Ncert Solutions Chemistry Class 11th

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New answer posted•a year ago
S
a year ago

Moles of NO =   6 0 0 × 1 0 − 3 × 1 0 0 3 0 = 2

∴ moles of NO2 = 2

3NO2 + H2O -> 2HNO3 + NO

0 . 2 M 0 . 4 3 M           o r         p H = − l o g 0 . 4 3 = 0 . 8 8        

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New answer posted•a year ago
New answer posted•a year ago
S
a year ago

q = 0, w = 0, U = 0 = f (V, T)

As volume changes, temperature must change to maintain constant internal energy.

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New answer posted•a year ago
S
a year ago

Zero acidic hydrogen present in anionic part of (R)

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New answer posted•a year ago
S
a year ago

Z = P V m R T ⇒ 0 . 5 = 2 4 × V m 2 4

V m = 0 . 5 l / m o l e

( P + a V m 2 ) ( V m − b ) = R T

( 2 4 + a 0 . 5 2 ) ( 0 . 5 − 0 . 1 ) = 2 4

2 4 + a 0 . 5 2 = 6 0

a 0 . 5 2 = 3 6

a = 3 6 × 0 . 5 2 = 9 a t m − l 2 m o l 2
 

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New answer posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
S
a year ago

Red phosphorus undergoes disproportionation on reaction with alkali to form H4P2O6 and PH3

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New answer posted•a year ago