Chemistry 12 April 2001 Provincial Examination
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1 Chemistry 1 April 001 Provincial Examination ANSWER KEY / SCORING GUIDE CURRICULUM: Organizers 1. Reaction Kinetics. Dynamic Equilibrium 3. Solubility Equilibria 4. Acids, Bases, and Salts 5. Oxidation Ð Reduction Sub-Organizers A, B, C D, E, F G, H, I J, K, L, M, N, O, P, Q, R S, T, U, V, W Part A: Multiple Choice Q K C S CO PLO Q K C S CO PLO 1. D K 1 1 A1 5. B U 1 4 K9. A U 1 1 A4 6. C U 4 K11 3. C K 1 1 B1 7. B U 1 4 L11 4. B U 1 1 B6 8. D U 4 L1 5. C U 1 1 B4 9. C U 1 4 M4 6. D H 1 1 C5 30. B U 1 4 N3 7. C U 1 D3 31. C U 1 4 N 8. A K D5 3. D H 1 4 O3; K6 9. B U 1 E 33. C U 1 4 P 10. D U E, B K 1 4 P1, B H F4 35. B U 1 4 Q 1. A K 1 F3 36. D K 1 4 R1 13. D U 1 F6 37. A K 1 4 R 14. D K 1 3 G1 38. C U 1 5 S1 15. D K 1 3 G 39. B U 5 S3 16. A U 1 3 G8 40. D U 1 5 S5 17. C U 1 3 H1 41. C H 1 5 S5 18. A H 3 H5 4. D U 1 5 S6 19. B U 3 H6 43. B U 1 5 T4 0. A K 1 3 I 44. A H 1 5 U5 1. C K 4 J 45. D K 1 5 U9. C U 1 4 J A K 5 U10 3. A H 4 K1, D K 1 5 W 4. A K 1 4 K6 48. B U 5 W4 Multiple Choice = 60 marks (48 questions) 014chk May 8, 001
2 Part B: Written Response Q B C S CO PLO 1. 1 U 5 1 A3. U 3 F U 4 F8; D U 3 3 I U 4 3 G5; I U 4 J U 5 4 P1, U 5 4 M U 5 5 T U 4 5 W8 Written Response = 40 marks Multiple Choice = 60 (48 questions) Written Response = 40 (10 questions) EXAMINATION TOTAL = 100 marks LEGEND: Q = Question Number K = Keyed Response C = Cognitive Level B = Score Box Number S = Score CO = Curriculum Organizer PLO =Ê Prescribed Learning Outcome 014chk - - May 8, 001
3 PART B: WRITTEN RESPONSE Value: 40 marks INSTRUCTIONS: Suggested Time: 50 minutes You will be expected to communicate your knowledge and understanding of chemical principles in a clear and logical manner. Your steps and assumptions leading to a solution must be written in the spaces below the questions. Answers must include units where appropriate and be given to the correct number of significant figures. For questions involving calculation, full marks will NOT be given for providing only an answer. 1. An Alka-Seltzer tablet is added to water to produce carbon dioxide gas. The gas was collected using water displacement. The following data is recorded: Time (s) Volume of CO (ml) chk May 8, 001
4 a) Calculate the average rate of reaction for the formation of CO gas for the times: i) 0 10s (1Êmark) vol 30. ml 00. ml rate = = t s 30. ml = = 030. ml s s ii) 10 0 s (1Êmark) vol rate = = t 0. 0 ml 3. 0 ml 0. 0 s s ml = s = 170. ml s b) Suggest a reason why the rate of reaction from 0 to 10. 0s is slower than the rate from to 0. 0 s? (1Êmark) The surface area of the tablet increases as the tablet crumbles. c) The rate of reaction is not constant during the entire interval from to s. Describe the change in rate and explain a reason for the change. (Êmarks) The rate decreases during the interval because the surface area decreases. marks 014chk May 8, 001
5 . A flask is initially filled with some HI. At equilibrium, the [ HI]= 080. mol L. What is the H [ ] at equilibrium? (3Êmarks) HI( g) H( g) + I( g) Keq = 0. 5 K eq = 05. [ H][ I] HI [ ] x x = ( )( ) ( 080. ) x = 040. [ H]= 040. M 014chk May 8, 001
