Successfully reported this slideshow.
We use your LinkedIn profile and activity data to personalize ads and to show you more relevant ads. You can change your ad preferences anytime.

Chemistry Perfect Score 2011 module answer

4,939 views

Published on

Chemistry Perfect Score 2011 module answer

Published in: Technology, Business
  • Be the first to comment

Chemistry Perfect Score 2011 module answer

  1. 1. BAHAGIAN PENGURUSANSEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER JAWAPANMODUL PERFECT SCORE 2011 CHEMISTRY [KIMIA]  Set 1  Set 2  Set 3  Set 4  Set 5 1
  2. 2. JAWAPAN SET 1PAPER 2 : STRUCTURED QUESTIONSection ANo. Answer Mark1 (a) The formula that shows the simplest whole number ratio of atoms 1 of each element in a compound. (b) H2SO4 + Zn → ZnSO4 + H2 2 (c) Heating, cooling and weighing are repeated until a constant mass is 1 obtained. (d) Element Copper Oxygen Mass, g 47.70 – 25.30 53.30 – 47.70 =22.40 =5.60 Mole atom 22.40 5.60 64 16 = 0.35 = 0.35 Simplest ratio 1 1 Empirical formula = CuO 4 (e) H2 + CuO → Cu + H2O 2 (f) To prevent the hot copper from being oxidized again. 1 (g) Magnesium ribbon Heat 2 TOTAL 13No. Answer Mark2 (a) (i) Al2CO3 1 (ii) Al2(CO3)3 Al2O3 + 3CO2 2 (iii) The number of mole of Al2 (CO3) 3 = 70.2/ 234 = 0.3 mol 1 Based on the balanced equation; Al2 (CO3)3 : Al2O3 1 : 1 0.3 : 0.3 1 Mass of Ag = 0.4 x 102 = 30.6 g 1 (iv) Based on the balanced equation Al2 (CO3)3 : CO2 1 : 3 0.3 : 0.6 1 Volume of CO2 = 0.9 x 24 = 21.6 dm3 1 = 21600 cm3 1 (b) (i) Zinc carbonate 1 (ii) Zinc oxide and carbon dioxide 1 2
  3. 3. (iii) ZnCO3 → ZnO + CO2 1 TOTAL 12No. Answer Mark3 (a) (i) The number of protons found in the nucleus of an atom 1 (ii) 7 1 (b) 1 33 Q 16 (c) P and S // Q and R 1 (d) (i) Q and R 1 (ii) Have same proton number but different nucleon number // 1 Have same number of protons but different number of neutrons (e) (i) Melting point : 63 OC [values & unit must be correct] 1 (ii) Section Physical state AB Solid DE Liquid and gas 1 (iii) the heat energy absorbed by the particles is used 1 to overcome the forces of attraction between particles 1 TOTAL 10No. Answer Mark4 (a) Sodium and magnesium // sodium and aluminium // magnesium and 1 aluminium (b) Halogen 1 (c) 2.8.3 1 (d) (i) Sodium, magnesium, aluminium, chlorine, argon 1 Atomic size decreases (ii) From left to right : The proton number // the positive charge increases from sodium to argon 1 The forces of attraction by the nucleus on the electrons (nuclei attraction) 1 in the first three occupied shells become stronger (e) (i) Sodium burnt rapidly and brightly with a yellow flame // 1 White fumes liberated // white solid formed (ii) 2Na + Cl2 → 2NaCl [Formula of reactants and product are correct] 1 [Balanced equation] 1 has high melting / boiling point // conduct electricity 1 (iii) in molten state or aqueous solution // soluble in water TOTAL 10 3
