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Mr = 2 ×14.4 = 28.8 When reactivity of the hydrocarbons and the value of Mr are taken into consideration then the mixture can only by formed from CH ≡ CH (M_r = 26) and CH₃ –CH ≡ CH (Mr = 40). 

2CrCl₃ + 3 Br₂ + 16 KOH → 2 K₂CrO₄ + 6 KBr + 6 KCl + 8 H₂O 

(a) Reaction with hydrochloric acid: 1.52 g – 0.56 g = 0.96 g of a metal reacted and 0.896 dm³ of hydrogen (0.04 mol) were formed. 0.96 

combining mass of the metal: 11.2 x 0.96/0.896 = 12 g

Possible solutions: 

Reaction: Mg + 2 HCl → MgCl₂ + H₂ 

(b) Reaction with sodium hydroxide: 

 1.52 g – 0.96 g = 0.56 g of an element reacted, 0.896 dm³ (0.04 mol) of hydrogen were formed. 

combining mass of the metal: 11.2 x 0.56/0.896 = 7 g

Possible solutions: 

Reaction: Si + 2 NaOH + H₂O → Na₂SiO₃ + 2 H₂ 

(c) Combining of both elements: 

0.96 g Mg + 0.56 g Si = 1.52 g of silicide Mg_xSi_y

w(Mg) = 0.96 g/1.52 g = 0.63

w (Si) = 0.56 g/1.52 g =0.37

x : y = 0.63/24 : 0.37/28 = 2:1

silicide: Mg₂Si 

(d) Reaction of the silicide with acid: 

Mg₂Si + 4 HCl → 2 MgCl₂ + SiH₄ 

n(Mg₂SI) =1.52 g/76 g mol-1 = 0.02 mol

n(SiH₄) = 0.448 dm3/22.4 dm3 mol-1 = 0.02 mol

(e) Reaction of silane with oxygen: 

SiH₄ + 2 O₂ → SiO₂ + 2 H₂O 

V = 1 dm³

On the assumption that T = const: p₂ =n₂/n₁ p₁

n₁(O₂) = 1dm³/22.4 dm³ mol-1 =0. 0446 mol

Consumption of oxygen in the reaction: n(O₂) = 0.04 mol 

The remainder of oxygen in the closed vessel: 

n₂(O₂) = 0.0446 mol – 0.04 mol = 0.0046 mol 

p₂ = 0 .0046 mol/0 .0446 mol x p₁ = 0.1 p₁

Compound A: 

K_xN_yO_z 

x : y : z = 38.67/39.1 = 13.85/14 = 47.48/16 = 0.989 : 0.989 : 2.968 = 1 : 1 : 3

A : KNO₃ 

Compound B: 

K_pN_qO_r

p : q r: 45.67/39.1 = 16.47/14 = 37.66/16 = 1.173 : 1.176 : 2.354 = 1 : 1 : 2

B : KNO₂ 

Equation: 2 KNO₃ → 2 KNO₂ + O₂ 

The reaction: IO₃^- + 5 I^- + 6 H⁺ = 3 I₂ + 3H₂O 

occurs to be quantitative both with respect to IO₃^- and H⁺ . Consequently the first titration (in the presence of sulphuric acid) is suitable for the determination of iodate, while the second one for the determination of the hydrochloric acid content.

CuO + H₂SO₄ → CuSO₄ + H₂O 

n(CuO) = m(CuO)/M(CuO) = 20 g/79.5 g mol^-1 = 0.2516 g

n(H₂SO₄) = n(CuSO₄) = 0.2516 mol 

Mass of the CuSO₄ solution obtained by the reaction: 

m(solution CuSO₄) = m(CuO) + m(solution H₂SO₄) = 

= m(CuO) + n(H₂SO₄) x M(H₂SO₄)/w(H₂SO₄) 

= 20 g + 0.2516 mol 98 g mo-1/ 0.20

m(solution CuSO₄) = 143.28 g 

Mass fraction of CuSO₄:  

(a) in the solution obtained:

w(CuSO₄) = m(CuSO₄)/m(solution CuSO₄) 

= n(CuSO ) (CuSO )/(solution CuSO₄) = 028

(b) in saturated solution of CuSO₄ at 20°C: 

w(CuSO₄) = 209 g/120.9 g = 0.173

(c) in crystalline CuSO₄ . 5 H₂O: 

w(CuSO₄) = M(CuSO₄)/M (CuSO₄.5H₂O) =0.639

Mass balance equation for CuSO₄: 

0.28 m = 0.639 m₁ + 0.173 m₂ 

m - mass of the CuSO₄ solution obtained by the reaction at a higher temperature. 

m₁ - mass of the crystalline CuSO₄ .5H₂O. 

m2 - mass of the saturated solution of CuSO₄ at 20 °C. 

0.28 × 143.28 = 0.639 m₁ + 0.173 × (143.28 - m₁) 

m1 = 32.9 g 

The yield of the crystallisation is 32.9 g of CuSO₄ . 5H₂O. 