6 3. Consider the following equilibrium system: NOCl ( g) NO( g) + Cl( g) Keq = A 1. 00L flask is filled with 00. mol NOCl, 0. 10mol NO and 0. 10mol Cl. State and show by calculation the direction in which the reaction proceeds to reach equilibrium. (4Êmarks) Direction: Reaction proceeds to the left. Calculations: NOCl( g) NO( g) + Cl( g) K K Trial Trial NO Cl = [ ] [ ] NOCl > K [ ] = ( ) 010. ( 010. ) ( 00. ) = eq 014chk May 8, 001
7 4. In a titration, ml of NaCl ( aq) reacts completely with 4. 0 ml of M AgNO 3. What is the [ Cl ] in the original solution? (3Êmarks) + aq aq ( s) Ag ( ) + Cl ( ) AgCl mol Ag mol L L mol Cl + = ( )( ) = mol + = mol Ag mol [ Cl ]= L = M ( Deduct 1 mark for incorrect significant figures. ) 014chk May 8, 001
8 5. The following data was obtained when 0. 0 ml of a saturated solution of PbI was evaporated to dryness. Mass of evaporating dish g Mass of evaporating dish and residue g Use this information to determine the Ksp of PbI. (4Êmarks) Mass of PbI = 30. 6g g = 0. 01g Moles PbI 1mol = 0. 01g = g 5 mol 1 mark 1 mark + [ Pb ]= mol L = M [ I ]= L 5 mol = M + ( ) ( ) = Ksp = [ Pb ][ I ] = chk May 8, 001
9 6. a) Write the equation for the predominant reaction of HC O 4 with HSO 3. (1Êmark) HCO4 + HSO 3 CO4 + HSO 3 base acid b) Identify a Br nsted-lowry conjugate acid base pair from the above reaction. (1Êmark) HC O acid base OR HSO 4 CO4 3 HSO3 base acid 1 mark 014chk May 8, 001
10 7. a) In the space below, sketch the titration curve for the reaction whenê 010. M HCl is added to 100. ml of 010. M NaOH. (3Êmarks) ph mark for general shape 1 mark for showing curve drop 1 mark for location of equivalence point Volume of added HCl (ml) See graph above. 3 marks b) Describe two changes in the titration curve that would result from using 010. M CH3COOH in place of the HCl. (Êmarks) Any two of the following for 1 mark each. The equivalence point is > 7. The vertical part of the curve is shorter. Buffer region. 014chk May 8, 001
11 [ ] 050 3( ) 8. Calculate the OH in. M NH aq. (5Êmarks) K b Kw = = K a = ÊÊÊÊÊ NH + H O NH + OH [] I [ C] x +x +x [ E] 050. x x x K b [ + NH ][ OH ] 4 = = NH [ 3] Use assumption that 050. x 050. or use the quadratic formula. x = [ OH ]= ( x)( x) 050. x ( ) = 3 M ( Deduct 1 mark for incorrect significant figures. ) 5 marks marks 014chk May 8, 001
12 9. Balance the following redox reaction in basic solution. (5Êmarks) ( ) MnO C O MnO CO basic + ( 4 ) + 3 ( HO + CO 4 CO3 + 4H + e ) 3e + 4H + MnO MnO + H O H O + MnO + 3C O MnO + 6CO + 4H marks (1 mark for each half cell) 1 mark for multiplication 1 mark for addition 1 mark for basic OH + MnO + 3C O MnO + 6CO + H O 014chk May 8, 001
13 10. a) Draw and label the parts of an operating electrolytic cell during the electrolysis of molten potassium chloride KCl ( l). (3Êmarks) + Ð DC Power Source Anode Cathode Cl Ð K + 1 mark for single container 1 mark for power supply 1 mark for electrodes b) Define the term oxidizing agent. (1Êmark) An oxidizing agent is a species which causes another to lose electrons. END OF KEY 014chk May 8, 001
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