Chemistry NCERT Exemplar Solutions Class 12th Chapter Five

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2 months ago

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alok kumar singh

Contributor-Level 10

xf (x) + 2 = λ (x – 2) (x – 3) (x – 4) (x – 5)

x = 0 λ = 1 6 0                

x = 10 -> 10f (10) = 26

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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alok kumar singh

Contributor-Level 10

  f ( x ) = { 2 x 2 + 3 , 2 < x < 1 x 2 , 1 x < 0 3 2 1 , 0 x < 1 x 2 3 + 1 2 , 1 x < 2

f ( 1 ) = f ( 1 + ) = f ( 1 ) = 1                

f ( 1 + ) = f ( 1 ) = 1 2 2                

Points of discontinuity

x = 0, 1

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alok kumar singh

Contributor-Level 10

v = l a + m b  

= ( 2 l + m , l + 2 m , 2 l = m )               

  v . ( 3 , 2 , 1 ) = 0 l = 4 m . . . . . . . ( i )              

  | v . a ^ | = 1 9 9 l 2 m = 5 7 . . . . . . . . ( i i )              

(i) & (ii) Þ l = 6, m =   3 2

2 v = ( 2 1 , 1 8 , 2 7 )              

| 2 v | 2 = 1 4 9 4            

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alok kumar singh

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l i m x 1 x n f ( 1 ) f ( x ) x 1  

= l i m x 1 9 x n ( x 6 + 2 x 4 + x 3 + 2 x + 3 ) x 1                

= 9n – 19 = 44 -> n = 7

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alok kumar singh

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Due to equal and similar charge particle will repel each other, hence will never precipitate

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alok kumar singh

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Due to equal and similar charge particle will repel each other, hence will never precipitate.

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alok kumar singh

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Due to common ion effect solubility of AgCl will decreases in KCl, AgCl and AgNO3 but in deionized water, no common ion effect will takes place so maximum solubility.

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alok kumar singh

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Covalent character is L i C l > N a C l > K C l > C s C l .  As the cationic size increases polarization decreases.

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alok kumar singh

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(A) Cr = [Ar]3d54s1

              (B) m =   l t o + l

              (C) According to Aufbau principle, orbital are filled in order of their increasing energies.

              (D) Total nodes = n – 1

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