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PRE-LEAVING CERTIFICATE EXAMINATION, 2012

MARKING SCHEME

CHEMISTRY

HIGHER AND ORDINARY LEVEL

*WMS12*

35 Finglas Business Park, Tolka Valley Road, Finglas, Dublin 11T: 01 808 1494, F: 01 836 2739, E: info@examcraft.ie, W: www.examcraft.ie

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Introduction

In considering the marking scheme the following should be noted.

1. In many cases only key phrases are given which contain the information and ideas that must appear in the candidate’s answer in order to merit the assigned marks.

2. The descriptions, methods and defi nitions in the scheme are not exhaustive and alternative valid answers are acceptable.

3. The detail required in any answer is determined by the context and the manner in which the question is asked, and by the number of marks assigned to the answer in the examination paper, and in any instance, therefore, may vary from year to year.

4. The bold text indicates the essential points required in the candidate’s answer. Words, expressions or statements separated by a solidus (/) are alternatives which are equally acceptable. A word or phrase in bold, given in brackets, is an acceptable alternative to the preceding word or phrase. Note, however, that words, expressions or phrases must be correctly used in context and not contradicted, and where there is evidence of incorrect use or contradiction, the marks may not be awarded.

5. In general, names and formulas of elements and compounds are equally acceptable except in cases where either the name or the formula is specifi cally asked for in the question. However, in some cases where the name is asked for, the formula may be accepted as an alternative.

6. There is a deduction of one mark for each arithmetical slip made by a candidate in a calculation.

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HIGHER LEVEL

Section A [At least two questions must be answered from this section]

1. (a) 5 x 3; (b) 3 + 3, 5; (c) 6 + 3; (d) 3 + 6 + 6.

2. (a) 4 + 4; (b) 3; (c) 6; (d) 3 + 3 + 3; (e) 3 x 4; (f) 3 + 3; (g) 6.

3. (a) 3 + 3; (b) 3, 4 x 3, 3, 3; (c) 5; (d) (i) 4 x 3; (ii) 3 + 3.

Section B

4. Eight items to be answered. Six marks are allocated to each item and one additional mark is added to each of the fi rst two items for which the highest marks are awarded. (a) 6; (b) 2 x 3; (c) 2 x 3; (d) 2 x 3; (e) 2 x 3; (f) 6; (g) 6; (h) 2 x 3; (i) 6; (j) 2 x 3; (k) A: 6 or B: 2 x 3.

5. (a) 5; (i) 6, 3; (ii) 3 x 3; (b) 3, 3, 6, 6; (c) 3 x 3.

6. (a) (i) 4, 4; (ii) 2 x 3; (iii) 3; (iv) 6; (b) 3 x 3; (c) 6; (d) 4 x 3.

7. (a) 3, 5 x 1; (b) 3, 6, 3; (i) 6; (ii) 3; (c) 6; (d) 6; (e) 3 x 3.

8. (a) 2, 3; (b) 3; (c) 6; (d) 9, 3 +3; (e) 3; (f) 6 + 3 +3; (g) 3 + 3.

9. (a) 7 x 1; (b) 3 x 3; (c) 6; (d) 3 x 6; (e) 6; (f) 4.

10. (a) 4, 3, 3, 3, 6, 6. (b) 6, 4, 6, 6, 3. (c) 4, 2 x 3, 6 + 3 x 3.

11. (a) 9, 6, 4, 3, 3. (b) 4, 2 x 3, 7, 8. (c) A: 7, 6, 3, 3, 6. B: 6, 2 x 3, 3, 6, 4.

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SECTION A

Eight questions. These must include at least two questions from Section A.All questions carry equal marks (50 marks).Note: (i) In many cases, most especially defi nitions, candidates’ answers must contain key phrases in order to get the assigned marks. In calculations, 3 marks are deducted for a mathematical error, 1 mark for a slip or missing units, but no further deduction is applied if the calculation, otherwise correct, is completed. (ii) Alternative valid answers are acceptable.

