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UNIT 1 ANSWERS the Water Is Called the Solvent and the Sodium Chloride Is the Solute

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UNIT 1 ANSWERS the Water Is Called the Solvent and the Sodium Chloride Is the Solute ANSWERS 345 5 ▶ Sodium chloride dissolves in water to form a solution. UNIT 1 ANSWERS The water is called the solvent and the sodium chloride is the solute. If the solution is heated to 50 °C some of the water evaporates until the solution becomes CHAPTER 1 saturated and sodium chloride crystals start to form. 1 ▶ a melting b freezing 6 ▶ a er 60 c subliming/sublimation d subliming/sublimation 50 g wat 2 ▶ a 40 30 20 10 0 solid liquid gas solubility/g per 100 0 20 40 60 80 100 120 Note: Solids should have regularly packed particles temperature/°C touching. Liquids should have most of the particles b 94 +/−1 g per 100 g touching at least some of their neighbours, but with The values obtained in this question and in c depend gaps here and there, and no regularity. Gases should on the line of best fit. In the exam there will always be have the particles well spaced. some tolerance – a range of values will be accepted. b Solids: vibration around a fixed point. Liquids: c From the graph, the solubility at 30 oC is 10 g per 100 g particles can move around into vacant spaces, but of water. with some difficulty because of the relatively close 40 ____ ​​ × 10 = 4 g packing. 100 c Evaporation: Some faster moving particles break away Therefore 4 g of sodium chlorate will dissolve. from the surface of the liquid. Boiling: Attractive forces d i 53 +/−1 °C are broken throughout the liquid to produce bubbles ii The solubility at 17 oC is 7 ±1 g per 100 g, therefore of vapour. 20 − 7 = 13 g must precipitate out of the solution. 3 ▶ a i A – gas; B – liquid; C – solid; D – liquid; E – solid Answers of 13 ±1 g are acceptable. ii A – gas; B – solid; C – solid; D – liquid; E – solid iii A – gas; B – liquid; C – solid; D – gas; E – solid CHAPTER 2 b A, because it is a gas. 1 ▶ Element Compound Mixture c It sublimes and therefore is converted directly from a solid to a gas without going through the liquid stage. hydrogen magnesium oxide sea water d D – because it has a lower boiling point the forces of calcium copper(II) sulfate honey attraction between particles will be weaker therefore it blood will also evaporate more easily than substance B (the mud only other substance that is a liquid at 25 °C). potassium iodide 4 ▶ a The ammonia and hydrogen chloride particles have solution to diffuse through the air in the tube, colliding with air 2 ▶ a mixture b mixture c element particles all the way. d element e compound f compound b i Its particles will move faster. ii It would take slightly longer for the white ring to 3 ▶ Substance X is the pure substance – it melts at a fixed form, because the gas particles would be moving temperature. Substance Y is impure – it melts over a more slowly at the lower temperature. range of temperatures. c Ammonia particles are lighter than hydrogen chloride 4 ▶ a crystallisation b (simple) distillation particles and so move faster. The ammonia covers c fractional distillation d chromatography more distance than the hydrogen chloride in the same e filtration time. 5 ▶ For example: Stir with a large enough volume of cold water d i Ammonium bromide. to dissolve all the sugar. Filter to leave the diamonds on ii The heavier hydrogen bromide particles would the filter paper. Wash on the filter paper with more water to move more slowly than the hydrogen chloride remove any last traces of sugar solution. Allow to dry. particles, and so the ring would form even closer 6 a M b R to the hydrobromic acid end than it was to the ▶ hydrochloric acid end. The ring will also take c 0.45 ±0.01 (measure to the centre of the spot and slightly longer to form because of the slower remember to measure from the base line and not from moving particles. the bottom of the paper) d G and T e P 346 ANSWERS CHAPTER 3 g 82, lead h 1 ▶ a the nucleus b electrons c proton d proton and neutron 2 ▶ a 9 b sum of protons + neutrons in the nucleus c 9 p, 10 n, 9 e d The protons and electrons have equal but opposite 6 ▶ Palladium is a metal and so is likely to have any of the charges. The atom has no overall charge, therefore following properties: there must be equal numbers of protons and • good conductor of electricity electrons. • forms a basic oxide 3 ▶ a 26 p, 30 n, 26 e • is shiny when polished or freshly cut b 41 p, 52 n, 41 e • is malleable c 92 p, 143 n, 92 e • is ductile • is a good conductor of heat 4 ▶ a Atoms with the same atomic number but different mass numbers. They have the same number of The first two are mentioned specifically on the syllabus. protons, but different numbers of neutrons. 