Chemistry NCERT Exemplar Solutions Class 12th Chapter Two

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2 months ago

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A
alok kumar singh

Contributor-Level 10

ΔTb=iKbm1

ΔTbΔTf=Kb*1Kf*236=12=Kbkf*12

KbKf=1x=1

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2 months ago

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A
alok kumar singh

Contributor-Level 10

 Number of neutrons = 26

Number of electrons = 25

% of extra electrons = 262525*100=4%

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2 months ago

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P
Payal Gupta

Contributor-Level 10

(c) 2CH3CH2Clether2NaC2H5C2H5+2NaCl (WurtzReaction)

(d) 2C6H5Clether2NaC6H5C6H5+2NaCl (WittingReaction)

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

 PT=PA+PB=0.8? ? atm

PA=PB=0.4atm

YA=0.5YB=0.5

XA=0.2

PAo=2atm.

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

 λ=h2mEE=KineticEnergy

E=h22mλ2

= (6.626*1034)22*9.1*1031* (3.3*1010)2

= 2.215 * 10-18

Eabsorbed – Erequired + K. E

EabsorbedErequired=1+K.EErequired

=1+2.215*101813.6*1.602*1019=2.016

 2

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A
alok kumar singh

Contributor-Level 10

PT=XAPA0-PB0+PB0

ATQ

550=14PA0-PB0+PB0

2200=PA0-PB0+4PB0

560=15PA0-PB0+PB0

2200=PA0-PB0+5PB0

PA0+3PB0=2200

PA0±4PB0=2800PB0=600

PA0=400mmHg

 

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A
alok kumar singh

Contributor-Level 10

Balmer series lies in the visible region.

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P
Payal Gupta

Contributor-Level 10

P A 0 = 5 0 t o r r

P B 0 = 1 0 0 t o r r

Mole fraction of A in liquid phase, xA = 0.3

Mole fraction of B in liquid phase, xB = 0.7

Now; PA=PA0xA

PB=PB0xB

= 100 * 0.7 = 70 torr

Mole fraction of B in vapour phase,  yB=PBPA+PB=7085

7085=x17

X = 14

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2 months ago

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P
Payal Gupta

Contributor-Level 10

Work function,  ? 0=6.63*1019J

Threshold wavelength λ0=?

Using ; ? 0=hcλ0

λ0=hc? 0

= 300 * 109 m = 300 nm

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

According to Henry's law

P = KH * mole fraction of solute

0.92=46.82*1000*molofO255.5

Mole of  O2 = 1.09 * 10-3

Millimole=1.091

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