Class 11th

Get insights from 8k questions on Class 11th, answered by students, alumni, and experts. You may also ask and answer any question you like about Class 11th

Follow Ask Question
8k

Questions

0

Discussions

28

Active Users

0

Followers

New answer posted

7 months ago

0 Follower 4 Views

A
alok kumar singh

Contributor-Level 10

6. Steam distillation is a special type of separation technique for temperature sensitive materials such as natural organic compounds. Few organic compounds tend to decompose at higher temperatures and normal distillation does not fit for this purpose. So, water is added to the apparatus and the temperature of the compounds is depressed, evaporating them at lower temperature. Once the distillation is accomplished, the vapours are condensed and  there occurs,   the separation of the constituents at ease.

New answer posted

7 months ago

0 Follower 10 Views

A
alok kumar singh

Contributor-Level 10

5. Whenever the difference in the boiling points of two liquids is not much than simple distillation technique cannot be used to separate them. The vapours of such liquids are formed within the same temperature range and are condensed simultaneously and hence the technique of fractional distillation is used in such cases. In this technique, vapours of a liquid mixture are passed through a fractionating column before condensation. The fractionating column is fitted over the mouth of the round bottomed flask. The vapours of the liquid with higher boiling point condense before the vapours of the liquid with lower boiling point. The vapour

...more

New answer posted

7 months ago

0 Follower 6 Views

A
alok kumar singh

Contributor-Level 10

4. Liquids which have different boiling points vaporize at different temperatures. The vapours are cooled and then liquids so formed are collected separately. Liquid A can be separated from B and C due to the large difference in boiling points. Liquid B and Liquid C have boiling points very close to each other and cannot be separated by simple distillation and are separated by fractional distillation. Liquid B distilled first since the order of boiling points of A, B and C is as follows: B < C < A.

New answer posted

7 months ago

0 Follower 7 Views

A
alok kumar singh

Contributor-Level 10

3. When the difference in boiling points of two liquids is not much then simple distillation cannot be used to separate them. Vapours of such liquids are formed within the same temperature range and are condensed simultaneously. The technique of fractional distillation is used in such cases. In this technique, vapours of a liquid mixture are passed through a fractionating column before condensation. The fractionating column is fitted over the mouth of the round bottom flask. The liquid (A) with low boiling point will distill first.

New answer posted

7 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

2.  Benzoic acid can be purified by hot water because of following characteristics.
(i) Benzoic acid is more soluble in hot water and less soluble in cold water.
(ii) Impurities present in benzoic acid are either insoluble in water or more soluble in water to such an extent that they remain in solution as the mother liquor upon crystallisation.

New answer posted

7 months ago

0 Follower 5 Views

A
alok kumar singh

Contributor-Level 10

1. The atomic orbitals combine to give rise to a new set of equivalent orbitals called hybrid orbitals and are used in bond formation. This phenomenon is known as hybridization which can be defined as the process of intermixing of the orbitals of slightly different energies so as to redistribute their energies, resulting in the formation of new sets of orbitals having equivalent energies and shape. In H2C=C=CH2 (allene) carbon atom l and 3 are sp2 hybridized as each one has 3σ bonds while carbon atom 2 has 2σ bonds so it is sp hybridized while the allene molecule as a whole is found to be non-planar. 

New answer posted

7 months ago

0 Follower 208 Views

P
Payal Gupta

Contributor-Level 10

15. [3x22ax+3a2]3

[3x2+a(2x+3a)]3

We know that by binomial theorem,

(a+b)3 = a3+b3+3ab(a+b)

a3+b3+3a2b+3ab2

Then,

[3x2+a(2x+3a)]3

= (3x2)3 + [a(2x+3a)]3 + [3(3x2)2 a(2x+3a)] + [ 3(3x2){a(2x+3a)}2]

= 27x6 + [a3(2x+3a)3] + [3(9x4)(2ax+3a2)] + [3(3x2){a2(3a2x)2}]

= 27x6 + [a3{8x3+27a3+3(4x2)(3a)+3(2x)(9a2)}] + [54ax5+81a2x4] + [ (9a2x2) (9a2+4x212ax) ]

