Chemistry Answers Chapter 1

Answers Chapter 1 Exercises 1 (a) CuCO3 → CuO + CO2 6 X has diffused more quickly, so it must be a lighter gas. Its particles have greater velocity than the particles of Y at the same temperature. (Note though that they will both have the same value for average kinetic energy.) 7 From the kinetic molecular theory we would expect a solid to be more dense than its liquid, and therefore that ice would sink in water. 8 Bubbles will be present through the volume of the liquid. A brown gas is visible above the brown liquid. As the two states are at the same temperature, the particles have the same average kinetic energy and are moving at the same speed. The inter-particle distances in the gas are significantly larger than those in the liquid. 9 At certain conditions of low temperature and low humidity, snow changes directly to water vapour by sublimation, without going through the liquid phase. 10 Steam will condense on the skin, releasing energy as it forms liquid at the same temperature (E–D on Figure 1.4). This is additional to the energy released when both the boiling water and the condensed steam cool on the surface of the skin. (a) 2KNO3(s) → 2KNO2(s) + O2(g) 11 B (b) CaCO3(s) + H2SO4(aq) → CaSO4(s) + CO2(g) + H2O(l) 12 AW 01_EX_12 (b) 2Mg + O2 → 2MgO (c) H2SO4 + 2NaOH → Na2SO4 + 2H2O (d) N2 + 3H2 → 2NH3 (e) CH4 + 2O2 → CO2 + 2H2O 2 (a) 2K + 2H2O → 2KOH + H2 (b) C2H5OH + 3O2 → 2CO2 + 3H2O (c) Cl2 + 2KI → 2KCl + I2 (d) 4CrO3 → 2Cr2O3 + 3O2 (e) Fe2O3 + 3C → 3CO + 2Fe 3 (a) 2C4H10 + 13O2 → 8CO2 + 10H2O (b) 4NH3 + 5O2 → 4NO + 6H2O (c) 3Cu + 8HNO3 → 3Cu(NO3)2 + 2NO + 4H2O (d) 6H2O2 + 2N2H4 → 2N2 + 10H2O + O2 (e) 4C2H7N + 15O2 → 8CO2 + 14H2O + 2N2 4 (a) Sand and water: heterogeneous (b) Smoke: heterogeneous (c) Sugar and water: homogeneous (d) Salt and iron filings: heterogeneous (e) Ethanol and water: homogeneous (f) Steel: homogeneous 5 (c) 2Li(s) + 2H2O(l) → 2LiOH(aq) + H2(g) (d) Pb(NO3)2(aq) + 2NaCl(aq) → PbCl2(s) + 2NaNO3(aq) (e) 2C3H6(g) + 9O2(g) → 6CO2(g) + 6H2O(l) In the answers, exercises and text there is inconsistency about whether the water formed in combustion reactions is gas or liquid. At room temperature it will ultimately finish up as a liquid, as shown here. 2 10:01 13 This is the same as the number of moles of carbon atoms used. These calculations have used L = 6.02 × 1023 (a) 7.22 × 1022 (b) 3.01 × 1024 (c) 1.20 × 1023 Therefore the number of carbon atoms used = moles of chalk × (6.02 × 1023 mol–1) = 6.02 × 1023 y 100.09 14 0.53 mol H 15 0.250 mol 16 (a) 262.87 g mol−1 (b) 176.14 g mol−1 (c) 164.10 g mol−1 (d) 248.22 g mol−1 36 (a) 2.50 mol (b) 5.63 mol (c) 665.5 g (a) 2C4H10 + 13O2 → 8CO2 + 10H2O 17 189.1 g 18 1.5 mol 19 0.0074 mol Cl− 38 4.355 kg 20 1.83 × 10 C atoms 39 (a) CaCO3 → CaO + CO2 21 171 g (integer value because no calculator) 22 10.0 g H2O 23 2.0 mol N2 > 3.0 mol NH3 > 25.0 mol H2 > 1.0mol N2H4 24 (a) CH (b) CH2O (c) C12H22O11 (d) C4H9 37 (b) 1.59 g 24 (e) C4H7 (f) CH2O (b) 92.8% (c) CaCO3 is the only source of CO2; all the CaCO3 undergoes complete decomposition; all CO2 released is captured; heating does not cause any change in the mass of the other minerals present. 