Ncert Solutions Chemistry Class 12th

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Vishal Baghel

Contributor-Level 10

In leaching suitable compound is used as solvent.

A l 2 O 3 + 2 N a O H + 3 H 2 O 2 N a [ A l ( O H ) 4 ]

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alok kumar singh

Contributor-Level 10

Stannane is the tetrahedral shape of molecule.

Hybridisation is- sp3, shape and structure are tetrahedral

 

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alok kumar singh

Contributor-Level 10

Liquation refining process is applicable for the metal having low m.p, but containing impurities have higher m.p

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alok kumar singh

Contributor-Level 10

Higher adsorbtion of gas is corresponds to higher liquefaction and higher liquefaction is directly proportional to the higher critical temperature.

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Vishal Baghel

Contributor-Level 10

(a) Nitric oxide formation -> Pt is used as catalyst

(b) Haber's process -> Fe is used as catalyst

(c) Hydrolysis of ester -> Acid (H2SO4) is used as catalyst

(d) SO3 formation -> NO is used as catalyst

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alok kumar singh

Contributor-Level 10

Total 3 streo- isomers

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alok kumar singh

Contributor-Level 10

Mole of polyhydric alcohol =  1 . 8 4 * 1 0 3 9 2 = 2 . 0 * 1 0 5 m o l e .

Mole of H2 gas produced = 1 . 3 4 4 2 2 4 0 0 = 6 . 0 * 1 0 5 m o l e .  

No of -OH gp present =  6 . 0 * 1 0 5 2 . 0 * 1 0 5 = 3

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alok kumar singh

Contributor-Level 10

Only H2S2O8 has per-oxo bond.

 

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alok kumar singh

Contributor-Level 10

λ t = l n N o N t

0 . 6 9 3 t 1 / 2 * t = l n 1 N t

 Or  0.693*10030=ln1Nt

  1 N t = 1 0

N t = 1 1 0 = 0 . 1              

Or Nt = 1 * 10-1   μ g

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