Class 10

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 4 Pair of Linear Equations in Two Variables Ex 4.3 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 4th Lesson Pair of Linear Equations in Two Variables Exercise 4.3

10th Class Maths 4th Lesson Pair of Linear Equations in Two Variables Ex 4.3 Textbook Questions and Answers

Question 1.
Solve each of the following pairs of equations by reducing them to a pair of linear equations.
i) \(\frac{5}{x-1}\) + \(\frac{1}{y-2}\) = 2
\(\frac{6}{x-1}\) + \(\frac{3}{y-2}\) = 1
Answer:
Given
\(\frac{5}{x-1}\) + \(\frac{1}{y-2}\) = 2
\(\frac{6}{x-1}\) + \(\frac{3}{y-2}\) = 1
Put \(\frac{1}{x-1}\) = a and \(\frac{1}{y-2}\) = b,
then the given equations reduce to
5a + b = 2 ……… (1)
6a – 3b = 1 ………. (2)

⇒ b = \(\frac{7}{21}\) = \(\frac{1}{3}\)
Substituting b = \(\frac{1}{3}\) in equation (1) we get

⇒ (x – 1) . 1 = 3 × 1
⇒ x – 1 = 3
⇒ x = 3 + 1 = 4
b = \(\frac{1}{y-2}\) ⇒ \(\frac{1}{3}\) = \(\frac{1}{y-2}\)
⇒ (y – 2) . 1 = 3 × 1
⇒ y – 2 = 3
⇒ y = 3 + 2 = 5
∴ Solution (x, y) = (4, 5)

ii) \(\frac{x+y}{xy}\) = 2;
\(\frac{x-y}{xy}\) = 6
Answer:
Given
\(\frac{x+y}{xy}\) = 2
⇒ \(\frac{x}{xy}\) + \(\frac{y}{xy}\) = 2
⇒ \(\frac{1}{y}\) + \(\frac{1}{x}\) = 2
\(\frac{x-y}{xy}\) = 6
⇒ \(\frac{x}{xy}\) – \(\frac{y}{xy}\) = 6
⇒ \(\frac{1}{y}\) – \(\frac{1}{x}\) = 6
Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b,
then the given equations reduces to

⇒ b = \(\frac{8}{2}\) = 4
Substituting b = 4 in equation (1) we get
a + 4 = 2 ⇒ a = 2 – 4 = -2
but a = \(\frac{1}{x}\) = -2 ⇒ x = \(\frac{-1}{2}\)
b = \(\frac{1}{y}\) = 4 ⇒ y = \(\frac{1}{4}\)
∴ Solution (x, y) = \(\left(\frac{-1}{2}, \frac{1}{4}\right)\)

iii) \(\frac{2}{\sqrt{x}}\) + \(\frac{3}{\sqrt{y}}\) = 2;
\(\frac{4}{\sqrt{x}}\) – \(\frac{9}{\sqrt{y}}\) = -1
Answer:
Given
\(\frac{2}{\sqrt{x}}\) + \(\frac{3}{\sqrt{y}}\) = 2 and \(\frac{4}{\sqrt{x}}\) – \(\frac{9}{\sqrt{y}}\) = -1
Take \(\frac{1}{\sqrt{x}}\) = a and \(\frac{1}{\sqrt{y}}\) = b,
then the given equations reduces to
2a + 3b = 2 …….. (1)
4a – 9b = – 1 …….. (2)

⇒ b = \(\frac{5}{15}\) = \(\frac{1}{3}\)
Substituting b = \(\frac{1}{3}\) in equation (1) we get
2a + 3\(\left(\frac{1}{3}\right)\) = 2
⇒ 2a + 1 = 2 ⇒ 2a = 2 – 1 ⇒ a = \(\frac{1}{2}\)

∴ Solution (x, y) = (4, 9)

iv) 6x + 3y = 6xy
2x + 4y = 5xy
Answer:
Given
6x + 3y = 6xy
⇒ \(\frac{6x+3y}{xy}\) = 6
⇒ \(\frac{6x}{xy}\) + \(\frac{3y}{xy}\) = 6
⇒ \(\frac{6}{y}\) + \(\frac{3}{x}\) = 6
2x + 4y = 5xy
⇒ \(\frac{2x+4y}{xy}\) = 5
⇒ \(\frac{2x}{xy}\) + \(\frac{4y}{xy}\) = 6
⇒ \(\frac{2}{y}\) + \(\frac{4}{x}\) = 6
Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b,
then the given equations reduces to
3a + 6b = 6 ……. (1)
4a + 2b = 5 ……. (2)

⇒ b = \(\frac{9}{18}\) = \(\frac{1}{2}\)
Substituting b = \(\frac{1}{2}\) in equation (1) we get
3a +6\(\left(\frac{1}{2}\right)\) = 6
⇒ 3a = 6 – 3
⇒ a = \(\frac{3}{3}\) = 1
but a = \(\frac{1}{x}\) = 1 ⇒ x = 1
b = \(\frac{1}{y}\) = \(\frac{1}{2}\) ⇒ y = 2
∴ Solution (x, y) = (1, 2)

v) \(\frac{5}{x+y}\) – \(\frac{2}{x-y}\) = -1
\(\frac{15}{x+y}\) + \(\frac{7}{x-y}\) = 10
where x ≠ 0, y ≠ 0
Answer:
Given
\(\frac{5}{x+y}\) – \(\frac{2}{x-y}\) = -1 and
\(\frac{15}{x+y}\) + \(\frac{7}{x-y}\) = 10
Take \(\frac{1}{x+y}\) = a and \(\frac{1}{x-y}\) = b, then
the given equations reduce to
5a – 2b = – 1 ……… (1)
15a + 7b = 10 ……… (2)

⇒ b = \(\frac{-13}{-13}\) = 1
Substituting b = 1 in equation (1) we get
5a – 2(1) = -1
⇒ 5a = -1 + 2
⇒ 5a = 1
⇒ a = \(\frac{1}{5}\)
but a = \(\frac{1}{x+y}\) = \(\frac{1}{5}\) ⇒ x + y = 5
b = \(\frac{1}{x-y}\) = 1 ⇒ x – y = 1
⇒ x = \(\frac{6}{2}\) = 3
Solving the above equations
Substituting x = 3 in x + y = 5 we get
3 + y = 5 ⇒ y = 5 – 3 = 2
∴ Solution (x, y) = (3, 2)

vi) \(\frac{2}{x}\) + \(\frac{3}{y}\) = 13
\(\frac{5}{x}\) – \(\frac{4}{y}\) = -2
where x ≠ 0, y ≠ 0
Answer:
Given
\(\frac{2}{x}\) + \(\frac{3}{y}\) = 13 and
\(\frac{5}{x}\) – \(\frac{4}{y}\) = -2
Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b, then
the given equations reduce to
2a + 3b = 13 ……… (1)
5a – 4b = -2 ……… (2)

⇒ b = \(\frac{69}{23}\) = 3
Substituting b = 3 in equation (1) we get
2a + 3 (3) = 13
⇒ 2a = 13 – 9
⇒ a = \(\frac{4}{2}\) = 2
but a = \(\frac{1}{x}\) = 2 ⇒ x = \(\frac{1}{2}\)
b = \(\frac{1}{y}\) = 3 ⇒ y = \(\frac{1}{3}\)
∴ Solution (x, y) = (\(\frac{1}{2}\), \(\frac{1}{3}\))

vii) \(\frac{10}{x+y}\) + \(\frac{2}{x-y}\) = 4
\(\frac{15}{x+y}\) – \(\frac{5}{x-y}\) = -2
Answer:
Given
\(\frac{10}{x+y}\) + \(\frac{2}{x-y}\) = 4 and
\(\frac{15}{x+y}\) – \(\frac{5}{x-y}\) = -2
Take \(\frac{1}{x+y}\) = a and \(\frac{1}{x-y}\) = b, then
the given equations reduce to
10a + 2b = 4 ……… (1)
15a – 5b = – 2 ……… (2)

⇒ b = \(\frac{16}{16}\) = 1
Substituting b = 1 in equation (1) we get
10a + 2(1) = 4
⇒ 10a = 4 – 2
⇒ a = \(\frac{2}{10}\) = \(\frac{1}{5}\)
but a = \(\frac{1}{x+y}\) = \(\frac{1}{5}\) ⇒ x + y = 5 ……. (3)
b = \(\frac{1}{x-y}\) = 1 ⇒ x – y = 1 …….. (4)
Adding (3) and (4)

⇒ x = \(\frac{6}{2}\) = 3
Substituting x = 3 in x + y = 5 we get
3 + y = 5 ⇒ y = 5 – 3 = 2
∴ Solution (x, y) = (3, 2)

viii) \(\frac{1}{3x+y}\) + \(\frac{1}{3x-y}\) = \(\frac{3}{4}\)
\(\frac{1}{2(3x+y)}\) – \(\frac{1}{2(3x-y)}\) = \(\frac{-1}{8}\)
Answer:
Given
\(\frac{1}{3x+y}\) + \(\frac{1}{3x-y}\) = \(\frac{3}{4}\) and
\(\frac{1}{2(3x+y)}\) – \(\frac{1}{2(3x-y)}\) = \(\frac{-1}{8}\)
Take \(\frac{1}{3x+y}\) = a and \(\frac{1}{3x-y}\) = b, then
the given equations reduce to

⇒ a = \(\frac{2}{8}\) = \(\frac{1}{4}\)
Substituting a = \(\frac{1}{4}\) in equation (1) we get

Solving (3) and (4)

⇒ x = \(\frac{6}{6}\) = 1
Substituting x = 1 in 3x + y = 4
⇒ 3(1) + y = 4
⇒ y = 4 – 3 = 1
∴ The solution (x, y) = (1, 1)

Question 2.
Formulate the following problems as a pair of equations and then find their solutions.
i) A boat goes 30 km upstream and 44 km downstream in 10 hours. In 13 hours it can go 40 km upstream and 55 km downstream. Determine the speed of the stream and that of the boat in still water.
Answer:
Let the speed of the boat in still water = x kmph
and the speed of the stream = y kmph
then speed in downstream = x + y
Speed in upstream = x – y
and time = \(\frac{\text { distance }}{\text { speed }}\)
By problem,
\(\frac{30}{x-y}\) + \(\frac{44}{x+y}\) = 10
\(\frac{40}{x-y}\) + \(\frac{55}{x+y}\) = 13
Take \(\frac{1}{x-y}\) = a and \(\frac{1}{x+y}\) = b, then
the given equations reduce to
30a + 44b = 10 ……… (1)
40a + 55b = 13 ……… (2)

⇒ b = \(\frac{1}{11}\)
Substituting b = \(\frac{1}{11}\) in equation (1) we get

⇒ x = 8
Substituting x = 8 in x – y = 5 we get
8 – y = 5
⇒ y = 8 – 5 = 3
∴ The solution (x, y) = (8, 3)
Speed of the boat in still water = 8 kmph
Speed of the stream = 3 kmph.

ii) Rahim travels 600 km to his home partly by train and partly by car. He takes 8 hours if he travels 120 km by train and rest by car. He takes 20 minutes more if he travels 200 km by train and rest by car. Find the speed of the train and the car.
Answer:
Let the speed of the train be x kmph
and the speed of the car = y kmph
By problem,

Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b, then
the given equations reduce to
15a + 60b = 1 ……… (1)
8a + 16b = \(\frac{1}{3}\) ⇒ 24a + 48b = 1 ……… (2)

⇒ a = \(\frac{-1}{-60}\) = \(\frac{1}{60}\)
Substituting a = \(\frac{1}{60}\) in equation (1) we get

but a = \(\frac{1}{x}\) = \(\frac{1}{60}\) ⇒ x = 60 kmph
b = \(\frac{1}{y}\) = \(\frac{1}{80}\) ⇒ y = 80 kmph
Speed of the train = 60 kmph and
speed of the car = 80 kmph

iii) 2 women and 5 men can together finish an embroidery work in 4 days while 3 women and 6 men can finish it in 3 days. Find the time taken by 1 woman alone and 1 man alone to finish the work.
Answer:
Let the time taken by 1 woman to complete the work = x days
and time taken by 1 man to complete the work = y days
∴ Work done by 1 woman in 1 day = \(\frac{1}{x}\)
Work done by 1 man in 1 day = \(\frac{1}{y}\)
By problem,

Take \(\frac{1}{x}\) = a and \(\frac{1}{y}\) = b,
then the above equations reduce to
2a + 5b = \(\frac{1}{4}\) and 3a + 6b = \(\frac{1}{3}\)
⇒ 8a + 20b = 1 …….. (1) and
9a + 18b = 1 ……… (2)

⇒ b = \(\frac{1}{36}\)
Substituting b = \(\frac{1}{36}\) in equation (1) we get

but a = \(\frac{1}{x}\) = \(\frac{1}{18}\) ⇒ x = 18 and
b = \(\frac{1}{y}\) = \(\frac{1}{36}\) ⇒ y = 36
∴ Time taken by 1 woman = 18 days
1 man = 36 days

AP State Board Syllabus AP SSC 10th Class English Textbook Solutions Chapter 3B Once Upon a Time Textbook Questions and Answers.

AP State Syllabus SSC 10th Class English Solutions Chapter 3B Once Upon a Time

10th Class English Chapter 3B Once Upon a Time Textbook Questions and Answers

Comprehension

I. Tick (✓) the option that will complete each of the following statements. In some cases more than one option may be possible.

1. In the first five stanzas the poet is talking about
a) the honest and innocent world of children.
b) the insincere world of adults.
c) the difference between the past and the present.
d) the old and the young.
Answer:
(b) ✓
(c) ✓

2. The last four lines of the poem suggest
a) hope.
b) regret.
c) a sense of loss.
d) eagerness to learn.
Answer:
(a) ✓
(b) ✓

3. The expression ice-cold-block eyes’ means
a) The eyes are wet with tears.
b) expressionless eyes,
c) a state of lack of feelings.
d) a dead man’s eyes.
Answer:
(b) ✓
(c) ✓

4. ‘They’ in line 4 of stanza 1 refers to
a) people in the past.
b) present day people.
c) all adults.
d) young children.
Answer:
(c) ✓

5. ‘Their …………. eyes search behind my shadow’ means
a) they avoid meeting his eyes.
b) they try to look at the darker side of the person.
c) they convey no emotions.
d) they try to see what is not there.
Answer:
(b) ✓

6. The poet has learnt
a) to shake hands.
b) the ways of the world.
c) to laugh.
d) to put on masks.
Answer:
(d) ✓

7. The poet wants to learn from his son because his son
a) is not corrupted by the ways of the world.
b) is more informed.
c) knows about good manners more than his father.
d) is more caring.
Answer:
(a) ✓

II. Answer the following questions in a sentence or two each.

Question 1.
When did people shake hands with their hearts?
Answer:
The people, in their childhood, when they didn’t know the falsehood and hypocrisies of the world, when they were not corrupted by the ways of the world, shook hands with their hearts.

Question 2.
What is the poet crying over? What help does he want from his son?
Answer:
The poet regrets for losing the traits of his own character such as honesty, modesty, and sincerity. He laments over getting corrupted by the ways of the world. He regrets for his character being influenced by hypocrisy and fallacies of the world. The poet requests his son to help him regain his sincere and heartful, innocent and child-like smile.

Question 3.
“Most of all, I want to relearn
how to laugh, for my laugh in the mirror
shows only my teeth like a snake’s bare fangs !”
What does the poet mean by these lines?
Answer:
The poet feels his smile as fictitious, insincere, and hypocritic. He feels such a smile is dangerous. The comparison of his teeth to snake’s fangs makes false, mask-like smile seem dangerous.

Question 4.
What is the tone of the poem?
Answer:
The tone of the poem is roughly equivalent to the mood it creates in the reader. In Once Upon a Time’ the tone of the poem in the earlier stanzas is abashed, regretful but in the last stanza the poet ends the poem in an optimistic and hopeful tone. Thus the poet begins the poem in a negative tone i.e. somber but ends positively i.e. opti¬mism.

Question 5.
“Now they shake hands without hearts :
while their left hands search
my empty pockets.”
Why do the left hands search empty pockets now? What does this indicate?
Answer:
The poet expresses his concern for the influence of the western world on age-old African custom. He feels that the once enthusiastic and friendly society of Africa now treated its own people like strangers and looked at each other with suspicion and hostility. The white imperialists always exploited and plundered the wealth of their colonies. So their left hands search the empty pockets of their subjects in an endeavour to rob them further.

Question 6.
The poet uses certain words to express frustration and sorrow. Identify these words.
Answer:
The phrases “ice-block-cold eyes”, “shake hands without hearts”, “doors shut on me”, “learned to wear many faces”, “teeth like a snake’s bare fangs” are used to express the poet’s regret. The phrases or lines such as “…believe me, son. I want to be what 1 used to be”, “unlearn these muting things”, “want to relearn how to laugh” are the lines used to express his frustration.

Once Upon a Time Summary in English

Once upon a time, the people used to laugh with their hearts. There used to be sincerity in their laugh. Their laugh came from their hearts. There was genuinety in their actions and feelings. But people laugh superficially, in present. Their laugh is ficticious, feelingless. The eyes are dead like feelingless, and unsympathetic/apathetical. Even people shake hands mechanically and wish the people artificially but not heartfully.

In the third stanza the poet explains more about the changes the man possesses as he grows in age. He has noticed falsehood, superfluous feelings and deteriorating human relations in present day society. The poet also says that the people lie when they say the positive phrases like “Feel at home” and “Come again.” When the poet visits their house for the third time thinking that their words are genuine, the doors are shut on his face. In this material and artificial world the poet has learnt many things especially wearing many faces, like putting on many dresses. That means he changes his expressions and feelings to suit the situations and needs of the people with whom he is to deal with.

In behaving like that he loses his own character and traits of his self. As this is the way of the world the poet has also learnt to laugh with teeth but not with heart. He also has learnt to shake the hands of others but not with heart. He has learnt to say ‘Goodnight’ when he means Good riddance’. He has learnt to say Glad to meet you,’ when he is not glad and he says, ‘It’s nice talking to you’ when he is bored of talking.

But the writer is fed up with the forcible hypocrisy and pretension of falsehood. He wants to regain his real spirit and character. He wants to abandon all this falsehood. He wants to laugh sincerely as the children do. His laugh reveals all the fallacies of the world. When he looks at himself in the mirror his teeth are exposed and they appear like the fangs of a snake.

In the last stanza the poet appeals to his son to show him how to smile whole-heartedly. The poet’s desire to regain his original traits of his character, sincerity and to give up his falsehood and hypocrisy reveals his yearning for the innocence, faithfulness and sincerity.

Once Upon a Time Glossary

cock-tail (n) : a drink usually made from a mixture of one or more alcoholic drinks

conform (v) : to be and thinking the same way as most other people in a group or society; normally acceptable

portrait (n) : a painting, drawing or photograph of a person especially of the head and shoulders

good-riddance (n) : a feeling of relief when an unwanted person leaves

muting (adj) : changing all the time; expressionless/not expressed in speech

fangs (n) : long, sharp teeth of some animals like snakes and dogs

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 14 Carbon and its Compounds.

AP State Syllabus SSC 10th Class Chemistry Important Questions 14th Carbon and its Compounds

10th Class Chemistry 14th Lesson Carbon and its Compounds 1 Mark Important Questions and Answers

Question 1.
Define Isomerism. (AP March 2016)
Answer:
The phenomenon of possessing same molecular formula but different properties by the compounds is known as “Isomerism”.

Question 2.
Give the names of the functional groups. (AP March 2018)
a) – COOR
b) – OH
Answer:
a) Ester
b) Alcohol

Question 3.
How do you explain the role of Oxygen in combustion process? (TS March 2015)
Answer:
Oxygen helps the combustion (or) No combustion will take place without oxygen.
Ex : C + O₂ → CO₂

Question 4.

Predict and write the products. (TS March 2016)
Answer:

Question 5.
Write two uses of nano tubes. (TS June 2017)
Answer:

1. Nano tubes are used as molecule wires.
2. In intigrated circuits nano tubes are used to connect the components together.
3. Nano tubes are used to incert Bio-molecules into the single cell.

Question 6.
Write two uses of Ethanol in day to day life. (TS March 2018)
Answer:
Ethanol is used in
i) Preparation of Alchoholic drinks
ii) Preparing tincture iodine
iii) Preparing cough syrup and tonics

Question 7.
Write the atomic structure of the following carbon compound. 3, 7-dibromo-4, -6 dichloro – oct-5-ene-l, 2-diol. (TS March 2019)
Answer:

Question 8.
Thanish added acetic acid along with concentrated sulphuric acid to ethanol what would be his observation during the experiment? (AP SCERT: 2019-20)
Answer:

1. He may observe that the resulting mixture is a sweet odoured substance.
2. The substance is ethyl acetate, an ester.

Question 9.
Why do the various micelles present in water do not come together to form a precipitate? Guess the reason. (TS June 2019)
Answer:
The various micelles present in water do not come together to form a precipitate as each micelle repels the other because of the ion-ion repulsion.

Question 10.
Mention any two uses of graphite in day to day life. (TS June 2019)
Answer:
Uses of graphite in day to day life :

1. Pencil lead.
2. Lubricant.

Question 11.
What is “Allotropy”?
Answer:
The property of an element to exist in two or more different forms due to the difference in their atomic arrangement is called “Allotropy” and the different forms are called allotropes.

Question 12.
‘Diamond is a bad conductor of heat.’ Why?
Answer:
Diamond is a bad conductor of heat due to lack of free electrons.

Question 13.
What is ‘cleavage’?
Answer:
Cleavage is a property of splitting of crystals of some minerals in certain directions to produce a flat, even surface.

Question 14.
“Diamond is the hardest natural substance but is brittle.” Why?
Answer:
Diamond is the hardest natural substance but is brittle and can be broken due to the property of cleavage.

Question 15.
Explain about high refractive index of diamond.
Answer:
Diamond has a high refractive index, due to which most of the light that enters the diamond gets reflected back internally. This internally reflected light is responsible for the brilliance of a diamond.

Question 16.
What is catenation?
Answer:
Catenation is the phenomenon in which atoms of same element join together to form long chains.

Question 17.
What is an alkyl group?
Answer:
If one hydrogen is removed from an alkane, it is called alkyl group.
Ex : CH₄ → methane
CH₃ → methyl group

Question 18.
What is polymerization?
Answer:
The reaction in which a large number of identical and simple molecules join together to form a large molecule is called ‘polymerization’.

Question 19.
What do you understand by a ‘Functional group’?
Answer:
A group of atoms in carbon compounds showing characteristic properties is called a functional group.

Question 20.
Name some functional groups.
ANswer:
Alcohol – OH, Aldehyde – CHO, Ketone – > C = O, Carboxylic acid (- GOOH), ester (-COOR), and amine – NH₂ are some important functional groups.

Question 21.
What is pyrolysis?
Answer:
Decomposition of a compound on heating in the absence of air is called pyrolysis.

Question 22.
What is hydrocarbon?
Answer:
Compounds containing only carbon and hydrogen are called ‘hydrocarbons’.
Ex : Alkanes (Saturated hydrocarbons),
Alkenes and Alkynes (Unsaturated hydrocarbons).

Question 23.
What is ‘Saturated hydrocarbon’? (Or) What is an alkane?
Answer:
The valency of carbon is 4, of all the valencies of carbon, are satisfied, the resultant hydrocarbons are referred to as ‘saturated hydrocarbons’ or alkanes. Their general formula is C_nH_2n+2.

Question 24.
What are ‘Unsaturated hydrocarbons’?
Answer:
The hydrocarbons containing one or more double bonds or triple bonds between two carbon atoms are called ‘unsaturated hydrocarbons’.
Ex : C₂H₆ and C₃H₆, etc.

Question 25.
What are alkenes?
Answer:
Alkenes are unsaturated hydrocarbons having at least one (C = C) double bond in their structures, Alkenes are also called olefins. Their general formula is C_nH_2n.
Ex : Ethylene (C₂H4) and propene (C₃H₆), etc.

Question 26.
What are alkynes?
Answer:
Alkynes are unsaturated hydrocarbons having at least one (\(C \equiv C\)) triple bond in their structures. Their general formula is C_nH_2n-2.
Ex: Acetylene (\(\mathrm{HC} \equiv \mathrm{HC}\))

Question 27.
Mention the natural sources of carbon compounds.
Answer:
Plants, wood, natural gas, coal, petroleum, etc. are the natural sources of carbon compounds.

Question 28.
Explain about methanol (or) methyl alcohol.
Answer:
Methanol is the simplest alcohol, It is the first member of the homologous series of alcohol. It is also known as wood alcohol, as it was initially obtained by the destructive distillation of wood.

Question 29.
What is organic chemistry?
Answer:
The chemistry of carbon compounds (excluding the carbonates, bicarbonates, carbides, cyanides, carbon dioxide, and carbon monoxide) is called organic chemistry. The large number of organic compounds necessitated their study in separate branch of chemistry, known as organic chemistry,

Question 30.
What is halogenation?
Answer:
Alkanes react with halogens in the presence of sunlight. For example, when a mixture of methane and chlorine is exposed to sunlight, a hydrogen atom of methane is replaced by a chlorine atom,

Question 31.
How many rings are there in buckminsterfullerene?
Answer:
In buckminsterfullerene, there are 32 rings, of them 12 are pentagonal rings and 20 are hexagonal rings.

Question 32.
Give example for homologous series.
Answer:
CH₄ and C₂H₆ → These differ by a – CH₂ unit.
and C₂H₆ and C₃H₈ → These differ by a – CH₂ unit.

Question 33.
What is hybridisation?
Answer:
The intermixing of orbitals to form equivalent new orbitals is called hybridisation.

Question 34.
What are nanotubes?
Answer:
Nanotubes are allotropic form of carbon.

Question 35.
What are homologous series?
Answer:
The ierles of carbon compound in which successive compounds differ by -CH₂ unit is called homologous series.

Question 36.
Write the molecular formula of the fourth member of the homologous series of alcohols.
Answer:
CH₃ – CH₂ – CH₂ – CH₂ – OH

Question 37.
What is a catalyst?
Answer:
The substance which does not take part in chemical reaction but changes the rate of reaction.

Question 38.
Why are oils liquids at room temperature?
Answer:
Oils are unsaturated compounds so they are in liquid state.

