oc.ffi rt-n-l CHEMISTRY Foundation Chemistry [Turn over OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary GCE TIME

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1 oc.ffi RECOGNISING ACHIEVEMENT OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary GCE CHEMISTRY Foundation Chemistry Wednesday I JUNE 2005 Morning 2811 t hour Candidates answer on the question paper. Additional materials: Data Sheet for Chemistry Scientific Calculator Candidate Name Centre Number Candidate Number rt-n-l TIME t hour INSTRUCTIONS TO CANDIDATES o Write your name in the space above. Write your Centre number and Candidate number in the boxes above. o Anwver all the questions.. Write your ansrrrers in ths spaces provided on the question paper. o Read each question carelully and make sure you know what you have to do before starting your answer. INFORMATION FOR CANDIDATES. The number of marks is given in brackets [ ] at the end of each question or part question. o You will be awarded marks for the quality of written communication where this is indicated the question. FOR EXAMINER'S USE o You may use a scientificalculator. Qu. Max. Mark. You may use the Data Sheet for Chemistry You are advised to show all the steps in any calculations I TOTAL 60 sp (cwslm) 2005 [N,V100/34231 This question paper consists of I printed pages. Registered Charity Number: [Turn over

2 2 Answer all the questions. Fu Examinefs Ug 1 The element titanium, Ti, atomic number 22, is a metal that is used in the aerospace industry for both airframes and engines. A sample of titanium for aircraft construction was analysed using a mass spectrometer and was found to contain three isotopes, 46Ti, 47Ti and 48Ti. The results of the analysis are shown in Table 1.1 below. Table 1.1 isotope 46Ti 47Ti 48Ti relative isotopic mass OO percentage composition (a) (i) Explain the term isotopes. (ii) Complete the table below for atoms of two of the titanium isotopes....t11 isotope protons neutrons electrons 46Ti 4zra l2l (b) Using theinformation Table 1.1, calculate the relative atomic mass of this sample of titanium. Give youranswer to three significant figures. (c) Complete the electroniconfiguration of a titanium atom. t2l 1s22s22p...t Jun05

3 (d) Titanium has metallic bonding. 3 For Examinels Use (i) Explain what is meantby metallic bonding. Use a diagram in your answer (ii) How does metallic bonding allow titanium to conduct electricity?..t1 I (e) A student reacted 1.44 g of titanium with chlorine to form 5.70 g of a chloride X. (i) How many moles of Ti atoms were reacted? t1l (ii) How many moles of CI atoms were reacted? l2l (iii) Determine the empirical formula of X. (iv) Construct a balanced equation for the reaction between titanium and chlorine. t1l...t1 l (v) At room temperature, X is a liquid which does not conduct electricity. What does this information suggest about the bonding and structure in X?...t21 [Total: 16] [Turn over

4 4 2 The Group 2 element radium, Ra, is used in medicine for the treatment of cancer. Radium was discovered in 1Bg8 by Pierre and Marie Curie by extracting radium chloride from its main ore pitchblende. For Examinefs Use (a) Predicthe formula of radium chloride....t11 (b) Pierre and Marie Curie extracted radium from radium chloride by reduction. Explain what is meant by reduction, using this reaction as an example..t2l (c) Radium reacts vigorously when added to water. Ra(s) + 2H2O(l) + Ra(OH)r(aq) + nr(o) (i) Use the equation to predict two observations that you would see during this reaction. (ii) Predict a ph value for this solution. (d) Reactions of the Group 2 metals involve removal of electrons. The electrons are removed more easily as the group is descended and this helps to explain the increasing trend in reactivity. (i) The removal of one electron from each atom in 1 mole of gaseous radium atoms is l2l t1l called the t21 The equation for this process in radium is: I2l (ii) Atoms of radium have a greater nuclear charge than atoms of calcium. Explain why, despite this, less energy is needed to remove an electron from a radium atom than from a calcium atom Jun05...t31 [Total: 13]

5 5 A student had a stomach-ache and needed to take something to neutralisexcess stomach acid. He decided to take some Milk of Magnesia, which is an aqueous suspension of magnesium hydroxide, Mg(OH)r. For Examinefs Use (a) The main acid in the stomach is hydrochloric acid, HCI(aq), and the unbalanced equation for the reaction that takes place with Milk of Magnesia is shown below. Mg(oH)r(s) +...HC(aq)-+...Mgcl(ac) +...H2o(l) Balance the equation by adding numbers where necessary in the unbalanced equation above. t1l (b) The student'stomach contained 5OO cm3 of stomach fluid with an acid concentration of 0.108moldm-3. The student swallowed some Milk of Magnesia containing Mg(OH)e. He wondered whether this dose was sufficient to neutralise the stomach acid. Assume that all the acid in the stomach fluid was 0.108moldm-3 hydrochloric acid. (i) How many moles of HCI were in the 500 cm3 of stomach fluid? (ii) Calculate the mass of Mg(OH), necessary to neutralise this stomach fluid. I1l t3l (iii) Determine whether the student swallowed too much, too little, or just the right amount of Milk of Magnesia to neutrafise the stomach acid. (c) Chewing chalk has been used for many years to combat excess stomach acid and indigestion tablets often contain calcium carbonate, CaCOr. Suggest, with the aid of an equation, how these tablets work...11 l.l2l [Total: 8] 2811 Jun05 [Turn over

