Ncert Solutions Chemistry Class 11th

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R
Raj Pandey

Contributor-Level 9

2 C ( s ) + 2 H 2 ( g ) ? ? C 2 H 6 ( g ) Δ H = Δ H f ( C 2 H 6 )

Δ H f ( C 2 H 6 ) = [ 2 * ( 3 9 4 ) ] + [ 3 * ( 2 8 6 ) ] [ 1 * ( 1 5 6 0 ) ] K J / m o l e

= (-788 – 858 + 1560) KJ/mole = -86 kJ/mole

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A
alok kumar singh

Contributor-Level 10

InA-O-H, if EN of ' A' is 2.1 then it will be neutral, as XA-X0=X0-XH. (where X is EN)

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V
Vishal Baghel

Contributor-Level 10

Kindly go through the answers

(1.25)

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V
Vishal Baghel

Contributor-Level 10

Moles of NO =   6 0 0 * 1 0 3 * 1 0 0 3 0 = 2

moles of NO2 = 2

3NO2 + H2O -> 2HNO3 + NO

0 . 2 M 0 . 4 3 M o r p H = l o g 0 . 4 3 = 0 . 8 8        

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V
Vishal Baghel

Contributor-Level 10

Kindly go through the answers

(7.00)

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V
Vishal Baghel

Contributor-Level 10

q = 0, w = 0, U = 0 = f (V, T)

As volume changes, temperature must change to maintain constant internal energy.

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V
Vishal Baghel

Contributor-Level 10

Zero acidic hydrogen present in anionic part of (R)

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V
Vishal Baghel

Contributor-Level 10

Z = P V m R T 0 . 5 = 2 4 * V m 2 4

V m = 0 . 5 l / m o l e

( P + a V m 2 ) ( V m b ) = R T

( 2 4 + a 0 . 5 2 ) ( 0 . 5 0 . 1 ) = 2 4

2 4 + a 0 . 5 2 = 6 0

a 0 . 5 2 = 3 6

a = 3 6 * 0 . 5 2 = 9 a t m l 2 m o l 2
 

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V
Vishal Baghel

Contributor-Level 10

The blue color of the solution is due to the ammoniated electron which absorbs energy in infrared region of light

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V
Vishal Baghel

Contributor-Level 10

(0) Path A to D is adiabatic.                                                                                                                                              

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