Question 1(a) Add deionised water to the salt and stir with rod until all solid has dissolved. (3 marks) Pour into clean volumetric fl ask using funnel. (3 marks) Wash beaker and rod and add washings to volumetric fl ask. (3 marks) Fill fl ask to 1cm3 of mark, remove funnel, use dropper to bring liquid to mark. (3 marks) Stopper and invert several times. (3 marks)(b) Methyl Orange / Methyl Red. (3 marks) Orange/yellow Peach/ red (3 marks) Indicator is a weak acid and too much would interfere with the reaction. (5 marks)(c) Na2CO3 HCl V1 = 20 V2 = 20.17 M1 = X M2 = 0.13 n1 = 1 nb = 2 M1 V1 = M2 V2 n1 n2 20 x M1 = 20.17 x 0.13 1 2 M1 = 0.06555 moles per litre [accept 0.064 – 0.067] (6 marks) 0.06555 x 106 = 6.948g/litre [accept 6.93 – 6.97g] (3 marks)(d) 4.69 x 4 = 18.76 grams/l 18.76 – 6.948 = 11.812 grams of water. (3 marks) Percentage 11.812/ 18.76 x 100 = 62.9% [accept 61% – 64%] (6 marks) 6.948 g of salt contain 11.812 g of water 106g (1 mole) of salt contains 180.2 g of water. 180.2/18 = 10 X = 10 (6 marks)

Question 2(a) X = water / H2O (4 marks) Y = calcium dicarbide / calcium carbide / carbide / calcium acetylide / CaC2 (4 marks)(b) black / dark / grey (off-white) / grey-black / dirty solid (3 marks)(c) CaC2 + H2O C2H2 + CaO / CaC2 + 2H2O C2H2 + Ca(OH)2 (6 marks)(d) Hydrogen sulphide H2S and phosphine PH3. Both gases removed by bubbling them through acidifi ed copper sulphate. (9 marks)(e) Unsaturated means the compound contains either a C = C double bond or a C ≡ C triple bond (4) Reagent is bromine water (4) Type of reaction is Addition Reaction (4)

(12 marks)(f) Common name is Acetylene (3) Major use oxyacetylene fl ame (torch – but not blowtorch) / cutting metals / welding metals /making ethanal / making propanone / making propan-2-ol / making polymers (plastics) / pesticides / fuel for lamps any one (3) (6 marks)(g) Soluble in methylbenzene because ethyne is non polar. (6 marks)

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Question 3(a) Volatile liquid is a liquid with a low boiling point. Or evaporates easily (3 marks) Relative molecular mass is the mass of a molecule relative to the mass of 1/12 mass of the carbon-12 isotope. (3 marks)(b) diagram (3) mass (4 x 3) volume (3) temperature (3) (21 marks) Method 1 Diagram: fl ask, sealed (covered) with foil with small hole (pinhole)* , immersed so that at least half is under water. * Accept if hole mentioned in account of experiment. Label required: any one correct label. (3) Mass: get mass of fl ask and foil (3) [add liquid and arrange as in diagram] heat until liquid gone / heat until fl ask appears empty / vaporised (3) cool (dry) and reweigh (3) get mass of sample by subtraction (Get difference) (3) Volume: fi ll fl ask and empty into graduated (measuring) cylinder (3) Accept method using mass & density Temperature: use thermometer (probe, sensor) to read temperature of water (or got from diagram). (3) Note: temperature of water or steam cannot be assumed to be 100 oC. Method 2. Diagram: gas syringe with self-sealing cap (septum cap, can be shown sealed), surrounded by heating device (oven, steam jacket, beaker of water). Label required: any one correct label. (3) Mass: get mass of hypodermic (syringe) containing liquid (3) inject some liquid into gas syringe (3) reweigh hypodermic (syringe) (3) get mass by subtraction (Get difference) (3) Volume: read from gas syringe. (3) Temperature: read from thermometer (probe, sensor) in heating device (or got from diagram). (3) Note: temperature of water or steam cannot be assumed to be 100 oC.(c) barometer / bourdon guage / barograph (barothermograph) / pressure sensor (not probe) (5 marks)(d) (i) Temperature = 97 + 273 = 370 (3) Volume = 132 × 10–6 m3 / 0.0000132 m3 (3) n = PV = 1 × 105 × 132 × 10–6 RT 8.3 × 370 (3) = 0.00429 / 0.0043 moles (3) (12 marks) (ii) Mr = 0.25 (3) 0.00429 = 58.2 (3) (6 marks)

