Ncert Solutions Chemistry Class 11th

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New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

C N - 14 electrons, so B . O = 3  and diamagnetic.

New answer posted

3 months ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

  V r m s > V a v > V m p

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

p K b 1 K b 1   Basic strength                                                                                                        

New answer posted

3 months ago

0 Follower 2 Views

V
Vishal Baghel

Contributor-Level 10

Theory based.                                                                                                                                                

New answer posted

3 months ago

0 Follower 1 View

V
Vishal Baghel

Contributor-Level 10

Theory based.                                                                                                                                                                             

New question posted

3 months ago

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New answer posted

3 months ago

0 Follower 6 Views

R
Raj Pandey

Contributor-Level 9

PV = nRT
n = PV/RT = (1 atm * 4*10? L) / (0.083 LatmK? ¹mol? ¹ * 300 K) = 1.6 * 10? mol
Mass = n * Molar Mass = 1.6 * 10? mol * 16 g/mol = 25.6 * 10? g ≈ 26 * 10? g

New answer posted

3 months ago

0 Follower 4 Views

R
Raj Pandey

Contributor-Level 9

Energy per second = 1000 J / 10 s = 100 J/s
Energy of one photon E = hc/λ = (6.626*10? ³? * 3*10? ) / (400*10? ) = 4.965 * 10? ¹? J
Number of electrons ejected = Total energy / Energy per photon = 100 / (4.965 * 10? ¹? ) = 20.14 * 10¹? ≈ 2 * 10²?

New answer posted

3 months ago

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R
Raj Pandey

Contributor-Level 9

A + B? 2C
Initial: 1, 1
At eq: 1-x, 1+2x
K = [C]²/ ( [A] [B]) = (1+2x)²/ (1-x)² = 100
(1+2x)/ (1-x) = 10
1+2x = 10-10x => 12x = 9 => x = 3/4
[C] = 1+2x = 1+2 (3/4) = 1+1.5 = 2.5M = 25 * 10? ¹M

New answer posted

3 months ago

0 Follower 5 Views

R
Raj Pandey

Contributor-Level 9

C? H? + 13/2 O? → 4CO? + 5H? O
1 mole C? H? (58 g) produces 5 mole H? O (90 g)
∴ 90 g H? O obtained from 58 g C? H?
∴ 72g H? O obtained from (58/90) * 72g = 46.4 g
= 464 * 10? ¹g

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