Chemical Ideas Answers 5w175a

Answers to AS problems in Chemical Ideas

Section 1.1   1 a 2.0 e 5.0   2 a 144  3

b 5.3 c 1.3 f 50.2 b Neodymium



d 10.0

Mass of sample/g

Amount of sample/mol

Number of atoms

– – 55.8 80.2 63.5

– 2.00 – – 0.50

6.02 × 1023 – 6.02 × 1023 12.04 × 1023 –

4 a The mass of the sample is needed to be sure that iodine and oxygen are the only elements in the compound. b The relative number of moles of iodine and oxygen. c To change the relative number of moles into the ratio of moles of oxygen relative to 1 mole of iodine. d In order to produce a ratio involving whole numbers e I2O5, I4O10, I6O15, etc. f The molar mass is needed.   5 a 1 e 0.25 i 2

b 0.5 f 0.5 j 5

c 0.25 g 0.25

d 0.1 h 0.1

Atoms of copper are approximately twice as heavy as atoms of sulfur. Thus the same mass contains only half as many moles of copper as it does of sulfur.   6 Black copper(II) oxide (CuO) contains equal numbers of copper and oxygen particles (Cu2+ and O2– ions). Red copper(I) oxide (Cu2O) contains twice as many copper particles as oxygen particles (Cu+ and O2– ions).   7 a H2O b CO c CS2 d CH4 e Fe2O3 f CuO g CaO h SO2 i MgH2   8 a 92.3 b 7.7 c CH   9 a SiH4 b CO c CO2 d MgO e C2H6O f CaCO3 g HClO3 h NaHCO3 10 a CH2 b P2O3 c AlCl3 d BH3 e C4H5 f C3H4 g CH2O h C12H22O11 11 a H2O2 b CO c C2H2 d C6H6 e C6H12 12 a 2 b 11 c 2 d 10 e 2 13 a 30.0 b 78.0 c 129.9 d 100.1 e 158.0 f 241.8 g 132.0 14 a 2 b 4 c 10 d 0.02 e 5 f 1 × 106

Section 1.2 1

a b c d e f g h i j

2Mg + O2 → 2MgO 2H2 + O2 → 2H2O 2Fe + 3Cl2 → 2FeCl3 CaO + 2HNO3 → Ca(NO3)2 + H2O CaCO3 + 2HCl → CaCl2 + CO2 + H2O H2SO4 + 2NaOH → Na2SO4 + 2H2O 2HCl + Ca(OH)2 → CaCl2 + 2H2O 2Na + 2H2O → 2NaOH + H2 CH4 + 2O2 → CO2 + 2H2O 2CH3OH + 3O2 → 2CO2 + 4H2O

2 3

a b c d e a b c d e

2Ca + O2 → 2CaO Ca + 2H2O → Ca(OH)2 + H2 C + CO2 → 2CO N2 + 3H2 → 2NH3 C3H8 + 5O2 → 3CO2 + 4H2O Zn(s) + H2SO4(aq) → ZnSO4(aq) + H2(g) Mg(s) + 2HCl(aq) → MgCl2(aq) + H2(g) MgCO3(s) → MgO(s) + CO2(g) 2C2H6(g) + 7O2(g) → 4CO2(g) + 6H2O(l) BaO(s) + 2HCl(aq) → BaCl2(aq) + H2O(l)

6 7 8

a b c d e f a b a b c d e f

56 tonnes S + O2 → SO2 64 g 64 tonnes 2 tonnes 112 tonnes 217 tonnes 0.58 tonnes Fe2O3 + 3CO → 2Fe + 3CO2 159.6 g 1.43 g 1.43 tonnes 2.86 tonnes 11.9 tonnes

Section 1.3 1 a All the magnesium reacts. b So that we know the number of moles of each substance involved in the reaction. c Mass of 1 mole of magnesium oxide. d Because 2 moles of magnesium oxide are produced. e To find the mass of magnesium oxide produced from 1 g of magnesium. f 80.6/48.6 would be multiplied by 50 rather than by 6. 2 20 g 3 a 2.8 g b 3.1 g c 2.5 g 4 3667 g (3.667 kg) 5 a C8H18 + 12.5O2 → 8CO2 + 9H2O b 175 kg c 154 kg

200 984_13_SAC SP_TT Ideas.indd 200

Salters Advanced Chemistry, Pearson Education Ltd 2008. © University of York. This document may have been altered from the original.