Question 39.
Why are fats solids at room temperature?
Answer:
They are saturated compounds so they are in solid state.

Question 40.
Do you know the police detect whether suspected drivers have consumed alcohol or not? Explain.
Answer:
Orange Cr₂O₇^2- changes bluish green Cr³⁺ during the process of the oxidation of alcohol. The length of die tube that turned into green is the measure of die quantity of alcohol that had been drunk.

Question 41.
What is pka?
Answer:
The negative value of logarithm of dissociation constant of an acid.

Question 42.
What is Saponification?
Answer:
Alkaline hydrolysis of triesters of higher fatty acids producing soaps is called saponification.

Question 43.
What is a soap?
Answer:
Sodium or potassium salt of fatty acid.

Question 44.
What is micelle?
Answer:
A spherical aggregate of soap molecules in water is called micelle.

Question 45.
What change will you observe if you test soap with litmus papers?
Answer:
Red litmus turns into blue.

Question 46.
Write the valency of carbon in CH₃ – CH₃, CH₂ = CH₂ and \(\mathrm{HC} \equiv \mathrm{CH}\)?
Answer:
The valency of carbon in CH₃ – CH₃ is 4.
The valency of carbon in CH₂ = CH₂ is 3.
The valency of carbon in \(\mathrm{HC} \equiv \mathrm{C}\) – H is 2.

Question 47.
Out of butter and groundnut oil which is unsaturated in nature?
Answer:
Groundnut oil is unsaturated in nature.

Question 48.
What are hydrophobic and hydrophilic parts in soap?
Answer:

Question 49.
Name the carboxylic acid used as preservative.
Answer:
Acetic acid is used as preservative.

Question 50.
Why does graphite act as a good conductor of electricity?
Answer:
Graphite is a good conductor of electricity because of delocalized x electron system.

Question 51.
Among objects made of glass and diamond, which one shines more? Why?
Answer:
Diamond shines more because of low conical angle of 24,4° and also high refractive

Question 52.
Write IUPAC names of the following compounds.

Answer:
a) 2, 2, 3, 3 – tetra methyl butane
b) 3-chloro butan-l-oic acid.

Question 53.
What is the difference between combustion and oxidation reaction?
Answer:
Combustion is an oxidation reaction where a compound is burnt in the presence of oxygen, whereas oxidation is addition of oxygen which does not require any burning.

Question 54.
Write the order of priority of functional groups for naming carbon compounds.
Answer:

Question 55.
What is glycerol?
Answer:
The trihydroxy alcohol is called glycerol.

Question 56.
What do you mean by CMC?
Answer:
CMC means Critical Micelle Concentration.

Question 57.
Name the simplest chloride of saturated hydrocarbon.
Answer:
Chloro methane or methyl chloride.

Question 58.
Write the IUPAC name of next homolog of CH₃CH₂CHO.
Answer:
The next homolog of CH₃CH₂CHO is CH₃CH₂CH₂CHO (its IUPAC name is butanol). Since homologs differ by – CH₂.

Question 59.
How do physical properties like boiling point and melting point vary as the number of carbon atoms increases in a homologous series?
Answer:
There is regular gradation in physical properties of homologous series. So the physical properties like boiling point and melting point vary.

Question 60.
What is meant by Hybridisation?
Answer:
Mixing two atomic orbitals with the same energy levels to give a degenerated new type of orbitals.

Question 61.
Write any two uses of Graphite.
Answer:
i) Conductor
ii) Lubricant

Question 62.
Write any two examples to Amorphous form of carbon.
Answer:
i) Coke
ii) coal
iii) charcoal.

Question 63.
Write any two examples to crystalline forms of carbon.
Answer:
i) Diamond
ii) graphite

Question 64.
What are the applications of Buckminster fullerene?
Answer:
i) Antioxidants
ii) Anti aging and damage agent in cosmetic sector.

Question 65.
What is meant by catenation?
Answer:
Binding of an element to itself through covalent bonds to form chain or ring molecules.

Question 66.
Write any one use of nanotubes.
Answer:
i) Used as molecular wires.
ii) Used in integrated circuits.

Question 67.
On which reason, graphite is used as lubricant and as the lead in pencils?
Answer:
Graphite has free electrons.

Question 68.
How many isotopes are there for C₄H₁₀, what are they?
Answer:
i) n – Butane
ii) Iso – Butane

Question 69.
CH₃ – CH = CH – CH₃, how many sigma bonds are present in the above compound?
Answer:
11

Question 70.
Write the IUPAC name of Ethyle alcohol.
Answer:
Ethanol.

Question 71.
Classify the following into alkanes, alkenes and alkynes.
C₁₂ H₂₂, C₁₀ H₂₂, C₁₁ H₂₂
Answer:
i) C₁₀ H₂₂ – Alkanes
ii) C₁₁ H₂₂ -Alkenes
iii) C₁₂ H₂₂-Alkynes

Question 72.
Hi ……… I am carboxylic acid. I am used in the making vinegar, who am I?
Answer:
Acetic acid.

Question 73.
What does IUPAC represent?
Answer:
International Union of Pure and Applied Chemistry.

Question 74.
Write any one example for esterification reaction.
Answer:
CH₃COOH + C₂H₅OH → CH₃COOC₂H₅ + H₂O

Question 75.
A compound with molecules formula C₂H₆O is used in cough syrup. Identify the compound.
Answer:
Ethyl Alcohol.

Question 76.
Which substance is added for the denaturation of ethyl alcohol?
Answer:
Pyridine.

Question 77.
What is the abbreviation of CMC?
Answer:
Critical Micelle Concentration.

Question 78.
Write the names of polar end and non-polar end in a soap.
Answer:
Polar end – COO^– Na⁺, Non-polar end – R.

Question 79.
Write the IUPAC name of the alcohol which one carbon atom.
Answer:
Methanol.

Question 80.
Write the chemical equation which indicates the preparation of ethanol industrially?
Answer:

Question 81.
What is the formula of chloroform? Write its one use.
Answer:
CHCl₃, Anesthetic.

Question 82.
Which type of hydrocarbons are participate in addition reaction?
Answer:
Unsaturate Hydrocarbons.

Question 83.
What are the oxidising agents used in oxidisation of C₂H₅?
Answer:
K₂Cr₂O₇, KMn0₄.

Question 84.
What is meant by catalyst?
Answer:
To change die rate of reaction without itself undergoing any permanent chemical change.

Question 85.
What are the main constituents of LPG?
Answer:
Butane, Methane.

Question 86.
What is the difference between saturated and unsaturated hydrocarbons?
Answer:
Saturated house single bonds, unsaturated have multiple bonds.

Question 87.
Describe a test for carboxylic acid.
Answer:
React with metals liberate hydrogen gas.

Question 88.
What is meant by denatured alcohol?
Answer:
Unfit for human consumption by adding one or more chemicals.

Question 89.
Complete the following equation.
CH₄ + 2O₂ →
Answer:
CH₄ + 2O₂ → CO₂ + 1H₂O

Question 90.

i) In the above substance, what is the hybridisation of 3^rd carbon?
Answer:
sp²

ii) What is the hybridisation of 4^th carbon?
Answer:
sp³

Question 91.
What is the main misuse of Ethanol?
Answer:
Drinking.

Question 92.
What is gasohol?
Answer:
10% Ethyl alcohol with gasoline.

Question 93.
Write any two uses of Ethyle alcohol.
Answer:
i) Good solvent
ii) Additive to automotive gasoline.

Question 94.
Write two IUPAC name

Answer:
3 – Chloro 1 – Butane

Question 95.
Name the following functional groups.
i) – COOR
ii) R – COOH
Answer:
i) – COOR (Ester)
ii) R – COOH (Carboxylic acid)

Question 96.
Name the crystalline allotrope of carbon which conducts electricity.
Answer:
Graphite.

Question 97.
Ravi gets confused while understanding the between R – COOH and R – OH functional groups, ask him one question to classify it.
Answer:
i) What is carboxylic acid?
ii) What is Alcohol?

Question 98.
Formic acid (HCOOH)
Farmaldehyde (HCHO)
Methanol (CH₃OH), then answer the following questions.
i) Which is present in ants?
Answer:
HCOOH (Formic acid).

ii) Which is used to preservation of dead bodies?
Answer:
HCHO (Farmaldehyde).

Question 99.
Write the symbolic representation showing the functional groups.
i) amine
ii) amide
Answer:
i) R – NH₂
ii) R – CONH₂

Question 100.
How many sigma and pi-bonds present in Acetylene?
Answer:
\(\mathrm{HC} \equiv \mathrm{CH}\) ; σ bonds – 3 ; π bonds – 2

Question 101.
Which of the following will give substitution reactions?
CH₄, C₃H₆, C₃H₄, C₅H₁₂, C₄H₈
Answer:
CH₄, C₅ H₁₂

Question 102.
Which of the following will give addition reactions?
CH₄, C₃H₆, C₃H₄, C₅H₁₂, C₄H₁₀
Answer:
C₃H₆, C₃H₄

Question 103.
What is a homologous series?
Answer:
Same functional group, difference between successive members is a simple structural unit – CH₂.

Question 104.
Name the hydrocarbon which is used in the artificial ripening of fruits?
Answer:
C₂H₄

Question 105.
Define fermentation process.
Answer:
Chemical break down of a substance by bacteria, yeast or other microorganisms.

Question 106.
Define functional group.
Answer:
They are specific substituents within molecules that are responsible for die characteristic chemical reactions.

Question 107.
Which hydrocarbons participate in sp² hybridisation?
Answer:
C₂H₄

Question 108.
Name the following compounds,
i) CH₃ – CH₂ – Br
Answer:
1 – Bromo Ethane

ii)

Answer:
Ethanol

iii)

Answer:
2 – Butanone

iv)

Answer:
2, 3 – dichloro Butane

Question 109.
Which constituents are present in tincture Iodine?
Answer:
i) Iodine
ii) Alcohol.

Question 110.
Write the uses of esters in daily life.
Answer:
i) Solvents
ii) Plasticizers

Question 111.
Name the gas evolved when acetic acid reacts with sodium hydrogen carbonate.
Answer:
The gas liberated is carbon dioxide.

Question 112.
Name the organic acid present in vinegar. Write its chemical formula.
Answer:
The acid present in vinegar is acetic acid. Its formula is CH₃COOH.

Question 113.
Why is graphite a good conductors’of electricity?
Answer:
Graphite has free electrons.

Question 114.
Why does carbon form compounds mainly by covalent bonding?
Answer:
Tbtravalency.

Question 115.
Why are alkanes called as paraffins?
Answer:
Low reactivity.

Question 116.
Draw two possible structures with formula C3HgO and what they are called?
Answer:

Question 117.
Draw structure of 3 – methyl pentan-3-ol.
Answer:

Question 118.
Draw the shape of soap molecule.
Answer:

Question 119.
Draw the shape of Micelle.
Answer:

Question 120.
Draw the shape of methane.
Answer:

Question 121.
Draw the structure of pentanoic acid.
Answer:

Question 122.
How do you appreciate the role of diamond in space probes?
Answer:
Since it has the ability to filter out harmful radiations, it is used in making protective windows for space probes.

Question 123.
How do you appreciate the role of acetic acid as a preservative?
Answer:

• Dilute acetic acid is used as a food preservative in the preparation of pickles and sauces,
• As vinegar, it is also used as an appetiser for dressing food dishes.

Question 124.
How do you appreciate the role of diamond in surgery?
Answer:
A sharp edged diamond is used as a tool to remove cataract in eye surgery.

10th Class Chemistry 14th Lesson Carbon and its Compounds 2 Marks Important Questions and Answers

Question 1.
Draw the simple figure of a soap molecule. (AP March 2016)
Answer:
Structure of soap molecule :

Question 2.
Draw the structure of the methane molecule. Write its bond angle. (TS June 2015)
Answer:
The bend angle in methane is 109°2 8′.

Question 3.
a) Why are vegetable oils healthy as compared to vegetable ghee? (TS March 2015)
b) Write the IUPAC name of

Answer:
a) Because vegetable oils contain unsaturated fatty acids or vegetable oils are easily digestible.
b) 3 – Mono chloro butene (or) 3 Chloro butene

Question 4.
What are alkenes? Write the general formula of alkenes. Give an example for alkenes. (TS June 2017)
Answer:

• Unsaturated hydrocarbons those are having carbon * carbon double bond are known as alkenes.
• The general formula of Alkenes is CHH2h.
• Example : Ethelene (C2H4).

Question 5.

Based on the diagram, answer the following.
1) Write the name of the compound.
2) Write the name of functional group in the structure. (AP March 2019)
Answer:

1. The compound is 2, 3-di ethyl-cycle hexan-1-ol.
2. Alcohol (OH) is the functional group in the structure.

Question 6.
Identify the functional groups in the following compounds and write IUPAC names.

Answer:

The IUPAC name of the compound Is 2 – Chloro-Butan 1-ol.

The IUPAC name of the compound is 3 – Methyl-2-Butan-one.

Question 7.
Draw the structure of butanoic acid C₃H₇COOH.
Answer:
Formula of butanoic acid is C₄H₅O₂.

Question 8.
What is ‘Isomerism’?
Answer:
Compounds having same molecular formula but different structures are called isomers, and the phenomenon is called isomerism.
Ex: C₄H₁₀ exists an n-hutane and iso-butane.

Question 9.
How do you detect leakage in the cylinder?
Answer:

• To detect any leakage of gas from die cylinder, a strong-smelling substance like ethyl mercaptan (C₂H₅ SH) is added to die gas.
• Then the leakage can be easily detected by the foul smell of die ethyl mercaptan.

Question 10.
How is LPG gas useful for environment?
Answer:

• Because of its heat producing capacity (calorific value), it is considered to be a good fuel.
• It bums without producing smoke. Hence, it does not cause any pollution.
• It is a dean fuel and can be conveniently handled.

Question 11.
How is ethanol useful in pharmaceutical industry?
Answer:

• Solutions in ethanol are often prepared in pharmaceutical industry, these solutions are known as tinctures.
• For example, a solution of Iodine and potassium iodide in ethanol is called tincture of iodine.
• It is also used as an important raw material for the synthesis of many organic compounds, for example, ethanol, ethanoic acid, ethanoie anhydride, esters, chloroform, etc.

Question 12.
How are synthetic detergents harmful for environment?
Answer:

• Some synthetic detergents resist biodegradation, i.e. they are not decomposed by micro-organisms such as bacteria.
• Hence, they cause water pollution in lakes and rivers.
• They tend to persist for a long time, making the water unfit for aquatic life.

Question 13.
Explain about allotropic forms of carbon.
Answer:

Question 14.
Diamond is considered to be the purest form of carbon. How can we prove it?
Answer:
When diamond is heated in oxygen alone, it bums at about 800° C and forms carbon dioxide leaving no residue. This proves that diamond to be the purest form of carbon.

Question 15.
Why does carbon not form C⁴⁺? Why?
Answer:

• Electronic configuration of carbon is 1s²2s²2p².
• If carbon loses four electrons from the outer shell, it will form C⁴⁺ ions.
• This requires huge amount of energy which is not available normally.
• Therefore C⁴⁺ formation is not possible.

Question 16.
Why does carbon form compounds mainly by covalent bonding?
Answer:
Carbon is unable to form C⁴⁺ ion as well as C^4- ion. So carbon has to satisfy its tetra- valency by sharing electrons with other atoms. So it mainly forms covalent bonding.

Question 17.
Define Allotropy. What are the allotropic forms of carbon?
Answer:
The property of an element to exist in two or more physical forms having more or less similar chemical properties but different physical properties is called allotropy. The allotropic forms of carbon are graphite, diamond, etc.

Question 18.
Identify the unsaturated compounds of the following.
a) CH₃ – CH₂ – CH₂
b) CH₃ – CH = CH₃
c)

Answer:
a) CH₃ – CH₂ – CH₂ saturated compound.
b) CH₃ – CH = CH₃ unsaturated compound.
c)

Question 19.
Define Isomers. Write structural formula of isomers of butane.
Answer:
Compounds having same molecular formula but different properties are called isomers.
Isomers of butane :
1) CH₃ – CH₂ – CH₂ – CH₃
Butane

Question 20.
What happens when a small piece of sodium is dropped into ethanol?
Answer:
When a small piece of sodium is dropped into ethanol it releases hydrogen gas and forms sodium ethoxide.
2C₂H₅OH + 2 Na → 2C₂H₅ONa + H₂

Question 21.
What type of reaction takes place between ethane and chlorine?
Answer:
Substitution reaction takes place between ethane and chlorine in die presence erf sunlight
CH₄ + Cl₂ → CH₃Cl + HCl
CH₃Cl +Cl₂ → CH₂Cl₂ + HCl
CH₂Cl₂ + Cl₂ → CHCl₃ + HCl
CHCl₃ + Cl₂ → CCl₄ + HCl

Question 22.
What are the two properties of carbon which lead to the huge number of carbon compounds we see around us?
Answer:

1. Catenation
2. Isomerism.

Question 23.
How could you name the following compounds?
a) CH₃ – CH₂ – CH₂ – Br
b) CH₃ – CH₂ – CH₂ – CH₂
Answer:
a) Bromo propane
b) Hexyne

Question 24.
Give examples for primary, secondary and tertiary amines.
Answer:
Primary amine – CH₃NH₂
Secondary amine – CH₃ – NH – CH₃

Question 25.
Write the following conversions.
1) Ethanol to Ethene
2) Ethene to Ethanol
3) Methane to carbon tetra chloride.
Answer:
1) Ethanol reacts with cone. H₂SO₄ at about 170°C to give ethene.

2) Ethanol is prepared from ethene by the addition of water vapour in the presence of catalyst P₂O₅.

3) Methane reacts with chlorine in the presence of sunlight. Hydrogen atoms of CH₄ are replaced by chlorine at

Question 26.
Name the following compounds and which one is saturated among them.
a) CH₃ – \(\mathrm{C} \equiv \mathrm{H}\) – CH₃
b) CH₃ – CH = CH – CH₃
c) CH₃ – CH₂ – CH₂ – CH₃
Answer:
a) 2-Butyne
b) 2 – Butene
c) Butane

Butane does not show any double or triple bonds. Its valency is completely satisfied with formation of single bond. So it is a saturated compound.

Question 27.
How do you identify the given organic compound contains carboxylic acid functional group?
Answer:

• On adding carbonates and bicarbonates the compound containing carboxylic acid group evolves carbon dioxide gas.
• When warmed with alcohol and cone. H₂SO₄ a pleasant fruity smell is produced due to formation of ester.

Question 28.
Explain briefly about the structure of “Diamond”.
Answer:

• In a diamond, each carbon atom is surrounded by four other carbon atoms.
• In these carbon atoms, each carbon atom undergoes in its excited state sp3 hybridisation.
• These are placed at the four corners of a regular tetrahedron.
• This results in a 3-dimensional network of carbon atoms.
• So diamond is in three dimensional structure.

Question 29.
Explain briefly about the structure of “Graphite”.
Answer:

• In graphite, each ‘C’ is surrounded by three other ‘C’ atoms.
• The ‘C’ atoms are arranged in layers.
• In the layer structure, the carbon atoms are in trigonal planar environment.
• Each layer consists of a 2-dimensional hexagonal network.

Question 30.
Diamond is an extremely bad conductor of electricity.” Why?
Answer:
1) In diamond, each carbon atom is covalently bonded with four other carbon atoms.
2) So, the four outermost electrons of a carbon atom are engaged or trapped in the covalent bonds, having no free electrons making it a bad conductor of electricity.

Question 31.
Why is diamond hard but graphite is smooth and slippery?
Answer:
Diamond has sp³ hybridisation with tetrahedral environment. As C – C bonds are very strong any attempt to distort the diamond structure requires large amount of energy. Hence diamond is one of the hardest material.

Whereas graphite has sp² hybridisation with layer structure with trigonal planar environment. The layers tend to slide on one another. So graphite is smooth and slippery.

Question 32.
An organic compound X with a molecular formula C₂H₆O undergoes oxidation within presence of alkaline KMnO₄ to form a compound Y. X on heating in presence of con. H₂SO₄ at 443 K gives Z. Which on reaction with Br₂ and decolorizes it? Identify X, Y, and Z and write the reactions involved.
Answer:
X is ethanol.

Question 33.
Complete the following reactions.

Answer:

Question 34.

What are A and B?
Answer:
1) Alkynes undergo addition reaction in the presence of nickel catalyst and hydrogen to form Alkene.

Question 35.
Draw the structure for the following compounds.
a) Propanoic acid
b) Chlorobutane
c) Hexanone
d) Pentanal
Answer:
a) CH₃CH₂COOH
b) CH₃CH₂CH₂CH₂Cl
c) CH₃CH₂CH₂CH₂COCH₃
d) CH₃CH₂CH₂CH₂CHO

Question 36.
Give IUPAC names of the following compounds. If more than one compound is possible, name all of them.
i) A chloride derived from butane.
ii) A ketone derived from pentane.
Answer:
i) The following chlorides are possible for butane.

ii) The following ketones are possible for pentane.

Question 37.
a) What are the various possible structural formulae of a compound having molecular formula C₃H₆?
b) Give IUPAC names of the above possible compounds and represent them in structure.
c) What is the difference between those
Answer:

b) The IUPAC names of compounds are propene and cycle propane.
c) The main difference Is that the first compound Is alkene-an unsaturated compound and second is cyclo alkane-a saturated compound.

Question 38.
Draw isomeric forms of C₆H₁₄.
Answer:
Isomers of hexane :

Question 39.
How do you appreciate the role of carbon in everyday life?
Answer:

• Major components of our daily food have carbohydrates, proteins, fats, etc. which are all made up of carbon compounds.
• The fibres of cloth are made up of cellulose and other types of materials, which are all carbon compounds.
• Cement and steel form the core of any of the modern buildings. Carbon bestows steel with hardness, while limestone (CaCO₃) a major constituent of cement also contains carbon.

Question 40.
How do you appreciate the role of oxygen in combustion process?
Answer:

• When the oxygen supply is insufficient, the fuels burn incompletely producing mainly a yellow flame.
• When the oxygen supply is sufficient, the fuels burn completely producing a blue flame.

Question 41.
How do you appreciate the role of Ethanol as a fuel?
Answer:

• A material which is burnt to obtain heat is called a fuel. Since ethanol burns with a clear flame giving a lot of heat, it is used as a fuel.
• Some countries add ethanol to petrol to be used as a fuel in cars. Thus ethanol is used as an additive in petrol.
• Ethanol alone can also be used as a fuel for cars.

Question 42.
What are the uses of fullerenes?
Answer:
Fullerenes are under study for potential medical use such as specific antibiotics to target resistant bacteria and even target cancer cells such as melanoma.

Question 43.
Write the HJPAC names of the following compounds.
i) CH₃ – CH₀ – CH₂ – CH₂ – CH₂ – CH₂ – CH₂ – CH₂ – CH₂OH
ii) CH₃ – CH₂ – CH = CH- CH₂ – \(\mathrm{C} \equiv \mathrm{CH}\)
iii) CH₃ – CH₂ – CH₂ – CH₂ – CHO
iv) CH₃ – CH₂ – CH₂ – CH₂ – COOH
Answer:

1. nananol
2. 4- ene – 1 heptyne
3. pentanal
4. pentanoic acid

Question 44.
What are the uses of alcohol?
Answer:

• Alcohols are goods solvent for resin and gums.
• Ethanol is used in the thermometers because of its low freezing point.
• One of the products of ethyl alcohol is chloroform, which is used as an aesthetic.
• 10% ethanol in gasoline is a good motor fuel.
• It is used in medicines such as tincture iodine, cough syrups and many tonics.

Question 45.
What are the uses of acetic acid?
Answer:

• 5 to 8% solution of acetic acid in water is called vinegar and is used widely as a preservative in pickles.
• Used as a laboratory reagent.
• Used in the production of perfumes, dyes, esters, etc.
• Used in medicine.

10th Class Chemistry 14th Lesson Carbon and its Compounds 4 Marks Important Questions and Answers

Question 1.
Write IUPAC names for the following carbon compounds.

Answer:
A) 2 – methyle pentane – 3 – ol
B) 3 – chloro, 4 – Methyle hexanoic acid
C) 2 Bromo – Bute – 2 – ene
D) 2, 5 Dimethyle hexane

Question 2.
(AP June 2017)
Observe the given carbon compound and answer the following questions.
a) Give numbering to the carbons in the given compound according to IUPAC rules.
b) Name the functional group present in the given compound.
c) Name the word root for the given carbon compound.
d) Write the IUPAC name of the given compound.
Answer:
a)

b) The given compound contains functional group – OH. It is an alcohol.
c) Word root: The number of carbon atoms present in the molecules is called word root. Here the word root is (C₅) – pent.
d) IUPAC name of the given compound is pent 4 – ene 2 – ol.

Question 3.
Alkanes are considered as Paraffins. So, they undergo substitution reactions but not addition reactions. Explain with suitable example. (AP March 2017)
Answer:

Question 4.

Observe the structure and answer the following.
a) Write the name of principal functional group present in the compound.
b) Identify the parental chain in the compound.
c) What are the substituents in the above compound?
d) Name the above compound as per IUPAC nomenclature. (AP June 2018)
Answer:
a) Ketone
b)

c) Methyl group ; Hydroxy group
d) 7 – hydroxy – S – methyl heptan – 2 – one

Question 5.
In the table given below, fill the information in the empty boxes and give answers to the following questions. (TS June 2015)
a Write the general formula of alkanes from the table.

b) How many a bonds are there in C₃H₆?
c) What sequential order did you notice in the molecular formulae?
d) There exist single bonds between carbon atoms of alkanes. Do you agree with this statement? Give reasons.
Answer:

a) The general formula of Alkanes is C_nH_2n+2
b) The number of o bonds in C₂H₆ are 7.
c) Two successive alkanes are differed by – CH₂ group.
d) Except Methane all other alkanes have single bonds between carbon atmos because it is a saturated hydro carbon.

Question 6.
Why do we call alkanes as paraffins? Explain the substitution reactions of alkanes. (TS June 2016)
Answer:
a) 1. Alkanes are saturated hydrocarbons with least reactivity.
2. Therefore they are called paraffins.
3. Parum = little and affins = affinity.

b) 1. A reaction in which one atom or a group of atoms in a given compound is replaced by other atom or group of atoms is called a substitution reaction.
2. Alkanes have single bonds and undergo substitution reactions.

