Ncert Solutions Chemistry Class 11th

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Vishal Baghel

Contributor-Level 10

Moles of N2O= 2 . 2 4 4 = 1 2 0

Δ H = n C p Δ T = 1 2 0 * 1 0 0 ( 4 0 ) = 2 0 0 J

Δ U = q p + w

w = P e x t . Δ V

w = 1 ( 1 6 7 . 7 5 2 1 7 . 1 ) 1 0 0 0 * 1 0 1 . 3 J = + 5 J

Δ U = 2 0 0 + 5 = 1 9 5 J

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Vishal Baghel

Contributor-Level 10

n R T P = 0 . 9 0 * 0 . 0 8 2 1 * 3 0 0 1 8 * 3 2 * 7 6 0 = 2 9 . 2 1 2 9

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Vishal Baghel

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B a C l 2 + S O 3 2 B a S O 3 W h i t e d i l . H C l S O 2 b u r n i n g s u l p h u r l i k e s m e l l

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alok kumar singh

Contributor-Level 10

Here, total meq of acetic acid = 50 * 0.1 = 5

And total meq of NaOH = 25 * 0.1 = 2.5

After neutralization process

Meq of left acetic acid = 2.5

And meq of formed CH3COONa = 2.5

p H = p K a + l o g 1 0 [ S ] [ A ]

p H = 4 . 7 6 + l o g 1 0 2 . 5 2 . 5 = 4 . 7 6 = 4 7 6 * 1 0 2  

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alok kumar singh

Contributor-Level 10

Hence, x = 727 (the nearest integer)

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Vishal Baghel

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Borax forms basic buffer solution in aqueous medium

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Vishal Baghel

Contributor-Level 10

Baking soda = NaHCO3

Washing soda = Na2CO3. 10H2O

Caustic soda = NaOH

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alok kumar singh

Contributor-Level 10

HOCl produce in the stratospheric cloud, by the hydrolysis reaction of ClONO2.

C l O N O 2 ( g ) + H 2 O ( g ) H O C l ( g ) + H N O 3 ( v a p )

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alok kumar singh

Contributor-Level 10

Due to H- bond in water, it has high melting point and melting point of other hydrides of the group are depending upon the molecular weight.

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alok kumar singh

Contributor-Level 10

Due to high crystallity Be has the highest M.P.

Be = 1560 K

Mg = 925 K

Ca = 1120 K

Sr = 1062 K

 

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