Chemistry

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New answer posted

9 months ago

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V
Vishal Baghel

Contributor-Level 10

Ammonical AgNO3   

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9 months ago

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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

(Cr, Mn) -> Thermite reduction

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9 months ago

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V
Vishal Baghel

Contributor-Level 10

A ( g ) ? 2 B ( g ) .

E m 2 m o l V l 4 m o l V l

N e w E q m 4 x V 4 + 2 x V

V * 2 4 x 2 V 4 2 x 2 V

New Eqm 4 x y 2 V 4 + 2 x + 2 y 2 V

= 4 Z 2 V = 4 + 2 Z 2 V ( Z = x + y )

Now, KC =  [ B ] 2 [ A ] = constant

o r , ( 4 V ) 2 ( 2 V ) = ( 4 + 2 Z 2 V ) ( 4 Z 2 V ) Z = 1 . 2 9

 mole of B at new Eqm = 4 + 2Z = 6.58

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9 months ago

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V
Vishal Baghel

Contributor-Level 10

Sucrose D (+) glucose + D (-) Fructose

  [ α ] = + 6 6 . 5 ° [ α ] = + 5 2 . 5 ° [ α ] = 9 2 . 8 °

Thus rotation changes from positive to negative after hydrolysis. Due to this reason hydrolysis of sucrose is known as inversion and mixture after hydrolysis is known as invert sugar.

Therefore option (B) is correct.

New answer posted

9 months ago

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V
Vishal Baghel

Contributor-Level 10

[ Z n ( g l y ) 2 ] is optically active compound

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V
Vishal Baghel

Contributor-Level 10

In H2O (polar solvent) dibromophenol derivative and in CS2 (non-polar solvent moneobromo phenol derivate is obtained.

New answer posted

9 months ago

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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
 

New answer posted

9 months ago

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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

New answer posted

9 months ago

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V
Vishal Baghel

Contributor-Level 10

Therefore option (b) is correct.

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