3. For example :
Methane (CH₄) reacts with chlorine in the presence of sunlight.

Question 7.
Write the types of Allotropes of Carbon. Give any three examples of each. (TS March 2016)
Answer:
The allotropes of carbon are classified into two types. They are
i) Amorphous forms,
ii) Crystalline forms.
Examples :
Amorphous forms :
Coal, coke, wood charcoal, animal charcoal, lamp black, gas carbon, petroleum coke, sugar charcoal, etc.

Crystalline forms :
Diamond, graphite, and buckminsterfullerene.

Question 8.
Write any 4 characteristic features of homologous series of Organic compounds. (TS March 2016)
Answer:
Homologous series :
The series of carbon compounds in which two successive compounds differ by – CH2 unit is called homologous series.

Characteristic features of homologous series :

1. They have one general formula.
   Ex : Alkane (C₄H_{2n + 2}), Alkene (C₄H_2n), Alkyne (C₄H_2n-2)
2. Successive compounds in their series possess a difference of (- CH₂) unit.
3. They possess similar chemical properties due to the same functional group.
4. They show a regular gradation in their physical properties.

Question 9.
List out the materials required to conduct the experiment to understand the esterification reaction. Explain the procedure of the experiment. How can you identify that an ester is formed in this reaction?(TS March 2017)
Answer:
Required Material :
Test tube, beaker, tripod* burner, water, wire guage, ethanol (absolute alcohol), glacial acetic acid, concentrated sulphuric acid.

Procedure :

1. Take 1 ml of ethanol and 1 ml of glacial acetic acid along with a few drops of concentrated sulphuric acid in the test tube.
2. Warm it in a water bath or in a beaker containing water for atleast five (5) minutes.
3. Pour the warm contents into a beaker containing 20-50 ml of water and observe the odour of the resulting mixture.
   If we smell sweet odour from the beaker, we can confirm that ester is formed.

Question 10.
Explain the Isomerism and Catenation properties of carbon. (TS March 2018)
Answer:
Catenation properties of carbon :
i) Carbon has ability to form longest chains with its own atoms. This special property of carbon is called catenation.
ii) Due to catenation property of carbon it can form largest chain containing millions of carbon atoms, branches and cyclic compounds.

Isomerism of carbon :
The phenomenon of possessing some molecular formula but different properties by the compounds is known as isomerism.

Molecular formula of above two molecules is C₄ H₁₀ but they have different structure. These two are isomers.

By there two special properties of carbon it can make number of compounds.

Question 11.

Observe the above table and answer the following questions. (TS March 2019)
1) Write the general formula of Alkanes.
2) Mention the names of unsaturated hydrocarbons.
3) Write the homologous series of Alkynes.
4) Write the formula of Hexyne.
Answer:
1) General formula for Alkanes : C_nH_2n+2.
2) Unsaturated Hydrocarbons in the list are :
Propene C₃H₆, Butene C₄H₆, Pentyne C₅H₈, Hexyne C₆H₁₀.

3) Homologous series of Alkynes is C₂H₂ (Ethyne), C₃H₄ (Propyne), C₄H6 (Butyne), C₅H₈ (Pentyne), C₆H₁₀ (Hexyne).

4) Formula of Hexyne is C₆H₁₀.

Question 12.
Complete the following table based on functional groups of organic compounds, their structural formulas and respective suffixes. (AP SCERT: 2019-20)

Answer:

Question 13.
Explain the occurrence of carbon.
Answer:
Carbon occurs in nature in free state as well as in combined state.

Question 14.
What is sp hybridisation? Explain.
Answer:

• Each carbon is only joining to two other atoms rather than four or three.
• Here the carbon atoms hybridise their outer orbitals before forming bonds, this time they only hybridise two of the orbitals.
• They use the ‘s’ orbital (2s) and one of the 2p orbitals, but leave the other 2p orbitals unchanged.
• The new hybrid orbitals formed are called sp-hybrid orbitals, because they are made by an s-orbital and a p-orbital reorganizing themselves.

Question 15.
Write the characteristics of homologous series of organic compounds.
Answer:
Characteristics of homologous series :

1. They have one general formula.
   e.g.: Alkanes (C_nH_2n+2)
2. Successive compounds in the series possess a difference of – CH₂ unit.
3. They possess similar chemical properties due to same functional group.
   e.g.: C – OH
4. They show a regular gradation in their physical properties.

Question 16.
What is sp³ hybridisation with diagram? Explain.
Answer:
The excited carbon atom allows its one s-orbital (2s) and three p-orbitals (2p_x, 2p_y, 2p_z) to intermix and reshuffle into four identical orbitals known as sp³ orbitals. Thus, carbon atom undergoes sp³ hybridization. The four electrons enter the new four identical hybrid orbitals known as sp³ hybrid orbitals, one each as per Hu nd’s rule.

1) Since carbon has four unpaired electrons, it is capable of forming bonds with four other atoms.

2) When carbon reacts with hydrogen, four hydrogen atoms allow their ‘s’ orbitals containing one electron each to overlap with four sp³ orbitals of carbon atom which are oriented at an angle of 109°. 28’.

3) Four orbitals of an atom in the outer shell orient along the four corners of a tetrahedron to have minimum repulsion between their electrons. ‘The nucleus of the atom is at the centre of the tetrahedron.

4) This leads to form four sp³ – s sigma bonds between carbon atom and four hydrogen atoms, All these bonds are of equal energy,

Question 17.
What is sp² hybridisation? Explain.
Answer:
Consider ethene molecule

• In the formation of CH₂ – CH₂ each carbon atom in its excited state undergoes sp² hybridisation by intermixing one s-orbital (2s) and two p-orbitals (say 2p_x and 2p_y) and reshuffling to form three sp² orbitals.
• Mow each carbon atom is left with one ‘p’ orbital (say 2p_z) unhybridised,
• The three sp² orbitals having one electron each get separated around the nucleus of carbon atoms at an angle of 120°.
• When carbon is ready to form bonds one sp² orbital of one carbon atom overlaps the sp² orbital of the other carbon atom to form sp² – sp² sigma (σ) bond,
• The remaining two sp² orbitals of each carbon atom get overlapped by ‘s’ orbitals of two hydrogen atoms containing unpaired electrons.
• The unhybridised p_z orbitals on the two carbon atoms overlap laterally as shown in figure to form a π (pi) bond.
• Hence, there exist a sigma (σ) bond and a pi π (pi) bond between two carbon atoms in ethene molecule. Hence, the molecule ethene (C₂H₄) is

Question 18.
The list of some organic compounds is given below.
Ethanol, ethane, methanol, methane, ethyne and ethene.
From the above list name the compound …………..
a) formed by the dehydration of ethanol by cone. H₂SO₄.
b) which forms methanoic acid on oxidation?
c) which forms chloroform on halogination in the presence of light?
d) which are unsaturated compounds?
e) which have compounds containing alcohol group?
Answer:
a) Dehydration ethanol in the presence of Cone. H₂SO₄ forms ethene,
b) Methanol on oxidation turns to methanoic acid,
c) Methane in the presence of light forms chloroform,
d) Unsaturated compounds are ethene and ethyne.
e) The compounds containing alcohol group are methanol, ethanol,

Question 19.
Give the IUPAC names of the following compounds.

Answer:

1. 1 – propyne
2. 3 – pentanel (or) pentamSml
3. 2 – methyl propane
4. 1, 2 dichloro ethane

Question 20.
Give one example of each of the following.
i) Saturated hydrocarbon
ii) Cyclic compounds
iii) Unsaturated hydrocarbon
iv) Functional group
v) Homologous series
Answer:
i) Saturated hydrocarbons are Alkanes, So the examples are methane (CH₄), Ethane, (C₂H₆).
ii) Cyclic compounds are cycle alkanes, eg : Cyclo propane (C₃H₆), Cycle butane (C₄H₆).
iii) Unsaturated hydrocarbons are Aikynes, eg : Ethene (C₂H₄), Propene
iv) The examples for functional groups are ‘ 1. Aldehyde – CHO, 2. Alcohol = OH
v) A series of carbon compounds that differ by – CH₂ with similar chemical properties is called homologous series.
eg: 1, Alkane, 2, Alkene

Question 21.
Write the differences between saturated and unsaturated hydrocarbons.
Answer:

Saturated hydrocarbons	Unsaturated hydrocarbons
1) All the four valencies of each carbon atom are satisfied by forming single covalent bonds with carbon and hydrogen atoms,	1) The valencies of at least two carbon atoms are not fully satisfied by the hydrogen atoms.
2) Carbon atoms are joined only by single bonds.	2) Carbon atoms are joined by at least one double bond or by a triple bond.
3) They are less reactive due to non­availability of electrons in the single covalent bond therefore they undergo substitution reactions,	3) They are more reactive because of the presence of electrons in the double or triple bond and therefore undergo addition reactions.

Question 22.
Answer the following.
a) What are the first three members of carboxylic acid series?
b) Name the compounds which can be oxidised directly or in stages to produce ethanoic acid.
c) Write one equation each when acetic acid reacts with a metal, a base, and a carbonate.
d) Name the organic compound formed when acetic acid and ethanol react together.
Answer:
a) The first three members of carboxylic acids are :
i) Methanoic acid – HCOOH
ii) Ethanoic acid – CH₃COOH
iii) Propanoic acid – CH₃CH₂COOH

b) Ethanol in stages oxidises to acetic acid whereas ethanol directly oxidises to ethanoic acid.

c) i) 2 CH₃COOH + 2 Na → 2CH₃COONa + H₂
ii) CH₃COOH + NaOH → CH₃COONa + H₂O
iii) CH₃COOH + Na₂CO₃ → CH₃COONa + H₂O + CO₂

d) When ethanol reacts with ethanoic acid it forms an ester namely ethyl acetate.

Question 23.
What are the rules to be followed to name a carbon compound?
Answer:
Rules to be followed
i) Longest carbon chain is selected,
ii) Chain is numbered in such a way that the branched chain or substituent gets the smallest number,
iii) If the functional group is present, it is given the. lowest number,
iv) Substituents are named in the alphabetical order,
v) The position of substituents are prefixed with hyphen,
vi) Multiple substituents are written with numerical prefixes such as di or tri,

Question 24.
Write suffixes and prefixes for some important characteristic functional group in a tabular form.
Answer:

Question 25.
Correct the following statements.
1) Alkenes undergo substitution reactions.
2) Alkanes are polar in nature.
3) When sodium piece is added to ethanol oxygen gas liberates.
4) On complete combustion of carbon compound it gives carbon monoxide and water.
Answer:

1. Alkenes are unsaturated hydrocarbons. So they undergo addition reactions.
2. Alkanes are covalent compounds. So they are non-polar in nature.
3. When sodium piece is added to ethanol it releases hydrogen gas.
4. On complete combustion of carbon compound it forms carbon dioxide and water.

Question 26.
Copy and complete the following table which relates to three homologous series of hydrocarbons.

Answer:

Question 27.
Draw the structures of isomers of butane.
Answer:
Isomers of butane are n-butane, iso butane and cyclo butane :
Structures :

Question 28.
Draw the structures of the following.
a) Ethanoic acid
b) Propanal
c) Propene
d) Chloro propene
Answer:
Structures:

Question 29.
Draw the structures of the following compounds
a) 2 – bromo pentane
b) 2 – methyl propane
c) butanal
d) 1 – hexyne
Answer:

Question 30.
Write the molecular formula of the first four compounds of the homologous series of aldehydes.
Answer:
Homologous series of aldehydes ate Formaldehyde, Acetaldehyde, Propionaldehyde and Butanaldehyde.

Question 31.
How many isomers can be drawn for pentane with molecular formula C-H(2? What are they? Draw their structures and mention theii common names.
Answer:
Isomers of pentane are three. These are
1) Pentane
2) Iso pentane
3) Neo pentane.
Structures :

Question 32.
Draw the Allotropes of Carbon. Diamond
Answer:

Question 33.
Draw the Graphite.
Answer:

Question 34.
Draw the Buckminsterfullerene (₆₀C).
Answer:

Question 35.
Draw the Nanotubes. A.
Answer:

Question 36.
Draw the structures of Methane :
Answer:
Methane :

Question 37.
Draw the structures of Ethyne :
Answer:

Question 38.
Draw structures of the Ethane and electron dot structure of Chlorine.
Answer:
Ethane:

Chlorine:

Question 39.
Draw the electron dot structures of Ethanoic acid arid Ethyne (Acetylene).
Answer:
Ethanoic acid (Acetic acid) :

Structure of Ethyne (Acetylene) :

Question 40.
Draw the electronic dot structure of ethane molecule.
Answer:

Question 41.
Write the structures of the following compounds.
a) prop-l-ene
b) 2, 3-dimethyl butane
c) 3-hexene
d) 2-methyl prop-l-ene
Answer:

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 13 Principles of Metallurgy.

AP State Syllabus SSC 10th Class Chemistry Important Questions 13th Principles of Metallurgy

10th Class Chemistry 13th Lesson Principles of Metallurgy 1 Mark Important Questions and Answers

Question 1.
Which method is suitable to enrich sulphide ores? (AP June 2016)
Answer:
Froth flotation method is suitable to enrich sulphide ores.

Question 2.
We use P.V.C. pipes for water supply instead of metal pipes. Why? (AP March 2017)
Answer:
PVC pipes do not rust. So they are used as water pipes instead of metal.

Question 3.
Arrange the metals Fe, Na, Ag and Zn in increasing order of their chemical reactivity. (TS March 2017)
Answer:
Ag < Fe < Zn < Na
(OR)
Ag, Fe, Zn, Na

Question 4.
Write the deferences between Roasting and Calcination. (TS June 2018)
Answer:
1) Burning of ore in the presence of air or oxygen is called “Roasting”.
So in the roasting air is present.

2) Burning of ore in the absence of air or oxygen is called “Calcination.”
So in the calcination air is absent.

Question 5.
What are the preventive methods do you take for rusting iron materials? (TS March 2018)
Answer:

1. Covering the surface of iron materials with paint or by some chemicals.
2. Electroplating.

Question 6.
Mention the application of thermite process in daily life. (AP SCERT: 2019-20)
Answer:
1) Joining railing of railway tracks,
2) Joining cracked machine parts.

Question 7.
What are the essential condition that iron articles get rust? (TS June 2019)
Answer:
The essential condition that iron articles get rust is presence of water and air both.

Question 8.
What is metallurgy?
Answer:
The process of extraction of metals from their ores is called metallurgy.

Question 9.
What is the Bronze an alloy of?
Answer:
Bronze is an alloy of copper and tin.

Question 10.
What are ores?
Answer:
The minerals from which the metals are extracted without economical loss are called ores.

Question 11.
What is the percentage of Aluminium oxide in Bauxite?
Answer:
50-70%.

Question 12.
Why is 16th group called chalcogen family?
Answer:
Chaleo means ore and genus means produce. We notice that the ores of many metals are oxides and sulphides. This is why oxygen-sulphur (16th group) group as chalcogen family.

Question 13.
Which metals form oxides, sulphides and carbonates?
Answer:
Moderate reactive metals.

Question 14.
Based on the reactivity arrange the metals.
Answer:
Based on reactivity we can arrange metals in descending order of their reactivity as shown below:

Question 15.
What is gangue?
Answer:
The unwanted material in the ore is called gangue.

Question 16.
What is activity series?
Answer:
Arrangement of the metals in decreasing order of their reactivity is known as activity series.

Question 17.
Why do we add some impurities to ore?
Answer:
We add some suitable impurities to ore in order to decrease its melting point.

Question 18.
What is roasting?
Answer:
Roasting is a pyrochemical process in which ore is heated in the presence of oxygen or air below its melting point.

Question 19.
What is thermite process?
Answer:
The reaction of metal oxides with aluminium is called thermite process.

Question 20.
Write the chemical equations involving thermite reaction.
Answer:
2 Al + Fe₂O₃ → Al₂O₃ + 2 Fe + heat
2 Al + Cr₂O₃ → Al₂O₃ + 2 Cr + heat

Question 21.
How do you convert cinnabar into mercury?
Answer:
When cinnabar (HgS) is heated in air, it is first converted into (HgO), then reduced to mercury on further heating.

Question 22.
What is distillation of metals?
Answer:
The extracted metal in the molten state is distilled to obtain the pure metal as distillate by distillation of metals. Here the impurities are high boiling point metals.

Question 23.
What is poling?
Answer:
The molten metal is stirred with logs (poles) of greenwood and impurities are removed either as gases or they get oxidized and form slag over surface of the molten metal is called poling.

Question 24.
What is liquation?
Answer:
A low melting metal can be made to flow on a slope surface to separate it from high melting impurities is called liquation.

Question 25.
What is calcination?
Answer:
Calcination is a pyrochemical process in which the ore is heated in the absence of air.

Question 26.
What is flux?
Answer:
Flux is a substance added to the ore to remove the gangue from it by reacting with ore.

Question 27.
What are the ores of iron?
Answer:
Haematite (Fe₂O₃), Magnetite (Fe₃O₄).

Question 28.
What are the ores of zinc?
Answer:
Zinc blende (ZnS), Zincite (ZnO).

Question 29.
What is the formula of gypsum and metal present in gypsum?
Answer:
The formula of gypsum is CaSO₄.2H₂O. The metal present in gypsum is calcium.

Question 30.
What is a furnace?
Answer:
The furnace is one which is used to carry out pyrochemical process in metallurgy.

Question 31.
Arrange the following chlorides in ascending order of reactivity of respective metals. MgCl₂, NaCl, PbCl₂, HgCl₂.
Answer:
The ascending order is HgCl₂, PbCl₂, MgCl₂, NaCl.

Question 32.
What are the fuel and flux for haematite ore?
Answer:
The coke is used as fuel and limestone (CaCO₃) is used as flux for haematite ore.

Question 33.
Why can copper not displace zinc from its compound?
Answer:
Copper is less reactive than zinc. So copper cannot displace zinc from its compound or salt.

Question 34.
How do various metals in activity series react with chlorine on heating?
Answer:

1. All the metals react with chlorine on heating to form their respective chlorides.
2. But the reactivity decreases from top to bottom.

Question 35.
How do you know the reactivity of metals with chlorine decreases from top to bottom?
Answer:
We know that the reactivity of metals with chlorine decreases from top to bottom the heat evolved when the metal reacts with one mole of chlorine gas to form chloride.

Question 36.
Give some examples for corrosion.
Answer:
Examples for corrosion :

1. The rusting of iron (Iron oxide)
2. Tarnishing of silver (Silver sulphide)
3. Development of green coating on copper (Copper carbonate) and bronze.

Question 37.
Why are potassium, sodium, calcium never found in free state?
Answer:
The metals potassium, sodium and calcium are so reactive that is why they never exist in free state.

Question 38.
How do you extract metals at the top of activity series?
Answer:
The metals at the top of activity series are extracted by electrolysis of their fused compounds.

Question 39.
What is meant by enrichment of ore?
Answer:
Some physical methods are useful in removing unwanted rocky material from ore. It is called enrichment of ore.

Question 40.
How do you extract metal from the crude metal?
Answer:
To extract metal from enriched ore it is converted into metallic oxide by reduction reaction. Then this metallic oxide is further reduced to get metal with certain impurities.

Question 41.
What are the impurities you get in the refining of copper?
Answer:
Antimony, selenium, tellurium, silver, gold and platinum.

Question 42.
What is slag?
Answer:
The substance formed due to reaction of gangue and flux,
Eg : CaSiO₃, FeSiO₃

Question 43.
What is meant by pyrochemical reactions?
Answer:
Pyre means heat. So the chemical reactions involving heat are called pyrochemical reactions.

Question 44.
Can you mention some articles that are made up of metals?
Answer:
Jewellery, conducting wires and utensils.

Question 45.
Why are we mixing small amount of carbon to iron?
Answer:
To make iron hard and strong.

Question 46.
What is the main difference between steel and stainless steel?
Answer:
Steel will rust whereas stainless steel will not rust.

Question 47.
Give some examples for corrosion.
Answer:
The rusting of iron, tarnishing of silver, development of green coating on copper.

Question 48.
Do metals exist in the same form as that we use in our daily life?
Answer:
No, they exist as ores and minerals and some may exist in the form of metals.

Question 49.
Do you know how metals are obtained?
Answer:
The metals are extracted from their ores.

Question 50.
Have you ever heard the words like ore, mineral and metallurgy?
Answer:
Yes, these words are related to extraction of metal.

Question 51.
Where do we carry out pyrochemical processes in metallurgy?
Answer:
Pyrochemical processes can be carried out inside the furnace.

Question 52.
Which process converts sulphide ore into oxide ore?
Answer:
Roasting is the process which converts sulphide ore into oxide ore.

Question 53.
Is silver mineral or ore? Justify your answer.
Answer:
Silver is neither mineral nor ore. It is a metal.

Question 54.
Give two examples for corrosion.
Answer:
1) Rusting of iron
2) Green coating on copper.

Question 55.
Name the form of carbon used in the blast furnace for the extraction of iron.
Answer:
Carbon is used in the form of coke to reduce iron in blast furnace.

Question 56.
Give name and formulae of sulphide ore of lead and mercury.
Answer:
a) Sulphide ore of lead is Galena. Its formula is PbS.
b) Sulphide ore of mercury is Cinnabar. Its formula is HgS.

Question 57.
What are the various pyrochemical processes used in metallurgy?
Answer:
The various pyrochemical processes used are
a) Smelting,
b) Roasting,
c) Calcination.

Question 58.
What is the gas released at anode when fused sodium chloride is electrolysed?
Answer:
When sodium chloride is electrolysed, sodium metal is formed at cathode and chlorine gas is formed at anode.
NaCl → Na⁺ + Cl^–

Question 59.
Which pyrochemical process is useful to convert zinc blende into oxide ore?
Answer:
Zinc blende is sulphide ore of zinc. Its formula is ZnS. So it can be converted into oxide ore by heating strongly in excess of air known as roasting.
2 ZnS + 3 O₂ → 2 ZnO + 2 SO₂

Question 60.
Which pyrochemical process is useful to convert Magnesite into oxide ore?
Answer:
Magnesite is carbonate ore of magnesium. Its formula is MgCO₃. So it can be converted into oxide ore by heating in the absence of air.
MgCO₃ → MgO + CO₂

Question 61.
What is the main impurity present in iron when it is removed from the blast furnace?
Answer:
The main impurity that can be removed is slag because it is formed when gangue in the ore reacts with flux.
CaO + SiO₂ → CaSiO₃

Question 62.
Name two metals normally manufactured by the electrolysis of fused compounds.
Answer:
Metals with high reactivity can be extracted by electrolysis of their fused compounds.
The examples are potassium, sodium and calcium.

Question 63.
What are the examples of corrosion?
Answer:

1. Rusting of iron.
2. Tarnishing of silver.
3. Development of green coating on copper (CuCO₃).
4. Green coating on Bronze.

Question 64.
What is importance of prevention of corrosion?
Answer:

1. Save the money.
2. Preventing accidents such as a bridge collapse.
3. Failure of a key component.

Question 65.
What is meant by galvanisation?
Answer:
Preventing the rust on metals by using layer of zinc. This phenomena is called galvanisation.

Question 66.
Write the example of electroplating in daily life.
Answer:

1. Rold gold.
2. Copper coating on cookware.

Question 67.
What is formula and name of iron rust?
Answer:
Iron rust is equal to hydrated ferric oxide.
Formula : Fe₂O₃ × H₂O

Question 68.
What is importance of alloying?
Answer:

1. Improving the properties of metal.
2. To avoid the rust.
3. To increase the hardness.

Question 69.
Which one used as flux extracting of iron from heamatite?
Answer:
Limestone or calcium carbonate (CaCO₃).

Question 70.
Marne the two metals which corrode easily?
Answer:
Iron and copper.

Question 71.
Atmospheric air always contains moisture. Then, how can you protect iron articles from the affect of atmosphere?
Answer:
By painting, oiling and greasing, etc.

Question 72.
Explain the terms gangue and flux.
Answer:
The impurity present in the ore is called gangue. The substance added to the ore to remove gangue from it is called flux.

Question 73.
What are the metals are present in carnallite?
Answer:
Potassium (K) and Magnesium (Mg).

Question 74.
Write the elements are present in high reactivity series.
Answer:
Na, Mg, Al, K, Ca
(11 12 13 19 20)

Question 75.
Write the elements that are in moderate reactivity series.
Answer:
Fe, Cu, Zn, Pb.

Question 76.
Name the two metals which do not corrode easily?
Answer:
Gold and platinum.

Question 77.
Mention some important methods of refining.
a) Distillation
b) Poling
c) Liquation
d) Electrolysis
Answer:
d) Electrolysis

Question 78.
What is the role of furnace in metallurgy?
Answer:
Furnace is the one which is used to carry out pyrochemical process in metallurgy.

Question 79.
What is meant by calcination?
Answer:
It is the process of heating the concentrated ore in the absence of air.

Question 80.
Write the equation of heating of one sulphide ore in the process of roasting.
Answer:

Question 81.
Mention two methods which produce very pure metals?
Answer:
a) Electrolytic reduction.
b) Smelting.

Question 82.
What are the applications of thermite reaction in daily life?
Answer:
a) To join railings of railway tracks.
b) To join cracked machine parts.

Question 83.
Arrange the metals Ag, Mg, K in reactivity series.
Answer:
K > Mg > Ag.

Question 84.
How do you extract highly reactive metals?
Answer:
Highlyreactive metals can be extracted by electrolysis of their fused compounds.

Question 85.
What is dressing of an ore?
Answer:
The process of removal of impurities from an ore is called dressing of the ore or concentration of the ore.

Question 86.
Write the equation of example of calcination.
Answer:

Question 87.
Write the some properties of metals.
Answer:
Malleability, Ductility, Sonarity and Electrical conductivity.

Question 88.
Mention the stages involved in extraction of a metal from its ore.
Answer:

1. Dressing or concentration.
2. Extraction of crude metal.
3. Refining or purification of the metal.

Question 89.
How do you extract moderately reactive metals?
Answer:
These metals are generally sulphides and carbonates. They are converted into oxides before reducing them to metals.

Question 90.
Give an example for reduction of metal oxide with carbon.
Answer:
The oxides are reduced by coke in a furnace which gives the metal and carbon monoxide.