Question 4Eight items to be answered. Six marks to be allocated to each item and one additional mark to be added to each of the fi rst two items for which the highest marks are awarded.(a) S = +2(b) Sulfate; white precipitate (cloudiness) with barium chloride (barium nitrate, barium ions) solution which is insoluble in dilute hydrochloric acid (HCl)(c) Ideal gas: perfectly obeys the gas laws (Boyle’s law, the kinetic theory, PV = nRT) under all conditions of temperature and pressure (d) Atoms. Moles = 560/22,400 = 0.025 moles: 0.025 x 6 x 10 23 x 5 = 7.5 x 10 22 atoms.(e) Sigma and Pi: sigma – “end-on” (“head-on”) overlap of orbitals / no nodal line (3) pi – “side-on” (“sideways”, “lateral”) overlap of p-orbitals or d- or f- orbitals (3) [Both can be got from diagrams.](f) AA; Atomic Absorption.(g) Copper: = 1s2 2s2 2p6 3s2 3p6 4s1 3d10(h) Isotope; atoms of same element (same atomic number, same Z, same number of protons) with different mass numbers (different A, different number of neutrons)(i) Structure: Propanone H O H

H C C C H

H H (6 marks)

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(j) Charles’s Law: volume varies directly with kelvin (absolute) temperature / V / T* = k / V1 / T1 = V2 / T2 / V T(k) A Photodissociation: (ii) O3 O2 + O

●

B Low density: LDP shorter chains (HDP longer chains) / LDP much branching (HDP little branching) /LDP molecules packed loosely (HDP molecules packed tightly) / LDP fl exible (HDP rigid)

Question 5(a) Electronegativity: relative (measure of) attraction / number expressing (giving) attraction (3) for shared electrons / for electrons in a covalent bond (5 marks) (i) Ammonia; x x H N H (6 marks) x H

The ammonia molecule has three atoms of hydrogen bonded to one atom of nitrogen. Nitrogen has 5 electrons in its outer level and needs 3 more to get 8 in its outer level. Each hydrogen shares its electron with the nitrogen to form three covalent bonds with 3 bonded pairs and one lone pair of electrons. (3 marks) (ii) Shape: Pyramidal 107o Explain: greater repulsion by lone pairs / l.p.– l.p. > etc. pushes bonds closer together / reduces (lowers) bond angle from 109o to 107o. (9 marks)(b) Energy level: discrete (fi xed, restricted) energy of electron / energy of electron in orbital / shell (orbit) which electrons of equal energy can occupy (3 marks) Atomic orbital: high probability region for electron / region in which electron is likely to be found (3 marks) Iron; 1s2 2s2 2p6 3s2 3p6 4s2 3d6 (6 marks) Fe 2+ 1s2 2s2 2p6 3s2 3p6 4s0 3d6 Fe3+ 1s2 2s2 2p6 3s2 3p6 4s0 3d5 More stable; half fi lled d orbital (6 marks)(c) Boiling Points: ethene is non-polar no intermolecular forces and has a lower boiling point; methanol has hydrogen bonds [intermolecular forces] and the molecule is polar making it have a higher boiling point. (9 marks)

Question 6(a) (i) Reference: 2,2,4-trimethylpentane (isooctane) // heptane (n-heptane) (4 marks)

H CH3 H CH3 H

H C C C C C H

H CH3 H H H

H H H H H H H

H C C C C C C C H

H H H H H H H (4 marks)

(ii) Structural Features: short chain length / branching / ring (cyclic) / aromatic. [Any two] (6 marks) (iii) Catalyst poison / destroys catalytic converter/poisonous emission (3 marks) (iv) Additive; oxygenate / alcohol / methanol / ethanol / ether (alkoxyalkane) / methyl-t- butyl ether (MTBE, 2-methoxy-2-methylpropane) (6 marks)