Calculations are made on the assumption that the following reactions take place:

2 NaCl → 2 Na⁺ + 2 Cl 

cathode: 2 Na⁺ + 2 e^– → 2 Na 

anode: 2 Cl^- – 2 e^– → Cl–  

Cl₂ + Cu → CuCl₂ 

Because the electrolyte solution is permanently being stirred the following reaction comes into consideration: 

CuCl2 + 2 NaOH → Cu(OH)₂ + 2 NaCl 

On the assumption that all chlorine reacts with copper, the mass of NaCl in the electrolyte solution remains unchanged during the electrolysis

m(NaCl) = n M = c V M = 2 mol dm^-3 × 0.2 dm³ × 58.5 g mol^-1 = 23.4

V(H₂) = 22.4 dm³ , i. e. n(H₂) = 1 mol  

The amount of water is decreased in the solution by:  

n(H₂O) = 2 mol 

m(H₂O) = 36 g 

Before electrolysis: 

m(solution NaCl) = V ρ = 200 cm³ × 1.10 g cm^-3 = 220 g 

% NaCl = 23.4 g/220 g x 100 = 10.64

After electrolysis: 

m(solution NaCl) = 220 g – 36 g = 184 g

% NaCl = (23.4 g/184 g x 100) = 12.72

(b) 5-bromo-2-hydroxy-1-hexanal 

(b) NH⁺₄ – NH₃

H₃AsO₄ + 4Zn + 8H⁺ → AsH₃ + 4Zn²⁺ + 4H₂O 

(e) A solution neutral to litmus contains Na⁺, Ca²⁺, and PO^3-₄ .

(c) 9.8 × 10^-9 ;

Cl₂ + Ca(OH)₂ → CaOCl₂ + H₂O 

1. HCOOH (aq) + Br₂ (aq) → CO₂ (g) + 2H⁺ (aq) +2Br⁻(aq) 

2. BrO^-₃ + 5Br− + 6H⁺ →3 Br₂ + 3H₂O 

The ions of the remaining seven solutions can easily be identified by mutual reactions. Out of the 21 possible reactions,12 are common positive reactions. Additional information is available from the colour of 4, and the smell of one solution. AgNO₃: reacts with all the six compounds; NH₃: with the exception of iodide it gives a characteristic reaction with all the others salts;  

Fe(SCN)₃: its colour and reaction with NH₃,I^-, Ag⁺ are characteristic;COCl₂: can be detected from its colour and by adding NH₃ or Ag⁺; KI: can be identified by its reaction with Ag⁺ and from the evolution of I₂ due to an addition of Fe³⁺ or Cu²⁺;

CuCl₂: can be detected from its colour and reaction with NH₃, I^- and Ag⁺ ; 

NiCl₂: has a characteristic colour and reacts with NH₃ and Ag⁺ . 

(a) [Cr(H₂O)₆]³⁺ 4OH^–→ [Cr(OH)4(H₂O)₂]^– + 4H₂O 

2[Cr(OH)₄(H₂O)₂]^– + 3H₂O₂ + 2OH^– → 2 CrO^2-₄+12H₂O 

(b) Equation is given in 2a. 

Peroxo compounds contain the functional group: O^2-₂

Examples: H₂O₂, Na₂O₂, BaO₂, H₂SO_5, H₂S₂O_8, K₂C₂O₆, CrO_5, [VO₂]³⁺

(a) Mass percentage composition: 54.56 % C; 9.21 % H; 36.23 % O 

(b) Empirical formula: C₂H₄O 

(c) Molar mass: 88 g mol^-1 Molecular formula: C₄H₈O₂ 

(d) Possible structures: 

1. CH₃-CH₂-CH₂-COOH 

2. CH₃-CH(CH₃)-COOH 

3. CH₃-O-CO-CH₂-CH₃ 

4. CH₃-CH₂-O-CO-CH₃ 

5. CH₃-CH₂-CH₂-O-CO-H

6. CH₃-CH(CH₃)-O-CO-H  

7. CH₃-CH₂-CH(OH)-CHO

8. CH₃-CH(OH)-CH₂-CHO 

9. CH₂(OH)-CH₂-CH₂-CHO

10. CH₃-C(OH)(CH₃)-CHO 

11. CH₂(OH)-CH(CH₃)-CHO 

12. CH₃-O-CH₂-CH₂-CHO  

13. CH₃-CH₂-O-CH₂-CHO 

14. CH₃-O-CH(CH₃)-CHO 

15. CH₃-CH₂-CO-CH₂-OH 

16. CH₃-CH(OH)-CO-CH₃

17. CH₂(OH)-CH₂-CO-CH₃

18. CH₃-O-CH₂-CO-CH₃  

(e) The possible structures are 3, 4, 5, 6. 

(f) The structure of the compound C is CH₃-CH₂-O-CO-CH₃.