Question 91.
Give an example for reduction of oxide ore with CO.
Answer:

Question 92.
What is flux?
Answer:
Flux is a substance added to the ore to remove the gangue from it by reacting with ore. If the impurity is acidic substance, basic substance is used as flux and vice – versa.

Question 93.
How do various metals in activity series react with chlorine on heating?
Answer:

1. All the metals react with chlorine on heating to form their respective chlorides.
2. But the reactivity decreases from top to bottom.

Question 94.
What are the substances to be added if the gangue is acidic or basic?
Answer:
If the gangue (impurity) is acidic substance like SiO₂, basic substance like CaO is used as flux and if the impurity is of basic nature like FeO acidic flux like SiO₂ is added to the gangue.

Question 95.
Why will stainless steel not rust?
Answer:
Stainless steel is prepared by mixing iron with nickel and chromium. Nickel and chromium are less reactive with oxygen. So stainless steel will not rust.

Question 96.
Why is sodium metal stored in kerosene?
Answer:
Sodium is highly reactive with both air (oxygen) and water. So it should be stored in kerosene.

Question 97.
Which metal gets covered with protective film of oxide when exposed to air?
Answer:
The metal is aluminium. When aluminium is exposed to air it forms a protective layer of aluminium oxide (Al₂O₃).

Question 98.
All ores are minerals, but all minerals need not be ores. Why?
Answer:
A mineral from which a metal can be extracted and economical loss is called ore.

Question 99.
Why is carbon not used for reducing aluminium from aluminium oxide?
Answer:
The oxide of Aluminium is very stable and can be reduced by electrolytic process.

Question 100.
Name few metals which occur in native state in nature. Why?
Answer:
Gold, Platinum, Silver are the metals which occur in native state, because of their low chemical reactivity.

Question 101.
Why do we call oxygen – sulphur group is chalcogen family?
Answer:
Chaleo means ore. We know that most of ores of many metals are oxides and sulphides. That’s why oxygen – sulphur group is called chalcogen family.

Question 102.
Aluminium occurs in combined state in nature whereas gold is in free state. Why?
Answer:
Gold has low reactivity and so occurs in free state. Aluminium is electropositive metal and high reactivity. So it is oxide or chloride.

Question 103.
What are the uses of thermite reaction?
Answer:
Thermite reaction is used to join railings of railway tracks or cracked machine parts.

10th Class Chemistry 13th Lesson Principles of Metallurgy 2 Marks Important Questions and Answers

Question 1.
Define mineral. Mention any two ores of ‘magnesium’. (AP June 2017)
Answer:
1) Minerals :
The elements or compounds of the metals which occur in nature in the earth crust are called ‘minerals’.

2) Two ores of magnesium :
Magnesite – MgCO₃
Carnalite – KCl MgCl₂ 6H₂O

Question 2.
Potassium, Sodium, Magnesium are high reactive metals and occur as chlorides in nature. Suggest and explain the suitable method for the extraction of the above metals from their ores. (AP March 2017)
Answer:

• The suitable method to extract these metals from their chlorides is electrolysis of their fused compounds.
• It is not feasible for method of reduction, electrolysis of their aqueous solutions.

Question 3.
Predict, what happens in the field of domestic use of metals if alloys were not discovered. (TS June 2016)
Answer:
If alloys were not discovered,

1. All the vessels/utensils made of single metal like iron, copper, aluminium, etc. may be used for cooking purpose.
2. We may face problems like rusting of iron, tarnishing of silver and copper, etc.
3. We may face the problems of corrosion of home appliances.
4. We may face difficulties in cleaning of the vessels due to rusting and tarnishing.
5. Cost of the utensils may be risen, because of less availability of the metals like copper.
6. Using of the plastic ware may be risen for storage due to lack of steel containers.
7. Brass, steel, bronze, etc. utensils are not available to use.
8. Making of jewellary is also difficult.

Question 4.
Give an example with the chemical equation for the reduction of ores using more reactive metals. (TS March 2017)
Answer:
The reaction of Iron oxide with aluminium.
Fe₂O₃ + 2Al → 2Fe + Al₂O₃ + Heat
(Or)
Reaction of Titanium Chloride with Magnesium.
TiCl₄ + 2Mg → Ti + 2MgCl₂
(Or)
Reaction of Titanium Chloride with Sodium.
HCl₄ + 4Na → Ti + 4NaCl
(Or)
Reaction of Cromium oxide with aluminium.
Cr₂O₃ + 2Al → 2Cr + Al₂O₃ + Heat

Question 5.
Write two precautions to prevent corrosion of metals in your daily life. (TS June 2018)
Answer:
Precautions to be taken to prevent corrosion of metals.
i) Painting the metals.
ii) By keeping the metals in the dry places.
iii) Cover the surface by other metals that are inert or non reactive to the atmosphere.
iv) Applying oil/grease to the metals.
v) Making of alloys.

Question 6.

High reactivity	Moderate reactivity	Low reactivity
K, Na, Ca, Mg, Al	Zn, Fe, Pb, Cu	Ag, Au

Observe the table and answer the following questions. 4jt*y (June 2019
i) Which of the above metals found even in free state in nature ?
ii) Which of the above metal’s ore are concentrated by using magnetic separation?
Answer:
i) Ag, Au.
ii) Fe.

Question 7.
Silicon is a metalloid. How do you support this?
Answer:
Silicon exhibits following properties, so I conclude that it is a metalloid.

1. It is metallic lustre in nature.
2. It exists in several metallic and non-metallic compounds.
3. It has brittle nature.
4. All metalloids usually occur in combined states both metals and non-metals.

Question 8.
Explain the reaction of various metals in activity with cold water.
Answer:
1) From potassium to magnesium displace hydrogen from cold water with decreasing reactivity. Potassium reacts with cold water violently but reaction of Magnesium is very slow. The reactivity order is given below.
Mg < Ca < Na < K

2) From aluminium to gold do not displace hydrogen from cold water.

Question 9.
How do various metals in activity series react with steam?
Answer:

• The metals from potassium to iron displace H₂ (Hydrogen gas) from steam with decreasing reactivity. That means the reaction of potassium with steam is voilent but the reaction of iron is very slow.
• The metals from lead to gold do not displace hydrogen from steam.

Question 10.
How do various metals in activity series react with dilute strong acids?
Answer:
1) The metals from potassium to lead displace hydrogen from dilute strong acids with decreasing reactivity.
a) The reaction of potassium is explosive.
b) The reaction of magnesium is vigorous.
c) The reaction of iron is steady.
d) The reaction of lead is slow.

2) The metals from copper to gold do not displace H₂ from strong dilute acids.

Question 11.
What are the preventive techniques used in corrosion of metals?
Answer:
Prevention of corrosion of metals :

• Covering the surface of metal with paint or by some chemicals like bisphenol which prevent the surface of metallic object to come in contact with atmosphere.
• Covering the surface of metal by other metals like tin or zinc that are inert or react themselves with atmosphere to save the metal.
• An electrochemical method in which a sacrificial electrode of another metal like magnesium and zinc, etc. corrodes itself to save the metal.

Question 12.
What are the chemical reactions that take place inside blast furnace?
Answer:
The chemical reactions that take place inside the blast furnace.

Question 13.
What are the various types of furnaces? Explain.
Various types of furnaces :
1) Blast furnace:
Blast furnace has both fire box and hearth combined in big chamber which accommodates both ore and fuel.

2) Reverberatory furnace :
It has both fire box and hearth separated, but the vapours (flame) obtained due to burning of the fuel touch the ore in the hearth and heat it.

3) Retort furnace :
In this furnace there is no direct contact between the hearth or fire box and even the flames do not touch the ore.

Question 14.
Why is alloying preferred for metals? Explain with examples.
Answer:

• Alloying is a method of improving properties of a metal. We can get desired properties by this method.
• For example, iron is the most widely used metal. But it is never used in its pure state.
• This is because pure iron is very soft and stretches easily when hot.
• But, if it is mixed with a small amount of carbon, it becomes hard and strong.
• When iron is mixed with nickel and chromium we get stainless steel which will not rust.

Question 15.
What is 22 carat gold? Why is it preferred for making jewellery?
Answer:

• Pure gold, known as 24 carat gold is very soft.
• So it is not suitable for making jewellery.
• It is alloyed with either silver or copper to make it hard.
• So they use 22 carat gold in which pure gold is alloyed with 2 parts of either silver or copper for making gold jewellery.

Question 16.
Write about electrolysis of NaCl.
Answer:
1) Fused NaCl is electrolysed with steel cathode and graphite anode.

2) The metal sodium (Na) will be deposited at cathode and chlorine gas liberates at the anode.
At Cathode : 2 Na⁺ + 2e^– → 2 Na
At Anode : 2 Cl^– → Cl₂ + 2e^–

Question 17.
Identify the metal present in the following ores.
i) Epsom Salt
ii) Horn Silver
iii) Cinnabar
iv) Galena
Answer:
i) Magnesium
ii) Silver
iii) Mercury
iv) Lead

Question 18.
What is meant by extraction of metals? Write the main stages of extraction of metals from its Ore.
Answer:
Separation of metals from ores is called extraction of metals. Extraction of metals involves mainly three stages.

1. Concentration or dressing
2. Extraction of crude metal
3. Refining or purification of the metal.

Question 19.
Write differences between roasting and calcination.
Answer:

Roasting	Calcination
1. Ore is heated in the presence of oxygen or air.	1 Ore is heated in the absence of air.
2. Sulphide ore is converted into oxide ore.	2. Carbonate ore is converted into oxide ore.

Question 20.
What are the differences between minerals and ores?
Answer:

Minerals	Ores
1) Minerals contain a low percentage of metal.	1) Ores contain a large percent of metal.
2) Metals cannot be extracted from minerals.	2) Ores can be used for the extraction of metals.
3) All minerals cannot be called ores.	3) All ores are minerals.

Question 21.
What are the different types of reduction?
Answer:
The different types of reduction are
a) Chemical reduction,
b)Auto reduction,
c) Electrolytic reduction.

Question 22.
How do potassium and sodium react with oxygen?
Answer:
a) Potassium and sodium form oxides of type M₂O in limited supply of oxygen.
4 K + O₂ → 2 K₂O
4 Na + O₂ → 2 Na₂O

b) In excess of oxygen they form peroxides of type M₂O₂.
2 Na + O₂ → Na₂O₂
2 K + O₂ → K₂O₂

Question 23.
How does reactivity of chlorine vary in the reactivity series?
Answer:

1. All metals react with chlorine on heating to form their respective chlorides but with decreasing reactivity in the reactivity series.
2. This is understood from the heat evolved when metal reacts with one mole of chlorine gas to form chloride.

Question 24.
Name two ores of calcium and give their formulae.
Answer:
The ores of calcium are

1. Gypsum (CaSO₄ • 2H₂O)
2. Limestone (CaCO₃)

Question 25.
Which method is useful to separate sand from iron? Explain.
Answer:

• Sand can be separated from iron by using magnetic separation method.
• This can be done by using electromagnet. Iron being a magnetic material is attracted by electromagnet whereas sand is not attracted by electromagnet.
• So these materials are separated.

Question 26.
Which metals do not displace hydrogen from dilute strong acids?
Answer:

• Copper, mercury, silver, platinum, gold do not displace hydrogen from dilute strong acids like HCl, H₂SO₄, etc.
• The reactivity of these metals are less than hydrogen. So, they are unable to displace hydrogen from dilute acids.

Question 27.
Which metals are not found in free state? Why?
Answer:

• The metals like potassium, sodium, calcium, magnesium and aluminium are never found in free state in nature.
• The reason is that these metals have high reactivity. So, they exist as compounds.

Question 28.
Why do silver and gold exist even in free state?
Answer:

• Silver and gold are least reactivity metals. So, they are also called noble metals.
• Due to least reactivity they are unable to react with other elements.

Question 29.
How do moderately reactive metals occur in nature?
Answer:

• The metals like zinc, iron, lead are moderately reactive.
• They are found in the earth’s crust mainly as oxides, sulphides and carbonates.

Question 30.
Mention the most important metals and non-metals from the following products.
a) Annapurna salt
b) Liquid used in thermometer
c) Lead of the pencil
d) Chlorophyll
e) Filament in electric bulb
f) Enamel layer on teeth
Answer:
a) Annapurna salt : Iodine, chlorine – Non-metals
b) Liquid used in thermometer : Mercury – Metal
c) Lead of the pencil : Graphite – Non-metal
d) Chlorophyll : Magnesium – Metal
e) Filament in electric bulb : Tungsten – Metal
f) Enamel layer on teeth : Calcium phosphate – Non-metal

Question 31.
What is a furnace? Explain various parts of furnace.
Answer:
Furnace :
Furnace is the one which is used to cany out pyrochemical processes in metallurgy.

Furnace has mainly three parts :
1) Hearth :
Hearth is the place inside the furnace where the ore is kept for heating.

2) Chimney:
Chimney is the outlet through which flue (waste) gases go out of the furnace.

3) Fire box :
Fire box is the part of the furnace where the fuel is kept for burning.

10th Class Chemistry 13th Lesson Principles of Metallurgy 4 Marks Important Questions and Answers

Question 1.
What is a furnace? Draw Reverberatory furnace and label it parts. (AP March 2018)
Answer:
1) Furnace :
Furnace is the one which is used to carry out pyrochemical processes in metallurgy.
2) Diagram of Reverberatory furnace.

Question 2.
What are the various techniques used in purification of the crude metals? Explain.
(OR)
State the methods used for the purification of crude metals. Explain in which context these methods are used. (TS June 2015)
Answer:
1) The process of obtaining the pure metal from the impure metal is called refining of the metal.
2) Some of the processes of refining are
i) Distillation
ii) Poling
iii) Liquation
iv) Electrolytic refining.

3) The process that has to be adopted for purification of a given metal depends on the nature of the metal and its impurities.

Various methods adopted in purification of metals :
1) Distillation :
This method is very useful for purification of low boiling metals like zinc and mercury containing high boiling metals as impurities. The extracted metal in the molten state is distilled to obtain the pure metal as distillate.

2) Poling :
The molten metal is stirred with logs (poles) of greenwood. The impurities are removed either as gases or they get oxidized and form slag (Scum) over the surface of molten metal.

3) Liquation :
Low melting metal like tin can be made to flow on a slopey surface to separate it from high melting impurities.

4) Electrolytic refining :

1. In this method, the impure metal is made to act as anode.
2. A strip of the same metal in pure form is used as cathode.
3. They are put in a suitable electrolytic bath containing soluble salt of the same metal.
4. The required metal gets deposited on the cathode in the pure form.
5. The metal constituting impurity, goes as the anode mud.

The reactions are :
At anode : M → Mn⁺ + ne^–
At cathode : Mn⁺ + ne^– → M. ; (M = pure metal, n = 1, 2, 3, …….)

Question 3.
Four metals A, B, C and D are in turn added to the following solutions one by one. The observations made are tabulated below. (TS March 2015)

Answer the following based on the given information.
i) Which is the most reactive metal? Why?
ii) What would be observed, if ‘B’ is added to a solution of Copper (II) sulphate and why?
iii) Arrange the metals A, B, C and D in order of increasing reactivity.
iv) Which one among A, B, C and D metals can be used to make containers that can be used to store any of the above solutions safely?
Answer:
i) Metal ‘B’ is more reactive.
– Metal ‘B’ is replacing iron from iron sulphate.
– Metal ‘A’ is replacing copper from copper sulphate.
– Metal ‘C’ is replacing silver from silver nitrate.

ii) Metal B displaces Cu from CuS04 solution. Because metal B is more reactive than Cu.

iii) D < C < A < B.

iv) The container made up of metal D can be used to store any solution mentioned above.

Question 4.
Write the physical methods used for the concentration of the ore. Explain the method used for concentration of the sulphide ore. (TS June 2017)
Answer:
Physical methods used for the concentration of the ore is,
i) Hand Picking
ii) Washing
iii) Froth floatation
iv) Magnetic Separation.

Concentration of sulphide ore :

• Sulphide ore is concentrated by using froth floatation Method.
• The ore with impurities is tinely powdered and kept in water taken in a flotation cell.
• Air under pressure is blown to produce froth in water.
• Froth so produced, taken the ore particles to the surface whereas impurities settle at the buttom.
• Froth is separated and washed to get ore particles.

Question 5.
Draw a neat diagram of froth floatation process for the concentration of sulphide ore why we add pine oil to the mixture in this process? (AP SCERT 7201 9-20)
Answer:

Froth floatation process for the concentration of sulphide ores

1.  The mineral particles in the ore are preferentially wetted by the oil and float on the top of the froath.
2. The gangue particles are wetted by water and settle down.
3. Thus, the minerals can be separated from the gangue by adding pine oil.

Question 6.
Describe the reaction of various metals in activity series with oxygen.
Answer:

• The metals which are at the bottom of activity series have very low reactivity and do not burn or oxidase even on surface.
  Eg : Ag, Pt, Au.
• The metals like Pb, Cu and Hg do not burn but only form a surface layer of oxide, i.e., PbO, CuO, HgO.
• The metals like Al, Zn, Fe react with oxygen to form respective oxides.
• The metals like Ca and Mg burn with decreasing vigorousity to form oxides.
• The metals like K, Na burn vigorously to form Na₂O, K₂O in limited supply of oxygen but form peroxides in excess of oxygen.

Question 7.
How do you reduce purified ore to the metal of the top of activity series? Explain.
Answer:
The reduction of ore to particular metal mainly depends on the position of metal in the activity series.

Extraction of metals at the top of activity series :

1. Simple chemical reduction methods like heating C, CO, etc. to reduce the ores of the metals are not possible with metals like K, Na, Ca, Mg and Al.
2. The temperature required for the reduction is too high and more expensive.
3. The only method available is to extract these metals by electrolysis of their fused compounds.

Question 8.
How do you extract metals in the middle of activity series?
Answer:
Extraction of metals in the middle of the activity series :
The ores of these metals are generally present as sulphides or carbonates. Therefore prior to reduction of ores of these metals, they must be converted into metal oxides.

The metal oxides are then reduced to the corresponding metals by using the following methods :
1) Reduction of metal oxides with carbon :
The oxides are reduced by coke in closed furnace which gives the metal and carbon monoxide (CO).

2) Reduction of oxide ores with CO :

3) Auto (Self) reduction of sulphide ores:
In the extraction of copper from its sulphide ore, the ore is subjected to partial roasting in air to give its oxide.
2 Cu₂S + 3O₂ → 2 Cu₂O + 2SO₂

When the supply of air is stopped and temperature is raised, it results in the reaction of rest of the sulphide ore with oxide to form metal and S02. ‘
2 Cu₂O + Cu₂S → 6 Cu + SO₂

4) Reduction of ores (compounds) by more reactive metals :
When highly reactive metals such as sodium, calcium, aluminium, etc. are used as reducing agents, they displace metals of low reactivity from the compound.

Question 9.
How do you extract metals at the bottom of the activity series?
Answer:
1) Metals at the bottom of the activity series are often found in free state.

2) The oxides of these metals can be reduced to metals by heat alone and sometimes by displacement from their aqueous solutions.

3) When cinnabar (HgS) is heated in air, it is converted into HgO, then reduced to mercury on further heating.

4) Displacement from aqueous solution :
When Ag₂S is dissolved in KCN solution, it forms dicyanoargentate ions. When these ions are treated with Zn dust powder then Ag is precipitated.
Eg : Ag₂S + 4 CN^– → 2 [Ag(CN)₂]^– + S^2-
2 [Ag(CN)₂]^–_(aq) + Zn_(s) → [Zn(CN)₄]^2-_(aq) + 2 Ag_(s)

Question 10.
Explain the process involved in corrosion.
Answer:
1) Corrosion is an electrochemical phenomenon.
2) In corrosion, a metal is oxidised by loss of electrons generally to oxygen and results in the formation of oxides.
3) During corrosion at a particular spot on the surface of an object made of iron, oxidation takes place and that spot behaves as anode.
Anode : 2 Fe_(s) → 2 Fe²⁺ + 4e^–
4) Electrons released at this anodic spot move through the metal and go to another spot and reduce oxygen at that spot in the presence of H+.

5) This spot behaves as cathode.
Cathode : O_2(g) + 4 H⁺_(aq) + 4e^– → 2H₂O_(l)
Net reaction : 2 Fe_(s) + O_2(g) + 4 H⁺_(aq) → 2 Fe²⁺_(aq) + 2 H₂O_(l)

6) The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide (Fe₂O₃ . XH₂O) and with further production of hydrogen ions.

Question 11.
Explain electrolytic refining with an example.
Answer:

• The impure metal is taken as anode and pure metal is taken as cathode.
• They are put in a suitable electrolytic bath containing soluble salts of same metal.
• The required metal gets deposited on the cathode in the pure form.
• The metal constituting the impurity goes as the anode mud.

Electrode reactions :
At Anode : M → Mⁿ⁺ + ne^–
At Cathode : Mⁿ⁺ + ne^– → M (M pure metal); where n = 1, 2, 3,………….

Examples :

1. In order to refine copper, impure copper is taken as anode and pure copper strips are taken as cathode.
2. The electrolyte is an acidified solution of copper sulphate.
3. As a result of electrolysis copper in pure form is transferred from anode to cathode.
   At Anode : Cu → Cu²⁺ + 2e^–
   At Cathode : Cu²⁺ + 2e^– → Cu
4. The soluble impurities go into the solution, whereas insoluble impurities from the blister copper deposited at the bottom of anode as anode mud.

Question 12.
What is activity series? Give two examples each of them.
i) Low reactivity metals
ii) Moderate reactivity metals
iii) High reactivity metals
Answer:
Arranging metals in descending order of their reactivity is called activity series,
e.g.:
i) Low reactivity : Ag (Silver), Au (Gold)
ii) Moderate reactivity : Zn (Zinc), Fe (Iron)
iii) High reactivity : K (Potassium), Na (sodium)

Question 13.
Write the balanced chemical equations, extraction of iron from haematite in the Blast furnace.
Answer:

Question 14.
How do you prevent corrosion of various metals?
Answer:
Prevention of corrosion :The corrosion can be prev ented by the following methods.
1) Barrier protection :
In this method the metal surface is not allowed to come in contact with moisture, oxygen and carbon dioxide. This can be achieved by the following.
a) By coating iron with oils, paints, coal tar, grease, pitch, etc.
b) By blowing steam over red hot iron to form protective coating of Fe₃O₄.
c) By alloying iron with Ni, Cr, Si, etc.

2) Sacrificial protection:
Sacrificial protection means covering the iron surface with a layer of metal which is more electropositive than iron thus prevents iron from losing electrons. It is done by following methods.
a) By galvanisation (by dipping iron in a bath of molten zinc).
b) By tinning (by dipping iron in molten tin).
c) By the coating of copper.
d) Decorative coating : By using Zn, Mg and A/ powders mixed with paints.

3) Electrical protection:
In this method, the iron object to be protected from corrosion is connected to more active metal eg. magnesium, zinc or aluminium directly or through a wire. The iron object acts as cathode and the protective metal acts as anode. The anode is gradually used up to the oxidation of metal to its ions due to loss of electrons. Hydrogen ions collect at cathode and prevent rust formation.

4) Using anti-rust solution:
On applying alkyl phosphates and alkyl chromates to the iron objects corrosion can be prevented.

Question 15.
What are the salient features of the activity series?
Answer:
Salient features:

1. Any metal which is placed higher up in the series can displace any metal below it in order to from the salt solution of the later metal.
2. The larger the difference in the position of metals in the series, the more rapidly does the displacement take place.
3. Metals which are placed above hydrogen in the series have the ability to reduce ions from dilute sulphuric acid to liberate hydrogen gas.
4. Oxides of metals K, Na, Ca and Mg cannot be reduced by H2, CO or C.
5. Oxides and nitrates of less reactive metals Hg, Ag and Au decompose to give metals on being strongly heated.
6. Metals below copper such as mercury, silver, platinum and gold do not rust easily.
7. Hydrogen though a non-metal, has been included in the series.
   It occupies the position based on its formation of positive ions.

Question 16.
How do you extract metals based on activity series?
Answer:

• Highly active metals like potassium, sodium, calcium, magnesium and aluminium are obtained by the electrolysis of their fused halides or oxides, that is, by electrolytic reduction because their oxides cannot be reduced by common reducing agents like carbon, carbon monoxide and hydrogen.
• Zinc is obtained only by heating its oxide with carbon.
• Iron, lead and copper are obtained by reduction of their oxides with carbon, carbon monoxide and hydrogen.
• Copper is obtained by reducing black copper oxide with carbon or by air reduction.
• Mercury and silver are obtained by heating their respective oxides to temperature above 300°C when they lose oxygen and are reduced to free metals.
• However, less active mercury can also be obtained by merely heating its sulphide in air.
• Silver and Gold are obtained by displacement from solutions containing their ions by more electropositive metal zinc.

Question 17.
X is an element in the form of a powder. X burns in oxygen and the product is soluble in water. The solution is tested with litmus.
Write down the answers for the following questions from the above information and give reasons.
i) If X is a metal, then which colour will litmus turn ?
ii) If X is a non-metal, then which colour will litmus turn ?
iii) If X is a reactive metal, what gas will be released with dilute sulphuric acid ?
Answer:
i) If X is a metal, then the litmus turns into blue because metal reacts with oxygen and forms metallic oxide and aqueous solution of metallic oxide ore basic in nature.

ii) Mf X is a non-metal, then the litmus turns into red because non-metal reacts with oxygen and forms non-metallic oxide and aqueous solution of non-metallic oxide ore acidic in nature.

iii) If X is a reactive metal, then it releases hydrogen gas from sulphuric acid because more reactive metal displaces hydrogen from acid.

Question 18.
Complete the missing statements and give reasons.
i) Metals are ……………………….., while non-metals are poor conductors of heat.
ii) Metals are malleable, while non-metals are ……………………….. .
iii) Metals form positive ions, while non-metals form ……………………….. .
iv) Non-metals form acidic oxides, while metals form ……………………….. .
Answer:
i) Good conductors.
Reason :
Metals containing free electrons are very good conductors of electricity whereas non-metals are bad conductors of electricity because they do not have free electrons.

ii) Non-malleable.
Reason :
Metals are hard. So, they can be beaten into sheets whereas non-metals are soft, so they are non-malleable.

iii) Negative ions.
Reason :
Metals are electropositive in nature. They easily lose electrons to form positive ions, whereas non-metals are electronegative in nature. So, they gain electrons to form negative ions.

iv) Basic oxides.
Reason :
Non-metallic oxide solutions turn blue litmus into red. They are acidic in nature. So, they are called acidic oxides whereas metallic oxide solutions turn red litmus into blue. They are basic in nature. So, they are called basic oxides.