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(b) Processes: Isomerisation, dehydrocyclisation, and catalytic cracking. (9 marks)(c) Heat of Combustion: heat change when 1 mole of a substance is burned in excess oxygen. (6 marks)

(d) 7C + 8H2 C7H16 ΔH = – 224.2 kJ mol

–1

C + O2 CO2 ΔH = – 394 kJ mol–1

H2 + ½O2 H2O ΔH = – 286 kJ mol–1

C7H16 7C + 8H2 ΔH = 224.2 kJ mol–1 (3)

7C + 7O2 7CO2 ΔH = 7 x – 394 / –2758 kJ mol–1 (3)

8H2 + 4O2 8H2O ΔH = 8 x – 286 / – 2288 kJ mol–1 (3)

C7H16 + 11O2 7CO2 + 8H2O ΔH = – 4821.8 kJ mol–1 (3)

OR Σ ΔHformation(products) – Σ ΔHformation(reactants) = ΔH [7 × – 394 / – 2758 (3) 8 × – 286 / – 2288 (3)] – [(–224.2) / + 224.2 ] (3) = – 4821.8 kJ mol–1 (3) (12 marks)

Question 7(a) Rate of reaction: change in concentration [Products/reactants] per unit time.(3) 5 Factors : Temperature, concentration, particle size, nature of the reactants and the presence of a catalyst (5) (8 marks)(b) GRAPH: labelled and scaled axes [Accept “time” or “minutes”; “mass” or “g”.] (3) points plotted correctly (6) [Allow 3 marks if six or more points plotted correctly; assume (0, 0) is plotted correctly] curve drawn [has to be drawn to (0, 0)] (3) Note: the (6) for points plotted correctly not given if graph paper not used. (12 marks) (i) Tangent drawn at 3 min on graph Rate Mass / Time (2.1 g / 3min) = 0.7g per min.[Allow + or – 0.2] (6 marks) (ii) 0.7/44g = 0.0159 moles/min (3 marks)(c) Slower Rate explain: acid less concentrated / rate decreases with concentration / fewer collisions at lower concentration (6 marks)(d) Effect: see candidate’s graph* [steeper at start; levels off sooner; reaches same height] (6) [Allow (3) if description is fully correct but not shown on the graph] [Allow (3) if two of the three conditions above are shown on the graph] (6 marks)(e) Dust Explosion: combustible dust particles // dryness // above certain concentration // source of ignition (light, spark, fl ame, static electricity) // oxygen (air, atmosphere) * // enclosed space ANY THREE: (3 x 3) (9 marks)

Question 8(a) Le Chatelier’s Principle: reactions at equilibrium (2) oppose (Accept ‘minimise’, ‘relieve’) applied stress(es)* (3) (5 marks) *If the word stress(es) is replaced by particular examples (e.g. pressure), all three (temp., pressure & conc.) must be given. (b) Name; Haber Process. (3 marks)(c) Kc = [NH3]

2 [Square brackets essential] (6 marks) [N2][H2]

3 (d) N2 + 3H2 2NH3 start: 4.5 mol 13.5 mol 0 equil 3 mol 9 mol 3 mol ÷ 2 1.5 mol/L 4.5 mol/L 1.5 mol/L [3 x 3] (9 marks) Kc = [1.5] 2 [1.5] x [4.5]3 (3 marks) = 0.01646 (3 marks)(e) Catalyst : Iron Fe (3 marks)(f) Ideal: Low Temperature and High Pressure (6 marks) Low temperature favour an exothermic reaction [forward to the right] (3 marks) High pressure favours the side with the least no of moles [to the right] (3 marks)

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(g) High Pressure 200 Atmosphere high temperature is used 673 K [/ compromise temperature / specifi ed high temperature] (3 marks) At low temperatures rate is too low (slow) / activation energy too high (3 marks)

