Ncert Solutions Chemistry Class 11th

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P
Payal Gupta

Contributor-Level 10

4HNOI3 (l)+3KCl (s)Cl2 (g)+NOCl (g)+2H2O (g)+3KNO3 (g)

4 moles of HNO3 produced 3 mol of KNO3

Here mole of produced KNO3110101

If 3 mol of KNO3 produced by 4 moles of HNO3

 1 mole of KNO3 produced by 43 moles of HNO3

and 110101 mole of KNO3 produced by 4*1103*101 moles of HNO3 = 1.45 mole of HNO3

Hence mass of HNO3 = mole * mol.wt = 145 * 63 = 91.48  91.5gm

New answer posted

3 months ago

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V
Vishal Baghel

Contributor-Level 10

Element              e- gain enthalpy

F                           -333

Cl                         -349

Te                        -190

Po             &nb

...more

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V
Vishal Baghel

Contributor-Level 10

Δ H = 3 ( Δ H c o m b ) C 2 H 2 ( g ) ( Δ H c o m b ) C 6 H 6 ( l )

= 1 3 0 0 * 3 ( 3 2 6 8 ) k J / m o l e

= ( 3 9 0 0 + 3 2 6 8 ) k J / m o l e

= 6 3 2 k J / m o l e

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A
alok kumar singh

Contributor-Level 10

C P , m = C v , m + R

C v , m = 2 0 . 7 8 5 8 . 3 1 4 = 1 2 . 4 7 1 J K 1 m o l 1

n = 5 0 0 0 1 2 . 4 7 1 * 2 0 0 = 2 5 1 2 . 4 7 1 2

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A
alok kumar singh

Contributor-Level 10

K O 2 , N O 2 , C l O 2 , N O are paramagnetic.

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Payal Gupta

Contributor-Level 10

For an acid- base titration, Methly orange exist at end point as quinonoid form.

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A
alok kumar singh

Contributor-Level 10

2 N O C l ( g ) ? 2 N O ( g ) + C l 2 ( g )

t = 0       2 mol/Lit             0            0

teq          2 – x                    x            x/2


K C = [ N O ] 2 * [ C l 2 ] [ N O C l ] 2 = x 2 * x ( 2 x ) 2 * 2 = x 3 ( 2 x ) 2 * 2      


= 0 . 4 * 0 . 4 * 0 . 4 1 . 6 * 1 . 6 * 2 = 0.0125 = 125 * 10-4

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3 months ago

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A
alok kumar singh

Contributor-Level 10

Moles of H C l = moles of N H 3

= 2 * moles of urea = 2 * 0.6 60 = 0.02

N . V = 0.02

New answer posted

3 months ago

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

Conceptual.

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