Question 19.
Answer the following questions.
a) i) Name two naturally occurring compounds of zinc other than carbonate and give their formulae.
ii) Give equations for the extraction of zinc from zinc carbonate.
b) Write equations for the action of zinc on the following.
i) dil. H₂SO₄
ii) Copper (II) sulphate solution.
Answer:
a) i) The ores of zinc other than carbonate ore are zinc blende (ZnS) and Zincite (ZnO).

b) i) Zn + dil. H₂SO₄ → ZnSO₄ + H₂
Zr_(s) + CuSO_4(aq) → ZnSO_4(aq) + Cu_(s)

Question 20.
i) The ore zinc blende is an important source of the metal zinc. What is the name of zinc compound in zinc blende?
ii) What is the compound obtained by roasting zinc blende?
iii) What is the type of chemical reaction carried out after roasting in order to obtain zinc?
iv) What is the name of the alloy formed between zinc and copper?
Answer:
i) The zinc compound in zinc blende is ZnS (zinc sulphide).
ii) By roasting zinc blende it converts into zinc oxide.
2 ZnS + 3O₂ → 2 ZnO + 2 SO₂
iii) The chemical reaction carried out to convert zinc oxide to zinc metal is reduction in the presence of coke.

iv) The alloy of copper and zinc is bronze.

Question 21.
The basic material used for the production of iron in the blast furnace are limestone, coke and air in addition to iron ore.
a) Name one iron ore and write its formula.
b) Hot air is blown at the base of furnace where it reacts with coke. Give the chemical equations for the reactions that take place.
c) Higher up in the furnace the iron ore is reduced to iron by one of the gases produced in the furnace. Give the chemical equation for the reaction by which the gas is produced and give a balanced equation to show how the ore is reduced to iron.
d) Which compound produced from limestone takes part in forming the slag?
Answer:
a) The iron ore is Haematite (Fe₂O₃).
b) Coke bums partially to produce carbon monoxide gas.
2 C_(s) +O_2(g) → 2 C0_(g)

c) Iron oxide reacts with carbon monoxide gas and reduces to iron.
Fe₂O₃ + 3 CO → 2 Fe + 3 CO₂

d) Calcium carbonate (limestone) undergoes calcination to produce calcium oxide which takes part in the reaction to form slag.
CaCO_3(s) → CaO_(s) + CO₂
CaC_(s) + SiO_2(s) → CaSiO_3(l)

Question 22.
What information do you get from metal activity series given below.
K > Na > Ca > Mg > Al > Zn > Fe > Pb > [H] > Cu > Ag > Pt > Au
Answer:

• Metals below hydrogen [H] cannot displace hydrogen from acids and above hydrogen can displace hydrogen from acids.
• Metals which are higher in the series, can displace metals below it from the salt solution.
• The higher the position, the more active is the metal.
• Hydrogen has electropositive character, so it is placed among the metals.

Question 23.
The results of reactions of metals A, B, C, D, E with different solutions are given in the table below. Observe the table and write answers.

1) Which is the highly reactive metal? Why?
2) Which is the least reactive metal? Why?
3) Which metals form brown layer?
4) Arrange the metals A, B, C, D, E in the order of their reactivity?
5) Among these identify the silver, copper, iron, zinc and aluminium.
Answer:

1. Metal ‘E’ is more reactive among all the metals given because it displaces all the elements from the compounds given in the table,
2. Metal ‘C’ is the least reactive metal because it does not displace any other metal from the compounds given in the table.
3. Metals B and E will form brown layer.
4. The ascending order is as follows C < A < D < B < E.
5. C is silver, A is copper, D is iron, B is zinc and E is aluminium.

Question 24.
Draw the diagram of blast furnace and label its parts.
Answer:

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.4

10th Class Maths 3rd Lesson Polynomials Ex 3.4 Textbook Questions and Answers

Question 1.
Divide the polynomial p(x) by the polynomial g(x) and find the quotient and remainder in each of the following:
i) p(x) = x³ – 3x² + 5x – 3, g(x) = x² – 2
ii) p(x) = x⁴ – 3×2 + 4x + 5, g(x) = x² + 1 – x
iii) p(x) = x⁴ – 5x + 6, g(x) = 2 – x²
Answer:
i) Given polynomials are
p(x) = x³ – 3x² + 5x – 3 and
g(x) = x² – 2
Here, dividend and divisor are both in standard forms.
So, we have

∴ The quotient is x – 3 and the remainder is 7x – 9.

ii) Given polynomials are
p{x) = x⁴ – 3x² + 4x + 5 and
g(x) = x² + 1 – x
Here, the dividend is already in the standard form and the divisor is not in the standard form. It can be written as x² – x + 1.
We have,

∴ The quotient is x² + x – 3 and the remainder is +8.

iii) Given polynomials are
p(x) = x⁴ – 5x + 6 and
g(x) = 2 – x²
Here, the dividend is already in the standard form and the divisor is not in the standard form. It can be written as -x² + 2.
So, we have

∴ The quotient is -x² – 2 and the remainder is -5x + 10.

Question 2.
Check in which case the first polynomial is a factor of the second polynomial by dividing the second polynomial by the first polynomial:
i) t² – 3, 2t⁴ + 3t³ – 2t² – 9t – 12
ii) x² + 3x + 1, 3x⁴ + 5x³ – 7x² + 2x + 2
iii) x³ – 3x + 1, x⁵ – 4x³ + x² + 3x + 1
Answer:
i) Given first polynomial is t² – 3.
Second polynomial is
2t⁴ + 3t³ – 2t² – 9t – 12.
Let us divide 2t⁴ + 3t³ – 2t² – 9t – 12 by t² – 3, we get

Since the remainder is 0, therefore, t² – 3 is a factor of 2t⁴ + 3t³ – 2t² – 9t – 12.

ii) Given first polynomial is x² + 3x + 1
Second polynomial is 3x⁴ + 5x³ – 7x² + 2x + 2
Let us divide 3x⁴ + 5x³ – 7x² + 2x + 2 by x² + 3x + 1, we get

Since the remainder is 0, therefore x² + 3x + 1 is a factor of 3x⁴ + 5x³ – 7x² + 2x + 2.

iii) Given first polynomial = x³ – 3x + 1
Second polynomial = x⁵ – 4x³ + x² + 3x + 1
Let us divide x⁵ – 4x³ + x² + 3x + 1 by x³ – 3x + 1, we get

Here, remainder is 2(≠ 0).
Therefore, x³ – 3x + 1 is not a factor of x⁵ – 4x³ + x² + 3x + 1.

Question 3.
Obtain all other zeroes of 3x⁴ + 6x³ – 2x² – 10x – 5, if two of its zeroes are \(\sqrt{\frac{5}{3}}\) and –\(\sqrt{\frac{5}{3}}\).
Answer:
Let the other two zeroes are α and β.
Now compare the given polynomial 3x⁴ + 6x³ – 2x² – 10x – 5 with the standard form ax⁴ + bx³ + cx² + dx + e we get a = 3, b = 6, c = -2, d = -10, e = -5

–\(\frac{5}{3}\)αβ = \(\frac{-5}{3}\) ⇒ αβ = 1
now (α – β)² = (α + β)² – 4αβ
= (-2)² – 4(1)
= 4 – 4 = 0
α – β = 0 …. (2)
Now solving (1) and (2) we get

⇒ α = -1, β = -1
Then the remaining the zeroes are -1 and -1.
Hence all zeroes of it = –\(\sqrt{\frac{5}{3}}\), \(\sqrt{\frac{5}{3}}\), -1, -1.

Question 4.
On dividing x³ – 3x² + x + 2 by a polynomial g(x), the quotient and remainder were x – 2 and -2x + 4, respectively. Find g(x).
Answer:
Given, p(x) = x³ – 3x² + x + 2
q(x) = x – 2 and
r(x) = -2x + 4
By division algorithm, we know that Dividend = Divisor × Quotient + Remainder
p(x) = q(x) × g(x) + r(x)
Therefore, x³ – 3x² + x + 2
= (x – 2) × g(x) + (- 2x + 4)
⇒ x³ – 3x² + x + 2 + 2x – 4 = (x – 2) × g(x)
g(x) = \(\frac{x^{3}-3 x^{2}+3 x-2}{x-2}\)
On dividing x³ – 3x² + x + 2, by x – 2, we get

First term of g(x) = \(\frac{\mathrm{x}^{3}}{\mathrm{x}}\) = x²
Second term of g(x) = \(\frac{-x^{2}}{x}\) = -x
Third term of g(x) = \(\frac{x}{x}\) = 1
Hence, g(x) = x² – x + 1.

Question 5.
Give examples of polynomials p(x), g(x), q(x) and r(x), which satisfy the division algorithm and
i) deg p(x) = deg q(x)
ii) deg q(x) = deg r(x)
iii) deg r(x) = 0
Answer:
Let q(x) = 3x² + 2x + 6, degree of q(x) = 2
p(x) = 12x² + 8x + 24, degree of p(x) = 2
Given degree p(x) = degree q(x)
i) Using division algorithm,
We gave, p(x) = q(x) × g(x) + r(x)
On dividing 12x² + 8x + 24 by 3x² + 2x + 6, we get

Since, the remainder is zero, therefore 3x² + 2x + 6 is a factor of 12x² + 8x + 24.
∴ g(x) = 4 and r(x)= 0

ii) Let p(x) = x⁵ + 2x⁴ + 3x³ + 5x² + 2
q(x) = x² + x + 1, degree q(x) = 2
Given degree q(x) = degree r(x)
On dividing x⁵ + 2x⁴ + 3x³ + 5x² + 2 by x² + x + 1, we get

Here, g(x) = x³ + x² + x + 1 and r(x) = 2x² – 2x + 1
degree of r(x) = 2.
∴ deg g(x) = deg r(x).

iii) Let p(x) = 2x⁴ + 8x³ + 6x² + 4x + 12, r(x) = 2
Here, degree r(x) = 0
On dividing 2x⁴ + 8x³ + 6x² + 4x + 12 by 2, we get

Here, g(x) = x⁴ + 4x³ + 3x² + 2x + 1 and r(x) = 10
so degree of r(x) = 0

AP State Board Syllabus AP SSC 10th Class English Textbook Solutions Chapter 4A Rendezvous with Ray Textbook Questions and Answers.

AP State Syllabus SSC 10th Class English Solutions Chapter 4A Rendezvous with Ray

10th Class English Chapter 4A Rendezvous with Ray Textbook Questions and Answers

Look at the picture and answer the questions that follow.

Question 1.
Is this picture (a) a poster, (b) an advertisement, or (c) a painting? Justify your option.
Answer:
It is a poster. The word/expression “RAJKAPOORS” tells us that it is directed by Raj Kapoor, a well-known Hindi actor. “SHREE 420” is the title of the movie. The persons in the picture are the actors of this film. Hence, it is evident that it is the poster of a movie.

Question 2.
Who do you think are the persons shown in this picture? Name them.
Answer:
The persons in this picture are cine actors. They are Raj Kapoor and Nargis.

Question 3.
What do you think is ‘SHREE 420?
Answer:
I think ’SHREE 420’ is a North Indian movie. It was directed and produced by Raj Kapoor. Raj Kapoor, Nargis, and Nadira were the main characters in this movie. It hit the screen on 6th September 1955.

Comprehension

Answer the following questions.

Question 1.
What did Ray’s detractors accuse him of? Did Roberge agree to their accusation? If not, why?
Answer:
Ray’s detractors accused him that he made his reputation selling India’s poverty to the West. Roberge didn’t agree to their accusation. He was not attracted by the material poverty depicted in Ray’s films. The thing struck him most was that the enormous spiritual poverty of some rich people is much more deplorable than material poverty.

Question 2.
‘I didn’t come here to convert. In fact, I am the one who got converted.’ Who said these words? What different shades of meanings do you find in the words of the speaker?
Answer:
Roberge, the French-Canadian priest said these words. Although he came from the West, he didn’t speak with the arrogance of the West. He told that he came to India on a quest to know the world and in the process know himself. He also told that he didn’t come to India to convert and in fact he was the one who got converted.

The word ‘convert’ means ‘change’. Usually, the West come to India to convert the Indian people. They are known for their arrogance and look down upon Indians. They think that they are superiors in terms of knowledge, culture, civilization, etc. In one sense they want to convert the Indians and make them refined. In the other sense, the Westerners want to convert the Indians to Christianity. As far as Roberge is concerned, he was the one who got converted with the newly acquired knowledge about the world and got a chance to know about himself.

Question 3.
Roberge took nine years to meet Ray in person after joining St. Xavier’s college. Why did he take so long time? What would you do if you were in his place?
Answer:
Roberge wanted to meet Ray in person but it took him nine years after reaching Kolkata and joining St. Xavier’s college, to meet him. Although he wanted to meet Ray right way, he didn’t want to just go and see him as he was a living museum piece. Roberge wanted to prepare himself, get to know his works more, so that when they met, there could be a worthwhile dialogue between them. If I were in Roberge’s place, I would do the same as he did.

Question 4.
How was Ray perceived by the outsiders? Was this perception true of Ray’s real character?
Answer:
The outsiders perceived Ray that Ray’s massive physical and intellectual stature might have made him come across as cold, aloof and even intimidating. This perception was not true of Ray’s real character. In reality Ray was a very simple and unassuming man with a subtle sense of humour.

Question 5.
What is meant by the line, ‘Ray took off where Tagore signed out.’ What was Ray searching for?
Answer:
Roberge feels that there is a striking comparison between Tagore and Ray. He also feels that philosophically too, Ray took off where Tagore signed out. This means that Ray began his work where Tagore ended. In a way Ray continued to walk on the path shown by Tagore. We find the analogy in Ray’s last three films ‘Ganashatru’, ‘Shakha Prashakha’ and ‘Agantuk’ with Tagore’s ‘Shabhyatar Sankat’. Both their works contain the message that it would be a sin to lose faith in man. Ray was an agnostic throughout his life. Even in the face of death, Ray was searching for answer about the existence of God. This was suggested by some of the music that he used in Shakha Prashaka.

Question 6.
How did Roberge try to take ‘Chitrabani’ forward? How did ‘Chitrabani’ help filmmaking in Bengali?
Answer:
Roberge founded ‘Chitrabani’, a communication and film institute in 1970. It is the first of its kind in West Bengal. He lent Ray’s name as cofounder as a token of their friendship. Ray was made the member of the first governing body and later, he became its adviser. Roberge arranged most of the initial funding from Canadian agencies. Thus, Roberge tried to take ‘Chitrabani’ forward. He acted as the director of Chitrabani for a period of 26 years. Chitrabani not only produced important documentary features, but also became breeding ground for local talent for film-making. Thus, the institute helped film-making in Bengali.

Question 7.
The theme of ‘Rendezvous with Ray’ is… (Tick any two options.)
a) To explain the efforts of Chitrabani.
b) To picturize the illustrious life of Ray.
c) To explain the experiences of Roberge with Ray.
Answer:
(b)To picturize the illustrious life of Ray. (✓)
(c) To explain the experiences of Roberge with Ray. (✓)

Vocabulary

1. Read the following passage and notice the underlined words.
It was a Sunday evening. It was already dark. We wanted to watch the film ‘Gajani’.We had a square meal and came out. There were no lights. There was a power cut. It was pitch dark. It was bitterly cold. We looked here and there …….

The word pairs ‘square meal’, ‘pitch dark’ and ‘bitterly cold’ are found together. That means they co-occur. Such co-occurring words or word combinations are called collocations(co + locate = collocate). These expressions are natural.

Let us look at another example.

We say	We don’t say
Ride a motor cycle	Drive a motor cycle

Read the following report and fill in the blanks with the word from the box below that collocates with the underlined words or phrases.

The venue of the celebration was ‘Ravindra Bharathi’, Hyderabad. It was the 100 days’ celebration of the film ‘Animals Forever’. Avinash, the hero, was full of life with his ……..(1)……… performance in the film. He was admired by everyone. In fact, he was considered to be the main reason for the success of the film. The hall resonated with ……..(2)……… clapping when he came onto the dais. The auditorium with packed audience honoured him with a ……..(3)……… ovation. The producer felicitated every one in the unit in a ……..(4)……… manner. The event was momentous and unforgettable.
Answer:

1. outstanding (outstanding performance)
2. thunderous (thunderous clapping)
3. standing (standing ovation)
4. fitting (fitting manner)

II. One-word substitutes

‘Rendezvous’ is a one-word substitute for ‘a meeting place’.
You can work in groups and pick out similar one-word substitutes from the text equivalent to the meanings given below:

1. A short stay between two places in one’s journey → stopover
2. A person who brings out new books → compiler
3. A group of three films that has the same characters or subject → trilogy
4. An impressive entrance to a building → portal
5. A person who tries to make something less good by criticising it → detractor
6. A person who is extremely important or large in size → colossus
7. A person who is responsible for a problem or a crime → culprit
8. A handwritten document → manuscript
9. A statement that expresses something people believe is true and is to be followed → dictum
10. Using more words than needed → verbose
11. Something designed to teach people some moral → didactic
12. A person who is not sure about the existence of God → agnostic
13. A branch of philosophy that studies the principles of beauty in art → aesthetics
14. A result of a situation or of an action → fallout
15. A film that gives facts about something → documentary

III. In the lesson ‘Rendezvous with Ray’ we come across certain words /expressions that are not from English, e.g.: en route. This means ‘on the way’. These expressions are taken from languages like Latin, Greek, and Portuguese and so on. There are certain instances where these expressions are used in English perhaps because of their precision in meaning. Some of them along with their meanings are given below…

1. en masse = all together, in large numbers.
2. viva voce = a spoken exam
3. in toto = totally
4. alma mater = mother of the soul (school or university) included
5. ex officio = because of the rank or job/by virtue of office
6. in absentia = in the absence
7. detour = a longer route we take to avoid a danger
8. verbatim = word for word, exactly as spoken or written
9. status quo = situation as it is now
10. ad hoc = not planned in advance
11. bona fide = genuine, real or legal
12. lingua franca = link language
13. magnum opus = the greatest work
14. sine die = indefinitely

Read the following paragraph and All in the blanks with appropriate expressions given in the list above.

Children for Films

On the 14th november, on the occasion of Children’s Day, the children across the state requested the officials to conduct Children’s Film Festivals more often. They submitted a memorandum to the Secretary to the Government in Hyderabad in this regard.

The Government conceded to the request and came out with a proposal to set up an ___(1)___ committee to serve the purpose before a permanent body is in place. It was proposed by the Government that the committee would be led by a de-partment official as an ___(2)___ president. The committee should conduct a written exam along with a ___(3)___ to identify student representatives at mandal, district and state level to strengthen the culture of film festivals among the children. The ___(4)___ of the students should be verified for such identification. The proposal made the screening of at least a ___(5)___ of a director mandatory every year. The children were thrilled to bits on this.
Answer:

1. ad hoc
2. ex officio
3. viva voce
4. en masse
5. magnum opus

IV. In the previous classes we learnt how the words form with the help of prefixes and suffixes. Now, we will learn another aspect of word formation i.e. through roots. In the lesson we have a word ‘agnostic’ which is formed out of the root ‘gnos’ that means ‘to know’. When the prefix ‘a-‘ is added to the root ‘gnos’, we have the word ‘agnostic’.

A. Given below are some prefixes, roots and suffixes. Form words using them and write their meanings.

B. Analyse the following words in terms of the prefixes, roots and suffixes and their meanings.

C. Think of some words that begin and end with the following prefixes and suffixes.
Prefixes: dis-, bi-, pro-, pre-

i) Words that begin with prefix a “dis-” :

ii) Words that begin with prefix “bi-“:

iii) Words that begin with prefix “pro-“:

iv) Words that begin with prefix “pre-” :

Suffixes:
i) Words that end with suffix “-cide” :

ii) Words that end with suffix “-tion” :

iii) Words that end with suffix “-ist” :

iv) Words that end with suffix “-logy”:

Grammar

I. Arrange the following sentences in proper order and write a paragraph.You may insert appropriate linkers wherever necessary.

Stream of Comedy

In every Indian language, a comedian is an essential character of films. In Telugu,Relangi and Ramanareddy provided comedy which made the audiences laugh heartily. Their appearance appealed to the film lovers.

Relangi was fat and short.
Ramana Reddy was lean and tall.
Their accent amused the film lovers.
Padmanabham and Allu Ramalingaiah followed their footsteps.
Rao Gopal Rao’s stint as villain-cum-comedian has been admired by all.
Mr. 101 Districts, Nutan Prasad left an indelible mark in the hearts of the audience.
Ali and Sunil are comedians.
They are also considered heroes.
Rajababu came later.
Brahmanandam has had a long career.
He has a world record.
His name found place in the Guinness Book of World Records.
He amused the people for long.
People have been laughing.
Sorrows of people are taking a back seat.
Answer:
Stream of Comedy :
In every Indian language, a comedian is an essential character of films. In Telugu, Relangi and Ramanareddy provided comedy which made the audiences laugh heartily. Their appearance appealed to the film lovers. Relangi was fat and short whereas Ramana Reddy was lean and tall. Their accent amused the film lovers. Then Padmanabham and Allu Ramalingaiah followed their foot steps. Raja Babu came later and amused the people for long. Rao Gopal Rao’s stint as villain-cum-comedian has been admired by all. Mr. 101 Districts, Nutan Prasad left an indelible mark in the hearts of the audience. Another familiar comedian, Brahmanandam has had a long career. He has a world record and his name found place in the Guinness Book of World Records. Ali and Sunil are comedians and they are also considered heroes. People have been laughing with the efforts of comedians and sorrows of people are taking a back seat.

II. Prepositions following ‘adjectives’ and ‘verbs’:

In ‘Rendezvous with Ray’ we come across certain verbs and adjectives followed by prepositions. For example, acquainted with, brought out, prevented from, explained to

Read the following sentences and observe the underlined words.

1. I am amazed at you, Victoria.
2. Victoria, a precocious girl of ten, was dressed in colours.
3. He is trying to adapt himself to the regrettable occasion.
4. He was very fond of Jimmy.
5. I am sure I am sorry for it.
6. You’ve been waiting for me to begin tea.
7. I see the little trifles that belonged to father lying around.
8. This always appealed to me.

In the above sentences, the underlined adjectives and verbs are usually followed by certain prepositions.

A. Here is a list of adjectives and verbs. Tick( ✓ ) the prepositions that follow the adjectives and the verbs. Later, use them in sentences of your own.

1. a) Mr. Bhargav is proud of his son’s achieving the title.
b) Miss Lalitha is the proud owner of her new flat.

2. a) Raju is married to Brahmam’s sister.
b) She got married to Sujan.

3. a) Mr. Williams is good at English.
b) She is very good with her neighbours.
c) It is good weather for going on a picnic.

4. a) Both the brothers are different from each other
b) His car is different to mine.

5. a) The district collector is keen to help them.
b) He is not keen on playing with them.
c) Our boss is keen for the work to resume.

6. a) Guntur is famous for tobacco and mirchi.
b) France is famous for its wine.

7. a) She is capable of looking after my child.
b) He is not capable of doing this job.

8. a) My brother is responsible for all this mess.
b) The police caught the man who was responsible for the theft.

9. a) An atheist is a person who doesn’t believe in God.
b) Believe in yourself; or you can’t achieve the success.

10. a) The HM is shouting at Gopal for his misdeed.
b) The boy is shouting for somebody’s help.

11. a) Mr. Teja is thinking of his new college.
b) Why don’t you think of your job seriously ?

12. a) I don’t agree with him in any case.
b) They don’t agree on everything.
c) She agreed to meet him at the airport.
13. a) She has to depend on her grandparents as she lost her parents.
b) They depend entirely on the funds given by the government.

14. a) He is recovering from his severe illness.
b) The gold has been recovered from the culprit.

15. a) This site belongs to Rama Rao.
b) These assets belong to Raju.

16. a) I have applied for a job of typist.
b) She applied to two foreign universities and was accepted by them.

B. Fill in the blanks with suitable prepositions.

1. All last winter Sharath suffered ________ conghs and colds.
2. Anand is unaccustomed _________ the heat.
3. Kumar was afraid _________ his enemies.
4. Sriram was always arguing ________ his brother.
5. Sindhu was dedicated _______ her job.
6. Priyanka was shocked _______ the hatred they had shown.
7. I said _______ you, “I am thinking going ___________ to America. I have actually dreamt ________ it.”
8. I want to talk ________ the group about their exams.
9. I was terrified ________ her.
10. I’ve always been terribly fond _______ you.
11. If you continue to support someone who is in trouble you are loyal _______ them.
12. If you don’t understand any of these words, you could refer ________ a dictionary.
13. It wasn’t his car, in fact I don’t know who it belongs ________ .
14. My problems are very similar _______ yours.
15. People started to shout ________ the driver.
16. She had always been bad ________ languages.
17. She listened me and then told me _______ her problems.
18. The accident sadly resulted ________ the death of a man.
19. The buses are often late, so you can’t depend _______ them.
20. They may feel jealous ________ your success.
Answer:

1. from
2. to
3. of
4. with
5. to
6. at
7. to; about/of; of
8. to
9. of
10. of
11. to
12. to
13. to
14. to
15. at
16. at
17. to ; about
18. in
19. on/upon
20. of

III. Read the following paragraph and notice the use of the past perfect and simple past.

A. As all the actors had taken their positions, the curtain rose. They started acting as thedirector had asked them to. The audience enjoyed the play very much. The hero kicked the comedian since the comedian had done mischievous things. The musician fell off his chair after the comedian had fallen on him. The power went off after the musician had landed on the cables. There was darkness and silence everywhere. After a while two persons in the audience started a conversation.