6 6 4 Chlorine is used in the preparation of many commercially important materialsuch as bleach and iodine. Fu Examinefs Us (a) Bleach is a solution of sodium chlorate(i), NaOCI made by dissolving chlorine in aqueous sodium hydroxide. Clr(O) + 2NaOH(aq) + NaoCt (aq) + NaCl (aq) + H2o(l) Determine the changes in oxidatio number of chlorine during the preparation of bleach and comment on your results...t31 (b) lodine is extracted commercially from seawater with chlorine gas. Seawater contains very small quantities of dissolved iodide ions, which are oxidised to iodine by the chlorine gas. (i) Write an ionic equation for the reaction that has taken place...t21 (ii) Use your understanding of electronic structure to explain why chlorine is a stronger oxidising agent than iodine. l2l [Total: 7]

7 7 5 In this question, one mark is available for the quality of use and organisation of scientific terms. For Examinefs Use Nitrogen and oxygen are elements in Period 2 of the Periodic Table. The hydrogen compounds of oxygen and nitrogen, HrO and NH' both form hydrogen bonds. (a) (i) Draw a diagram containing two HrO molecules to show what is meantby hydrogen bonding. On your diagram, show any lone pairs present and relevant dipoles. t3l (ii) State and explain two anomalous properties of water resulting from hydrogen bonding. (b) The 'dof-and-cros.d diagram of an ammonia molecule is shown below...t4i Predict, with reasons, the bond angle in an ammonia molecule Junos [Turn over

8 (c) The atomic radii of nitrogen and oxygen are shown below. 8 For Examine/s Use element nitrogen oxygen atomic radius /nm Explain why a nitrogen atom is larger than an oxygen atom....t41 Quality of Written Gommunication [1] [Total: 16J END OF QUESTION PAPER OCFI has made every effort to trace the copyright holders of items used in this Question Paper, but il we have inadvertently overlooked any, we apologise Jun05

2811 Mark Scheme Abbreviations, annotations and conventions used in the Mark Scheme = alternative and acceptable answers for the same marking point ' = separates marking points NOT = answers which

9 2811 Mark Scheme Abbreviations, annotations and conventions used in the Mark Scheme = alternative and acceptable answers for the same marking point ' = separates marking points NOT = answers which are not worthy of credit ( ) = words which are not essential to gain credit = (underlining) key words which must be used to gain credit ecf = error carried forward AW = alternative wording ora = or reverse arqument Question Expected Answers Marks 1 (a) (i) atoms of same elemenusame atomic number... with different numbers of neutrons/different masses y' IlI (ii) isotope protons neutrons electrons ooti { o'ta / tzl (b) A, = t / = 47.7 / t2i (c) 1 s'2s'2po3s=3po3 d' 4s' { tll (d) (e) (i) ( ii) o-o-o positive ions y' electronr / lmust be labelled) electrons move y' moles Ti = = mol/0.03 mol (accept use of answer from (b)) tzl tlt tlt (i i) mass of Cl = = 4.26 g / moles Cf = = mol { = mol gets 1 mark t2l (iii) Ti:Cl = : O.12 = 1.4. Empiricaformula = TtClq / : mol gives TiCls for 1 mark t{l (iv) (v) Ti +2Clz+TaCl4/ (ecf possible from (iii) covalent / simple molecutar y' 11I tzl Total: 16

10 I Mark Scheme June 2005 porcviations, fnotatione and feventions rd [' in the Mark heme = altemative and acceptable answers for the same marking point ' = separates marking points NOT = answers which are not worthy of credit ( ) = words which are not essential to gain credit = (underlining) key words which must be used to gain credit ecf = error carried forward AW = alternative wording ora = or reverse arqument testion Expected Answers Marks (a) RaCl2 r' tlt (b) Reduction is gain of electrons/decrease in oxidatio number { Ra2* gains 2 electrons -+ Ra/ Oxidation state goes from +2 in Raclz 0 in Ra / t2l (c) (i) effervescence/bu b bles r/ Ra disappears/dissolves / tzl (ii) 8-14 / tll (d) (i) Firsi y' ionisation (eneryfi{ l2l Ra(g) + Ra*(g) + e- {/ 1 mark for equation 1 mark for state symbols '-' not required on 'e' Jzl (ii) atomic radii of Ra > atomic radii of Cal Ra has electrons in shell further from nucleus than Cal Ra has more shells / Ra has more shielding than Ca y' :'more'ls essenfia/ Ra electron held less tightly/less attraction on electron/ t3l Total: 13

