Chemistry NCERT Exemplar Solutions Class 12th Chapter Eight

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R
Raj Pandey

Contributor-Level 9

496.00

P P M =   Mass of Iron     Mass of wheat   * 10 6

Mass of 10 =   Mass of Iron   100 * 10 3 * 10 6

Mole of F e = 1 g m

F e = 1 56  contain F e S O 4 7 H 2 O  mole of  atom = 1

  F e in 1 mole

56 g  in  1 g  mole

  1 56 Mass

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R
Raj Pandey

Contributor-Level 9

48.00

| w | = 1 2 ( 6 + 10 ) = 48 J

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R
Raj Pandey

Contributor-Level 9

1.52

E = E 0 - 0.0591 4 l o g ? H + 4

E = 1.23 + 0.0591 * p H

E = 1.23 + 0.0591 * ( 5 )

E = 1.52

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R
Raj Pandey

Contributor-Level 9

Kindly consider the following Image

 

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R
Raj Pandey

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C o N H 3 6 C l 3 + 3 A g N O 3 ? 3 A g C l

  Mole of   C o N H 3 6 C l 3 1 =   Mole of   A g N O 3 3

0.3 267.46 = 0.125 * V * 10 - 3 3

V = 26.92 m L

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R
Raj Pandey

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Order of I.E:           N a < A l < M g < S i

496 577 737 786

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Raj Pandey

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Number of bond between sulphur and oxygen = 8

Number of bond between sulphur and sulphur = 8

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R
Raj Pandey

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v = 1 λ = R z 2 ? 1 n 1 2 - 1 n 2 2

For H atom z = 1

v = R 1 n 1 2 - 1 n 2 2

For Balmer series:  n 1 = 2

If n 2 = 3 1 λ = R z 2 1 4 - 1 9

1 λ = R 5 36

λ m a x = 36 5 R

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R
Raj Pandey

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Aq. N a O H  in a burette and aqueous oxalic acid in a conical flask.

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Raj Pandey

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Without catalyst: K = A e - E a / R T

  Presence catalyst:   10 6 K = A e - E c / R T E q ( B ) - E q ( A ) 10 6 = e - ( E - E c ) / R T E - E c R T = 2.303 * 6 Δ E = E c - E = ( - 2.303 ) * 6 R T

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