B. Read the following conversation and fill in the blanks with appropriate verb forms,
i. e. past perfect/simple past.

1. Sarath : Oh ! What ________ (happen)? Everything ________ (be) disturbed before the play ________ (come) to an end.
2. Bharath : Damn it! The play ________ (be) very interesting. Someone on the stage ________ (do) something when the hero ________ (throw) him off.
3. Sarath : I too ________ (see) it. It was the comedian. The hero ________ (hurl) him since he ________ (do) a mischievous thing.
4. Bharath : How disgusting! I (pay) one hundred rupees before I ________ (enter) this theatre. Everything has become a chaos.
5. Sarath : Where ________ (be) the director? Had he ________ (try) to set things right before the audience ________ (start) leaving, it would have been nice.
6. Bharath : The electrician ________ (restore) the power before the audience ________ (leave). Thank God! At last the play resumed.
Answer:

1. happened, had been, came
2. was, had done, threw
3. saw, hurled, had done
4. had paid, entered
5. was, tried, started
6. had restrored, left

IV. Given below is a paragraph with ten errors in the areas of ‘concord,tense, prepositions, punctuation and articles’. Edit the paragraph.

The Indian film industry has witnessed sweeping changes in the past hundred years. It started of with mute (mooki) films. Even then, people liked this new form of entertainment.There was several intervals in a film show because of a single projector. Later, the technology changes made a talkie films possible. A theatre of those days is like a rice mill. This type of theatre were called Touring Talkies’. But these didn’t tour. There were bamboo screens to served the purpose of walls of the modern theatre. ‘Cut Shows’ were a luxury of those days. Have you ever watch them Now, the modern theatres is completely different. Multiplexes with dts, 3D and 4D are a present reality.
Answer:
The Indian film industry has witnessed sweeping changes for the past hundered years. It started off with mute (mooki) films. Even then, people liked this new form of entertainment. There were several intervals in a film show because of a single projector. Later, the technological changes made the talkie films possible. A theatre of those days was like a rice mill. This type of theatres were called ‘Touring Talkies”. But these didn’t tour. There were bamboo screens to serve the purpose of walls of the modern theatre. ‘Cut Shows’ were a luxury of those days. Have you ever watched them? Now, the modern theatres are completely different. Multiplexes with dts, 3D, and 4D are a present reality.

Study Skills

Read the following passage and rewrite it replacing all idioms, without changing the meaning of the passage.

Tragedy in Comedy

“Cut!…. Shot OK”, shouted the director. The funny face of the comedian suddenly wore a worried look. He said to the director, “sir,….” The director did not allow him to
say anything. He said, “Yes, you can go ” Where should he go? The comedian really did not understand what the director had meant. He could not make out what he said. As per the original plans, the comedian had to attend the schedule for 15 more days continuously. This put the comedian in soup as his wife was seriously ill. He was referring to the break he needed to visit his wife and maybe the director was referring to the sets. Perhaps both of them were talking to cross purposes. He had always been loyal to the firm that brought him into the limelight. But he could not help his wife. He was in a tight corner then. Yes! He had to face difficulties, yet he had to make people laugh. This was the paradox of his life. Wasn’t it a ‘tragedy in comedy’?
Answer:
“Cut! Shot OK”, shouted the director. The funny face of the comedian suddenly wore a worried look. He said to the director, “Sir. ” The director did not allow him to say anything. He said, “Yes, you can go ” Where should he go ? The comedian really did not understand what the director had meant. He could not understand what he said. As per the original plans, the comedian had to attend the schedule for 15 more days continuously.

This put him in a difficult situation as his wife was seriously ill. He was referring to the break he needed to visit his wife and maybe the director was referring to the sets. Perhaps both of them understood the other’s words incorrectly- He had always been loyal to the firm that made him catch the public attention. But he could not help his wife. He was in a trouble then. Yes ! He had to face difficulties, yet he had to make people laugh. This was the peculiar situation of his life. Wasn’t it a ‘tragedy in comedy’?

Listening

Listen to the radio programme and tick (✓) the right answer to the following questions.

Jewels of India

This is All India Radio, Hyderabad.
Welcome to ‘Jewels of India Programme’ presented by Meenakshi.
In our programme today, we will know about the most famous Indian filmmaker.

We know that Bengal is a beehive of cultural activities. It has witnessed cultural reforms. It is the birthplace of many writers and intellectuals.

To continue this tradition, another great son of India took birth in Calcutta on 2nd May 1921. Who ioiiicl that be? It is none other than Satyajit Ray, the son of a writer… a grandson ol an amateur astronomer. His home provided him with an ambiance to think and feel

Graduating from Presidency College, Calcutta, Ray moved to the open air university, ’Shantiniketan1. A keen observer that he was, Ray imbibed the spirit of Shantiniketan. There he read extensively and developed interest in painting.

Back in Calcutta, Ray started working for a London-based advertising firm. Wasn’t it a right place for a creative man like Ray? He worked there for some time and moved to London to work there.

There he watched films. Film after film. He liked the film ‘Bicycle Thieves’.That kindled interest in Young Ray in film-making. Slowly his interest led him to a world of creativity. That was the birth of a great film maker. And Ray became a film maker.

His films ‘Pather Panchali’, ’Aparajito’, ’Agantuk’ were some of his master pieces. His vision for Bengali films was matchless.

Along with the Oscar Award, he received the Bharata Ratna. The legendary film maker passed away in 1992. But the film world cherishes his memories forever.

Next week we will know about another great personality.
Till then, good bye.

1. What was the programme about?
a) Ray
b) Jewels of India
c) Film-making
Answer:
b) Jewels of India

2. Which film inspired Ray?
a) Agantuk
b) Bicycle Thieves
c) Shantiniketan
Answer:
b) Bicycle Thieves

3. The highest civilian award Ray received was
a) Oscar Award
b) Bharath Ratna
c) Padma Vibhushan
Answer:
b) Bharath Ratna

Oral Activity

Make a short speech of 2-3 minutes about the film you have seen recently. You
should include the following points in your speech.
– What the film was.
– Why you liked it so much (You can talk about the direction, actors, costumes, music, songs, etc.)
Answer:
Dear Sir and my dear friends,
I am very glad to take this chance of speaking on the film (movie) ‘Satamanam Bhavathi”, which I have seen recently along with my family members. I like it so much as it is a family sentiment movie. All the family members can sit and watch it pleasantly.

The director of this film, Satish Vegesna presented it in a nice manner. We don’t find the scenes of violence, more fights, horror, sex, etc. in this movie. The famous music director Mickey J Meyer composed beautiful songs for this movie. Audience can easily be mesmerised by the melodious songs written by lyricists Sri Mani and Ramajogaiah Sastry. In the characters of Raju and Nitya, Sarvanand and Anupama Parameswaran acted su-perbly. Prakash Raj and Jayasudha’s action is wonderful. Sameer Reddy’s cinematography is one of the attractions of this film. I conclude that it is a super hit movie which attracts the attention of the spectators.

Thank you one and all for giving me this opportunity.

Rendezvous with Ray Summary in English

Rendezvous with Ray’ is the news report published by Frontline on July 29 and on August 11, 2006. This is all about Satyajit Ray, the distinguished cine director from Calcutta (Kolkata). Gaston Roberge, a French-Candian priest presents his experiences with Ray in this lesson.

In 1961, when Fr. Gaston Roberge was 26 years old, he was acquainted with the works of Satyajit Ray through the Apu Trilogy. He saw all the three films in one sitting -the incident which made him love the people of India and Bengali cinema and culture. Roberge gives a scholarly, original analysis of Ray’s works in his latest book, “Satyajit Ray, Essays : 1970-2005”. In his youth, Roberge knew of Bengal through Mircea Eliade’s La Nuit Bengalie, some of Tagore’s poems and a Reader’s Digest article on Mother Teresa. He was haunted by the harsh image of poverty shown in “Saint of the Slums” and Apu’s world came as a reassurance. Ray’s critics accused him that he made his reputation selling India’s poverty to the West. But Roberge was struck by the enormous spiritual poverty of some rich people which is worse than material poverty.

He wanted to meet Ray and make friends with him but it took him nine years after reaching Kolkata. He wanted to get to know Ray’s works more so that there could be a worthwhile dialogue between them when they met. Their friendship lasted about 22 years until Ray’s death in 1992. Their quiet friendship developed over the years. Ray’s friends called him Manikda. He was shy and discreet about showing his emotions. He was very simple and modest with a subtle sense of humour. Both the friends used to meet on Sundays at 9 a.m. at Ray’s residence on Bishop Lefroy Road, Kolkata. Ray used to show Roberge the private screenings and welcome his comments. Ray often addressed Roberge as the French- speaking priest in Bengali and Roberge felt it was Ray’s appreciation for him. Ray wrote his manuscripts in Bengali, with notes in English for his set-designer, with sketches and staff notation of music. Once, he missed the Charulatha screenplay. Though he knew who the culprit was, he didn’t want to take any action with humane concern.

We find striking comparison between Tagore and Ray in their works. There is a philosophical analogy too between them. We find all the characters – the rich, the poor, the powerful, the humble, the peasants, the city persons, children, teenagers, adults, old people, men, women, etc. in Ray’s movies.

Rendezvous with Ray Glossary

rendezvous : a meeting place

unique (adj) : usually good and special

en route (adv) : on the way

stopover (n) : a halt in a long journey

acquainted (v) : deliberately found out about something

trilogy (n) : a set of three films with the same artists or characters [The Apu Trilogy comprises three Begali fims directed by Satyajit Ray: Pather Panchali (1955), Aparajito (1956), and Apur Sansar (1959)]

fascinating (v) : being attracted

path-breaking (adj) : totally new

compilation (n) : book, list, record, etc., which consists of different pieces of information, songs, etc.

insight (n) : a sudden clear understanding of something especially a complicated situation or idea

portal (n) : an entrance

haunted (v) : obsessed

reassurance (n) : something that is said or done which makes someone feel calmer and less worried or frightened about a problem

accusation (n) : the act of charging somebody

detractors (n) : people who criticise

deplorable (adj) : very bad, unpleasant and shocking

arrogance (n) : showing pride

quest (n) : a long search for something that is difficult to find

muster up confidence (idiom) : gain confidence

right away (idiom) : immediately

worthwhile (adj) : important or useful

discreet (adj) : careful about what you say or do, so that you do not offend, upset, embarrass people or tell secrets

stature (n) : the degree to which someone is admired or regarded as important

aloof (adj) : unfriendly and deliberately not talking to other people

intimidating (adj) : making you feel worried and not confident

unassuming (adj) : showing no desire to be noticed or given special treatment, modest

subtle (adj) : not easy to notice or understand

screening (n) : the showing of a film or television programme

cemented (v) : made a relationship between people, countries or organizations firm and strong

elegance (n) : the state of being beautiful, attractive or graceful

manuscript (n) : a book or piece of writing before it is printed

notation (n) : a system of written marks or signs used to represent something such as music, mathematics or scientific ideas

culprit (n) : the person who is guilty of a crime or doing something wrong

reputation (n) : the opinion that people have about someone or something because of what has happened in the past

humane (adj) : treating people or animals in a way that is not cruel and causes them as little suffering as possible

colossus (n) : someone or something that is extremely big or extremely important

sign out (phr.v.) : to write one’s name in a book when one leaves a place such as a hotel, an office or a club

dictum (n) : a statement that is believed to be true and followed

analogy (n) : a feature that is similar

didactic (adj) : intended to teach people a moral lesson

verbose (adj) : using or containing too many words

impulse (n) : a sudden strong desire to do something without thinking about whether it is a sensible thing to do

aesthetics (n) : the art of judging beauty

denying (v) : saying that something is not true

agnostic (n) : a person who does not know whether God exists

frail (adj) : weak and thin

fallout (n) : result

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 5 Quadratic Equations Ex 5.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 5th Lesson Quadratic Equations Exercise 5.1

10th Class Maths 5th Lesson Quadratic Equations Ex 5.1 Textbook Questions and Answers

Question 1.
Check whether the following are quadratic equations.
i) (x + l)² = 2(x-3)
Answer:
Given: (x + l)² = 2(x – 3)
⇒ x² + 2x + 1 = 2(x – 3) = 2x – 6
⇒ x² + 2x + l – 2x + 6 = 0
⇒ x² + 7 = 0 is a Q.E.

ii) x² – 2x = (-2) (3 – x)
Answer:
Given: x² – 2x = -2(3 – x)
⇒ x² – 2x = -6 + 2x
⇒ x² – 4x + 6 = 0 is a Q.E.

iii) (x-2) (x + 1) = (x- 1) (x + 3)
Answer:
Given: (x – 2) (x + 1) = (x – 1) (x + 3)
⇒ x (x + 1) – 2 (x +1)
= x (x + 3) – 1 (x + 3)
Note : Compare the coefficients of x2 on both sides. If they are equal it is not a Q.E.
⇒ x² + x – 2x – 2 = x² + 3x – x -3
⇒ x² – x – 2 = x² + 2x – 3
⇒ 3x – 1 = 0 is not a Q.E.

iv) (x – 3) (2x + 1) = x(x + 5)
Answer:
Given: (x – 3) (2x + 1) = x(x + 5)
⇒ x (2x + 1) – 3 (2x + 1) = x . x + 5 . x
⇒ 2x² + x – 6x – 3 = x² + 5x
⇒ 2x² – 5x – 3 – x² – 5x = 0
⇒ x² – 10x – 3 = 0 is a Q.E.
(or)
Comparing the coefficients of x2 on both sides.
x . 2x and x . x
⇒ 2x² and x²
2x² ≠ x²
Hence it’s a Q.E.

v) (2x – 1) (x – 3) = (x + 5) (x – 1)
Answer:
Given: (2x – 1) (x – 3) = (x + 5) (x – 1)
⇒ 2x (x – 3) -1 (x – 3) = x (x – 1) + 5(x – 1)
⇒ 2x² – 6x – x + 3 = x² – x + 5x – 5
⇒ 2x² -7x + 3 – x² – 4x + 5 = 0
⇒ x² – 11x + 8 = 0
Hence it’s a Q.E.
(or)
Co.eff. of x² on L.H.S. = 2 × 1 = 2
Co.eff. of x² on R.H.S = 1 × 1 = 1
LHS ≠ RHS Hence it is a Q.E.

vi) x² + 3x + 1 = (x – 2)²
Answer:
Given: x² + 3x + 1 = (x – 2)²
⇒ x² + 3x + 1 = x² – 4x + 4
⇒ 7x – 3 = 0 is not a Q.E.

vii) (x + 2)³ = 2x (x² – 1)
Answer:
Given: (x + 2)³ = 2x(x² – 1)
⇒ x³ + 6x² + 12x + 8 = 2x³ – 2x [∵ (a + b)³ = a³ + 3a²b + 3ab² + b³]
⇒ -x³ + 6x² + 14x + 8 = 0
is not a Q.E. [∵ degree = 3]

viii) x³ – 4x² – x + 1 = (x – 2)³
Answer:
Given : x³ – 4x² – x + 1 = (x – 2)³
⇒ x³ – 4x² – x + 1 = x³ – 6x² + 12x – 8
⇒ 6x² – 12x + 8 – 4x² – x + 1 = 0
⇒ 2x² – 13x + 9 = 0 is a Q.E.

Question 2.
Represent the following situations in the form of quadratic equations:
i) The area of a rectangular plot is 528 m². The length of the plot (in meters) is one more than twice its breadth. We need to find the length and breadth of the plot.
Answer:
Let the breadth of the rectangular plot be x m.
Then its length (by problem) = 2x + 1.
Area = l . b = (2x + 1) . x = 2x² + x
But area = 528 m² (∵ given)
∴ 2x² + x = 528
⇒ 2x² + x – 528 = 0 where x is the breadth of the rectangle.

ii) The product of two consecutive positive integers is 306. We need to find the integers.
Answer:
Let the consecutive integers be x and x + 1.
Their product = x(x + 1) = x² + x
By problem x² + x = 306
⇒ x² + x – 306 = 0
where x is the smaller integer.

iii) Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360 years. We need to find Rohan’s present age.
Answer:
Let the present age of Rohan be x years.
Then age of Rohan’s mother = x + 26
After 3 years:
Age of Rohan would be = x + 3
Rohan’s mother’s age would be = (x + 26) + 3 = x + 29
By problem (x + 3) (x + 29) = 360
⇒ x(x + 29) + 3(x + 29) = 360
⇒ x² + 29x + 3x + 87 = 360
⇒ x² + 32x + 87 – 360 = 0
⇒ x² + 32x – 273 = 0
⇒ x² + 39x – 7x – 273 = 0
⇒ x (x + 39) – 7 (x + 39) = 0
⇒ (x – 7) (x + 39) = 0
⇒ x = 7 or x = -39 ‘x’ being age cannot be negative.
∴ x = Present age of Rohan = 7 years.

iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Answer:
Let the speed of the train be x km/h.
Then time taken to travel a distance of distance of 480 km = \(\frac{\text { distance }}{\text { speed }}\) = \(\frac{480}{x}\)
If the speed is 8km/h less, then time needed to cover the same distance would be \(\frac{480}{x-8}\)

⇒ x² – 8x = 1280
⇒ x² – 8x – 1280 = 0
where x is the speed of the train.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 7 Coordinate Geometry Ex 7.4 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 7th Lesson Coordinate Geometry Exercise 7.4

10th Class Maths 7th Lesson Coordinate Geometry Ex 7.4 Textbook Questions and Answers

Question 1.
Find the slope of the line joining the two given points.
i) (4,-8) and (5,-2).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-2+8}{5-4}\)
= 6

ii) (0, 0) and (√3, 3)
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{3-0}{\sqrt{3}-0}\)
= \(\frac{3}{\sqrt{3}}\)
= \(\frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}}\)
= √3

iii) (2a, 3b) and (a, -b).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-b-3b}{a-2a}\)
= \(\frac{-4b}{-a}\)
= \(\frac{4b}{a}\)

iv) (a, 0) and (0, b).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{b-0}{0-a}\)
= \(\frac{-b}{a}\)

v) A (-1.4, -3.7), B (-2.4, 1.3).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{1.3+3.7}{-2.4+1.4}\)
= \(\frac{5.0}{-1}\)
= -5

vi) A (3, -2), B (-6, -2).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{-2+2}{-6-3}\)
= 0

vii) A (-3\(\frac{1}{2}\), 3), B (-7, 2\(\frac{1}{2}\)).
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{2 \frac{1}{2}-3}{-7+3 \frac{1}{2}}\)
= \(\frac{-\frac{1}{2}}{-3 \frac{1}{2}}\)
= \(\frac{1}{2}\) × \(\frac{2}{7}\)
= \(\frac{1}{7}\)

viii) A(0, 4), B(4, 0)
Answer:
Slope = \(\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\)
= \(\frac{0-4}{4-0}\)
= \(\frac{-4}{4}\)
= -1

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 6 Progressions Ex 6.5 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 6th Lesson Progressions Exercise 6.5

10th Class Maths 6th Lesson Progressions Ex 6.5 Textbook Questions and Answers

Question 1.
For each geometric progression find the common ratio ‘r’, and then find a_n.
i) 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\), …….
Answer:
Given G.P.: 3, \(\frac{3}{2}\), \(\frac{3}{4}\), \(\frac{3}{8}\), …….

ii) 2, -6, 18, -54, …….
Answer:
Given G.P. = 2, -6, 18, -54, …….
a = 2, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{-6}{2}\) = -3
a_n = a . r^n-1 = 2 × (-3)^n-1
∴ r = -3; a_n = 2(-3)^n-1

iii) -1, -3, -9, -27, ……
Given G.P. = -1, -3, -9, -27, ……
a = -1, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{-3}{-1}\) = 3
a_n = a . r^n-1 = (-1) × 3^n-1
∴ r = 3; a_n = (-1) × 3^n-1

iv) 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\), …….
Given G.P. = 5, 2, \(\frac{4}{5}\), \(\frac{8}{25}\), …….
a = 5, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{2}{5}\)
a_n = a . r^n-1 = 5 × \(\left(\frac{2}{5}\right)^{n-1}\)
∴ r = \(\frac{2}{5}\); a_n = 5\(\left(\frac{2}{5}\right)^{n-1}\)

Question 2.
Find the 10^th and nth term of G.P.: 5, 25, 125,…..
Answer:
Given G.P.: 5, 25, 125,…..
a = 5, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{25}{5}\) = 5
a_n = a . r^n-1 = 5 × (5)^n-1 = 5^1+n-1 = 5ⁿ
a₁₀ = a . r⁹ = 5 × 5⁹ = 5¹⁰
∴ a₁₀ = 5¹⁰; a_n = 5ⁿ

Question 3.
Find the indicated term of each geometric progression.
i) a₁ = 9; r = \(\frac{1}{3}\); find a₇.
Answer:
a_n = a . r^n-1

ii) a₁ = -12; r = \(\frac{1}{3}\); find a₆.
Answer:
a_n = a . r^n-1

Question 4.
Which term of the G.P.
i) 2, 8, 32,….. is 512?
Answer:
Given G.P.: 2, 8, 32,….. is 512
a = 2, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{8}{2}\) = 4
Let the n^th term of G.P. be 512

512 = 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2 × 2
= 2⁹
∴ 2n – 1 = 9
[∵ bases are equal, exponents are also equal]
∴ 2n = 9 + 1 = 10
n = \(\frac{10}{2}\) = 5
∴ 512 is the 5^th term of the given G.P.

ii) √3, 3, 3√3, …….. is 729?
Answer:
Given G.P.: √3, 3, 3√3, …….. is 729
a = √3, r = \(\frac{a_{2}}{a_{1}}\) = \(\frac{3}{\sqrt{3}}\) = √3
now a_n = a . r^n-1 = 729
⇒ (√3)(√3)^n-1 = 729
⇒ (√3)ⁿ = 3⁶ = (√3)¹²
⇒ n = 12
So 12th term of GP √3, 3, 3√3, …….. is 729.

iii) \(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\), ……. is \(\frac{1}{2187}\)?
Answer:
Given G.P.: \(\frac{1}{3}\), \(\frac{1}{9}\), \(\frac{1}{27}\), ……. is \(\frac{1}{2187}\)

Let \(\frac{1}{2187}\) be the n^th term of the G.P., then
a_n = a . r^n-1 = \(\frac{1}{2187}\)

[∵ bases are equal, exponents are also equal]
7^th term of G.P is \(\frac{1}{2187}\).

Question 5.
Find the 12^th term of a G.P. whose 8 term is 192 and the common ratio is 2.
Answer:
Given a G.P. such that a₈ = 192 and r = 2
a_n = a . r^n-1
a₈ = a . (2)^8-1 = 192

= 3 × 2¹⁰ = 3 × 1024 = 3072.

Question 6.
The 4^th term of a geometric progression is \(\frac{2}{3}\) and the seventh term is \(\frac{16}{81}\). Find the geometric series.
Answer:
Given: In a G.P.

Now substituting r = \(\frac{2}{3}\) in equation (1)
we get,

Question 7.
If the geometric progressions 162, 54, 18, ….. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),….. have their n^th term equal, find the value of n.
Answer:
Given G.P.: 162, 54, 18, ….. and \(\frac{2}{81}\), \(\frac{2}{27}\), \(\frac{2}{9}\),……

Given that n^th terms are equal
a_n = a . r^n-1

⇒ 3^n-1+n-1 = 81 × 81
⇒ 3^2n-2 = 3⁴ × 3⁴
⇒ 3^2n-2 = 3⁸ [∵ a^m . aⁿ = a^m+n]
⇒ 2n – 2 = 8
[∵ bases are equal, exponents are also equal]
2n = 8 + 2
⇒ n = \(\frac{10}{2}\) = 5
The 5^th terms of the two G.P.s are equal.

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 3 Polynomials Ex 3.2 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 3rd Lesson Polynomials Exercise 3.2

10th Class Maths 3rd Lesson Polynomials Ex 3.2 Textbook Questions and Answers

Question 1.
The graphs of y = p(x) are given in the figure below, for some polynomials p(x). In each case, find the number of zeroes of p(x).

Answer:
i) There are no zeroes as the graph does not intersect the X – axis.
ii) The number of zeroes is one as the graph intersects the X – axis at one point only.
iii) The number of zeroes is three as the graph intersects the X – axis at three points.
iv) The number of zeroes is two as the graph intersects the X – axis at two points.
v) The number of zeroes is four as the graph intersects the X – axis at four points.
vi) The number of zeroes is three as the graph intersects the X – axis at three points.

Question 2.
Find the zeroes of the given polynomials,
(i) p(x) = 3x
(ii) p(x) = x² + 5x + 6
(iii) p(x) = (x + 2) (x + 3)
(iv) p(x) = x⁴ – 16
Answer:
i) Given p(x) = 3x
Let p(x) = 0
So, 3x = 0
x = \(\frac{0}{3}\) = 0,
Zeroes of p(x) = 3x is zero.
∴ No. of zeroes is one.

ii) Given p(x) = x² + 5x + 6 is a quadratic polynomial.
It has atmost two zeroes.
To find zeroes, let p(x) = 0
⇒ x² + 5x + 6 = 0
⇒ x² + 3x + 2x + 6 = 0
⇒ x(x + 3) + 2 (x + 3) = 0
⇒ (x + 3) (x + 2) = 0
⇒ x + 3 = 0 or x + 2 = 0
⇒ x = -3 or x = -2
Therefore the zeroes of the polynomial are -3 and -2.

iii) Given p(x) = (x + 2) (x + 3)
It is a quadratic polynomial.
It has atmost two zeroes.
Let p(x) = 0
⇒ (x + 2) (x + 3) = 0
⇒ (x + 2) = 0 or (x + 3) = 0
⇒ x = -2 or x = -3
Therefore the zeroes of the polynomial are -2 and – 3.

iv) Given p(x) = x⁴ – 16 is a biquadratic polynomial. It has atmost two zeroes.
Let p(x) = 0
⇒ x⁴ – 16 = 0
⇒ (x²)² – 4² = 0
⇒ (x² – 4) (x² + 4) = 0
⇒ (x + 2) (x – 2) (x² + 4) = 0
⇒ (x + 2) = 0 or (x – 2) = 0 or (x² + 4) = 0
⇒ x = -2 (or) x = 2 (or) x² = -4
Therefore the zeroes of the polynomial are 2, – 2, we do not consider √-4 since it is not real.

