CHEMISTRY 2811/01 Foundation Chemistry
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1 THIS IS A LEGACY SPECIFICATION ADVANCED SUBSIDIARY GCE CHEMISTRY 2811/01 Foundation Chemistry *CUP/T56917* Candidates answer on the question paper OCR Supplied Materials: Data Sheet for Chemistry (inserted) Other Materials Required: Scientific calculator Friday 9 January 2009 Afternoon Duration: 1 hour * * INSTRUCTIONS TO CANDIDATES Write your name clearly in capital letters, your Centre Number and Candidate Number in the boxes above. Use black ink. Pencil may be used for graphs and diagrams only. Read each question carefully and make sure that you know what you have to do before starting your answer. Answer all the questions. Do not write in the bar codes. Write your answer to each question in the space provided, however additional paper may be used if necessary. INFORMATION FOR CANDIDATES The number of marks is given in brackets [ ] at the end of each question or part question. The total number of marks for this paper is 60. You will be awarded marks for the quality of written communication where this is indicated in the question. You may use a scientific calculator. A copy of the Data Sheet for Chemistry is provided as an insert with this question paper. You are advised to show all the steps in any calculations. This document consists of 12 pages. Any blank pages are indicated. FOR EXAMINER S USE Qu. Max. Mark TOTAL 60 [L/100/3428] SP (SLM) T56917/3 OCR is an exempt Charity Turn over
2 2 Answer all the questions. 1 The answer to each part of this question is a number. (a) (i) How many neutrons are there in an atom of chlorine-37?... [1] (ii) How many electrons are needed to fill one orbital?... [1] (iii) How many electrons are there in the 3p sub-shell of a chlorine atom?... [1] (b) (i) Calculate the relative atomic mass of a sample of gallium containing 60% 69 Ga and 40% 71 Ga. Give your answer to three significant figures. [1] (ii) Calculate the relative formula mass of (NH 4 ) 2 CO 3. [1] (iii) Calculate the number of grams of NaNO 3 in 200 cm 3 of a mol dm 3 solution. [2] (iv) Calculate the number of molecules in 100 cm 3 of SO 2 gas at room temperature and pressure mol of SO 2 molecules occupies 24.0 dm 3 at room temperature and pressure. L = mol 1. [1] (c) Determine the oxidation number of chlorine in NaClO [1] [Total: 9]
3 2 Element A is in Period 3, Na Ar, of the Periodic Table. Some of the successive ionisation energies of element A are shown below. 3 ionisation energy/kj mol 1 1st 2nd 3rd 4th 5th 6th 7th (a) Define the term first ionisation energy.... [3] (b) Identify element A from the elements in Period 3, Na Ar. Explain how you decided on your answer. element A:... explanation: [3] (c) Elements in the same group in the Periodic Table have different ionisation energies. Explain why there is a trend in first ionisation energies for elements in the same group.... [3] [Total: 9] Turn over
4 4 3 Calcium chloride, CaCl 2, is used for dust control on roads and in car parks. Calcium chloride is made up of Ca 2+ and Cl ions. (a) Complete the table below. species protons electrons Ca 2+ Cl [2] (b) Draw a dot-and-cross diagram of CaCl 2. Show outer electron shells only. (c) Solid calcium chloride does not conduct electricity. An aqueous solution of calcium chloride does conduct. Explain the different conductivities of solid and aqueous calcium chloride.... [2] (d) Calcium chloride can be made by reacting limestone with hydrochloric acid. [2] CaCO 3 (s) + 2HCl(aq) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) In the laboratory, a student carries out this reaction using 4.85 g CaCO 3 and 1.50 mol dm 3 HCl. She then evaporates water from the solution to obtain solid CaCl 2. (i) How many moles of CaCO 3 were reacted? answer =...mol [2]
5 (ii) What mass of CaCl 2 is formed by the reaction of 4.85 g of CaCO 3? 5 answer =... g [1] (iii) Calculate the volume, in cm 3, of 1.50 mol dm 3 HCl that reacts with 4.85 g of CaCO 3. answer =... cm 3 [2] (e) Choose another chemical that could be reacted with hydrochloric acid to make calcium chloride. Write a balanced equation for the reaction.... [2] (f) Compound B is a calcium compound used in making paper. Compound B is manufactured by passing SO 2 gas through a solution of calcium hydroxide. Compound B has the following percentage composition by mass: Ca, 19.82%; H, 0.99%; S, 31.74%; O, 47.45%. (i) Determine the empirical formula for compound B. [2] (ii) Construct a balanced equation for the manufacture of compound B from calcium hydroxide by this method.... [1] [Total: 16] Turn over
