Chemistry August 000 Provincial Examination ANSWER KEY / SCORING GUIDE CURRICULUM: Organizers. 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. C U A 5. B K 4 L. B U A3 6. B U 4 K 3. C K B 7. A U 4 L4, 6 4. D U B 8. D U 4 L 5. B U B9 9. D K 4 L 6. D K C4 30. A K 4 M 7. D K D4 3. A H 4 M4 8. D H D6 3. A U 4 M4 9. B U D7 33. C U 4 N3 0. A U E 34. B U 4 O4. A U F 35. B U 4 P5. B U F3 36. B K 4 R3 3. B U F7 37. A K 5 S 4. C K 3 G 38. B U 5 S 5. A U 3 G 39. A U 5 S, 6. B U 3 G7 40. C U 5 S3 7. D U 3 H 4. A K 5 S6 8. C U 3 H 4. D U 5 T3 9. B U 3 H5 43. D H 5 T5, 6 0. C K 3 H7 44. D K 5 U. B H 3 I3 45. B U 5 U5. C U 3 I4 46. C U 5 U9 3. A U 4 J7 47. D U 5 W3, 4 4. D H 4 K 48. D K 5 W5 Multiple Choice = 48 marks 008chk - - September 6, 000
Part B: Written Response Q B C S CO PLO. U 3 A3, 4. U 3 E3 3. 3 U 4 F6 4. 4 U 3 3 H3, I6 5. 5 K 4 J0, 6. 6 U 3 4 J3, K8 7. 7 U 3 4 M3 8. 8 U 3 4 P4 9. 9 H 4 5 P4, W7 0. 0 U 4 5 T6, U Written Response = 3 marks Multiple Choice = 48 (48 questions) Written Response = 3 (0 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 008chk - - September 6, 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.. Consider the following reaction: CaCO3 ( s) + HCl( aq) CaCl( aq) + CO( g) + HO( l) A sample of some HCl was added to a sample of CaCO 3 in an open beaker. The total mass was monitored and the following data obtained: HCl (aq) CaCO 3(s) Time (min) 0.00.00.00 3.00 4.00 5.00 6.00 7.00 Total mass of beaker and contents (g) 50.00 48.50 47.50 46.95 46.60 46.4 46.33 46.30 8.00 46.30 008chk - 3 - September 6, 000
a) Find the average rate of production of CO in mol s during the first 5 minutes. ( marks) ( ) = 50. 00 g 46. 4g 500. min 0. 78g min mark mark for subtraction for division 0. 78gCO mol CO = 0. 063 mol 44. 0 gco mark rate in mol s 0. 063 mol 4 = = 7. 0 mol 60 s s mark b) Give one reason why the rate of production of CO would decrease with time. Explain, using collision theory. (Êmark) For Example: Reason: [ HCl] decreases Explanation: Less frequent collisions OR Reason: Surface area of CaCO 3 decreases Explanation: Less frequent collisions mark 008chk - 4 - September 6, 000
. Consider the following reaction: 3+ + aq aq aq Fe ( ) + SCN ( ) FeSCN ( ) yellow colourless red When a few drops of 60. M NaOH is added to 5. 0 ml of the above system, a precipitate ofê Fe OH ( ) 3 forms and the solution turns pale yellow. a) Explain this colour change in terms of Le Ch telierõs Principle. (Êmarks) The reduced [ Fe 3+ ] causes a shift to the left to offset theêstress. } marks b) Describe the effect on the rate of the reverse reaction as the colour change occurs. (Êmark) The rate of the reverse reaction decreases. ÊÊÊÊ mark 008chk - 5 - September 6, 000
3. Consider the following equilibrium: 3I( g) + 3F ( g) IF( g) + I4F ( g) Initially,. 00 0 mol of I and 3. 00 0 mol of F are put into a 0. 00 L flask. 3 At equilibrium, I F is 00. 0 M. Calculate the K eq. (4Êmarks) [ 4 ] 3I + 3F IF + I F 4 [] I 0. 000 0. 0300 0 0 [ C] 0. 0060 0. 00600 + 0. 00400 + 0. 0000 [ E] 0. 040 0. 040 0. 00400 0. 0000 mark for division by 0 marks for ICE table Ê ÊÊÊÊÊÊ Ê K K eq eq IF I F = ( ) ( ) 4 3 3 I F ( ) ( ) = ( ) 0. 00400 ( 0. 0000 ) 3 3 ( 0. 040) ( 0. 040) = 844. 0 marks 008chk - 6 - September 6, 000