  4. 4. No. Answer Mark5 (a) (i) X 1 (ii) 8 valence electron // electron arrangement 2.8 // achieve octet electron 1 arrangement (b) Covalent 1 (c) (i) VW4 ( (b) 1 (a) i i (ii) ) ( 1+1 W W V W W (iii) has low melting / boiling point // cannot conduct electricity 1 in molten and solid state . // insoluble in water// soluble in organic solvent. (d) (i) Ionic compound 1 (ii) Atom U donate one electron to form U+ ion 1 Atom W accept one electron to form W- ion 1 U+ ion and W- ion attracted to each other by strong electrostatic force / 1 ionic bond. (iii) 1 1 U W [Number of electron each shells are correct] [Number of charge symbol are correct] TOTAL 13 4
  5. 5. PAPER 2: ESSAY QUESTIONSection BNo. Answer Mark6 (a) Group 17 1 Period 3 1 Has seven valence electrons. 1 Has three shells occupied with electron 1 (b) (i) Between Y and X 1.Atom Y has 1 valence electron and atom X has 7 valence electron 1 2. to achieve octet electron arrangement 1 3. Atom Y loses/donates/transfers 1 electron to form ion Y+ 1 4. Atom X gains/receives 1 electrons from atom Y to form ion X- 1 5 Y+ ion and X- ion are attracted by a strong electrostatic force / ionic bond 1 6. Diagram + Y X 1 (ii) Between W and X 1. Atom W has 4 valence electrons and atom X has 7 valence electrons. 1 2. Each atom W contributes 4 electrons whereas each atom X contributes one electron for sharing. 3. to achieve octet electron arrangement 1 4. Four atoms of X share a pair of electrons with one atom W to form a 1 WX4 molecule / Diagram 1 X W X X W W W X W Molecules WX4 5
  6. 6. (c) Compound P : ionic bond 1 Compound Q : Covalent bond 1 Melting Point Compound P Ions are held by strong electrostatic forces. More energy is needed to overcome these forces. 1 Compound Q 1 Molecules are held by weak intermolecular forces. Only a little energy is required to overcome the forces. Or 1 Electrical conductivity 1 Compound P In molten state or aqueous solution , there are free moving ions 1 Ions carry charge 1 Compound Q 1 In molten and solid states , no free moving ions 1 exist as molecule TOTAL 20No. Answer Mark7 (a) (i) 2.8.7, Chlorine 1+1 (ii) 2Fe + 3Cl2 → 2FeCl3 Correct formulae of reactants and product 1 Balanced 1 (b) (i) Z,Y,X 1 Z more reactive than X 1 Atomic size of Z bigger than atomic size X 1 Valence electron become further away from nucleus 1 Valence electron to be more weakly pulled by the nucleus 1 Valence electron can be released more easily in atom Z 1 (ii) same/similar 1 Same valence electron 1 (c) X : 2.4 1 Y : 2.6 1 to achieve octet electron arrangement one X atom contributes four electron and each two Y atoms 1 contributes two electrons for sharing 1 Group 16 1 Period 2 1 6 valence electron 1 2 shells occupied with electrons 1 TOTAL 20 6
  7. 7. PAPER 2: ESSAY QUESTIONSection CNo. Answer Mark8 (a) (i) Dilute acid: Hydrochloric acid / Sulphuric acid/ Nitric acid 1 Metal N: Magnesium / zinc 1 (ii) Anhydrous calcium chloride 1 To dry the hydrogen gas 1 (iii) Example: Copper(II) oxide 1 Copper ion is reduced// reduction process 1 Because oxidation number of copper decrease from +2 to 0 Hydrogen is oxidised// oxidation process 1 Because oxidation number of hydrogen increase from 0 to +1 1 Hydrogen is reducing agent 1 