Question 9(a) Names A = Ethane B = Chloroethane. C = Ethene. D = Ethanol E = Ethanoic Acid. F = 1,2 dibromoethane. G = Ethanal (7 marks)(b) Three compounds. C Ethene E Ethanoic acid G Ethanal (9marks)(c) Chlorine (Cl2) and UV light (6 marks)(d) Elimination = 3 [C2H5OH to C2H4] Reagent Aluminium Oxide (Al2O3) (6 marks) Oxidation = 6 [ C2H5OH to CH3CHO] and 4 [C2H5OH to CH3COOH] (6 marks) Reagent for oxidation Sodium dichromate (Na2Cr2O7 /sulfuric acid (H2SO4) (6 marks)(e) Hydrogen (H2) and a nickel (Ni) catalyst (6 marks)(f) Ionic Addition = 5 (4 marks)

Question 10Answer any two of the parts (a), (b), and (c).(a) 3 C2H5OH + 2Cr2O7

2- + 16 H+ 4Cr+3 + 3 CH3COOH + 11 H2O Na2Cr2O7. 2H2O 46 + 104 + 112 + 36 = 298. No. of moles = 11.92/ 298 = 0.04 moles (4 marks) C2H5OH 24 + 5 + 16 + 1 = 46. No. of moles = 2.3/46 =0.05 moles (3 marks) Ratio of Ethanol to dichromate = 3:2 (3 marks) But actual ratio is 0.05 : 0.04. Dichromate is in excess. Therefore the ethanol is the limiting factor. (3 marks) CH3COOH = 60 grams. Theoretical yield = 60 x 0.05. = 3.0 grams (6 marks) Percentage Yield = 1.32/3.0 = 44% yield. (6 marks)

(b) BOD The amount of dissolved oxygen in p.p.m used up by a sample of water over a period of fi ve days at 20o C in the dark. (6 marks) The enrichment of water with nutrients [ nitrates or phosphates] (4 marks) (i) BOD: when sample kept in the dark for fi ve days at 20 oC (293 K) (2) (6 marks) (ii) 9.8 – 2.6 = 7.2 x 20 [dilution factor] = 144 p.p.m (6 marks) (iii) p.p.m = mg/litre (3 marks)(c) The indicator itself dissociates according to the equation HX H+ + X¯ (4) In base (alkali / high pH) equilibrium lies on the right (shifts forward) giving colour of X- ion / in base (alkali / high pH) indicator is dissociated (X-) giving colour of ions dissociated = present as ions (ionised) (3) In acid (low pH) equilibrium lies on the left (shifts backward) giving colour of (HX) molecule / in acid (low pH) indicator is associated giving colour of molecule (HX) Associated = present as molecules (3) (10 marks)

pH [H+] = √ 0.2 x 5.0 x 10-4 (6) [H+] = 0.01 (3) pH = 2 (3) Nitric Acid pH = 0.699 (3) (15 marks)FO

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Question 11Answer any two of the parts (a), (b), and (c)(a) Test for Phosphate ion. Place a sample of the test solution in a test tube. Add ammonium molybdate reagent and heat gently (9 marks) Yellow precipitate forms and is soluble in ammonia solution. (6 marks) Flame test (4 marks) Copper Green fl ame (3) Sodium Yellow/orange fl ame (3) (6 marks)(b) Reducing agent is oxidized so the oxidation number will increase (4 marks) SO3

2- is the reducing agent. C2O42- is the reducing agent (6 marks)

Cl2 + SO3 2– + H2O 2Cl

– + SO4 2– + 2H+ (7 marks)

16H+ + 2 MnO4− + 5C2 O4

2− 10 CO2 + 2Mn2+ + 8 H2O (8 marks)

