Chemistry

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New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

Kindly go through the answers

(13.00)

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New answer posted

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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.

New answer posted

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V
Vishal Baghel

Contributor-Level 10

Ammonical AgNO3   

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V
Vishal Baghel

Contributor-Level 10

Zero acidic hydrogen present in anionic part of (R)

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Vishal Baghel

Contributor-Level 10

(Cr, Mn) -> Thermite reduction

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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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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.

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3 months ago

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