6 4 The boiling points of the halogens chlorine, bromine and iodine are shown below. 6 halogen boiling point/ C chlorine 35 bromine 59 iodine 184 (a) Explain this trend in the boiling points of the halogens.... [3] (b) Iodine reacts with water as shown below. I 2 (s) + H 2 O(l) IO (aq) + 2H + (aq) + I (aq) Determine the changes in oxidation number of iodine in this reaction and comment on your answers.... [3]
7 (c) A student carries out the following investigation. 7 Step 1: The student adds an excess of chlorine gas to an aqueous solution of potassium bromide. Step 2: The student adds aqueous silver nitrate to the resulting solution. (i) In step 1, what would the student observe? Write an ionic equation for the reaction that takes place. observation:... ionic equation:... [2] (ii) For step 2, write an ionic equation, including state symbols, for the reaction that takes place.... [2] (d) Many covalent compounds of the halogens, such as CCl 4, have polar bonds. Polarity can be explained in terms of electronegativity. (i) Explain what is meant by the term electronegativity [2] (ii) Molecules of the covalent compound CCl 4 have polar bonds. Draw a diagram to show the shape of a molecule of CCl 4. On your diagram, show the polarity of the bonds. [2] (iii) A molecule of CCl 4 is non-polar. Explain why [1] [Total: 15] Turn over
8 5 In this question, one mark is available for the quality of spelling, punctuation and grammar. 8 Many physical properties can be explained in terms of bonding and structure. The table below shows some properties of magnesium, diamond and ice. substance magnesium diamond ice electrical conductivity of solid good poor poor melting point 649 C 3550 C 0 C Explain these properties in terms of bonding and structure
9 [10] Quality of Written Communication [1] END OF QUESTION PAPER [Total: 11]
10 10 BLANK PAGE PLEASE DO NOT WRITE ON THIS PAGE
11 11 BLANK PAGE PLEASE DO NOT WRITE ON THIS PAGE
12 12 PLEASE DO NOT WRITE ON THIS PAGE Permission to reproduce items where third-party owned material protected by copyright is included has been sought and cleared where possible. Every reasonable effort has been made by the publisher (OCR) to trace copyright holders, but if any items requiring clearance have unwittingly been included, the publisher will be pleased to make amends at the earliest possible opportunity. OCR is part of the Cambridge Assessment Group. Cambridge Assessment is the brand name of University of Cambridge Local Examinations Syndicate (UCLES), which is itself a department of the University of Cambridge.
13 2811 Mark Scheme January Foundation Chemistry Question Expected Answers Marks Additional Guidance 1 (a) i 20 1 ii 2 1 iii 5 1 (b) i ii Allow 96 iii moles of NaNO 3 = 0.05 mass = 0.05 x 85 = 4.25 (g) g worth 1 (wrong M r ) Accept 4.3 but not 4.2 (ecf for calculated moles x 85) iv 2.51 x Allow 2.5 x Calc: x Allow calc value and any degree of correct rounding down to 2.5 x (c) (+)7 1 Sign not required but do not credit 7 Accept VII Total 9 1
14 2811 Mark Scheme January 2009 Question Expected Answers Marks Additional Guidance 2 (a) Energy change when each atom in 1 mole of gaseous atoms loses an electron (to form 1 mole of gaseous 1+ ions) 3 Not element alone Compensate missed marks from correct equation (b) Si 3 Not consequential Sharp rise in successive ionisation energy between 4th and 5th IE (c) marking a change to a new shell / there are 4 electrons in the outer shell atomic radii increases/ there are more shells/atoms get bigger there is more shielding/ more screening ionisation energy decreases because the Increased shielding and distance outweigh the increased nuclear charge / the nuclear attraction decreases Not sub-shell 3 USE annotations with ticks, crosses, con, ecf, etc for this part. down the group not required more is essential allow more electron repulsion from inner shells Allow nuclear pull ignore any reference to effective nuclear charge Total 9 2