(c) A Blocking (preventing escape, absorption or reabsorption) of radiation (heat, energy, infrared) (4) by gases in the atmosphere (7 marks) Carbon dioxide (CO2) / methane (CH4) / chlorofl uorocarbon (CFC) / oxides of nitrogen (NOx) water (H2O) [Accept ozone] any three (6 marks) carbon dioxide: fossil fuel combustion / respiration / deforestation / aerosols / car exhausts methane: paddy fi elds (rice growing) / ruminants (cows, sheep, etc., anaerobic digestion) landfi ll (dumps) / natural gas leakage chlorofl uorocarbon: refrigeration / aerosol propellants / foams / fi re extinguishers nitrogen oxides: car exhausts / nitrogenous fertilisers water: [Accept fossil fuel combustion / evaporation / transpiration / perspiration (sweating) / excretion /respiration / car exhausts] ozone: [Accept welding /photocopiers] ANY ONE: [Must correspond with answer given in (ii) any one source 3 (3 marks) Nitrogen (N2) / oxygen (O2) / any identifi ed noble gas (name or formula) (3 marks) Role of oceans: Controls the levels of CO2. Dissolves CO2 and converts it to carbonates and hydrogenates, used for photosynthesis by marine plants, currents carry CO2 to lower colder regions in the sea where it remains dissolved. (6 marks) B Calcium chloride reduces the melting point of the mixture 1070 K to 870 K (6 marks) Anode = graphite; cathode = steel (2 x 3 marks) Na+ + e- Na (3 marks) Prevent sodium and chlorine (products) from reacting to reform NaCl. (6 marks) Street light/ coolant in nuclear reactors (4 marks)

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ORDINARY LEVEL

Section A [At least two questions must be answered from this section]

1. (a) 2 x 4; (b) 3 x 3; (c) 6; (d) 3, 6; (e) 3 x 3; (f) 2 x 3; (g) 3.

2. (a) 3 x 5; (b) 5; (c) 6; (d) 3, 6 (e) 6; (f) 3; (g) 6

3. (a) 6; (b) 5; (c) 6, 6; (d) 3; (e) 2 x 3; (f) 6; (g) 6; (h) 6.

Section B

4. Eight items to be answered. Six marks are allocated to each item and one additional mark is added to each of the fi rst two items for which the highest marks are awarded. (a) 2 x 3; (b) 6; (c) 6; (d) 2 x 3; (e) 6; (f) 6; (g) 6; (h) 6; (i) 6; (j) 6; (k) A 6 or B 6.

5. (a) 2 x 3, Protons/neutrons 2 x 3, Electrons 2 x 3; (b) 8 x 4.

6. (a) 2 x 3, 6, 6, 3 x 3; (b) 2 x 4; (c) 6, 6, 3.

7. (a) 6, 5; (b) (i) 6; (ii) 6; (iii) 4 x 6; (c) 3.

8. (i) 7 x 2; (ii) 3; (iii) 3; (iv) 6; (v) 3, 6; (vi) 3; (vii) 3 x 3; (viii) 3.

9. (a) 3, 6; (b) 8 x 4; (c) 3 + 6.

10. (a) (i) 7; (ii) 4 x 3, 6. (b) 5, axes 5, points 6, curve 3, oxygen 6; (c) 7, 6, 6, 6;

11. (a) (i) 6; (ii) 6; (iii) 6; (iv) 7. (b) 7, Volume 3 x 3, moles 6, molecules 3; (c) A: (i) 6, 3, Fixed 5, Variable 4, Factors 4, 3. B: (i) 4, 3 x 3, 6, 2 x 3.

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SECTION A

Eight questions. These must include at least two questions from Section A.All questions carry equal marks (50 marks).Note: (i) In many cases, most especially defi nitions, candidates’ answers must contain key phrases in order to get the assigned marks. In calculations, 3 marks are deducted for a mathematical error, 1 mark for a slip or missing units, but no further deduction is applied if the calculation, otherwise correct, is completed. (ii) Alternative valid answers are acceptable.

Question 1(a) A solution whose exact concentrate is known (4 marks) Substance from which a solution of exact concentration can be made //Substance which is stable, available pure and water soluble (4 marks)(b) Pipette (3) Wash Pipette with deionised water (3) and then wash with Sodium Hydroxide solution (3) (solution that it will contain) (9 marks)(c) Wash bottle used to wash down the sides of conical fl ask/Washes splashes into fl ask. (6 marks)(d) Indicator: Methyl orange/methyl red (3) Yellow/orange to Pink [Accept peach]/red (6) (9 marks)(e) M1xV1 = M2xV2 n1 n2

19.5 x 0.10 = 20 x M2 (3) 1 1

1.95/20 = M2 (3)

0..0975mol/l = M2 (3) (9 marks)