2811 Mark Scheme Ju Abbreviations, annotations and conventions used in the Mark Scheme = alternative and acceptable answers for the same marking point ' = separates marking points NOT = answers which

11 2811 Mark Scheme Ju Abbreviations, annotations and conventions used in the Mark Scheme = alternative and acceptable answers for the same marking point ' = separates marking points NOT = answers which are not worthy of credit ( ) = words which are not essential to gain credit = (underlining) key words which must be used to gain credit ecf = error carried forward AW = alternative wording ora = or reverse arqument Question Expected Answers Marks 3 (a) *Mg(OH)z(s) HCl(aq) -)...MgClz(aq) H2O(l) tll (b) (i) ( ii) mofes HCI = x 500/1000 = { mofes Mg(OH)2 = Tz x moles HCI = / mofar mass of Mg(OH)z = x2 = 58.3 / (do not penalise 24) Ilt (iii) mass Mg(OH)z = 58.3 x0.027 = 1.57 g g / (accept ans from (ii) x = ) (mass Mg(OH)z of 3.15 g would score 2 marks as 'ecf as molar ratio has not been identified) Too much it 2.42 g (dose) > ans to (ii) r' (ff answer to (ii) t 2.42 g then 'correct' response here would be 'Not enough' t3l Ill (c) CaCOa reacts with (or neutralises) HCI t/ (or CaGOr + HCI in an equation) CaCOs + 2HCl -+ CaClz + HzO + COz / (correct equation would score both marks) t2l Total: 8

12 1 lllark Scheme June 2005 pbreviations, nnotations and pnventions bcc in the ilark pheme i = alternative and acceptable answers for the same marking point ' = separates marking points NOT = answers which are not worthy of credit ( ) = words which are not essential to gain credit = (underlining) key words which must be used to gain credit ecf = error carried fonrard AW = alternative wording ora = or reverse arqument Uestion Expected Answers Marks (a) Clz(9) + NaOCI(aq) : CI(O) Ct(+17 t, Clz(g) ---+ NaCl(aq) : Cl(O) + Ca(-1) { Cl is both oxidised (in forming NaOCI) and reduced (in forming NaCl)/d isproportionation Cl reduces Cl to form NaCl AND Cl oxidises Cl in forming NaOCI r' T3I (b) (i) Cf2+21-+lz+2Cl-{{ 1 mark for species. 1 mark for balancing tzl (ii) Cl atom is smaller/has less shells / electron to be captured will be attracted more / t2l Total: 7

2811 Mark Scheme June Abbreviations, annotations and conventions used in the Mark Scheme = alternative and acceptable answers for the same marking point ' = separates marking points NOT = answers

13 2811 Mark Scheme June Abbreviations, annotations and conventions used in the Mark Scheme = alternative and acceptable answers for the same marking point ' = separates marking points NOT = answers which are not worthy of credit ( ) = words which are not essential to gain credit = (underlining) key words which must be used to gain credit ecf = error carried forward AW = alternative wording ora = or reverse arqument Question Expected Answers Marks 5 (a) (i) H bonding from O of 1 H2O molecule to H of another { dipoles shown y' with lone pair involved in bond / (ii) Two properties from: lce is lighter than water/ max density at 4'C { explanation: H bonds hold HzO molecules apart / open lattice in ice / H-bonds are longer / Higher melting/boiling point than expecte 6 { explanation: strength of H bonds that need to be broken / must imply that intermolecular bonds are broken t3l High surface tension/viscosity / explanation strength of H bonds across surface y' (b) NHo: 107" / (range ") efectron pairs repel other etectron pairs { fone pair has more repulsion { efectron pairs get as far apart as possible { (c) N has less protons than O (ora)/ electrons are in same shell /have same or similar shieldin g '/ weaker nuclear attraction in N (ord { sheff drawn in less by nuclear charge in N (ora\/ watch for distinction between nuclear attraction and nuclear charge in candidates' scripts. t4l t4l t4i QoWC: links together two statements in at least two of the sections (a)(ii),(b) and (c)/ tlt Total: 16 6

CHEMISTRY Foundation Chemistry. OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary GCE

CHEMISTRY Foundation Chemistry. OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary GCE OXFORD CAMBRIDGE AND RSA EXAMINATIONS Advanced Subsidiary GCE CHEMISTRY 2811 Foundation Chemistry Wednesday 8 JUNE 2005 Morning 1 hour Candidates answer on the question paper. Additional materials: Data