Class 11th

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New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

New answer posted

2 months ago

0 Follower 9 Views

V
Vishal Baghel

Contributor-Level 10

S 1 reaction depends on carbocation stability and cation form in 3 will be most stable.

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

On adding reaction equilibrium constant will get multiplied.

New answer posted

2 months ago

0 Follower 4 Views

V
Vishal Baghel

Contributor-Level 10

H can easily gain electron to form its anion.

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

4Li + O? → 2Li? O oxide
2Na + O? → Na? O? peroxide
K + O? → KO? superoxide

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

Due to resonance C – Cl bond in option B is the shortest.

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

U and H are temperature dependent
Cp, m – Cv, m = R (for 1 mole of ideal gas)
dU = CvdT

New answer posted

2 months ago

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V
Vishal Baghel

Contributor-Level 10

T = constant
P = constant

PV = nRT

PdV = nRdT

PdV + VdP = 0

ΔV = nRΔT/P

dV = (-)VdP/P
|ΔV| = V (ΔP/P)
V/P ΔP = nRΔT/P
ΔT = V/nR ΔP
C = V/nR T/P = 300/2 = 150

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