Answers_p1022-1061 - 1022 ANSWERS TO EXAMINATION QUESTIONS Answers to examination questions 1 Quantitative chemistry Q19 D Paper 1 IB questions and IB

Answers_p1022-1061 - 1022 ANSWERS TO EXAMINATION QUESTIONS...

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1 Quantitative chemistry Paper 1 IB questions and IB style questions Q1 D One mole of a chemical element is the relative atomic mass expressed in grams. Q2 D Mass of one molecule = (12 r 3) + (8 r 1) g mol - 1 6 r 10 23 mol - 1 = 7.33 r 10 - 23 g Q3 B 28 g mol - 1 r 2 mol = 56 g Q4 D A: 0.5 mol H 2 molecules; 1 mol H atoms, B: 0.03125 mol O 2 molecules; 0.0625 mol oxygen atoms; C: 3.9 r 10 - 3 mol S 8 molecules; 0.0312 mol S atoms; D: 6.258 r 10 - 3 mol Br 2 molecules; 0.0125 mol Br atoms Q5 C Amount = mass(g) molar mass (g mol - 1 ) = 180 g 18 g mol - 1 = 10 mol Number of molecules = 6 r 10 23 r 10 = 6 r 10 24 Q6 D Multiply the equation through by four: 12 moles of hydrogen peroxide would produce 8 moles of water. Q7 C 2H 2 S + 3O 2 l 2SO 2 + 2H 2 O Q8 D M r BH 3 = 14, molecular formula could be B 4 H 12 ( M r = 56); B: M r = 56; C: M r = 56; D: M r = 36.5 Q9 C The molecular formulas simplify to NF, CH 2 O, C 2 H 4 O (unchanged) and CN. Q10 A 2Cl 2 represents two chlorine molecules: Cl - Cl and Cl - Cl. Q11 C 2Ca(s) + O 2 (g) l 2CaO(s) (balanced with correct formulas). Q12 D The charge is not balanced. Q13 A Fe 2 (SO 4 ) 3 is composed of three SO 4 2 - and two Fe 3+ . Q14 B Amount of M atoms = (52 - 32) 40 = 0.50 mol; amount of S atoms = 32 32 = 1.0 mol. Multiplying by two gives MS 2 . Q15 D M + O 2 l MO 2 Amount of oxygen molecules = 16 32 = 0.5 mol O 2 molecules; 0.5 mol M atoms. Hence relative atomic mass of M = 36 0.5 = 72 g mol - 1 . Q16 B 2H 2 + O 2 l 2H 2 O; 8 mol H 2 and 2 mol O 2 ; stoichiometric molar ratio is 2 : 1, hence the gas mixture contains unreacted hydrogen and water – no oxygen, since it is the limiting reagent. Q17 B 3Mg(s) + Fe 2 O 3 (s) l 2Fe(s) + 3MgO(s) Amount of magnesium = 2.4 g 24 g mol - 1 = 0.1 mol; amount of iron = 0.1 mol r 2 3 . Mass of iron = 0.066 mol r 56 g mol - 1 = 3.733 g. Q18 A Molecular mass = 63 r 12 55.74 r 100 = 1356.3 g mol - 1 Q19 D Amount of NH 3 = 8.5 17 = 0.5 mol; amount of NO = 15 30 = 0.5 mol; percentage yield = 0.5 0.5 r 100 = 100% Q20 D There is range of kinetic energies for gas molecules in a sample at fixed temperature. Q21 C Avogadro’s hypothesis: the number of particles in a gas is directly proportional to volume at the same temperature and pressure. Q22 D Average kinetic energy of gas particles is directly proportional to the absolute temperature in kelvin. Q23 D High pressure will bring the gas molecules very close together. This causes more collisions and also allows the weak attractive forces to operate. With low temperatures, the gas molecules do not have enough kinetic energy to continue on their path to avoid that attraction. Q24 A P and V vary inversely at constant temperature. This relationship is termed Boyle’s law. Q25 D The average kinetic energy of the gas molecules is directly proportional to absolute temperature (a result of thermodynamics). At a given temperature the molecules of all species of gas, no matter what size, shape or mass, have the same average kinetic energy. Q26 A V 2 = 350 cm 3 r 99.3 kPa r 273 K 101.3 kPa r 295 K V 2 = 317.5034 cm 3 = 318 cm 3 Q27 B Amount of CO 2 = 88.0 44 = 2 mol, hence amount of carbon atoms = 2 mol; amount of H 2 O = 27 18 = 1.5 mol hence amount of hydrogen atoms = 3 mol; empirical formula is C 2 H 3 , hence possible molecular formulas include C 4 H 6 , C 6 H 9 , etc.
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