Class 12th

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New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

CoCl3.NH3 + AgNO3 A g C l ( 2 m o l )

[ C o ( N H 3 ) 5 C l ] C l 2 + A g N O 3 A g C l ( 2 m o l )  

x = 5

New answer posted

2 months ago

0 Follower 2 Views

R
Raj Pandey

Contributor-Level 9

kindly consider the following Image 

 

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

M n 2 + t 2 g 1 1 1 e g 1 1

(Number of unpaired electron = 5)

μ S = 3 5 = 5 . 9 1 6

New answer posted

2 months ago

0 Follower 2 Views

A
alok kumar singh

Contributor-Level 10

K = A . e E a / R T = ( 6 . 5 * 1 0 2 ) e 2 6 0 0 0 K / T  

E a 8 . 3 1 4 = 2 6 0 0 0

Ea = 216.164kJ/mol 216

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

In thin layer chromatography, the spots of colourless compounds, which are invisible to eyes can be detected by putting the plate under UV light.

Another detection technique is to place the plate in a covered jar containing I2 (s). Sometimes an appropriate reagent may also be sprayed on the plate, ninhydrin in case of amino acids.

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

H2 (g) + Cu2+ (aq)  2H+ (aq) + Cu (s)

E c e l l = E c e l l 0 2 . 3 0 3 R T n f l o g Q  

0.31 = 0.34 -  0 . 0 6 2 l o g [ H + ] C u 2 +  

[ C u 2 + ] = 1 0 7 ( ? [ H + ] = 1 0 3 )  

x = 7

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

DDT, organophosphates and dieldrin are pesticides.

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

[ N i ( C N ) 4 ] 2  

CN- is strong field ligand


N i 2 + = 4 s 0 3 d 8  

 Here; [ N i ( C N ) 4 ] 2 is square planar and diamagnetic.

[ N i ( C O ) 4 ]  

Ni = 4s23d8         Co is strong field ligand.

Here ;  [ N i ( C O ) 4 ]  is tetrahedral and diamagnetic

[ N i ( C N ) 4 ] 2  has 3d8 configuration while [ N i ( C O ) 4 ]  has 3d10 configuration.

New answer posted

2 months ago

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A
alok kumar singh

Contributor-Level 10

Δ T b = i K b m

Δ T f = i K f m

4 4 = K b * 1 . 5 K f * 4 . 5

K b K f = 3

New answer posted

2 months ago

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R
Raj Pandey

Contributor-Level 9

Valence shell electronic configuration of Eu in +2 O.S is,

  E u 2 + = 6 s 0 4 f 7  

It has half filled f-sub shell and stable.

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