Chemistry 1 June 000 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 CO PLO Q K C CO PLO 1. A K 1 A1 5. A U 4 K9. D K 1 A 6. B U 4 L4, N3 3. C U 1 A3 7. D K 4 L6 4. C H 1 A5 8. C U 4 L11 5. B K 1 B5 9. B K 4 M1 6. B U 1 C5 30. C K 4 M 7. B U D6 31. A U 4 M4 8. C K D9 3. A K 4 O 9. B U E 33. D U 4 O4 10. D K E, 4 34. B U 4 P3 11. C K F 35. C U 4 P5 1. C H F4 36. A U 4 P6 13. A U F6 37. B H 4 Q5 14. D U 3 G5 38. D K 5 S1 15. D K 3 G8 39. B K 5 S 16. A U 3 H 40. A K 5 S 17. A U 3 H4 41. C U 5 S3 18. B K 3 I 4. A U 5 S6 19. C H 3 H, I4 43. C K 5 T1 0. D H 3 I5 44. D U 5 T5 1. D K 4 J11 45. B H 5 U7. C K 4 K7 46. D U 5 U11 3. D U 4 J8 47. B U 5 V4 4. A U 4 K8 48. A U 5 W Multiple Choice = 48 marks 006chk - 1 - July 5, 000
Part B: Written Response Q B C S CO PLO 1. 1 U 3 1 C, C5. U 3 B6, E 3. 3 U 3 F6 4. 4 U 3 3 I5 5. 5 H 3 3 E, H5, K5 6. 6 U 4 4 M5 7. 7 U 5 4 N1, N, N3 8. 8 U 4 5 S4, U7 9. 9 U 5 U, U11 10. 10 U 5 W8 Written Response = 3 marks Multiple Choice = 48 (48 questions) Written Response = 3 (10 questions) EXAMINATION TOTAL = 80 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 006chk - - July 5, 000
PART B: WRITTEN RESPONSE Value: 3 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. a) Complete the steps in the following mechanism. (1 1 Êmarks) Step 1 NO + Pt NOPt 1 mark Step NOPt + NO N + OPt 1 mark Step 3 Ê Ê OPt O + Pt Overall NO N + O b) Define the term reaction intermediate and give an example from the completed mechanism above. (1 1 Êmarks) Definition: A substance which forms in one step of a mechanism and is used up in a later step. 1 mark Example: NOPt or OPt 1 mark 006chk - 3 - July 5, 000
. Consider the observations for the following equilibrium: Trial NO 4( g) NO( g) colourless brown ( ) ( ) Temperature C Colour I. 10 light brown II. 50 dark brown a) Sketch the potential energy curve on the graph below for this equilibrium. (1Êmark) PE progress of the reaction 1 mark b) Explain the colour change using Le Ch telierõs Principle. (1Êmark) An increase in temperature causes the reaction to shift to the right and the NO [ ] increases. 1 mark c) Other than changing temperature, what could be done to cause a shift to the left? (1Êmark) To cause a shift to the left add NO or remove NO 4 or decrease the volume. 1 mark 006chk - 4 - July 5, 000
3. Consider the data obtained for the following equilibrium: 3+ + aq aq aq Fe ( ) + SCN ( ) FeSCN ( ) [ Fe 3+ ] [ SCN ] [ FeSCN + ] Experiment 1 391. 10 Experiment 67. 10 3 80. 10 5 365. 10 4 9. 10 4? Calculate the [ FeSCN + ] in experiment #. (3Êmarks) K eq [ + FeSCN ] = 3+ Fe SCN = [ ][ ] 9. 10 3. 91 10 8. 0 10 5 ( )( ) = 94. 10 4 1 1 mark 94. 10 = x 6. 7 10 3. 65 10 3 4 ( )( ) [ FeSCN ]= x = 673. 10 + 4 M 1 1 mark ( Deduct 1 mark for incorrect significant figures. ) 006chk - 5 - July 5, 000
( ) is combined with 4. At 5 C, will a precipitate form when 5. 0 ml of 0. 010 M Pb NO3 75. 0 ml of 0. 010 M NaI? Support your answer with calculations. (3Êmarks) PbI + ( s) Pb ( aq) + I ( aq) + 5. 0 ml [ Pb ]= 0. 010 M = 0. 0050 M 100. 0 ml 75. 0 ml [ I ]= 0. 010 M = 0. 00750 M 100. 0 ml + Trial K Pb I sp = [ ][ ] = ( 0. 0050)( 0. 00750) = 14. 10 7 1 mark 1 mark 7 9 Since Trial Ksp ( 14. 10 ) > Ksp 85. 10 a precipitate does form. ( ) } 1 mark 006chk - 6 - July 5, 000