40 (a) 85.2 g 41 5.23 g C2H4Cl2 (b) 1.3 g H2 25 Na2S2O3 42 253.2 g theoretical CaSO3; 77.9% 26 CoSO4.7H2O 43 3.16 g ester 27 C17H25N 44 107 g of C6H6 needed NH3 45 (a) 2.40 mol 28 29 6.94 Li 30 CdS 31 empirical formula CH; molecular formula C6H6 32 empirical formula H2PO3; molecular formula H4P2O6 33 C10H16N5P3O13 for both empirical and molecular formulas 34 C 3 H 8O 35 Let y = mass of chalk in grams. moles of chalk used = yg y 100.09 g mol–1 (b) 0.0110 mol (c) 44 mol mass used × Mr(CaCO3) 46 (a) 35.65 dm3 (b) 5.7 dm3 47 0.652 dm3 48 0.138 mol Br2 and 0.156 mol Cl2, so more molecules of Cl2 49 0.113 dm3 50 0.28 dm3 51 90 kPa 52 16 °C 53 3.0 dm3 54 2.8 dm3 3 10:01 Answers 55 M = 133 g mol−1 so gas is Xe 56 90.4 g mol 57 Helium 58 311 dm3 59 empirical formula and molecular formula = SO3 60 At higher altitude the external pressure is less. As the air in the tyre expands on heating (due to friction with the road surface), the internal pressure increases. 61 73 52% NH3 by mass; assuming no side reactions occur and gases behave as ideal gases 74 3.20 × 105 kg 75 0.225 mol dm−3 (or round to 0.23 mol dm−3) 76 [Na2CO3] = 77 100 mol O2; 3.2 kg −1 (a) Particles are in constant random motion and collide with each other and with the walls of the container in perfectly elastic collisions. The kinetic energy of the particles increases with temperature. There are no inter-particle forces and the volume of the particles is negligible relative to the volume of the gas. (b) At low temperature, the particles have lower kinetic energy, which favours the formation of inter-particle forces and reduces gas PV pressure. <1 nRT YP mol dm−3 X Y(Q − P) [NaHCO3] = mol dm−3 X Challenge yourself 1 In cold climates, temperature may approach or go below the boiling point of butane so the butane stays liquid even when it is released from the pressure it is under when stored in its canister. This makes it ineffective as a fuel. 2 FeCl3.6H2O, CuSO4.5H2O, Co(NO3)2.6H2O 3 N = 18%, P = 22%, K = 17% 4 Many reactions with ‘useless’ by-products could have high stoichiometric yield under optimum conditions, but low atom economy, for example, methanoic acid production: 62 NH3 shows greater deviation than CH4 due to stronger intermolecular attractions, especially at low temperature. 63 B 64 2.81 g 65 4.93 g 66 0.0100 mol 67 0.400 mol dm−3 68 3.1 cm3 10Na(s) + 2KNO3(s) → K2O(s) + 5Na2O(s) + N2(g) 69 0.178 mol dm−3 70 0.0220 mol dm−3, 0.0802% HCl K2O(s) + Na2O(s) + SiO2(s) → Na2K2SiO4 (alkaline silicate glass) 71 0.106 mol dm−3 Na2SO4 and 0.115 mol dm−3 Pb(NO3)2; assume no side reactions, all PbSO4 precipitates 72 1217 tonne 2NaCOOH + H2SO4 → 2HCOOH + Na2SO4 For 100% conversion with stoichiometric reactants, the yield = 100%. 2 × 46.03 × 100% = atom economy = (2 × 68.01) + 98.08 39% 5 6 2NaN3(s) → 2Na(s) + 3N2(g) As NaOH dissolves, the separated Na+ and OH− ions become hydrated, i.e. they are surrounded by H2O molecules. This involves breaking the hydrogen bonds between the H2O molecules in pure water and allows closer packing, which reduces the volume. 