= 27x6 + [ 8a3x3+27a6+36a4x254a5x ] + [ 54ax5+81a2x4 ] + [ (81a4x2+36a2x4108a3x3 ]

= 27x6  8a3x3+27a6+36a4x254a5x  54ax5+81a2x4 + 81a4x2+36a2x4108a3x3

= 27x6– 54ax5 + 117a2x4  116a3x3 + 117a4x2  54a5x + 27a6

New answer posted

7 months ago

0 Follower 19 Views

P
Payal Gupta

Contributor-Level 10

14. For (a – b) to be a factor of anb nwe need to show (anbn) = (ab)k as k is a natural number.

We have, for positive n

an = (ab+b)n = [(ab)+b]n

=>an = nC0(ab)n + nC1(ab)n -1b + nC2(ab)n – 2b2 + ………… +nCn-1 (ab) bn1 + nCnbn

=>an= (ab)n + nC1 (ab)n1 b + nC2 (ab)n2 b2 + …………….…+ nCn-1 (ab) bn1 + bn [Since, nC0 = 1 and nCn = 1]

=> anbn = (ab)n +nC1 (ab)n1 b + nC2 (ab)n2 b2 + ……………… + nCn-1 (ab) bn1

=> anbn = (ab) [ (ab)n1 + nC1(ab)n2 b + nC2 (ab)n3 b2 +……….…… + nCn-1  bn1 ]

=> anbn = (ab) k where k = [ (ab)n1 + nC1 (ab)n2 b + nC2 (ab)n3 b2 +……….…… + nCn-1  bn1 ] is a natural number.

Therefore (a – b) is a factor o

...more

New answer posted

7 months ago

0 Follower 3 Views

P
Payal Gupta

Contributor-Level 10

13. The general term of the expansion (3+ax)9 is

Tr+1 = 9Cr 3(9r) (ax)r

= 9Cr 3(9r)arxr

At r = 2,

T2+1 = 9C2 3(92)a2x2

9!2!(92)! 37a2x2

9′8′7! /2′1′7! 37a2x2

= 36 *37a2x2

At r = 3,

T3+1 = 9C3 3(93)a3x3

9!3!(93)! 36a3x3

= 9'8'7'6!/3'2'1'6! 36a3x3

= 84 *36a3x3

Given that,

Co-efficient of x2 = co-efficient of x3

=> 36 * 37a2 = 84 * 36a3

=> a3a2 = 36' 3/84'36

=> a = 3′3/7

97

New answer posted

7 months ago

0 Follower 2 Views

P
Payal Gupta

Contributor-Level 10

1.The general term of the expansion (a + b)n is given by

Tr +1 = nCran–rbr

So, T1 = nC0an = an

T2 = nC1an-1b = n!1!(n1)! an-1 b = n*(n1)!(n1)! an-1b = nan-1b

T3 = nC2an-2b2 = n!2!(n2)[! an-2b2 = n *(n1)*(n2)!2*1*(n2)! an-2b2 = n(n1)2 an-2b2

Given,

T1 = 729

=>an = 729 ------------------ (1)

T2 = 7290

=>nan–1b = 7290 ------------- (2)

T3 = 30375

=> n(n1)2 an–2b2 = 30375 ------------------- (3)

Dividing equation (2) by (1) we get,

nan1ban = 7290729

=> nba = 10

Similarly dividing equation (3) by (2) we get,

n(n1)2 an–2b2 ÷ nan–1b = 303757290

=> n(n1)2 an–2b21nan1b = 303757290

=> n(n1)an2b2 * 1nan1b = 303757290 * 2

=> (n1)ba = 25′ *26

=> nba ba = 253

=> 10 – ba = 253 [since, nba&

...more

Get authentic answers from experts, students and alumni that you won't find anywhere else

Sign Up on Shiksha

On Shiksha, get access to

  • 66k Colleges
  • 1.2k Exams
  • 681k Reviews
  • 1800k Answers

Share Your College Life Experience

×
×

This website uses Cookies and related technologies for the site to function correctly and securely, improve & personalise your browsing experience, analyse traffic, and support our marketing efforts and serve the Core Purpose. By continuing to browse the site, you agree to Privacy Policy and Cookie Policy.