Question 3.
Draw the graphs of the given polynomial and find the zeroes. Justify the answers,
i) p(x) = x² – x – 12
ii) p(x) = x² – 6x + 9
iii) p(x) = x² – 4x + 5
iv) p(x) = x² + 3x – 4
v) p(x) = x² – 1
Answer:
i) Given polynomial p(x) = x² – x – 12.
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-3, 0) and (4, 0).
So, the zeroes of the polynomial are -3 and 4.
Justification:
Given p(x) = x² – x – 12 = 0
⇒ x² – 4x + 3x – 12 = 0
⇒ x(x – 4) + 3(x – 4) = 0
⇒ (x – 4) (x + 3) = 0
⇒ x – 4 = 0 and x + 3 = 0
x = 4 and x = – 3

ii) Given polynomial p(x) = x² – 6x + 9
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (3, 0).
So, the zeroes of the given polynomial are same i.e., 3.
Justification:
Given p(x) = x² – 6x + 9
⇒ x² – 3x – 3x + 9 = 0
⇒ x(x – 3) – 3(x – 3) = 0
⇒ (x – 3) (x – 3) = 0
⇒ x – 3 = 0 and x – 3 = 0
x = 3 and x = 3

iii) Given polynomial p(x) = x² – 4x + 5
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph does not cut the X – axis at any point.
So, the quadratic polynomial p(x) has no zeroes.
Justification: For the given p(x) = x² – 4x + 5 not possible to split in factors.

iv) Given polynomial p(x) = x² + 3x – 4.
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-4, 0) and (1, 0).
So, the zeroes of the polynomial are -4 and 1.
Justification:
Given p(x) = x² + 3x – 4 = 0
⇒ x² + 4x – x – 4 = 0
⇒ x(x + 4)- 1(x + 4) = 0
⇒ (x + 4) (x – 1) = 0
⇒ x + 4 = 0 and x – 1 = 0
x = – 4 and x = 1

v) Given polynomial p(x) = x² – 1
List of values of p(x):

Now, let’s locate the points listed above on a graph paper and draw the graph.

Result: We observe that the graph cuts the X – axis at (-1, 0) and (1,0).
So, the zeroes of the polynomial are – 1 and 1.
Justification:
Given p(x) = x² – 1 = 0
⇒ p(x) = (x + 1) (x – 1) = 0 [∵ a² – b² = (a + b) (a – b)]
⇒ x + 1 = 0 and x – 1 = 0
x = -1 and x = 1

Question 4.
Why are \(\frac{1}{4}\) and -1 zeroes of the polynomial p(x) = 4x² + 3x – 1 ?
Answer:
Given polynomial p(x) = 4x² + 3x – 1
Given zeroes are \(\frac{1}{4}\) and -1
Let x = \(\frac{1}{4}\)

Let x = -1
⇒ p(-1) = 4(-1)² + 3(-1)-1 = 4 – 3 – 1 = 4 – 4 = 0
∴ P(\(\frac{1}{4}\)) = 0 and p(-1) = 0
So these values are zeroes of the polynomial p(x).

AP State Board Syllabus AP SSC 10th Class Maths Textbook Solutions Chapter 10 Mensuration Ex 10.1 Textbook Questions and Answers.

AP State Syllabus SSC 10th Class Maths Solutions 10th Lesson Mensuration Exercise 10.1

10th Class Maths 10th Lesson Mensuration Ex 10.1 Textbook Questions and Answers

Question 1.
A joker’s cap is in the form of right circular cone whose base radius is 7 cm and height is 24 cm. Find the area of the sheet required to make 10 such caps.
Answer:

Radius of the cap (r) = 7 cm
Height of the cap (h) = 24 cm
Slant height of the cap (l) = \(\sqrt{r^{2}+h^{2}}\)
= \(\sqrt{7^{2}+24^{2}}\)
= \(\sqrt{49+576}\)
= √625
= 25
∴ l = 25 cm.
Lateral surface area of the cap = Cone = πrl
L.S.A. = \(\frac{22}{7}\) × 7 × 25 = 550 cm².
∴ Area of the sheet required for 10 caps = 10 x 550 = 5500 cm².

Question 2.
A sports company was ordered to prepare 100 paper cylinders without caps for shuttle cocks. The required dimensions of the cylinder are 35 cm length / height and its radius is 7 cm. Find the required area of thin paper sheet needed to make 100 cylinders.
Answer:

Radius of the cylinder, r = 7 cm
Height of the cylinder, h = 35 cm
T.S.A. of the cylinder with lids at both ends = 2πr(r+h)
= 2 × \(\frac{22}{7}\) × 7 × (7 + 35)
= 2 × \(\frac{22}{7}\) × 7 × 42 = 1848 cm².
Area of thin paper required for 100 cylinders = 100 × 1848
= 184800 cm²
= \(\frac{184800}{100 \times 100}\) m²
= 18.48 m².

Question 3.
Find the volume of right circular cone with radius 6 cm. and height 7 cm.
Answer:

Base radius of the cone (r) = 6 cm.
Height of the cone (h) = 7 cm
Volume of the cone = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 7
= 264 c.c. (Cubic centimeters)
∴ Volume of the right circular cone = 264 c.c.

Question 4.
The lateral surface area of a cylinder is equal to the curved surface area of a cone. If their base be the same, find the ratio of the height of the cylinder to slant height of the cone.
Answer:
Base of cylinder and cone be the same.

CSA / LSA of cylinder = 2πrh
CSA of cone = πrl
The lateral surface area of a cylinder is equal to the curved surface area of cone.
∴ 2πrh = πrl
⇒ \(\frac{h}{l}=\frac{\pi r}{2 \pi r}\)
⇒ \(\frac{h}{l}\) = \(\frac{1}{2}\)
∴ h : l = 1 : 2

Question 5.
A self help group wants to manufacture joker’s caps (conical caps) of 3 cm radius and 4 cm height. If the available colour paper sheet is 1000 cm², then how many caps can be manufactured from that paper sheet?
Answer:

Radius of the cap (conical cap) (r) = 3 cm
Height of the cap (h) = 4 cm
Slant height l = \(\sqrt{r^{2}+h^{2}}\)
(by Pythagoras theorem)
= \(\sqrt{3^{2}+4^{2}}\)
= \(\sqrt{9+16}\)
= √25
= 5 cm
C.S.A. of the cap = πrl
= \(\frac{22}{7}\) × 3 × 5
≃ 47.14 cm²
Number of caps that can be made out of 1000 cm² = \(\frac{1000}{47.14}\) ≃ 21.27
∴ Number of caps = 21.

Question 6.
A cylinder and cone have bases of equal radii and are of equal heights. Show that their volumes are in the ratio of 3 : 1.
Answer:
Given dimensions are:
Cone:
Radius = r
Height = h
Volume (V) = \(\frac{1}{3}\)πr²h

Cylinder:
Radius = r
Height = h
Volume (V) = πr²h

Ratio of volumes of cylinder and cone = πr²h : \(\frac{1}{3}\)πr²h
= 1 : \(\frac{1}{3}\)
= 3 : 1
Hence, their volumes are in the ratio = 3 : 1.

Question 7.
A solid iron rod has cylindrical shape. Its height is 11 cm. and base diameter is 7 cm. Then find the total volume of 50 rods?
Answer:

Diameter of the cylinder (d) = 7 cm
Radius of the base (r) = \(\frac{7}{2}\) = 3.5 cm
Height of the cylinder (h) = 11 cm
Volume of the cylinder V = πr²h
= \(\frac{22}{7}\) × 3.5 × 3.5 × 11 = 423.5 cm³
∴ Total volume of 50 rods = 50 × 423.5 cm³ = 21175 cm³.

Question 8.
A heap of rice is in the form of a cone of diameter 12 m. and height 8 m. Find its volume? How much canvas cloth is required to cover the heap? (Use π = 3.14)
Answer:

Diameter of the heap (conical) (d) = 12 cm
∴ Radius = \(\frac{d}{2}\) = \(\frac{12}{2}\) = 6 cm
Height of the cone (h) = 8 m
Volume of the cone, V = \(\frac{1}{3}\)πr²h
= \(\frac{1}{3}\) × \(\frac{22}{7}\) × 6 × 6 × 8
= 301.71 m³.

Question 9.
The curved surface area of a cone is 4070 cm² and its diameter is 70 cm. What is its slant height?
Answer:
C.S.A. of a cone = πrl = 4070 cm²
Diameter of the cone (d) = 70 cm
Radius of the cone = r = \(\frac{d}{2}\) = \(\frac{70}{2}\) = 35 cm
Let its slant height be ‘l’.
By problem,
πrl = 4070 cm²
\(\frac{22}{7}\) × 35 × l = 4070
110 l = 4070
l = \(\frac{4070}{110}\) = 37 cm
∴ Its slant height = 37 cm.

AP State Board Syllabus AP SSC 10th Class Physical Science Important Questions Chapter 9 Classification of Elements- The Periodic Table.

AP State Syllabus SSC 10th Class Chemistry Important Questions 9th Classification of Elements- The Periodic Table

10th Class Chemistry 9th Lesson Classification of Elements- The Periodic Table 1 Mark Important Questions and Answers

Question 1.
What is modern periodic law? (AP June 2015)
Answer:
Modern periodic law :
The physical and chemical properties of the elements are periodic functions of their electronic configurations.

Question 2.
Define Moseley’s periodic law. (AP June 2015)
Answer:
Moseley’s periodic law: The physical and chemical properties of elements are periodic functions of their atomic numbers.

Question 3.
Which group elements are called Carbon family? (AP Mareh 2016)
Answer:
14 (or) IVA Group of elements are called Carbon family.

Question 4.
Which atom is bigger in size, Ne or Ar? Why? (AP June 2018)
Answer:
Ar. In groups as we go down number of shells increases due to the formation of new shell.

“O Group”

He

Ne

Ar

Kr

Xe

Rn

Question 5.
A and B are two elements. The compound formed with A and B is A₂ B. What are the valencies of A and B. (TS March 2018)
Answer:
The valency of A is 1 and B is 2.

Question 6.
A teacher asked to give an example for Dobereiner’s triad. Ramu wrote them as “Li, Na, Mg”. In these three, identify which element does not belongs to this triad? (AP March 2019)
Answer:
Mg or Magnesium do not belongs to this triad.

Question 7.
Write the difference between Mendeleeff’s periodic law and modern periodic law. (AP SCERT: 2019-20)
Answer:
Mendeleeff’s periodic table is prepared based on atomic mass whereas modem periodic table is prepared based on atomic number (electronic configuration).

Question 8.
What is Dobereiner Triad? Give two examples to it.
Answer:
A group of three elements in which atomic weight of middle element is average of first and third element is called Dobereiner triad with similar propertion.
Eg: 1) U, Na, K
2) Cl, Br, I

Question 9.
What is Newlands’ law of octaves?
Answer:
When elements are arranged in the ascending order of their atomic weights, every eighth element starting from a given element resembles in its properties to that of starting element. This is called Newlands’ law of octaves.

Question 10.
What is MendeleefFs periodic law?
Answer:
MendeleefFs periodic law:
The physical and chemical properties of the elements are the periodic functions of their atomic weight.

Question 11.
What is the name given to horizontal rows and vertical columns in MendeleefFs periodic table?
Answer:
Horizontal rows are periods and vertical columns are groups.

Question 12.
What is the property on which MendeleefFs periodic table depends upon?
Answer:
Mendeleeff’s periodic table depends upon atomic weight.

Question 13.
What is the name given to I(A) group elements?
Answer:
Alkali metal family, because aliquili = plant ashes. Na, K, etc. were obtained from plant ash.

Question 14.
Why are VI A group elements called chalcogens?
Answer:
Chalcogeneous = Ore product. As the elements in group 16 (VI A) form ores with metals, they are called chalcogeneous family.

Question 15.
Why are VII A group elements called halogens?
Answer:
Halos – sea salt, genus – produced. So VII A (17) are obtained from nature as sea salt. So they are called halogen family.

Question 16.
What are halogens?
Answer:
Fluorine, Chlorine, Bromine, Iodine, and Astatine of VIIA group elements are called halogens, which are obtained from sea salt.

Question 17.
What are noble gases? What is the general electronic configuration of noble gases?
Answer:
The elements of group VIII A (18) are chemically least reactive so they are called noble gases. Their group electronic configuration is ns²np⁶ (except) for helium it is 1s².

Question 18.
What are Lanthanides?
Answer:
Elements acquiring same properties are called lanthanides, i.e. 4f elements. They are from ₅₈Ce (Cerium) to ₇₁Lu (Lutetium).

Question 19.
What are Actinides?
Elements acquiring different properties are called actinides, i.e. 5f elements. They are from ₉₀Th (Thorium) to ₁₀₃Lr (Lawrensium).

Question 20.
What are metals and non-metals?
Answer:
The elements with three or less electrons in the outer shell are considered to be metals and the ore with five or more electrons in the outer shell are considered to be non-metals.

Question 21.
What are metalloids?
Answer:
The properties of elements which are intermediate between the properties of metals and non-metals are called metalloids.

Question 22.
Which will behave like semi-conductors?
Answer:
Metalloids or semi-metals behave like semi-conductors.

Question 23.
What is valency?
Answer:
The combining power of element with respect to hydrogen, oxygen or indirectly any other element through hydrogen and oxygen is called valency.

Question 24.
What is the latest definition of valency?
Answer:
The number of electrons lost or gained or shared during a chemical reaction.

Question 25.
How do we measure atomic radius of solids?
Answer:
It is half of the distance of radius of each atom.

Question 26.
What is covalent radius?
Answer:
Half of the distance between length of covalent bond is called covalent radius.

Question 27.
In which units is atomic radius measured?
Answer:
Atomic radius is measured in pico meter (pm) units.
1 pm = 10^-12 m.

Question 28.
What is the method given by Milliken to calculate electronegativity of an element?
Answer:
According to Milliken, electronegativity of element is average value of its ionization energy and electron affinity.

Question 29.
What is electropositive character?
Answer:
The tendency of metals to remain positive ions in compounds is called electropositive character. (OR) The tendency of an atom to lose electrons to form positive ions.

Question 30.
What is screening effect or shielding effect?
Answer:
More the shells with electrons between the nucleus and the valence shell, they act as screens to decrease nuclear attraction over valence electron. This is called screening effect or shielding effect.

Question 31.
What do you mean by negative or positive electron gain enthalpy?
Answer:
The negative sign indicates that energy is liberated or lost, and the positive sign indicates that the energy is gained or absorbed.

Question 32.
What is a triad?
Answer:
A group of three elements with similar properties in which atomic weight of middle element is average of other two elements.

Question 33.
Chlorine, bromine, iodine are Dobereiner’s triads. How do you justify?
Answer:
Chlorine, bromine and iodine have similar properties and atomic weight of bromine is average of chlorine and iodine.

Question 34.
Why are lanthanides and actinides placed separately at the bottom of the periodic table?
Answer:
Lanthanides and actinides belong to f – block elements with different properties so they are placed at the bottom of periodic table.

Question 35.
Lithium, Sodium, and Potassium were put in one group on the basis of their similar properties.
1) What are those similar properties?
2) What is the common name of this group of family?
Answer:

1. They have same number of valence electrons that is 1 and valency 1. So they have similar chemical properties.
2. They are called alkali metals.

Question 36.
What are the following groups known as?
1) group VIIA elements
2) Zero group elements.
Answer:

1. Group VII A elements are called Halogens.
2. Zero group elements are called Noble gases.

Question 37.
An element Barium lies in 2^nd group; then answer the following.
1) What is its valency?
2) What will be the formula of its Phosphate?
Answer:

1. The element lies in second group. So its valency is 2.
2. The formula of Phosphate is Ba₃(PO₄)₂ [since the valency of Phosphate is 3],

Question 38.
A, B, C are three elements having their atomic numbers equal to 2,10 and 5 respectively.
a) Which of these elements belong to same period?
b) Which of these elements belong to same group?
Answer:
The electronic configurations of A, B, C are as follows
A – 2, B – 2, 8, C – 2, 3.
a) So, B and C belong to same period because valence electron enters same orbit.
b) A and C belong to same group because both are noble gases.

Question 39.
Which element of 3^rd period will form a chloride of Cl₄?
Answer:
It would be Silicon because its electronic configuration is 2, 8, 4. So, it lies in third period and its valency is 4.

Question 40.
Which two elements of 3^rd period will form a covalent compound?
Answer:
The two elements are phosporous and chlorine.

Question 41.
An element has atomic number 12. State whether it is metal or non-metal. Why?
Answer:
Its electronic configuration is 2, 8, 2. It lies in 2^nd group. The elements towards left of periodic table are generally metals. So the element is metal.

Question 42.
Elements X, Y, and Z belong to IA group of the periodic table. Their atomic radii are as follows.
X → 1.33 Å,
Y → 0.95 Å,
Z → 0.60 Å.
Arrange the elements in the increasing order of atomic number by giving reason.
Answer:
As we move from top to bottom in a group, atomic size increases and atomic number also increases.
So the correct increasing order is Z, Y, X.

Question 43.
An element has an atomic number 16. State
i) period to which it belongs
ii) the number of valence electrons.
Answer:
Its electronic configuration is 2, 8, 6.
i) So it belongs to 3^rd period (orbit number).
ii) The number of valence electrons is 6.

Question 44.
Why is energy absorbed when electron is added to uni-negative ion?
Answer:
It is difficult to add an electron to uni-negative ion. In order to overcome the repulsion between the electrons, actually energy should be supplied to add another electron to uni-negative ion.

Question 45.
When do you observe liberation of energy?
Answer:
Atoms of some elements gain electrons while forming ionic compounds. An atom is able to gain electron when the electron is attracted by the nucleus. Attraction involves the liberation of energy.

Question 46.
Why does nitrogen have less electron affinity value compared to oxygen?
Answer:
The electron affinity of nitrogen is less than oxygen because of stable configuration of nitrogen (i.e., 2p³ configuration).

Question 47.
Which one between Na and Na⁺ would have more size? Why?
Answer:
Na has more size because when one electron is removed from Sodium atom the nucleus attraction over outermost electron increases so atomic size decreases.

Question 48.
Second ionization energy of an element is higher than its first ionization energy. Why?
Answer:
It is difficult to remove an electron from unipositive ion when compared with neutral atom. So second ionization energy is always greater than first ionization energy.

Question 49.
Hydrogen can be placed in group’1 and group 7 periodic table. Why?
Answer:
Hydrogen has both +1 as well as -1 oxidation states. So still there is some ambiguity in position of hydrogen.

Question 50.
Why do inert gases have zero valency value?
Answer:
Inert gases show zero valency because they do not take part in chemical reactions due to stable configuration.

Question 51.
Element ’Z’ belongs to (second) 2nd group in the periodical table. Write the formula of oxide.
Answer:
The formula of oxide of the element is ZO.

Question 52.
Do the atom of an element and its ion have same atomic size?
Answer:
No, generally cation has smaller size and anion has greater size.

Question 53.
The electronegativities of the elements in period 3 of the periodic table are as. follows.

Arrange the elements in which they occur in the periodic table from left to right.
Answer:
Na Mg Al Si P S Cl

10th Class Chemistry 9th Lesson Classification of Elements – The Periodic Table 2 Marks Important Questions and Answers

Question 1.
An element has atomic number 17. Where would you expect this element in the Periodic Table? Why? (AP June 2018)
Answer:

• Electronic configuration of the given element is 1s² 2s² 2p⁶ 3s² 3p⁵.
• So, it is in 3^rd period and 17^th group of periodic table.
• Due to the valency electronic configuration of 3s² 3p⁵ it belongs to 3^rd period and 17^th group.

Question 2.
How do you appreciate the special nature of inert gases?
Answer:
I appreciate the special nature of inert gases because it helps us in explaining the formation of chemical bonds among the atoms of elements and their stability.

Question 3.
The atomic number of an element is 35. Where would you expect the position of this element in the periodic table? Why? (TS June 2015)
Answer:

• The Electronic configuration of element with atomic number 35 is 2, 8, 18, 7.
• So it has seven valence electrons.
• That’s why it is present in 17^th group or VII A group and 4^th period.
• The element is Bromine.

Question 4.
Why were Dobereiner, Newlands and Mendeleeff not 100% successful in their classification of elements? Why is the modern table relatively a better classification? (TS March 2016)
Predict the reason.
Answer:

• All the known elements at the time of Dobereiner could not be arranged in the form of triads.
• Newlands’ periodic table was restricted only for 56 elements.
• As Mendeleeffs classification is based on atomic weight, his classification led to two defects like anomalous pair of elements and dissimilar elements placed together.
• Modern periodic table was prepared on the basis of atomic number. So the periods and groups are clearly defined.

Hence Dobereiner, Newlands, and Mendeleeffs classifications were not 100% successful, but modern classification is successful.

Question 5.
Observe the electronic configurations given below and write the group and period numbers of those elements. (TS March 2016)

Answer:
a) The period number is 3 and group number is 1.
b) The period number is 3 and group number is 15.

Question 6.
Observe the information provided in the table and answer the questions given below it. (TS June 2017)

i) What are the s-block elements in the table?
ii) What are the ‘p’ block and ‘d’ block elements in the table?
Answer:
i) s-block elements : Na, Ca

ii) p-block elements : C, P
d-block elements : Ti, Ni.

Question 7.
Imagine, which one in each of the following pairs is large in size relatively with other? Explain. (AP March 2019)
(X) Na, Al (Y) Na, Mg⁺²
Answer:
(X) 1. Na is large in size than Al.
2. Atomic size gradually decreases from left to right in a period.

(Y) 1. Na is large in size than Mg²⁺.
2. Na is larger than Mg and Mg is larger than Mg²⁺.

Question 8.
What are the limitations of Dobereiner triad?
Answer:

• All the known elements could not be arranged in the form of triads.
• The law failed for very low mass or for very high mass elements.
  Eg : In case of F, Cl, Br the atomic mass of Cl is not an arithmetic mean of atomic masses of F and Br.
• As the techniques improved for measuring atomic masses accurately, the law was unable to remain strictly valid.

Question 9.
Distinguish between electron affinity and electronegativity.
Answer:

Electron affinity	Electronegativity
1. It is the property of an isolated gaseous atom.	1. It is the property of a bonded atom.
2. It is the energy released and is measured in ev/atom or kJ/mole.	2. It is relative quantity and has no units.
3. It is the attraction of an atom for a single electron.	3. It is the attraction of an atom for a pair of electrons.

Question 10.
What is electronegativity? What are the various methods used to determine electronegativity? Explain.
Answer:
Electronegativity :
The electronegativity of an element is defined as the relative tendency of its atom to attract electrons towards itself when it is bounded to the atom of another elements.

Various methods to calculate Electronegativity :
1) Milliken Scale :
According to Milliken, the electronegativity of an element is the average value of its ionization energy and electron affinity.

2) Pauling Scale :
Pauling scale is based on bond energies. The electronegativity of hydrogen is assumed as 2.20. Electronegativity of other elements is calculated with respect to hydrogen.

Question 11.
Give the electronic configurations of following elements. What do say about these elements by writing their electronic configurations?
a) Na
b) Al
c) Sc
d) Ce
Answer:
a) Na : 1s² 2s² 2p⁶ 3s¹
b) M : 1s² 2s²2 2p⁶ 3s² 3p¹
c) Sc : 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹
d) Ce : 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s² 4p⁶ 4d¹⁰ 5s² 5p⁶ 6s² 4f²

Inference :
The valence electron enters different orbitals. So these elements belong to different blocks in modern periodic table, i.e. s, p, d, and f respectively.

Question 12.
Do you think that Newlands’ law of octaves is correct? Justify.
Answer:
No, Newlands’ law of octaves was restricted to only 56 elements and did not leave any room for new elements. Elements that were discovered later could not be filled into Newlands’ table in accordance with properties.

Question 13.
Why did Mendeleeff have to leave certine blank spaces in his periodic table?
Answer:
Mendeleeff predicted that some elements were missing in the table so he left blank spaces at the appropriate places in the table.

Question 14.
Give reason for the need of classification of elements.
Answer:
Classification is necessary because it is difficult to remember the properties of all the elements separately. It is easy to identify the properties of elements by making them groups with similar properties.

Question 15.
x, y, and z are the elements of a Dobereiner’s triad. If the atomic mass of ‘x’ is 7 and that of ‘z’ is 39, what should be the atomic mass of ‘y’?
Answer:
The atomic mass of x = 7 ;
The atomic mass of z = 39
x, y, z form Dobereiner triad
∴ Atomic mass oi y = average of x and z = \(\frac{7+39}{2}\) = \(\frac{46}{2}\) = 23

Question 16.
Name the two elements that would expect to have chemical properties similar to element with atomic number 11. What is the base for your choice?
Answer:
The element with atomic number 11 is sodium and its electronic configuration is 1s² 2s² 2p⁶ 3s¹ or 2, 8, 1.

So it has one valence electron, i.e. present in I group. We know that the elements present in same group have same valence electrons. So they show similar properties.

Therefore the other two elements are Lithium and Potassium.

Question 17.
An element X belongs to 3^rd period and group 14 of the periodic table. State
a) the number of valence electrons
b) the valency
c) the name of the element.
Answer:
a) The number of valence electrons are 4.
b) Its valency = 8 – 4 = 4.
c) The electronic configuration of element is 2, 8, 4.
(Because 3^rd period means third orbit, group 14 has 4 valence electrons). So, the element with atomic number 14 is Silicon.

Question 18.
Why is it easier to remove 4f electron than 4s?
Answer:
Orbitals belonging to the same main shell have different penetration power towards the nucleus. In fourth main shell the order of penetration is like this 4s > 4p > 4d > 4f. So, it is easier to remove 4f electron than 4s.

Question 19.
How is screening effect responsible for low ionization of cesium?
Answer:

• More the shells with electrons between the nuclear and the valence shell, they act as screens and decrease nuclear attraction over valence electron. This is called the screening effect.
• More the screening effect, less is the ionization energy.
• Cesium with more inner shells has less ionization energy.

Question 20.
Why does Boron have less ionization energy when compared with Beryllium?
Answer:

• The electronic configuration of Be and B are 1s² 2s² and 1s² 2p² 2p¹.
• The element Boron has less ionization energy due to less penetration power of 2p compared to 2s.

Question 21.
We know that as we move from left to right ionization energy increases. But ionization energy Nitrogen is more than Oxygen. Why?
Answer:

• It is easier to remove an electron from Oxygen when compared to Nitrogen.
• This is because Nitrogen has stable 1s² 2s² 2p³ electronic configuration which contains half filled 2p orbitals whereas Oxygen has 1s² 2s² 2p⁴ configuration.

