Chemistry NCERT Exemplar Solutions Class 12th Chapter Eight

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New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

Except O 2 , remarginal C O 2 , O 3 , H 2 O , C F C s  are green house gases

New answer posted

2 months ago

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P
Payal Gupta

Contributor-Level 10

Enamines are inter conversible and have low stability with respect to imine. Among all C is most stable due to steric factor.

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2 months ago

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P
Payal Gupta

Contributor-Level 10

According to Ellingham diagram, the metal oxide with lower ΔGo is more stable than metal oxide of higher ΔGo at same temperature.

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

Correct electronic configuration of Pt is 78Pt = [Xe]4f145d96s1

It is an exceptional electronic configuration

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2 months ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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A
alok kumar singh

Contributor-Level 10

MnF4+4 MnF3+3

E.C = [Ar] 3d3                    [Ar]3d4

MnF2+2 [Ar]3d5 [EMn3+|Mn2+0=1.57VEMn4+|Mn2+0=1.27V]

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2 months ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution 
2MnO4+5H2O2+6H+2Mn2++8H2O+5O2

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

 IO4+H2O2IO3+O2

Oxidation number of l in,  IO4x+ (2)*4=1x=+7

Oxidation number of in,  IO3x+ (2)*3=1x=+5

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2 months ago

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A
alok kumar singh

Contributor-Level 10

eq of H2O2= eq of KMnO4

=0.316158*5=0.01=0.01*170.2*100=85%

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2 months ago

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V
Vishal Baghel

Contributor-Level 10

 No. of electron = 62 – 2 = 60

68Er+3 No. of electron = 68 – 3 = 65

70Yb+2 No. of electron = 70 – 2 = 68

71Lu+3 No. of electron = 71 – 3 = 68

65Tb+4 No. of electron = 65 – 4 = 61

63Eu+2 No. of electron = 63 – 2 = 61

65Tb+2 No. of electron = 65 – 2 = 63

69Tm+4 No. of electron = 69 – 4 = 65

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