17/4/08 08:26:38



Answers to AS problems in Chemical Ideas

Section 1.4 1 The particles in a gas are much further apart than in a liquid or solid. In a gas, therefore, the volume of the particles is a very small part of the total volume and does not significantly affect it. In a liquid or solid the particles are close together and their volumes must be taken into when deciding on the total volume. 2 a CH4(g) + 2O2(g) → CO2(g) + 2H2O(g) b Volume of oxygen is twice that of methane. c The volume of water vapour formed is twice the volume of methane burnt. 3 a H2(g) + Cl2(g) → 2HCl(g) b Volumes of hydrogen and chlorine are the same. Volume of hydrogen chloride is twice the volume of hydrogen or chlorine.

a 2H2(g) + O2(g) → 2H2O(l) b 5 cm3 a 3 b 3 c 4 d 2 e C3H4 (probably contains a triple bond or a benzene ring) 6 2.4 dm3 7 1.2 dm3 8 a 0.25 b 2 c 48 dm3 d 240 dm3 3 e 30 dm

4 5

Section 1.5 1 a i 0.02 dm3 ii 1.5 dm3 3 b i 220 000 cm ii 1600 cm3 2 There is 0.4 mole of sodium hydroxide dissolved in every dm3 of the solution. 3 a 0.5 b 0.4 c 1 d 0.2 e 0.05 f 0.002 4 a 2 b 2 c 5 d 0.2 e 4 f 0.2 5 a 0.25 b 0.2 c 5 d 0.4 e 0.5 f 0.125 6 a 40 g b 4 g c 20 g d 0.4 g e 800 g f 1 g 7 a 117 g b 3.95 g c 1.4 g d 9936 g e 0.00239 g f 2.385 g g 0.0126 g h 0.1825 g i 24.96 g j 13.895 g 8 Concentration/g dm–3

Concentration/mol dm–3

– – 31.5 13.4 – – 0.6

5.15 1.49 – – 0.174 0.065 –

9 Na+ Cl– + Na CO32– + Ag NO3– 2+ Mg Br– H+ SO42– (or HSO4–) 10 a 1 mol dm–3 b 0.02 mol dm–3 c 0.3 mol dm–3 d 0.4 mol dm–3 11 a 0.0019 mol b 0.0019 mol c 0.076 mol d 0.076 mol dm–3 or 2.77 g dm–3 12 a 0.0022 mol b 0.0044 mol c 0.176 mol d 0.176 mol dm–3 13 a 7.75 × 10–4 (0.000 775)  mol dm–3 b 0.0574 g dm–3 14 a 0.0025 b 0.0025 c 25 cm3 15 a 0.02 mol b 0.04 mol c 1 dm3 d 500 cm3 3 e 20 cm

Section 2.1 1 Isotope

Symbol

Atomic number

Mass number

Number of neutrons

carbon-13

13 6

 6

–

–

oxygen-16

–

–

–

 8

strontium-90

–

–

  90

–

iodine-131

–

53

131

78

iodine-123

–

53

123

70

C

3 4 5 6

a Ar(Br) = 79.9 b Ar(Ca) = 40.1 a 100 – x b 193x c 191(100 – x) d 193x + 191(100 – x) e [193x + 191(100 – x)] ÷ 100 f 60% iridium-193, 40% iridium-191 40% antimony-123; 60% antimony-121 25% rubidium-87

2 Protons

Neutrons

Electrons

a

35

44

35

b

35

46

35

c

17

18

17

d

17

20

17

Salters Advanced Chemistry, Pearson Education Ltd 2008. © University of York. This document may have been altered from the original.

984_13_SAC SP_TT Ideas.indd 201

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