Class 12th

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New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

(Cr, Mn) -> Thermite reduction

New answer posted

6 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

6 months ago

0 Follower 31 Views

V
Vishal Baghel

Contributor-Level 10

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

New answer posted

6 months ago

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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

6 months ago

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Vishal Baghel

Contributor-Level 10

Therefore option (b) is correct.

New answer posted

6 months ago

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Vishal Baghel

Contributor-Level 10

i = m t h e o r m a p p = 3 0 m a p p

Δ T b = i k b . m ] H A m x m = H + x + A x

0 . 0 1 5 6 = ( m * x m ) * 0 . 5 2 * m

m + x = 0.03

x = 0 . 0 1 i = 1 . 5 = 3 0 m a p p m a p p p = 2 0

 

New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

Addition on triple bond takes place by the addition of hydrogen

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Vishal Baghel

Contributor-Level 10

N = qvB

μ q v B = m d v d t = m v d v d S

μ q B S = m v

S = 3 * 1 0 6 * 4 0 . 3 * 1 0 6 * 0 . 2 = 2 0 0 m

New answer posted

6 months ago

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V
Vishal Baghel

Contributor-Level 10

1 u + 1 1 0 = 1 3 5

1 u = 1 3 5 1 1 0 = 2 7 7 0 = 1 1 4 1 u = 1 3 5 1 2 9 = 6 4 3 5 * 2 9 u = 3 5 * 2 9 6 4 = 1 5 . 8 5 t o 1 5 . 8 6

New answer posted

6 months ago

0 Follower 15 Views

V
Vishal Baghel

Contributor-Level 10

k Q e 0 . 1 2 + 1 2 m u 2 = k Q e 0 . 1 5 + 1 2 m u 2

k Q e * 0 . 0 3 0 . 1 2 * 0 . 1 5 = 1 2 m * ( v 2 u 2 )

Q = 9 . 1 * 9 9 * 1 0 1 1 1 2

= 7 5 . 0 7 5 * 1 0 1 2 4 π * ( 0 . 1 ) 2 = 5 . 9 7 n C / m 2

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