Class 12th

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New answer posted

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V
Vishal Baghel

Contributor-Level 10

Ammonical AgNO3   

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a year ago

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Vishal Baghel

Contributor-Level 10

(Cr, Mn) -> Thermite reduction

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a year 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

a year 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

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V
Vishal Baghel

Contributor-Level 10

Therefore option (b) is correct.

New answer posted

a year ago

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V
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

 

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a year ago

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Vishal Baghel

Contributor-Level 10

Addition on triple bond takes place by the addition of hydrogen

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V
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

a year 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

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