Ncert Solutions Chemistry Class 12th

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3 months ago

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alok kumar singh

Contributor-Level 10

l + H 2 O 2 + 2 H + l 2 + 2 H 2 O

  (-1)                 oxidation       (0)

Here, I- is reducing agent

H 2 O 2   behaves like oxidizing agent

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Payal Gupta

Contributor-Level 10

Cu+ = 3d9 4s0

n = 1

μ = 1 * ( 1 + 2 ) = 3 = 1 . 7 3 B M

the nearest integer is 2.

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alok kumar singh

Contributor-Level 10

In Ce & Eu (+3) oxidation state is more stable

? C e + 4 + e ? ? C e + 3 ( R e d u c t i o n )     so CeO2 is oxidising agent

E u + 2 ? E u + 3 + e ? ( o x i d a t i o n ) so EuSO4 is reducing agent

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alok kumar singh

Contributor-Level 10

Kindly go through the solution 

 

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Payal Gupta

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Density = M*zNAa3

d=63.5*46.02*1023* (3.596*1010)3*1000=9076kg/m3

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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

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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Payal Gupta

Contributor-Level 10

Stable carbocation formation is the important factor.

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Payal Gupta

Contributor-Level 10

α-anomer of maltose is 1,4 combination. (C1-C4) glycosidic linkage

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