(f) repeat without indicator or repeat with volumes determined by titration (3 marks) evaporate solution to small volume or allow pure salt to crystallise out (be produced) (3 marks)(g) Yellow (3 marks)

Question 2(a) X = Water//H2O (5) Y = Calcium dicarbide //CaC2 (5) Grey sandy solid (5) (15 marks)(b) Alkynes (5 marks)(c) H C C H (6 marks)(d) Bubble the gas through a sample of bromine water. The liquid will be decolourised / become paler. [Change from brown to colourless] (9 marks)(e) lots of soot / dark smoke / highly luminous fl ame (6 marks)(f) Oxygen O2 (3 marks)(g) Used for cutting metals / oxyacetylene blow torch (6 marks)

Question 3(a) Heat of Reaction is the heat change //energy change //when 1 mole is reacted / no of moles react according to balanced equation. (6 marks)(b) exothermic is a reaction in which heat is given out//off (5 marks)

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(c) Polystyrene/plastic//Insulated beaker. (6) Negligible heat capacity/good insulator (6) (12 marks)(d) burette/pipette (3 marks)(e) use accurate thermometer (short range / 0.1 deg calibrated) //mix (stir) //measure quickly // plot temperatures //record highest temperature reached. {any one} (6 marks)(f)

(6 marks)

(g) 0.05 moles of hydrochloric acid. (6 marks)(h) Neutralisation. (6 marks)

Question 4Eight items to be answered. Six marks are allocated to each item and one additional mark is added to each of the fi rst two items for which the highest marks are awarded.(a) High melting and boiling point/ dissolve in water/conduct electricity when molten or in solution//solids at room temperature. [any two](b) Marie Curie(c) Tetrahedral(d) A substance which alters the rate of a chemical reaction and is not changed by the reaction.(e) To kill bacteria.(f) Liquid Petroleum Gas(g) Bomb calorimeter(h) Increasing nuclear charge (more protons in nucleus)(i) Helium Nuclei. [ Two protons and two neutrons 4He2] (j) Niels Bohr(k) A – co-product is any other product formed along with the main product. B – a mixture of two or more elements one of which is a metal.

Question 5(a) Atomic Number: is the number of protons in // The nucleus of / an atom (3 marks) Relative atomic mass; Average mass of atoms of an element relative to // 1/12 of the mass of the carbon isotope C-12 (3 marks) 20 Protons 20 Neutrons (6 marks) 2 8 8 2 (3) 2 Two (3) (6 marks)(b) 1 = Positively 2 = Negatively 3 = Cations 4 = Anions 5 = Electrons 6 = Oppositely 7 = High 8 = Solids (8 x 4 marks)

Question 6(a) Homologous Series is a group of organic compounds with the same general formula, same functional group, same chemical properties, differ by CH2 [any 2] (6 marks) Saturated contain single C – C bonds (6 marks) Substitution reaction is where an atom or group of atoms are replaced/substituted by another atom or group of atoms. (6 marks) Structural isomers; same molecular formula different structural formula[different arrangement of atoms] (3) Correct drawing of n-butane (3) and 2methyl propane (3) (9 marks)

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(b) Residue – Road Surfacing/roofi ng Naphtha – Petrol/plastics Kerosene – Home heating Light Gasoline – Petrol. Refi nery gas – heating/cooking Gas oil – Cars /trucks/heating Any 2 (2 x 4 marks)(c) (i) Measure of the ability of a fuel to resist knocking//measure of a tendency of a fuel to cause autoignition. (6 marks) (ii) Lead. Poisoned catalytic converter//Poisonous / toxic / environmental reasons / health / pollution (6 marks) (iii) Catalytic cracking, //Isomerisation,// Dehydrocyclisation. Any 1 (3 marks) Question 7(a) Le Chatelier’s Principle: If a stress is applied to a system at equilibrium the system will alter to minimise the effect of the stress. (6 marks) A state of chemical equilibrium exists when the forward and reverse reactions occur simultaneously. (5 marks)(b) (i) Exothermic (6 marks) (ii) Kc = [SO3]

2

[SO2]2[O2] (6 marks)