5. When HCl is added to a saturated solution of CuC O 4, some precipitate dissolves. However, when HCl is added to a saturated solution of PbCl, additional precipitate forms. HCl HCl CuC O 4 PbCl Explain these observations. Support your explanation with chemical equations. (3Êmarks) CuC O + 4( s) Cu ( aq) + CO ÊÊÊ 4 ( aq) 1 mark H + from the acid reacts with the CO 4 to form HC O reducing the [ CO 4 ] and causing a shift to the product side. 4 1 mark + Pb ( aq) + Cl ( aq) PbCl( s) ÊÊÊ 1 mark The common ion effect causes a shift to the right. 1 mark 006chk - 7 - July 5, 000
6. A 0. 100 M solution of an unknown weak acid, HX, has a ph = 1. 414. What is the Ka for HX? (4Êmarks) + ÊÊÊÊÊÊÊ[ HO 3 ]= 0. 03855 M ÊÊ Ê 1 mark + HX + H O H O + X 3 [] I 0100. 0 0 [ C] 0. 03855 + 0. 03855 + 0. 03855 [ E] 0. 061 0. 03855 0. 03855 marks K ÊÊÊÊÊÊÊ K a a [ + HO ][ X ] 3 = HX [ ] = ( 0. 03855 )( 0. 03855 ) = 0. 04 0. 061 1 mark ( Deduct 1 mark for incorrect significant figures. ) 006chk - 8 - July 5, 000
7. Consider the salt ammonium acetate, NH4CH3 COO. a) Write the equation for the dissociation of NH CH COO 4 3. (1Êmark) + NH4CH3COO( s) NH4 ( aq) + CH3COO 1 mark ( aq) b) Write equations for the hydrolysis reactions which occur. (Êmarks) + + 4 3 3 NH + H O H O + NH 3 3 and CH COO + H O CH COOH + OH marks c) Explain why a solution of NH4CH3 COO has a ph = 700.. Support your answer with calculations. (Êmarks) K K a b for NH for CH COO the K for NH = K for CH COO a + 10 4 = 56. 10 3 10. 10 = 18. 10 14 5 + 4 b 3 = 56. 10 10 the acidic cation is completely neutralized by the basic anion. marks 006chk - 9 - July 5, 000
8. The metals Rh, Ti, Cr and Pd are individually placed in 1. 0Msolutions + + + + of Ê Rh, Ti, Cr and Pd and the cell voltages of the spontaneous reactions are determined. METAL ION Rh + Ti + Pd + Cr + Rh no reaction 0.35 V no reaction Ti.3 V.58 V? Pd no reaction no reaction no reaction Cr 1.51V no reaction 1.86 V a) Arrange the metals in order of increasing strength as reducing agents. (Êmarks) marks weakest reducing agent Pd, Rh, Cr, Ti strongest reducing agent b) Determine the cell voltage for Ti in a 10. M solution of Cr +. (Êmarks) + Ti + Rh = 3. V + Cr + Rh = 151. V + Cell voltage of Ti + Cr = 3. V 151. V = 07. V marks 006chk - 10 - July 5, 000
9. Consider the following reactions for a fuel cell : cathode: O + H O + 4e 4OH ( g) ( l) anode:? overall: H + O H O ( aq) ( g) ( g) ( l) a) Write the reaction at the anode. (1Êmark) H ( g) + 4 OH ( aq) 4 H O ( l) + 4 e 1 mark b) Discuss the advantage of a fuel-cell powered vehicle over an internal combustion powered vehicle by comparing the products formed. (1Êmark) The NO x produced by internal combustion cars is a source of acid rain. The H Ofrom a fuel-cell car is non-polluting. 1 mark 006chk - 11 - July 5, 000
10. Draw and label an electrochemical cell using a copper anode and having an E value > 100. V. (Êmarks) Volts anode Salt Bridge cathode 1 mark for suitable cathode Ñ Au for example. 1 3 mark for suitable ions Ñ Au + and Cu + for example. 1 mark for diagram being an electrochemical cell, not an electrolytic cell. END OF KEY 006chk - 1 - July 5, 000