7 ▶ They have a full outer shell (energy level) and so they b 35Cl: 17 p, 18 n, 17 e; 37Cl: 17 p, 20 n, 17 e have no tendency to form compounds by losing/gaining 6 × 7 + 93 × 7 electrons or sharing electrons. 5 ▶ ​​ _____________ ​​ = 6.93 100 8 ▶ Argon and potassium OR iodine and tellurium. 24 × 78.99 + 25 × 10.00 + 26 × 11.01 6 ▶ ​​ __________________________________ = 24.32 The elements would then be in a different group in the 100 Periodic Table. They would not have the same number 204 × 1.4 + 206 × 24.1 + 207 × 22.1+ 208 × 52.4 7 ▶ ​​ __________________________________________ = 207.241 of electrons in the outer shell as other members of the 100 group and would react in a completely different way. For 8 ▶ a 77 protons, 114 neutrons, 77 electrons example, potassium would be in Group 0 with the noble b Iridium-193 has 2 more neutrons in the nucleus. gases, and argon, which is very unreactive, would be in c More iridium-193 because the relative atomic mass is Group 1, with the highly reactive alkali metals. closer to 193 than 191. 9 ▶ This statement is true – it only applies to 1 element, CHAPTER 5 hydrogen (1H). 1 ▶ a Fe + 2HCl → FeCl2 + H2 CHAPTER 4 b Zn + H2SO4 → ZnSO4 + H2 c Ca + 2H2O → Ca(OH)2 + H2 1 ▶ a i strontium ii chlorine iii nitrogen d 2Al + Cr2O3 → Al2O3 + 2Cr iv caesium v neon e Fe2O3 + 3CO → 2Fe + 3CO2 b metals: caesium, molybdenum, nickel, strontium, tin f 2NaHCO3 + H2SO4 → Na2SO4 + 2CO2 + 2H2O non-metals: chlorine, neon, nitrogen g 2C8H18 + 25O2 → 16CO2 + 18 H2O 2 ▶ abc h Fe3O4 + 4H2 → 3Fe + 4H2O i Pb + 2AgNO3 → Pb(NO3)2 + 2Ag j 2AgNO + MgCl Mg(NO ) + 2AgCl Na Si S 3 2 → 3 2 k C3H8 + 5O2 → 3CO2 + 4H2O 2 ▶ a 44 b 60 c 142 d 132 3 ▶ a 2, 7 b 2, 8, 3 c 2, 8, 8, 2 e 286; (The common mistake would be not to multiply the whole water molecule by 10. So the mass of the 4 ▶ a 5 b 7 c 4 d 8 10H2O is 180. Students will commonly and wrongly 5 ▶ a A, F b A come up with 36 for this by multiplying the H2 by 10 c C and D because 5 shells are occupied but not the O as well. Work out the mass of the whole H O first and then multiply it by the number in front. d C because there are 7 electrons in the outer shell 2 That way you won’t make this mistake.) (energy level) f 392 e B, D f Calcium – it has 20 electrons and therefore must 3 ▶ a 13.9% b 35% have 20 protons in the nucleus. The atomic number c 21.2%; (Be careful of the cases where there are two is therefore 20. Calcium is the element with atomic nitrogen atoms in the fertiliser (all except KNO3). The number 20. masses of the nitrogen in those cases will be 28 and not 14.) ANSWERS 347 4 ▶ In each case, work out the Mr by adding up the relative c Na S O atomic masses (Ar values), and then attach the unit “g” to give the mass of 1 mole. Combining 3.22 g 4.48 g 3.36 g a 27 g mass b 331 g No. of 3.22 4.48 3.36 c 4.30 × 16 = 68.8 g moles of _____ = 0.14 _____ = 0.14 _____ = 0.21 d 0.70 × 62 = 43.4 g atoms 23 32 16 e 0.015 × 85 = 1.275 g Ratio of 1 1 1.5 f 0.24 × 286 = 68.64 g Don’t forget the water of moles simplifies crystallisation (divide by to 2 2 3 Strictly speaking the answers to d), e) and f) shouldn’t smallest be quoted to more than 2 significant figures, because number) the number of moles is only quoted to that precision.) Empirical formula is Na S O 5 ▶ In each case, work out the mass of 1 mole as above, and 2 2 3 then work out how many moles you’ve got in the stated d Carbon Hydrogen Bromine mass. You can use the equation: Given % 22.0 4.6 73.4 mass number of moles = ______________​​ ​​ No. of mass of 1 mole 22.0 4.6 73.4 20 moles of _____ = 1.833 ___ = 4.6 _____ = 0.9175 a ​​ ___ ​​ = 0.5 mol 12 1 80 40 atoms 3.20 b ​​ _____ ​​ = 0.0200 mol Ratio of 2 5 1 160 moles 2000 (divide c ​​ _____ ​​ = 25.2 mol; Don’t forget to convert kg to g! 79.5 by 50 smallest d ​​ ______ ​​ = 0.2 mol 249.5 number) 1 000 000 e ​​ _________ ​​ = 17 900 mol (or 17 857, although this is precise Empirical formula is C2H5Br 56 to more significant figures than the A r).
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