Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

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A
alok kumar singh

Contributor-Level 10

At equilibrium rate of forward and backward reaction becomes equal.

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V
Vishal Baghel

Contributor-Level 10

 2NO (g)+O2 (g)? 2NO2 (g)

Initial moles2mol1 mol

At equilibrium (2 – 2x) mole (1 – x) mol2x mol

no2 (1x)=0.6

x=0.4

nNO=22x=22*0.4

= 2 0.8 = 1.2

nNO2=2x=2*0.4=0.8

Kp= (0.82.6)2 (1.22.6)2 (0.62.6)

= 1.925

the nearest integer = 2.

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Vishal Baghel

Contributor-Level 10

I 2 + 1 0 H N O 3 ( c o n c ) 2 H l O 3 + 1 0 N O 2 + 4 H 2 O

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Vishal Baghel

Contributor-Level 10

I 2 + 1 0 H N O 3 ( c o n c ) 2 H l O 3 + 1 0 N O 2 + 4 H 2 O

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Payal Gupta

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C r 2 O 7 2 + 6 e + 1 4 H + 2 C r 3 + + 7 H 2 O  

Here, 6F electricity is required to reduce 1 mol  C r 2 O 7 2 to Cr3+.

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P
Payal Gupta

Contributor-Level 10

Using ; ksp = 108 S5

1.1 * 10-23 = 108 S5

S = ( 1 1 0 1 0 8 * 1 0 2 5 ) 1 / 5 M 1 0 5 M  

Specific conductance,

λ m 0 = κ 5 * 1 0 0 0 S m 2 m o l 1                              

= 3 * 10-3 Sm2 mol-1

So; x = 3

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Payal Gupta

Contributor-Level 10

Pentavalent oxides of group – 15 elements, E2O5 is more acidic than trivalent oxides, E2O3 of the same element.

Acidic strength of trivalent oxides decreases down the group as metallic strength increases.

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Payal Gupta

Contributor-Level 10

N2 gas is obtained by thermal decomposition of Ba (N3)2 as,

B a ( N 3 ) 2 Δ B a + 3 N 2

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V
Vishal Baghel

Contributor-Level 10

 S=Ksp=8*1028=2.82*1014

Or solubility of PbS = 282 * 10-16

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Vishal Baghel

Contributor-Level 10

HNO3    +            NaOH    ®          NaNO3      +        H2O

t = 0,     (600 * 0.2)meq    (400 * 0.1)meq    40 meq               40 meq

120 meq             40 meq

t =     80 meq               -

Total heat

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