Chemistry

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New answer posted

6 months ago

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V
Vishal Baghel

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

6 months ago

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V
Vishal Baghel

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

6 months ago

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V
Vishal Baghel

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

6 months ago

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Vishal Baghel

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

6 months ago

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Vishal Baghel

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

6 months ago

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V
Vishal Baghel

Contributor-Level 10

Moles of N2O= 2 . 2 4 4 = 1 2 0

Δ H = n C p Δ T = 1 2 0 * 1 0 0 ( 4 0 ) = 2 0 0 J

Δ U = q p + w

w = P e x t . Δ V

w = 1 ( 1 6 7 . 7 5 2 1 7 . 1 ) 1 0 0 0 * 1 0 1 . 3 J = + 5 J

Δ U = 2 0 0 + 5 = 1 9 5 J

New answer posted

6 months ago

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A
alok kumar singh

Contributor-Level 10

Kindly go through the image 

 

New answer posted

6 months ago

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A
alok kumar singh

Contributor-Level 10

Intermolecular H- bonding and intra-molecular H- bonding producing compound may be the phenol derivatives.

 

New answer posted

6 months ago

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Vishal Baghel

Contributor-Level 10

n R T P = 0 . 9 0 * 0 . 0 8 2 1 * 3 0 0 1 8 * 3 2 * 7 6 0 = 2 9 . 2 1 2 9

New answer posted

6 months ago

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alok kumar singh

Contributor-Level 10

Complete combustion of compound produces 0.2 gm CO2

Hence wt of carbon in 0.2 gm CO2

= ( 1 2 4 4 * 0 . 2 ) g m  

Therefore % of carbon in compound

= w t o f c a r b o n * 1 0 0 w t o f c o m p o u n d = 1 2 * 0 . 2 * 1 0 0 4 4 * 0 . 3 = 2 4 0 0 4 4 * 3 = 1 8 . 1 8 1 8 %  

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