Question 22.
Why is it difficult to remove an electron from Mg⁺ when compared with Mg?
Answer:

• The energy required to remove the first electron outermost orbit of a neutral gaseous atom of the element is called first ionization energy.
• The energy required to remove from unipositive ion of the element is called second ionization energy.
• Second ionization energy is always more than first ionization energy because it is difficult to move electron from unipositive ion due to greater nuclear attraction.
• So it is difficult to remove an electron from Mg⁺ when compared with Mg.

Question 23.
Using the periodic table predict formula of compound formed between an element ‘X’ of group 2 and another element of group 17.
Answer:

• The element X belongs to group 2. So, the number of valence electrons are 2 and its valency is 2.
• The element Y belongs to group 17 or VII. So, the number of valence electrons are 7 and its valency = 8 – 7= 1.

During formation compound elements exchange their valencies.
∴ The formula of compound is XY₂.

Question 24.
How do electronegativity values vary in period and group?
Answer:
Period :
When we move from left to right in period, the electronegativity increases due to decrease in atomic size.

Group:
When we move from top to bottom in a group, the electronegativity decreases due to increase in atomic size.

Question 25.
How does metallic and non-metallic characters vary in a period and group?
Answer:
Period:
When we move from left to right in a period, the metallic character decreases and non-metallic character increases.

Group :
When we move form top to bottom in a group, non-metallic character decreases and metallic character increases.

Question 26.
How do valency vary in period and group?
Answer:
Period :
When we move from left to right in a period, the valency does not follow a regular trend. For example, in second period the valency starts from 1 and increases to 4, then thereafter decreases to ‘O’.

Group :
When we move top to bottom in a group, the valency remains the same because in a group the valence electrons are same.

Question 27.
How does electron affinity vary in a period and group?
Answer:
Period :
When we move from left to right in period, electron affinity increases due to greater nuclear attraction over electron.

Group :
When we move from top to bottom, the electron affinity decreases in atomic size. As the size of the atom increases, the nuclear attraction over outermost electron decreases. So electron affinity decreases.

Question 28.
An element has atomic number 19. Where would you expect this element in the periodic table anti why?
Answer:
The electronic configuration of element is 1s²2s²2p⁶3s²3p⁶4s¹. So the element is in 4^th period and I group.

Question 29.
The electronic configuration of the element X, Y, and Z are given below,
a) X = 2, 5
b) Y = 2, 8, 1
c) Z = 2, 8
i) Which element belongs to 18^th group?
ii) Which element belongs to 15^th or V group?
iii) Which element belongs to third period?
Answer:
i) Z belongs to 18^th group because it is a noble gase (i.e. Ne).
ii) X belongs to 15^th or V group because it has 5 valence electrons.
iii) Y belongs to 3^rd period because the valence electron is present in 3^rd orbit.

Question 30.
Referring the part of periodic table given below answer the questions that follow.

1) What happens to the atomic size if moved from left to right? Support your answer.
2) What changes do you observe in the metallic properties of the elements when moved from left to right?
Answer:
1) When we move from left to right in a periodic table atomic radii of elements decrease, as a result the size of the,atom decreases,

2) When we move from left to right in a periodic table electronegativity values of elements increase, as a result the metallic properties of the elements decrease.

Question 31.
State the name of element, number of valence electrons, valency, the group number and the period number of each element given in the following table.

Answer:

10th Class Chemistry 9th Lesson Classification of Elements- The Periodic Table 4 Marks Important Questions and Answers

Question 1.
What is Ionization Energy? Explain the factors that effect Ionization Energy. (AP June 2017)
(OR)
What is ionization energy? What are the factors which influence ionization energy? Explain.
(OR)
Write the factors that influence ionization energy and explain any three of them. (TS March 2019)
Answer:
Ionization energy :
The energy required to remove an electron from the outermost orbit or shell of a neutral gaseous atom is called ionization energy.
Factors influencing ionization energy :
1) Nuclear charge :
As nuclear charge increases, ionization energy increases.

2) Screening effect or shielding effect:
More the screening effect, less is the ionization energy.

3) Penetrating power of the orbitals :
If the orbitals have less penetrating power, then the ionisation energy is less. Generally, the penetrating power of orbits are like this : s > p > d > f.

4) Stable configuration :
The elements having half-filled or completely filled orbitals have more stability. So the ionization energy is more when the element has stable configuration.

5) Atomic size :
As the atomic size increases, the nucleus attraction over outermost electron decreases. So ionization energy decreases.

Question 2.
Elements of one short period of the Periodic Table are given below in the order from left to right. (AP March 2017)
Li, Be, B, C, N, F, Ne
Answer the following:
(i) To which period, do these elements belong?
(ii) One element of this period is missing. Which is the missing element and where it should be placed?
(iii) Which of the above elements belong to the family of halogens? What is its electronegativity value?
(iv) How does the metallic character varies in the Period?
Answer:
(i) 2nd period.
(ii) Oxygen.
It should be placed between Nitrogen and Flourine.

(iii) Flourine
Electro negativity 4.0

(iv) Decreases from left to right.

Question 3.
In the table given below, names of some elements of families are given. Based on this, fill the information in the empty boxes. (TS June 2015)

Answer:

Question 4.
Two elements X and Y belong to Groups 1 and 2 respectively in the same period of the Periodic Table. Compare these elements with respect to : (TS March 2015)
i) number of electrons in their outermost orbit.
Answer:
The number of electrons in the outermost orbit of element X = 1
The number of electrons in the outermost orbit of element Y = 2

ii) their atomic size and their valencies.
Answer:
The atomic size of the Y is lesser than X
Valence of X = 1 ; Valence of Y = 2

iii) their ionisation energy and metallic character.
Answer:
The ionization energy of Y is greater than X, X has higher metallic character than Y.

iv) formulae of their chlorides and sulphates.
Answer:
Chloride of X …. XCl
Chloride of Y …. YCl₂
Sulphate of X …. X₂SO₄
Sulphate of Y …. YSO₄

Question 5.
How are the elements arranged into groups and periods in the Modern Periodic Table? Elements in a group possess similar properties, but elements in a period do not show similarities in their properties. Why? (TS June 2017)
Answer:

• The Modern periodic table is arranged in groups and periods based on the electronic configuration of the atoms of elements.
• Physical and Chemical Properties of elements are related to their electronic con-figurations particularly the outershell configurations.
• The atoms of the elements in a group posses similar electronic configurations.
• The elements in a group should have similar chemical properties and there should be regular gradation in their physical properties from top to bottom.
• Across the table from left to right in any period, elements gets an increase in the atomic number by 1 unit between any two successive elements.
• Therefore the electronic configuration of valence shell of any two elements in a given period is not same.
• Due to this reason elements along a period posses different chemical properties with regular gradation in their physical properties from left to right.

Question 6.
Explain any four factors which influence the electron affinity (Electron Gain Enthalpy). (TS March 2017)
Answer:
Factors effecting of electron affinity
1. Nuclear Charge :
If nuclear charge increases electron affinity increases, similarly it decreases if nuclear charge decreases.

2. Screening effect:
If screening effect value increases electron affinity increases, if it decreases electron affinity decreases.

3. Penetration power of the orbitals :
If penetration power of the orbitals increases electron affinity increases. If it decreases electron affinity decreases.

4. Stable configuration :
If an atom has stable electron configuration electron affinity will decreases.

5. Atomic radius :
If atomic radius increases electronic affinity will be increases. If atomic radius decreases electron affinity decreases.

6. Metallic property :
If metallic property increases electron affinity decreases.

7. Non-Metallic property :
Non-Metallic property increases electron affinity value increases.

Question 7.
Observe the information and answer the following questions. (TS June 2018)

Name of the Element	Atomic Number	Electronic Configuration
Sodium	11	[Ne] 3s1
Magnesium	12	[Ne] 3s2
Potassium	19	[Ar] 4s1
Calcium	20	[Ar] 4s2

1) What is valency of Magnesium?
Answer:
Valency of magnesium is two.

2) Which element has more electro-positivity?
Answer:
Potassium (K) has more electro-positivity.

3) Write the elements which belongs to (third) 3^rd Period.
Answer:
The elements which belongs to 3^rd period are Sodium (Na), Magnesium (Mg).

4) Write the elements which belongs to 1^st Group.
Answer:
Sodium (Na), Potassium (K) belong to 1^st Group.

Question 8.

Answer the following from the above in brmation. (TS March 2018)
i) Which element posses the higher atomic radius in the above table?
Answer:
The element having higher atomic radius is ‘K’ (Potassium)

ii) Mention two plair of element which forms ionic bond.
Answer:
Na, Cl Mg, CL

iii) Name the two elements having valency 2.
Answer:
Elements having valency 2 are Be, Mg, Ca, 0, S, Se.

iv) Which element has electronic configuration of 1s² 2s² 2p⁴.
Answer:
Oxygen.

Question 9.
Explain the significance of three quantum numbers in predicting the position of an electron in an atom. (AP SCERT: 2019-20)
Answer:
Each electron in an atom is described by a set of three quantum numbers n, 1 and ml.
1. Principal quantum number (n):
The principal quantum number is used to describe the size and energy of the main shell. It is denoted by ‘n’. ‘n’ has positive integer values of 1, 2, 3, It is used to know the number of orbitals (n²) and electrons in an orbit. (2n²).

As ‘n’ increases the shells becomes larger and the electrons in those shells are farther from the nucleus and their energies increases.

2. The angular – momentum quantum number (l) :
‘l’ has integer values from O’ to n – 1, for each value of ‘n’. Each ‘l’ value represents one sub-shell. It is used to describe the shape of an orbit.

3. The magnetic quantum number (m_l) :
The magnetic quantum number (m_l) has integer values between -l and +l including zero.

If l = 0, the possible ml value is 1.
l = 1, the possible ml value is -1, 0 and 1.
Thus for a certain value of 1, there are (2l + 1) integer values of ml.

These values describe the orientation of the orbital in space relative to the other orbitals in the atom.

Ex: When l = 1, (2l + 1) = 3, that means ml has 3 values namely -1, 0, 1 or three p orbitals, with different orientations along x, y, z axes, labelled as p_x, p_y and p_z orbitals.

Predicting the position of an electron in an atom :
If the values of n, l, and ml are 2, 1,-1 respectively the electron is present in 2px orbital in L – shell.

Question 10.
Answer the following question based on the values of the atomic radii of the elements of one of the periods in modern periodic table (AP SCERT: 2019-20)
Li (152), Be (111), B (88), C (77), N (74), O (66) and F (64)
a) What is the trend of atomic radii of given elements?
b) In the numerical listing of periods in the modern periodic table, what number was given to above elements?
c) Mention the unit of atomic radius.
d) Why the values of atomic radius varied along the period?
Answer:
a) Atomic radii of elements decrease while going left to right in the periodic table.

b) Period – 2
c) Unit of atomic radius is ‘pm’ (picometer).
d) There should be no change in distance between nucleus and outer most shell for the elements in one period.

But, nuclear charge increases because of the increase in the atomic number of elements in a period.

Hence, the nuclear attraction on the outer shell electrons increases.

As a result the size of the atoms decreases while going left to right in a period.

Question 10.
Mendeleeff classified the then known 63 elements in the form of a periodic table. Mention any two things that benefitted study of chemistry, to support the above statement.
Answer:

• Mendeleeff accepted minor inversions in the order of increasing atomic weights as these inversions resulted in elements being placed in the correct group.
• It was the extraordinary thinking of Mendeleeff that made the chemists to accept the periodic table and recognise Mendeleeff more than anyone else as the originator of the periodic law.
• At the time when Mendeleeff introduced his periodic table even electrons were not discovered.
• Even then the periodic table was prepared to provide a scientific base for the study of chemistry of elements.
• In his honour the 101 element was named “Mendelevium”.

Question 11.
How do these properties vary in period and group?
1) Valency
2) Atomic radius
3) Ionisation energy
4) Electron affinity
5) Electronegativity
6) Electropositivity
7) Metallic nature
8) Non-Metallic nature.
Answer:

Question 12.
Explain the salient features and achievements of the Mendeleeffs periodic table.
Answer:
Mendeleeffs periodic table is based on atomic weight.
1) Periodic law:
The physical and chemical properties of the elements are the periodic functions of their atomic weights.

2) Groups and sub-groups :
The vertical columns in Mendeleeffs periodic table are called groups. There are eight groups and elements in each group have similar properties. Each group is divided into sub-groups A and B.

3) Periods :
The horizontal rows are called periods. There are ‘seven’ periods in Mendeleeffs periodic table.

4) Predicting the properties of missing elements :
Based on the arrangement of elements in table, Mendeleeff predicted that some elements were missing and left blank spaces at appropriate places in the table. Later they were discovered.

5) Correction of atomic weight :
It is useful in correcting atomic weights of elements.

6) Anomalous series :
More atomic weight element like Tellurium (Ti) is placed before the less atomic weight element like Iodine in order to place these elements in the correct group.

Question 13.
How does atomic radius vary la period and group? Explain.
Answer:
Period :

1. As we move from left to right the atomic radius decreases because the electrons enter the same main shell.
2. The nuclear charge increases because of increase in atomic number of elements in period.
3. Hence, the nuclear attraction on the outer shell electron increases. As a result, the size of atom decreases.

Group:

1. Atomic radius increases from top to bottom in a group of the periodic table.
2. As we go down in a group, the atomic number of element increases. In order to accommodate more number of electrons, there are more additional shells.
3. As a result, the distance between the nucleus and the outer shell of atom increases.
4. So atomic size increases.

Question 14.
What is electron affinity? What are the factors which influence electron affinity?
Answer:
Electron affinity :

1. The electron affinity of an element is defined as the energy liberated when an electron is added to its neutral gaseous atom.
2. Electron affinity of an element is also called electron gain enthalpy of that element.
3. M_(g) + e^– → M^–_(g) + EA₁ (M = Atom of element, EA₁ = First Electron affinity)
   M^–_(g) + e^– → M^-2_(g) + EA₂ (EA₂ = Second Electron affinity)

Factors influencing Electron affinity :
1) Nuclear charge :
Greater the nuclear charge, greater the electron affinity value because of greater attraction for incoming electron.

2) Atomic size :
As the atomic size increases, the attractive force of the nucleus on the electron decreases. So electron affinity decreases.

3) Electronic configuration :
The elements having stable electronic configurations of half filled or completely filled valence sub-shells show very small tendency to accept additional electron. So the electron affinity is low or almost zero for these elements.

4) Penetrating power of the orbitals :
As the penetrating power of the orbitals increases, the electron affinity increases.

5) Screening effect or shielding effect:
More the screening effect of orbitals, less is the electron affinity value.

Question 15.
How did Mendeleeff correct atomic weights of various elements?
Answer:

• Atomic weight = Equivalent weight x Valency
• By using the formula, the atomic weight of Beryllium was calculated as 13.5 (Equivalent weight of Be = 4.5, valency = 3)
• With this atomic weight the element should be placed in wrong group.
• So Mendeleeff predicted its valency is only 2. From that he calculated the atomic weight of Beryllium as 9.
• Now it fitted into correct group.
• Similarly, Mendeleeff corrected atomic weights of Indium and Gold.

Question 16.
Answer the following questions if atomic number of element is 15.
1) What is the name of the element?
2) What is the electronic configuration of element?
3) Which period and group does it belong to?
4) How many valence electrons are there in the element?
5) What is the valency of the element?
Answer:

1. The element is phosporous,
2. The electronic configuration of element is 1s² 2s² 2p⁶ 3s² 3p³ or 2, 8, 5.
3. It belongs to 3^rd period (orbit number is 3) and V or 15 group (Number of electrons in valence orbit is 5.)
4. Number of valence electrons are 5.
5. Its valency is 8 – 5 = 3.

Question 17.
If an element belongs to 3^rd period and 17^th group, then answer the following questions.
1) What is its electronic configuration?
2) How many valence electrons are there in the element?
3) What is the valency of element?
4) What is atomic number of element?
5) What is the name of the element?
6) Give two more elements which have similar properties as this element?
Answer:

• The element belongs to 3rd period and 17^th group. So the valence orbit is 3^rd and number of valence electrons in that orbit is 7. So its electron configuration is 2, 8, 7.
• The number of valence electrons are 7.
• The valency of element = 8 – 7 = 1.
• The atomic number of element is 17.
• Name of the element is chlorine.
• Chlorine belongs to Halogen family. So Fluorine, Bromine, Iodine, and Astatine have similar properties as chlorine.

Question 18.
The elements of a periodic table are given below in the order from left to right.
Li Be B C O F Ne
1) To which period do these elements belong?
2) One element of this period is missing. Which is the missing element and where should it be placed?
3) Which one of the elements in this period shows the property of catenation?
4) Place the three elements fluorine, beryllium, and oxygen in the order of increasing electronegativity.
5) Which one of the above elements belongs to halogen series?
Answer:

1. The elements belong to 2^nd period.
2. The element which is missing is Nitrogen which is placed in between carbon and oxygen.
3. Carbon shows the property of catenation.
4. The ascending order of electronegativity for these element is Beryllium < Oxygen < Fluorine.
5. Fluorine belongs to halogen family.

Question 19.
A group of elements in periodic table is given below.
Boron, Aluminium, Gallium, Indium, and Thallium.
(Boron is the first element and Thallium is the last element)
Answer the following questions in relation to the above group of elements.
1) Which element has the most metallic character?
2) Which element would be expected to have the highest electronegativity?
3) If the electronic configuration of Aluminium is 2, 8, 3, how many electrons are there in outer shell of thallium?
4) The atomic number of Boron is 5. Write the chemical formula of the compound formed when Boron reacts with Chlorine.
5) Do the elements in the group to the right of this Boron group have more metallic or less metallic character? Justify your answer.
Answer:
1) Thallium has the most metallic character because as we move from top to bottom in a group the metallic character increases.

2) Boron has the highest electronegativity because as we move from top to bottom in a group electronegativity decreases.

3) Thallium is in the same group as Boron. So, the number of electrons in outermost shell of Thallium is 3.

4) The atomic number of Boron is 5. So, its electronic configuration is 2, 3. Therefore its valency is 3.

Whereas the atomic number of Chlorine is 17. So, its electronic configuration is 2, 8, 7. Therefore its valency is 1.

The formula of compound formed between Boron and Chlorine is BCl₃.

5) The elements in the group right to Boron group have lesser metallic character because as we move from left to right in a period metallic character decreases.

Question 20.
The following questions refer to the periodic table.
1) Name the first and the last element in period 2.
2) What happens to the atomic size of elements moving from top to bottom of a group?
3) Which of the elements has the highest electron affinity among the halogens?
4) What is common feature of the electronic configurations of the elements in group 16?
Answer:
1) The first and the last elements of 2^nd period or Lithium and Neon.

2) The atomic size decreases as we move from top to bottom in a group because there is an addition of shell each time as we move down the group.

3) Chlorine has the highest electron affinity. We know as move from top to bottom the electron affinity values decrease. But due to small size of fluorine there would be more electron-electron repulsions, if we add electron. So Chlorine has more electron affinity.

4) All these have same general outermost electronic configuration that is ns² np⁴.

Question 21.
Answer the following.
1) Elements of which groups have low ionization energy?
2) What is your guess about atomic size of an element with seven electrons among all the elements in the same period?
3) Which element has the highest electronegativity? Why?
4) Which element has the highest electropositivity? Why?
Answer:
1) Group IA, IIA elements have lower ionization energy values because they have metallic character.

2) As we move from left to right in a period atomic size decreases. So element with seven outermost electrons has least size among all the elements in the same period.

3) Fluorine has the highest electronegativity because when we move from left to right in a period atomic size decreases and electronegativity values increase. So Fluorine has the highest electronegativity.

4) Cesium has the highest electropositivity or positive character because when we move from top to bottom in a group atomic size decreases. So electropositive character increases. Therefore Cesium has the highest electropositive character.

Question 22.
Given below is the electronic configuration of A, B, C, D.

A) 1s2 2s2 2p¹	a) Which are the elements coming within the same period?
B) 1s2 2s2 2p6	b) Which are the elements coming within the same group?
C) 1s2 2s2 2p6 3s2 3p6	c) Which are the noble gas elements?
D) 1s2	d) Which group and period does the element C belong to?

Answer:
a) A and B belong to same period because the valence electrons of both the elements lie in the same orbit.
b) Elements A and C and elements B and D.
c) B and D are noble gas elements.
d) C belongs to 3^rd period (orbit number) and III group (Number of valence electrons).

Question 23.
Write down the characteristics of the element having atomic number 16.
i) Electronic configuration
ii) Period number
iii) Group number
iv) Element family
v) Number of valence electrons
vi) Valency
vii) Metal or non-metal
viii) Name of the element
Answer:
i) Electronic configuration of element is 1s² 2s² 2p⁶ 3s² 3p⁴ or 2, 8, 6.
ii) Period number is 3 because valence electron lies in 3^rd orbit.
iii) Group number is 6 because the number of valence electrons are 6.
iv) Element belongs to chalcogen family.
v) Number of valence electrons are 6.
vi) Valency = 8 – 6 = 2.
vii) It is a non-metal because in a period when we move from left to right non-metallic character increases.
viii) Name of the element is sulphur.

Question 24.
The second period element ‘F has electron gain enthalpy than the third period elements of same group ‘Cl’. Why?
Answer:

• In a group of elements, the electron gain enthalpy decreases from top to bottom.
• But in general the second element in a group, i.e. 3^rd period element has greater electron gain enthalpy than the first element, i.e. 2^nd period element.
  Ex : E.A of F < E.A of Cl.
• This is because Fluorine atom is smaller in size than Chlorine atom.
• F₂ also has strong inter electronic repulsions.
• In the addition of an electron to fluorine atom, the electronic repulsions overcome at the expense of a part of the energy liberated.

Hence the overall energy liberated is less than that of Chlorine atom.

Question 25.
Differentiate the metals and non-metals.
Answer:

Metals	Non-Metals
1. Metals have lustrous surface.	1. Non-metals do not have lustrous surface.
2. They show malleability.	2. They do not show malleability.
3. They show ductility.	3. They do not show ductility.
4. They produce sonorous sound.	4. They do not produce sonorous sound.
5. Generally they are hard.	5. Generally they are soft.
6. They are good conductors of electricity.	6. They are bad conductors of electricity.
7. Generally they liberate hydrogen gas when they are treated with acids.	7. They do not liberate hydrogen gas.

Question 26.
The electronic configuration of atom A is 2, 8, 6.
a) What is the atomic number of element A?
b) State whether the atomic size of element A is bigger or smaller than the atom having atomic number 14. Why?
c) Which of the elements exhibits similarity in chemical properties as element A 0(8), C(6), N(7), AV(18). Why?
d) How does the element form inert gas configuration?
Answer:
The electronic configuration of atom – A is 2, 8, 6.
a) Atomic number of element ‘A’ is 16, i.e. Sulphur.

b) The atom which has atomic number – 14 is Silicon (Si).

Atomic size of element decreases across period from left to right. So the atomic size of element ‘A’ is smaller than the atom having atomic number 14.

c) Element oxygen O₈ – exhibits similarity in chemical properties as element A, because they belong to the same group.

d) Given element – A becomes inert gas, i.e. Argon configuration by gaining ‘2’ electrons.

Question 27.
Select the correct answers from the choices A, B, C, D which are given with reference to the variation of properties in the periodic table. Which of the following is generally true?
A : Atomic size increases from left to right across a period.
B : Ionisation energy increases from left to right across a period.
C : Electropositive character increases going down a group.
D : Electronegativity increases going down a group.
Answer:
1) A is wrong because when we move from left to right the atomic number increases. So, the nuclear attraction over outermost orbital increases. Therefore the atomic size decreases.

2) B is correct but it does not follow a regular trend in a period.

3) C is correct. As move from top to bottom in a group atomic size increases. Therefore it is easy to lose electrons. So electropositive character increases.

4) D is wrong because as we move from top to bottom in a group atomic size increases. So electronegativity decreases.

Question 28.
Some elements belonging to second period of periodic table, and their atomic radii are given below. Observe them and write answers.

1) Write the elements in the ascending order of their atomic radii.
2) Which of the 2nd period elements closer to the configuration of inert gas?
3) Which is the outermost orbit of all these elements?
4) Which element’s atomic size is bigger, Beryllium or Carbon? Why?
Answer:

1. The ascending order of atomic sizes is O, N, C, B, Be and Li.
2. Lithium has closest inert gas configuration, i.e. 1s² 2s¹. Its nearest inert gas is Helium.
3. The outermost orbit for all these elements is second orbit.
4. Beryllium has more atomic size than Carbon. Because when we move across a period the atomic number increases. So nuclear attraction over outermost orbit increases and atomic size decreases. So carbon has lesser size than Beryllium.

Question 29.

Refer the above part of periodic table and answer the following questions.
a) Element with the least atomic size.
b) Write the electronic configuration of the elements B and E.
c) Identify the elements that have similar physical and chemical properties as the element Y.
d) Arranged elements increasing order of their electronegativity values.
Answer:
a) The element with least atomic size is E. Because when we move from left to right in a period the atomic size decreases,

b) Electronic configuration of B is 1s² 2s² 2p⁶ 3s² 3p¹ Because the element belongs to 13th group its general configuration is ns² np¹ and the element belongs to third period and its atomic number is 13. Similarly electronic configuration of E is 1s² 2s² 2p⁶ 3s² 3p¹. Because the element belongs to 16th group. Its general configuration is ns2np4 and it is in third period. So its atomic number is 16.

c) The elements which have similar physical and chemical properties with Y are X and Z. Because they lie in a single group, i.e. 1st group. In a group, elements are having similar physical and chemical properties.
d) Z < Y < X < B < C < D < E.

Question 30.
Consider the section of the periodic table given below.

1) Which is the most electronegative?
2) How many valence electrons are present in G?
3) Write the formula of the compound between B and H.
4) Which element has similar properties as J?
5) Which element has greater size-either D or E?
Answer:

1. J is the most electronegative. In a period electronegative values increase.
2. G is present in V group. So the number of valence electrons is 5.
3. B is present in first group. So its valency is 1 and hydrogen also has valency 1. Therefore the compound is BH.
4. K lies in same group as J. Elements belonging to same group have similar properties. So, K has similar properties as J.
5. E has greater size because as we move from top to bottom the atomic size increases.