Copper(II) ion// Copper(II) oxide is oxidising agent 1 (b) (i) Relative Molecular mass of (CH2)n = 56 1 (12 + 2) n = 56 n=4 Molecular formula = C4H8 1 (ii) Unglased porcelain chips Glass wool soaked in butanol Heat Water 2 Procedure: 1. A small amount of glass wool soaked in butanol is placed in a boiling tube. 1 2. The boiling tube is clamped horizontally 1 3. The unglazed porcelain chips are placed in the middle section of the boiling tube. 1 4. The boiling tube is closed with a stopper fitted with a delivery tube 1 5. The unglazed porcelain chips are heated strongly. Then, the glass wool is warmed gently to vaporize the propanol. 1 6. The gas released is collected in a test tube. 1 TOTAL 20 7
  8. 8. No. Answer Mark9 (a) (i) Formula that shows the simplest ratio of the number of atoms for each 1 element in the compound. (ii) Copper(II)oxide // lead(II)oxide 1 CuO + H2  Cu + H2O // PbO + H2  Pb + H2O 1+1 (b) (i) Magnesium oxide / zinc oxide 1 (ii) Procedure: 1. Clean magnesium / zinc ribbon with sand paper 1 2. Weigh crucible and its lid 1 3. Put magnesium ribbon into the crucible and weigh the crucible with its lid 1 4. Heat strongly the crucible without its lid 1 5. Cover the crucible when the magnesium starts to burn and lift/raise 1 the lid a little at intervals 6. Remove the lid when the magnesium burnt completely 1 7. Heat strongly the crucible for a few minutes 1 8. Cool and weigh the crucible with its lid and the content 1 9. Repeat the processes of heating, cooling and weighing until a constant 1 mass is obtained 10. Record all the mass 1 Result: Description Mass/g 1 Crucible + lid x Crucible + lid + magnesium y Crucible + lid + magnesium oxide z Calculation: Element Mg O Mass, g y-x z-y Mole y-x z-y 24 16 1 =0.1 =0.1 Simplest ratio 1 1 1 Empirical formula: MgO Max 10 (c) Element C H Mass (%) 84.6 15.4 Number of moles 84.6/12 15.4/1 1 =7.05 =15.4 1 Mole ratio 1 2 Empirical formula : CH2 1 RMM of (CH2)n = 70 1 [ 12 + 2]n = 70 14 n = 70 n = 5 1 Molecular formula : C5H10 1 20 8
  9. 9. JAWAPAN SET 2PAPER 2 : STRUCTURED QUESTIONSection A No. Answer Mark1 (a) Cell II 1 (b) (i) Magnesium electrode 1 (ii) e V 1 Magnesium Copper electrode electrode (iii) Copper electrode thicker // Brown solid deposited 1 (c) 1. Correct formulae of reactant and product 1 2. Balanced equation 1 Cu2+ + 2e → Cu (d) (i) Electrical energy to chemical energy 1 (ii) Blue colour remain unchange 1 (iii) 1. Concentration / Number of mole of Cu2+ ion remain unchanged 1 2. Rate of Cu2+ ion discharge at cathode is the same as rate of Cu atom ionize at anode 1 TOTAL 10 No. Answer Mark2 (a) (i) Iodine 1 r: formula/iodide/iodine gas (ii) MnO4 - + 8 H+ + 5 e → Mn2+ + 4 H2O 1 (iii) +7 → +2 1 reduction 1 (iv) Potassium chloride // iron(II) sulphate // [any reducing agent] 1 (b) (i) Zinc 1 (ii) 1. Correct formulae of reactant and product 1 2. Balanced equation 1 2 Zn + O2 → 2 ZnO a: 2 J + O2 → 2 JO (iii) K,J, L 1 (iv) Predict : no changes 1 r: no reaction Reason : L is more reactive than J/zinc 1 r: more electropositive TOTAL 11 9
