Class 11th

8.6. Write formulas for the following compounds:
(a) Mercury (II) chloride, (b) Nickel (II) sulphate, (c) Tin (IV) oxide, (d) Thallium
(I) sulphate, (e) Iron (III) sulphate, (f) Chromium (III) oxide.

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(a) HgCl₂ (b) NiSO₄ (c)SnO₂ (d) Tl₂SO₄ (e) Fe₂ (S0₄)₃ (f) Cr₂O₃.

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8.5. Calculate the oxidation number of sulphur, chromium and nitrogen in H₂SO₅, Cr₂O₇^2-and NO₃^-. Suggest structure of these compounds. Count for the fallacy.

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(a) H_2?SO₅? by conventional method.
Let x be the oxidation number of S
2 (+1) + x + 5 (−2) = 0
x = +8
+8 oxidation state of S is not possible as S cannot have an oxidation number more than 6. The fallacy is overcome if we calculate the oxidation number from its structure HO−S (O₂)−O−O−H.
−1+X+2 (−2)+2 (−1)+1=0
x=+6

(b) Dichromate ion
Let x be the oxidation number of Cr in dichromate ion
2x+7 (−2)=−2
x=+6
Hence the oxidation number of Cr in dichromate ion is +6. This is correct and there is no fallacy.

(c) Nitrate ion, by conventional method
Let x be the oxidation number of N in nitrate ion.
x+3 (−2)=−1
From the

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8.4. Fluorine reacts with ice and results in the change:

H₂O(s) + F₂ (g) ——> HF (g) + HOF (g)
Justify that this reaction is a redox reaction.

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Writing the O.N. of each atom above its symbol, we have,

here, the O.N. of F decreases from 0 in F₂ to -1 in HF and increases from 0 in F₂ to +1 in HOF. Therefore, F₂ is both reduced as well as oxidised. Thus, it is a redox reaction and more specifically, it is a disproportionation reaction.

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8.3. Justify that the following reactions are redox reactions:
(a) CuO (s) + H₂(g) → Cu(s) + H₂O(g)
(b) Fe₂O₃(s) +3CO(g) → 2Fe(s) + 3CO₂(g)
(c) 4BCl₃(g) +3LiAlH₄(s) → 2B₂H₆(g) + 3LiCl(s) + 3AlCl₃(s)
(d) 2K(s) +F₂(g) → 2K⁺F^–(s)

(e) 4 NH₃(g) + 5 O₂ → 4NO (g) + 6H₂O (g)

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(a) Here, O is removed from CuO, therefore, it is reduced to Cu, while O is added to H₂ to form H₂O, therefore, it is oxidised. Further, O.N. of Cu decreases from +2 in CuO to 0 in Cu but that of H increases from 0 in H₂ to +1 in H₂0. Therefore, CuO is reduced to Cu but H₂ is oxidised to H₂O. Thus, this is a redox reaction.

(b) Here O.N. of Fe decreases from +3 in Fe₂O₃ to 0 in Fe while that of C increases from +2 in CO to +4 in CO₂. Further, oxygen is removed from Fe₂O₃ and added to CO, therefore, Fe₂O₃ is reduced while CO is oxidised. Thus, this is a redox reaction.

(c) Here, O.N. of B decreas

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5.19. A mixture of dihydrogen and dioxygen at one bar pressure contains 20% by weight of dihydrogen. Calculate the partial pressure of dihydrogen.

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5.19. As the mixture H₂and O₂ contains 20% by weight of dihydrogen, therefore,

If H₂ = 20g, then O₂ = 80g

No. of moles of H₂ = 20/2 = 10 moles

No. of moles of O₂ = 80/32 = 2.5 moles

Partial pressure of H₂= [No. of moles of H₂/ (No. of moles of H₂+ No. of moles of O₂)] x

P_total= [10 (10 + 2.5)] x 1 = 0.8 bar

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8.2. What is the oxidation number of the underlined elements in each of the following and how do you rationalise your results?

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(a) In Kl₃, since the oxidation number of K is +1, therefore, the average oxidation number of iodine = -1/3. But the oxidation number cannot be fractional. Therefore, we must consider its structure, K⁺ [I —I < I]^–. Here, a coordinate bond is formed between I₂molecule and I^– ion. The oxidation number of two iodine atoms forming the I₂ molecule is zero, while that of iodine forming the coordinate bond is -1. Thus, the oxidation number of the three I atoms, atoms in Kl₃ is 0, 0 and -1, respectively.

(b) By conventional method O.N. of S in H₂S₄O₆is calculated as:

2 (+1) +4x + 6) (-2) = 0

Or x = +2.5

But all the four

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8.1. Assign oxidation number to the underlined elements in each of the following species:

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Let x be the oxidation number to the underlined elements in the given species:

(a) NaH₂PO₄

(+1) + 2 (+1) + x + 4 (-2) = 0

x + 3 – 8 = 0

x = +5

(b) NaHSO₄

(+1) + (+1) + x + 4 (-2) = 0

x – 6 = 0

x = +6

(c) H₄P₂O₇

4 (+1) + 2x + 7 (-2) = 0

2x -10 =0

x = +5

(d) K₂MnO₄

2 (+1) + x + 4 (-2) = 0

x – 6 = 0

x = +6

(e) CaO₂

2 + 2x = 0

x = -1

(f) NaBH₄

1 + x + 4 (-1) = 0 (Since H is present as hydride ion.)

x = +3

(g) H₂S₂O₇

2 (+1) + 2x + 7 (-2) = 0

x = +6

(h) KAl (SO₄)₂.12H₂O

+1 + 3 + 2x + 8 (-2) + 12 (2 x 1 - 2) = 0

x = +6

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5.18. 2.9 g of a gas at 95°C occupied the same volume as 0.184 g of hydrogen at 17°C at the same pressure. What is the molar mass of the gas?

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5.18. Given, P₁ = P₂ and V₁ = V₂

We know that P₁V₁ = P₂V₂

Or, n₁RT₁ = n₂RT₂

i.e., n₁T₁ = n₂T₂

Substituting n = w/M, we get

(W₁/M₁) x T₁ = (W₂/M₂) x T₂

(2.9/M₁) x (95 + 273) = (0.184/2) x (17 + 273)

M₁ = (2.9 x 368 x 2) / (0.184 x 290) = 40 g mol^-1

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5.17. Calculate the volume occupied by 8.8 g of CO₂ at 31.1 °C and 1 bar pressure. R = 0.083 bar LK^-1 mol^-1

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5.17. No. of moles of CO₂ = Given mass of CO₂ / Molar mass

= 8.8g / 44g mol^-1 = 0.2 mol

Pressure of CO₂ = 1 bar

RR = 0.083 bar dm³ K^–1 mol^–1

T = 273 + 31.1 K = 304.1 K

According to ideal gas equation,

PV = nRT

Therefore, V = nRT/P

= (0.2 x 0.083 x 304.1) / 1 bar = 5.048 L

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5.16. Pay load is defined as the difference between the mass of the displaced air and the mass of the balloon. Calculate the pay load when a balloon of radius 10 m, mass 100 kg is filled with helium at 1.66 bar at 27°C (Density of air = 1.2 kg m^-3 and R = 0.083 bar dm³ K^-1 mol^-1).

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