Chemistry NCERT Exemplar Solutions Class 12th Chapter Twelve

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3 months ago

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Payal Gupta

Contributor-Level 10

Moles of Br2= moles of C5H10560+10=570

w160=570

w=5*16070=807g

=1142.8*1021143*102g

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Payal Gupta

Contributor-Level 10

 Δt=hcλ=6.63*1034*3.08*108600*109

=6.63*3.08*1017600 J

=765.765*1021J

? 766*1021 J

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Payal Gupta

Contributor-Level 10

d [C]dt= (201010)=1mmoldm3

+d [D]dt= {d (B)dt}*1.5=32 {d [B]dt}

{d [B]dt}=2* {d [A]dt}

16 {d [B]dt}=13 {d [A]dt}=19 {+d [D]dt}=d [C]dt

 rate of reaction = +d [C]dt = 1m.m dm-3 S-1

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Payal Gupta

Contributor-Level 10

F e + 3 + l I 2 + F e + 2

Fe+2 undergoes reduction & I2 undergoes oxidation.

E c e l l o = E c a t h o d e o E a n o d e o

= (0.77 – 0.54) V = 0.23 V

= 23 * 10-2 V

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Payal Gupta

Contributor-Level 10

N2O4? 2NO2

(10.5)=0.5mol 2*0.5=1mol

kP= (11.5*1)2 (0.51.5*1)

=43=1.33

=710.15J/mol

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Payal Gupta

Contributor-Level 10

1000 ml solution contains 0.02 milli mole (mm)

500 ml solution contains 0.02 m.m

Solution made 1000 ml with H2O

m.m in final solution = 0.01 mm

Solution (A) + 0.01 m. m H2SO4

= 0.01 + 0.01

= 0.02 m.m

= 0.00002 * 103 mm

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Payal Gupta

Contributor-Level 10

ΔH=165kJ/mole

T =?

ΔS=550JK1

At equilibrium ; ΔG=0

T=ΔHΔS=165*1000550K

=3*100K=300K

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Payal Gupta

Contributor-Level 10

E1H=2.2*1018J

LiLi+2+2e, n=2

ELi+2=E1H*Z2n2=2.2*1018*3222

λ=4*108 m

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Payal Gupta

Contributor-Level 10

Molar mass of C7H5N3O6 = 12 * 7 + 1 * 5 + 14 * 3 + 16 * 6 = 84 + 42 + 96 = 227 g/mol

Number of moles = 6 8 1 2 2 7  = 3 mol

number of N-atoms

=3*6.02*1023*3=9*6.02*1023

=5418*1021

x=5418 

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