Chemistry NCERT Exemplar Solutions Class 11th Chapter Eight

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A
alok kumar singh

Contributor-Level 10

MnO? + H? C? O?2H? O - (H? )-> Mn²? + CO?
nf = (5) nf = (2)
Milli equivalent of C? O? ²? = mili equivalent of MnO?
2 x M x 10 = 5 x 0.05 x 10
M = 0.125 M
M = Strength / M.M of H? C? O?2H? O
Strength = 0.125 x 126g/L
= 15.75g/L

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a month ago

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alok kumar singh

Contributor-Level 10

H? SO? + 2NaOH → Na? SO? + 2H? O
Milli mole: 80 60 (limiting reagent)
H? + OH? → H? O
Milli mole: 160 60
Milli mole: 160-60 0 60
Total heat produced = (60/1000) x 57.1 x 1000 = 3426 J
Total heat absorbed = msΔT
ΔT = Totalheat / (m x s)
= 3426J / (1000 x 4.18) = 0.819K
= 81.9 x 10? ² K
Ans. = 82 (the nearest)

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alok kumar singh

Contributor-Level 10

3 [Co (en)? Cl? ]Cl + 3AgNO? (excess) → 3AgCl↓ (ppt) + 3 [Co (en)? Cl? ] NO?
Secondary valency of complex = 6

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a month ago

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alok kumar singh

Contributor-Level 10

E.C = σ1s² σ1s² σ2s² σ2s² σ2p? ² π2p? ² = π2p? ² π2p? ² = π2p? ²
Total electrons in BMO = 10

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

E_cell = E? _cell - (0.059/2)log { [Cu²? ]/ [Ag? ]²}
E? = E? _cell - (0.059/2)log [0.1/ (0.01)²] = 0.3095
E? _cell = 0.3095 + 0.0885 V
E? _cell = 0.398

Again, E? = E? _cell - (0.059/2)log [10? ²/ (10? ³)²]
E? = 0.398 - (0.059/2) x 4 V

E? = 0.28 V
E? = 28 x 10? ² V

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a month ago

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alok kumar singh

Contributor-Level 10

Kp = (Po? )¹/² = 4
Po? = 16 atm

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a month ago

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alok kumar singh

Contributor-Level 10

2SO? (g) + O? (g)? 2SO? (g)
t=0: 250 m bar 750 m bar -
0 625 m bar 2 x 250 = 500
P_total = 625 + 250m bar
= 875mbar

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a month ago

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alok kumar singh

Contributor-Level 10

A → 2B
t = 0: 1 mole 0
after 100 min: (1 - 0.1) mol 0.2 mol
K = (1/t)ln [a/ (a-n)]
K = (2.303/100)log (1/0.9) min? ¹
0.693/t? /? = (2.303/100) [log10 - log9] = (2.303/100) x 0.046
t? /? = 69.3 / (2.303 x 0.046) = 654.15 min
Ans. = 654 (the nearest integer)

New answer posted

a month ago

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alok kumar singh

Contributor-Level 10

i = total no. of particles after dissociation/association / total no. of particle before dissociation/association
HA? H? + A? (Dissociation)
0.5a
2HA → (HA)? (Association)
0.5a/2
i = (a + 0.25a)/a = 125 x 10? ²
Ans. = 125

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