  10. 10. PAPER 2 : ESSAY QUESTIONSection B No. Answer Mark3 (a) 1. Propanone is a covalent compound 1 2. Propanone exist as molecule // No freely moving ion in propanone 1 3. Sodium chloride is an ionic compound 4. Sodium chloride solution has freely moving ion 1 1 (b) (i) Properties Cell X Cell Y 1. Type of cell Voltaic cell Electrolytic cell 2. Energy Chemical → electrical Electrical → chemical change 1 3. Electrodes Anode: A Positive terminal: C // Copper 1 Cathode: B Negative terminal: D // Zinc 4. Ions in Cu2+, SO42-, H+ and OH- Cu2+, SO42-, H+ and OH- ions 1 electrolyte ions 5. Half Anode: Positive terminal: 1 equation Cu → Cu2+ + 2e Cu2+ + 2e → Cu Cathode: Negative terminal 1 Cu2+ + 2e → Cu Zn → Zn2+ + 2e 6. Observation Anode: Positive terminal: 1 Copper ecomes thinner Copper plate becomes thicker 1 Cathode: Negative terminal: 1….6 Copper becomes Zinc becomes thinner thicker (c) 1. Ag, M, L 1 2. L is more electropositive than silver 1 3. L displace silver from silver nitrate solution 1 4. M is more electropositive than silver 1 5. M displace silver from silver nitrate solution 1 6. M is less electropositive than L 1 7. M cannot displace L from L nitrate solution 1 (i Copper // Cu 1 i) TOTAL 20 No. Answer Mark4 (a) (i) 1. Correct formulae of reactant and product 1 2. Balanced equation 1 Zn + 2e → Zn2+ 3. Correct formulae of reactant and product 1 4. Balanced equation 1 Pb2+ + 2e → Pb (ii) 1. Zinc is oxidized 1 2. Zinc atom donates / losses electrons 1 3. Lead(II) nitrate / Pb2+ is reduced 1 4. Lead(II) nitrate / Pb2+ receives electrons 1 (b) (i) 1. Green colour of iron(II) sulphate change to brown 1 2. Correct formulae of reactant and product 1 3. Balanced equation 1 Cl2 + 2Fe2+ → 2Cl- + 2Fe3+ 4. Colourless solution of potassium iodide change to brown 1 5. Correct formulae of reactant and product 1 6. Balanced equation 1 Cl2 + 2I- → 2Cl- + I2 10
  11. 11. (ii) Test tube P : Cl- ion and Fe3+ ion 1+1 Test tube Q : Cl- ion and I2 1+1 (iii) 1. Add starch solution 1 2. Dark blue precipitate formed 1 TOTAL 20PAPER 2 : ESSAY QUESTIONSection C No. Answer Mark 2+5 (a) (i) 1. Cu // copper(II) ion 1 Equation 2. Correct formula of reactant and product 1 3. Balance 1 Cu2+ + 2e → Cu 4. Copper 1 (ii) 1. Oxygen 1 2. Insert glowing splinter into the test tube 1 3. Glowing splinter relights 1 (iii) 1. NO3- // nitrate ion 1 2. Oxygen 1 3. OH- ion is discharge 1 4. OH- ion is place lower than NO3- ion in the electrochemical series 1 Equation 5. Correct formula of reactant and product 1 6. Balance 1 4 OH- → 2 H2O + O2 + 4 e (b) Diagram 1. Functional apparatus 1 2. Label 1 Impure Pure copper copper Copper(II) sulphate solution 3. Pour [50 – 200 cm3] copper(II) sulphate solution into a beaker 1 4. Connect pure copper as cathode and impure copper as anode 1 5. Dip both pure and impure copper into copper(II) sulphate solution 1 6. Anode : Cu → Cu2+ + 2e 1 7. Cathode : Cu2+ + 2e → Cu 1 TOTAL 20 11