4 10:01 ( ( (( ) 1 D 2 D 3 B 4 C 5 A 6 B 7 D 8 D 9 D 10 D 11 A 12 C 13 D 14 C 15 D 16 A 17 (a) temperature: 4 ( ( ( mass: 3 pressure: 3 [1] so molecular formula is C2HCl3 [4] Award [3] (max) for 0.0341 dm3 or 22.7 dm3. Allow whole numbers for atomic masses. 19 3 Award [2] (max) for 22.9 dm . 3 Award [2] (max) for 0.0227 dm . Award [2] (max) for 0.034 dm3. (a) (( ) ) (( ) ) (( ) ) 2 × 1.01 (0.089) = 1.0 × 10−2 g H and 18.02 12.01 (0.872) = 2.38 × 10−1 g C 44.01 0.238 (100) = 18.3% C 1.30 1.0 × 10−2 (100) = 0.77% H [3] 1.30 Award [3] for correct final answer of 18.3%C and 0.77% H without working. Allow whole numbers for molar masses. ( ) NH3/ammonia in excess, by 10 dm3 volume of N2 produced = 25.0 dm [1] [2] 3 (a) n(HCl) = 0.200 mol dm−3 × 0.02720 dm3 = 0.00544 or 5.44 × 10−3 (mol) [1] (b) n(HCl) excess is 0.100 mol dm−3 × 0.02380 [1] dm3 = 0.00238 or 2.38 × 10−3 mol Penalize not dividing by 1000 once only in (a) and (b). (c) n(HCl) reacted = 0.00544 – 0.00238 = 0.00306 or 3.06 × 10−3 (mol) Award [3] (max) for 34.4 dm3. 18 [3] Award [3] for correct final answer without working. 20 Award [4] for correct final answer. ) ) ) Mr = (24.02 + 1.01 + 106.35) = 131.38 R = 8.31 J K−1 mol−1 (from IB Data booklet) Use PV = nRT (from IB Data booklet) nRT 1.50 × 8.31 × 298.15 V= = = 1.08 × 1.01 × 105 P 3 3 0.034.1 m = 34.1 dm ) Empirical formula = C2HCl3; Award [2] for correct empirical formula without working. (b) 0.0650 kg = 65.0 g 65.0 n= = 1.00 (mol) [1] 65.02 No penalty for using whole number atomic masses. 3 (c) n(N2) = × 1.00 = 1.50 (mol) 2 T = 25.00 + 273.15 = 298.15 K or 25.00 + 273 = 298 K P = 1.08 × 1.01 × 105 Pa or 1.08 × 1.01 × 102 kPa or 1.09 × 105 Pa or 1.09 × 102 kPa ) ) 35.45 = 0.433 g (Cl) and 143.32 0.433 (100) = 80.9% (Cl) [1] 0.535 Allow whole numbers for molar masses. 18.3 (c) = 1.52 mol C and 12.01 0.77 = 0.76 mol H and 1.01 80.9 = 2.28 mol Cl 35.45 Allow whole numbers for atomic masses. (b) (1.75) Practice questions [1] (d) 2HCl(aq) + CaCO3(s) → CaCl2(aq) + H2O(l) + CO2(g) or 2H+(aq) + CaCO3(s) → Ca2+(aq) + H2O(l) + CO2(g) [2] Award [1] for correct reactants and products. Award [1] if the equation is correctly balanced. Award [1] (max) for the following equations: 2HCl(aq) + CaCO3(s) → CaCl2(aq) + H2CO3(aq) 2H+(aq) + CaCO(s) → Ca2+(aq) + HCO(aq) 5 10:01 Answers (g) only CaCO3 reacts with acid or impurities are inert or non-basic or impurities do not react with the acid or nothing else in the eggshell reacts with acid or no other carbonates. [1] Ignore state symbols. (e) n(CaCO3) = 12 n(HCl) = 12 × 0.00306 = 0.00153 or 1.53 × 10−3 mol [2] Award [2] for correct final answer. Do not accept ‘all calcium carbonate reacts with acid’. (f) Mr(CaCO3) = 40.08 + 12.01 + 3 × 16.00 = 100.09 or 100.1 g mol−1 Accept 100. 21 NaCl 61.96%,CaCl238.04% m(CaCO3) = n × M = 0.00153 mol × 100.09 g mol−1 = 0.153 g 0.153 × 100 = 81.4% or %CaCO3 = 0.188 81.5% 22 (a) 0.115 mol H2O (b) 0.0574 mol K2CO3 (c) K2CO3.2H2O (d) Heat to constant mass – when further heating does not lead to further decrease in mass. [3] Ignore state symbols. Accept answers in the range 79.8% to 81.5%. Award [3] for correct final answer. 23 (a) NH3is in excess (b) HCl is limiting (c) 1.64 g ammonium chloride forms 10:01