(iii) Increase Pressure Increase yield [side with the least number of moles favoured] Addition of Oxygen Increase yield [system will remove excess and move to the right] Increase in Temperature Decrease yield [favours the endothermic reaction] Removal of SO2 Decrease yield [system will move to the left to replace it] (4 x 6marks)(c) Will not favour either reaction but equilibrium will be reached faster. (3 marks)

Question 8(i) A = Ethene B = Ethanol C = Ethanoic acid D = Chloroethane E = Benzene F = Ethanal G = Butane (7 x 2marks)(ii) G Butane (3 marks)(iii) D Chloroethane (3 marks)(iv) C2H5OH + 3O2 2CO2 + 3H2O (6 marks)(v) Elimination.(3) Aluminium Oxide (Al2O3) (6) (9 marks)(vi) E Benzene (3 marks)(vii) Sodium Ethanoate, Carbon Dioxide and Water. (9 marks)(viii) F Ethanal (3 marks)

Question 9(a) – log10 [H

+]//– log10[H3O+] / negative log to base ten of hydrogen ion concentration (3 marks)

Use a pH meter/ Use Universal Indicator. (6 marks)(b) 1 = Magnesium 2 = Temporary 3 = Hydrogen carbonate. 4 = Boiling 5 = Permanent 6 = Sulfates 7 = Ion Exchange 8 = Sodium (8 x 4 marks)(c) eutrophication is the rapid growth of plants and algae caused by an excess of nutrients in the water. Nitrates and Phosphates cause eutrophication. (9 marks)

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Question 10(a) (i) 2.16 – 1.82 = 0.34 g in 500 cm3. 0.64g / litre 640 mg/litre 640 p.p.m

(7 marks) (ii) Describe: Sample of solid / paste of solid // Preparatory step (clean wire / soak splint) // On nickel (nichrome, steel, wire) probe / wood splint // Placed in Bunsen fl ame // Colour imparted to fl ame (any 4 × 3) Colour: lilac (6) (18 marks)(b) Catalyst (5 marks) (i) Graph = Correctly labelled and scaled axes [Allow 3 marks only if axes are reversed] (5 marks) All points plotted correctly [Allow 3 marks if 6 to 8 points plotted correctly]

(6 marks) Curve drawn correctly (3 marks) (ii) 11 cm3 of oxygen produced in 1.4 to 1.6 min. [accept 1min 24 sec to 1 min 36sec]

(6 marks)(c) Oxidation = Loss of electrons (7 marks) (i) Blue solution decolourises / brownish precipitate formed [Allow 3 for ‘colour change’] (6 marks) (ii) Magnesium (Mg) is oxidised (6 marks) (iii) Magnesium is higher in the electrochemical series than copper (6 marks)

Question 11(a) (i) 1.6 / 16 = 0.1 moles (6 marks) (ii) 0.2 moles of oxygen (6 marks) (iii) 4.48 litres (dm3) (6 marks) (iv) CO2 = 44g (3) 0.1 mole = 4.4g (4) Ans 4.4g (7 marks)(b) Charles Law; At constant Pressure the volume of a fi xed mass of gas is directly proportional to the temperature. (7 marks) Volume: 2 x 10 5 x 6400 ÷ 312 = 1 x 10 5 x V ÷ 273 = 11,200 [LHS – 3 marks, RHS – 3 marks, calculation of V – 3 marks] (9 marks) How many moles = 0.5moles (6 marks) How many molecules = 3 x 10 23 molecules (3 marks)(c) A (i) feedstock = Reactants in industrial process /prepared raw materials (6 marks) (ii) Raw material = materials required (3 marks) Fixed Costs= must be paid regardless of the rate of production (5 marks) Variable costs = depend directly on the level of plant output. (4 marks) Factors = proximity of source of raw materials/water //good transport system// proximity of market //availability of suitable workforce [any two] (7 marks) B (i) Carbon atoms (4 marks) (ii) Diamond (3) and Graphite (3) . Diamond is covalently bonded (3). Graphite is also covalently bonded but also has van der Waals forces (6) (15 marks) Diamond = jewellery (3) Graphite = lubricant//pencil lead (3) (6 marks) FO

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