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CIE Jun 2014 v2 Paper 2 Q1
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![Answer all the questions in the spaces provided. (a) Explain what is meant by the term nucleon number. PMT[1] (b) Bromine exists naturally as a mixture of two stable isotopes, 79Br and 81Br, with relative isotopic masses of 78.92 and 80.92 respectively. (i) De ne the term relative isotopic mass.[2] (ii) Using the relative atomic mass of bromine, 79.90, calculate the relative isotopic abundances of 79Br and 81Br. (c) Bromine reacts with the element A to form a compound with empirical formula ABr3. The percentage composition by mass of ABr3 is A, 4.31; Br, 95.69. Calculate the relative atomic mass, Ar, of A. Give your answer to three signi cantgures. [3] Ar of A =[3]UCLES 2014 9701/22/M/J/14<br />
PMT (d) The elements in Period 3 of the Periodic Table show different behaviours in their reactions with oxygen. (i) Describe what you would see when separate samples of magnesium and sulfur are reacted with oxygen. Write an equation for each reaction. magnesiumsulfur[4] (ii) Write equations for the reactions of aluminium oxide, Al 2O3, with sodium hydroxide,hydrochloric acid.[2] (e) Phosphorus reacts with chlorine to form PCl 5. State the shape of and two different bond angles in a molecule of PCl 5. shape of PCl 5bond angles in PCl 5[2] [Total: 17]UCLES 2014 9701/22/M/J/14 [Turn over<br />](https://learnah.org/wp-content/uploads/2018/08/June_2014_v2_QP_-_Paper_2_CIE_Chemistry_A-level_1-0.jpg)
![Page 2 Mark Scheme Syllabus Paper GCE AS/A LEVELMay/June 2014 9701 22 Question Answers Mark Total PMT (a) The (total) number of protons and neutrons (in the nucleus of an atom) (b) (i) Mass of an atom(s) or isotope relative to (the mass) of (an atom of) carbon12 12 OR relative to carbon12 which is (exactly) 12 (units) allow a correct expression (ii) 79 Br 81 Br 78.92x 80.92(100x) where x = % abundance of Br 79 78.92 80.92(100 so = 79.9 100 x = 51 hence 79 Br : 81 Br = 51 : 49Cambridge International Examinations 2014<br />
Page 3 Mark Scheme Syllabus Paper GCE AS/A LEVELMay/June 2014 9701 22 Question Answers Mark Total PMT (c) Br 4.31 95.69 79.9 = 1 : 3 95.69 / 79.9 So = 3 4.31 rA/ 4.31 79.9 r = = 10.796 = 10.8 to 3 s.f. 95.69 3 sig figs allow alternative correct methods (d) (i) Mg: bright / white light / flame OR white solid / smoke Mg + O2MgO allow correct multiples S: blue flame OR white / steamy fumes OR yellow solid disappears S + O2SO2 allow correct multiples (ii) Al Al Al Al Al Al Al 2O3 + 2NaOH + 7H2O2NaAl (OH)4(H2O)2 OR 2O3 + 2NaOH + 3H2O2NaAl (OH)4 OR 2O3 + 2NaOH2NaAl O2 + H2O OR 2O3 + 2OH + 7H2O2[Al (OH)4(H2O)2] 2O3 + 2OH + 3H2O2[Al (OH)4] OR OR 2O3 + 2OH2Al O2 + H2O 2O3 + 6HCl2Al Cl 3 + 3H2O allow correct ionic equationsCambridge International Examinations 2014<br />
Page 4 Mark Scheme Syllabus Paper GCE AS/A LEVELMay/June 2014 9701 22 Question Answers (e) shape of PCl 5 =(trigonal) bipyramid(al) bond angles in PCl 5 = 120 and 90 Mark Total 17 PMT (a) (i) (The C2O4 2 ions) lose electrons owtte / ora (ii) 2MnO4 (aq) + 5C2O4 (aq) + 16H (aq)2Mn (aq) + 10CO2(aq) + 8H2O(l) 1+1+1 2 2+ (b) (i) 20.00.100 = 2(.00)10 (mol) 3 1000 (ii) 2t MnO4 : C2O4 = 2 : 5 so amount of C2O4 2t = (5 / 2)2.0010 t3 = 5(.00)10 (mol) t3 ecf from (b)(i) (iii) 5.0010 t3250 / 25 = 0.05(0) (mol) ecf from (b)(ii) (iv) amount = mass / M r so M r = mass / amount = 6.30 / 0.05 = 126 (v) ecf from (b)(iii) 12690 = 36 36 / 18 = 2.00 x = 2 Ecf from (b)(iv) if suitableCambridge International Examinations 2014<br />](https://learnah.org/wp-content/uploads/2018/08/June_2014_v2_MS_-_Paper_2_CIE_Chemistry_A-level_1-0.jpg)