4. a) Write the complete ionic equation for the reaction between NaBrO and AgNO 3 3. (Êmark) + + Na ( aq ) + BrO3 ( aq) + Ag ( aq) + NO3 ( aq) Na ( aq) + NO3 ( aq) + AgBrO3 ( s) mark + b) What is the maximum NaBrO 3 [ ] that can exist in equilibrium with 0 3. M AgNO? (Êmarks) AgBrO ( s) ÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊ Ê Ag + BrO 0. x + 3 3 K Ag BrO 5 ÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊÊ 53. 0 = ( 0. )( x) sp + = [ ][ 3 ] [ 3 ]= [ 3]= 5 x= BrO NaBrO 6. 0 M mark marks 008chk - 7 - September 6, 000
5. a) Define the term Br nsted-lowry conjugate acid-base pair. (Êmark) For Example: A pair of chemical species which differ by the presence of a proton on the acid which is absent from the base. mark b) Give an example of a conjugate acid-base pair. (Êmark) For Example: Acid: HO Base: OH mark 008chk - 8 - September 6, 000
6. Consider the acids HCl and HF. a) Only one of the following reactions occurs. Complete the equation for the reaction which does occur. ( mark) i) HCl + F? ii) HF + Cl? HCl + F HF + Cl mark for selection of correct reaction and mark for completing equation b) For the reaction that occurs, are reactants or products favoured? Explain. ( mark) For Example: Products are favoured mark ( ) thanê HF mark ( ) because HCl is stronger c) Explain why the other reaction will not occur. ( mark) For Example: Cl will not accept a proton. mark 008chk - 9 - September 6, 000
+ 7. Calculate the [ HO 3 ] of 00. M HNO. (3Êmarks) + 3 HNO + H O H O + NO [] I 00. M 0 0 [ C] x + x + x [ E] 00. x x x marks assume K a 00. x 00. [ + HO ][ NO ] 4 3 = 46. 0 = HNO = ( x)( x) 00. x ( ) x x 00. x= [ HO 3 ]= 68. 0 = ( )( ) + 3 [ ] M mark mark mark 008chk - 0 - September 6, 000
8. Write the formula equation and the net ionic equation for the reaction between 00. M H SO and 00. M Sr OH Formula equation: 4 H SO Sr OH SrSO H O 4( aq) ( aq) 4( s) ( ) Net ionic equation: + ( ). (3Êmarks) + ( ) + l marks + H ( aq) + SO4 ( aq) + Sr ( aq) + OH ( aq) SrSO4( s) + HO( l) marks 008chk - - September 6, 000
9. During the production of magnesium metal from sea water, magnesium ions are first precipitated from sea water as magnesium hydroxide. a) The magnesium hydroxide is then neutralized by hydrochloric acid, producing magnesium chloride. Write the neutralization reaction. ( mark) ( ) + + Mg OH ( s) HCl( aq) MgCl( aq) HO( l) mark b) The salt produced, magnesium chloride, is dried, melted and undergoes electrolysis. Write the reaction occurring at each electrode. (Êmarks) Anode:Cl Cl + e mark + Cathode: Mg + e Mg mark c) It is not possible to use electrolysis to remove Mg from a. 0M MgCl solution. Why? ( mark) For Example: Water is a stronger oxidizing agent than Mg +. mark 008chk - - September 6, 000
0. In the process of extracting tin from a sample of ore, the tin is removed as Sn + ions. A titration requires. 43 ml of 0. 070 M KCrO 7 to reach the equivalence point with the Sn + in a 0. 750 g sample of the ore. + + 4+ 3+ 3Sn ( aq) + CrO 7 ( aq) + 4H ( aq) 3Sn ( aq) + Cr ( aq) + 7HO l ( ) Using the reaction, calculate the percent mass of tin in the ore sample. (4Êmarks) 7 7 4 7 mol Cr O = ( 0. 070 mol L Cr O )( 0. 043 L) = 3. 643 0 mol Cr O mol Sn mol Sn mol CrO 7 + 3 = 3. 643 0 =. 093 0 mol Cr O + 4 7 3 + mol Sn mark mark + 3 mol Sn = mol Sn =. 093 0 mol Sn gsn=.093 0 8. 7 gsn mol Sn =. 97 0 mol Sn 3 gsn mark. 97 0 gsn % Sn = 0. 750 g Sn ore 00% = 7. 3% mark END OF KEY 008chk - 3 - September 6, 000