AUGUST 1995 CHEMISTRY 12 PROVINCIAL EXAMINATION ANSWER KEY / SCORING GUIDE

Size: px

Start display at page:

Download "AUGUST 1995 CHEMISTRY 12 PROVINCIAL EXAMINATION ANSWER KEY / SCORING GUIDE"

Delphia Lane
5 years ago
Views:

1 GST 99 EMISTRY PROVINIL EXMINTION NSWER EY / SORING GIE TOPIS. inetics. Equilibrium. Solubility. cids, ases, Salts. Oxidation Reduction PRT : MLTIPLE-OIE Q T S GR Q T S GR I-- I-- I-- I-- I-- I-F- I-- II-- II-- / E- II-E- II-- II-J- II-I- III-- / E- III-- III--8 III-- III-- III--6 III-E- IV-- IV-- IV-- IV IV-- IV-G- IV-F- IV-F- IV--9 IV-F-8 IV-J- IV-J- IV-I- IV-J-6 IV-- IV-L- V-- V--6 V-- V-- V-- V-E- V-G- V-G- V-G- V-G-8 V-- V-J- 98chpk - - pril, 996

2 PRT : WRITTEN-RESPONSE Q T S GR Q T S GR I--, I-- II-E- II-J- III-- III IV-I- IV-- IV-L- IV-J- V-E- V-I- Multiple-choice = 8 (8 questions) Written-response = ( questions) Total = 80 marks LEGEN: Q = Question = ognitive level T = Topic = eyed response S = Score GR = urriculum Guide Reference = Score box number 98chpk - - pril, 996

3 PRT : WRITTEN-RESPONSE. onsider the following reaction: ( s) +O ( g) O ( g) State one factor that would increase the rate of the above reaction. se collision theory to explain the increase in rate. ( marks) For example: catalyst: offers second reaction path of lower activation energy increase in temperature: increases fraction of particles with sufficient energy to react increase in surface area: increases probability of collisions increase in concentration of oxygen: increases probability of collisions ( mark for one of the above factors and mark for appropriate explanation.). efine and give an example of a homogeneous reaction. ( marks) reaction in which all reactants are in the same phase. mark e.g. ( g) +I ( g) I ( g) mark 98chpk - - pril, 996

4 . onsider the following system: O ( g) O ( aq) + heat List two ways in which more O could be dissolved in water. ( marks) i) ii) For example: i) decrease temperature ii) increase pressure iii) add water iv) add more O any two for mark each. In an experiment, 0.00 mol of O ( g) and 0.00 mol of O ( g) are placed in a.00 L container and the following equilibrium is achieved: O ( g) + O ( g) O ( g) t equilibrium, the [ O ] is found to be 0.60 mol L. alculate the value of eq. ( marks) O ( g) + O ( g) O ( g) [ I] [ ] [ E] marks eq = [ O ] O [ ] O =.0 0 ( ) ( ) ( 0.0) [ ] = marks 98chpk - - pril, 996

5 . In an experiment, a student pipettes a sample of saturated Mgr solution into a beaker and evaporates the sample to dryness. e recorded the following data: Volume of saturated Mgr ( aq) Mass of beaker Mass of beaker and residue.00 ml 89.0 g 9.7 g alculate the solubility of Mgr in moles per litre. ( marks) Mass of Mgr = 9. 7 g 89.0 g. g mark Molar mass of Mgr = 8. g mol Mol Mgr =. g mol 8. g =.0 0 mol mark Solubility =.0 0 mol L = mol L mark 98chpk - - pril, 996

6 6. What is the solubility of ao in g L? ( marks) ao a + + O Solubility = sp = mol L mark mark = g mol mark = gl mark = 7. 0 gl 7. Neutral red, Ind, is an acid-base indicator. a) Write an equation to represent the equilibrium of this indicator in water. ( mark) Ind + O O + + Ind b) What colour would this indicator be in 0. M NaO? ( mark) The indicator would be amber. 98chpk pril, 996

7 8. alculate the p of 0.0 M OO. ( marks) OO + O O + + OO [ I] [ ] x +x +x [ E] 0.0 x 0.0 x x marks [ ][ OO ] a = O+ [ OO] = ( x) x 0.0 ( ) =.8 0 x =. 9 0 [ O + ]=. 9 0 M p =.7 marks 9. Explain why normal rain water is slightly acidic. se an equation to support your answer. ( marks) O dissolves in water to make an acidic solution. O + O O or O + O O + + O mark mark 98chpk pril, 996

8 0. alculate the mass of solid NaO needed to neutralize 0.0 ml of 0. M O. ( marks) mol O = 0.00 L 0. mol L = 0.0 mol mol NaO = 0.0 mol O mol NaO mol O = 0.06 mol mark mark g NaO = 0.06 mol NaO 0.0 g NaO mol NaO =.0 g NaO mark. alance the following redox reaction occurring in an acidic solution. ( marks) s + lo so + lo Oxidation Method: (decrease ) mark for each oxidation number change mark for electron balance coefficient s+lo O so + lo mark for + and O (increase ) or alf-cell Method: mark ( ) mark ( ) mark s + O so e lo e lo + O s + 6 O + lo + + so + lo mark 98chpk pril, 996

9 . In the electrolysis of.0 M LiF, the products are oxygen gas and hydrogen gas. a) Write the anode half-reaction and include the E value. ( mark) O O e 0.8 V b) Write the cathode half-reaction and include the E value. ( mark) O + e + O ( 0 7 M) 0. V c) alculate the minimum voltage required for this electrolysis. ( mark) Minimum voltage is >. volts. EN OF EY 98chpk pril, 996

Chemistry 12 April 1996 Provincial Examination

Chemistry 12 April 1996 Provincial Examination hemistry pril 996 Provincial Examination NSWER EY / SORING GIE Topics:. inetics. Equilibrium. Solubility. cids, ases, Salts. Oxidation Reduction Part : Multiple hoice Q T S GR Q T S GR..... 6. 7. 8. 9.