15 2811 Mark Scheme January 2009 Question Expected Answers Marks Additional guidance 3 (a) Ca 2+ : 20 protons; 18 electrons Cl : 17 protons; 18 electrons 2 (b) cation shown with either 8 or 0 electrons AND anion shown with 8 electrons AND correct number of crosses and dots Correct charges on both ions 2 For 1st mark, if 8 electrons shown around cation then extra electron(s) around anion must match symbol chosen for electrons in cation. Circles not required Ignore inner shell electrons For charges, Allow: 2[Cl ] 2[Cl] last one) [Cl ] 2 (brackets not required except for Do not allow: for CaCl 2, [Cl 2 ] 2 [Cl 2 ] [2Cl] 2 [Cl] Max 1 if only one Cl - (c) solid: ions are fixed (AW) aqueous: ions are free (to move) (AW) (d) i molar mass CaCO 3 : (g mol 1 ) 4.85/100.1 = mol 2 If charge carriers are wrong but comparison is given, then award one mark, e.g. solid: electrons are fixed in lattice AND aqueous: electrons are free to move (1 mark 2 Not 100 for molar mass calc Allow rounding of calculator value back to 2 sig figs allow ECF If working shown for an incorrect molar mass, then the 2nd mark can be awarded as 4.85/calculated molar mass ii 5.38 or 5.39 g or 5.4 g 1 For information: x = x =
16 2811 Mark Scheme January 2009 Question Expected Answers Marks Additional guidance iii or or ECF: moles from (i) x or For information: 2 x = mol 2 x = ECF moles from (i) x 2 volume = 64.7 or 64.6 cm 3 or 65 For information (0.0970/1.50) x 1000 = 64.7 cm 3 (0.0969/1.50) x 1000 = 64.6 cm 3 ECF (moles above/1.50) x 1000 (e) Ca/CaO/Ca(OH) 2 Ca + 2HCl CaCl 2 + H 2 / CaO + 2HCl CaCl 2 + H 2 O / Ca(OH) 2 + 2HCl CaCl 2 + 2H 2 O 2 Ignore state symbols Allow any other suitable alternatives (f) i Ca : H : S : O = 19.82/40.1 : 0.99/1 : 31.74/32.1 : 47.45/16 or 1 : 2 : 2 : 6 empirical formula = CaH 2 S 2 O 6 2 Using atomic numbers gives CaHS 2 O 6 worth 1 Allow Ca(HSO 3 ) 2! ii Ca(OH) 2 + 2SO 2 CaH 2 S 2 O 6 1 If you see it, allow Ca(HSO 3 ) 2! Total 16 4
17 2811 Mark Scheme January 2009 Question Expected Answers Marks Additional Guidance 4 (a) down group/from Cl to I/, number of electrons/shells increases 3 Answers involving ionisation energies score 0 more/ stronger/ van der Waals / intermolecular forces/ induced dipoles/ instantaneous dipoles greater forces to break/more energy has to be put in to break forces (b) I 2 IO, 0 to +1 : oxidised I 2 I, 0 to 1 : reduced correct oxidised and reduced above/i is both oxidised and reduced / disproportionation (c) i goes orange/red/yellow 3 Sign not required but do not credit 1 Sign required here 2 Ignore brown Cl 2 + 2Br Br 2 + 2Cl Ignore spectator ions ii Ag + (aq) + Cl (aq) AgCl(s) correct equation correct state symbols (d) i attraction of an atom for electrons in a (covalent) bond/ bonding pair 2 Allow state symbols for (slightly) incorrect equations 2 ii iii correct 3-D tetrahedral shape shown showing one outward wedge and 1 inward wedge; 3 bonds below horizontal correct dipoles: δ+ on C and δ on each Cl Cl δ Cδ+ δ δ Cl Cl Cl δ the polarities/ dipoles cancel out / the molecule is symmetrical 2 For bond into paper, accept: 1 Total 15 Allow correct shape with no atom labels: Only need to show one dipole Cl Cl Cl Cl 5
18 2811 Mark Scheme January 2009 Question Expected Answers Marks Additional Guidance 5 Magnesium structure/bonding: giant metallic conducts by delocalised/free/mobile electrons melting point high because of the electrostatic attraction / attraction between (positive) ions and electrons 10 USE annotations with ticks, crosses, con, ecf, etc for this part. Credit information if given in annotated diagrams Watch out for contradictions, especially of bonding type Allow: positive ions with a sea of electrons for both structure and bonding marks if labelled, one if not. Diamond does not conduct/poor conductor: no mobile charge carriers/electrons/ions structure/bonding: giant ( ) covalent Giant only awarded if not given above melting point: high because strong/ lots of (covalent) bonds are broken Ice does not conduct: no mobile charge carriers/electrons/ions structure/bonding: H-bonds/intermolecular forces/ simple molecular melting point: Low because H bonds/intermolecular/ weak forces between molecules (are broken)/ higher than expected because H-bonds broken/ H-bonds stronger than other (named) intermolecular forces must refer to bonds being broken once QWC At least two sentences that show legible text with accurate spelling, punctuation and grammar so that the meaning is clear. 1 QWC mark must be indicated with a tick or cross through the Quality of Written Communication prompt at the bottom of page 9. Then scroll up to start of (b), counting ticks. Total 11 6
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