  12. 12. No. Answer Mark6 (a) (i) Metal P : Tin // Lead // Copper 1 Metal Q : Magnesium // Aluminium // Zinc 1 (ii) Exp I 1. Metal P is less electropositive than iron 1 2. Iron is oxidized 1 3. Iron losses electron // Fe → Fe2+ + 2e 1 4. Dark blue precipitate indicates the presence of Fe2+ ion 1 Exp II 5. Metal Q is more electropositive than iron 1 6. Metal Q is oxidized // Metal Q losses electron 1 7. Water and oxygen receive electron // 2H2O+O2 + 4 e → 4OH- 1 8. Pink colouration indicates the presence of OH- ion 1 (b) (i) 1. Bromine is reduced 1 2. Bromine molecule receives electron // Oxidation number of bromine 1 decrease / 0 → -1 3. Iron(II) sulphate / Fe2+ is oxidized 1 4. Fe2+ losses electron // Oxidation number of iron increases / +2→ +3 1 5. Correct formula of reactant and product 6. Balanced equation 1 Br2 + 2Fe2+ → 2Br- + 2Fe3+ 1 7. Brown colour of bromine decolourise 8. Green colour of iron(II) sulphate change to brown 1 1 (ii) 1. Add sodium hydroxide solution 1 2. Brown precipitate formed 1 TOTAL 20 12
  13. 13. JAWAPAN SET 3 PAPER 2 : STRUCTURED QUESTION Section A No. Answer Mark1 (a) (i) Solution in test tube C 1 (ii) Solution in test tube A 1 (b) 1. Higher than pH value of 0.1 moldm-3 HCl // The pH is 3/4/5/6 1 2. Ethanoic acis is a weak acid// Etanoic acid ionizes partially in water to produce 1 low concentration oh hydrogen ion 3. The lower the concentration, the lower the pH value 1 (c ) (i) Magnesium chloride 1 (ii) Mg + 2 H+ → Mg2+ + H2 1. Correct formula of reactant and product 1 2. Balanced equation 1 (iii) No of mole, HCl = 0.1 x 5 / 1000 = 0.0005 mol 1 Based on balanced equation, 2 mol of HCl : 1 mol of H2 0.0005 mol of HCl : 0.00025 mol of H2 // mol of H2 = 0.005/2 = 0.0025 Volume of hydrogen gas = 0.00025 x 24 dm3 1 = 0.006 dm3 // 6 cm3 1 (d) White precipitate 1 TOTAL 12 No. Answer Mark2 (a) (i) Solvent P: Water 1 Solvent Q: methyl benzene / propanone / suitable organic solvent 1 (ii) Effervescence / gas released // magnesium ribbon dissolved 1 (iii) 1. In the presence of solvent P/water , ethanoic acid ionize to form H+ ion. 1 2. H+ ion causes the ethanoic acid to show its acidic properties 1 3. In solvent Q, ethanoic acid exist as molecule// hydrogen ion does not present 1 (b) (i) 1. pH value increase / bigger 1 2. The lower the concentration of acid the higher the pH value 1 (ii) (0.5)(V) = (0.04)(250) // 1 V = 20 cm3 1 3 Alkali that ionize/dissociate completely in water to produce high concentration of hydroxide 1 1 (a) ions. (b) Alkaline / alkaline solution 1 1 (c) P: ion 1 Q: molecule 1 2 (d) No 1 Because there are no hydroxide ions in the solution// ammonia exist in the form of molecule. 1 2 13
  14. 14. (e) (i) 1 1. Colourless gas bubbles are released.// efeervesence 1 (ii) Mg + 2HCl  MgCl2 + H2 1. Correct formula 1 2. Balanced equation 1 2. Mol of Mg = 2.4/24 // 0.1 mol 1 4 3. Volume of H2 = 0.1  24 dm3 = 2.4 dm3 1 Total 11NO ANSWER MARK 4 (a) (i) Green 1 (ii) Double decomposition reaction 1 (b) (i) carbon dioxide 1 (ii) CuCO3 → CuO + CO2 1. Reactants and products are correct 1+ 1 2. Equation is balanced (iii) Copper(II) carbonate Heat Lime water - Labelled diagram 2 - Functional (c) 1 mol CuCO3 = 12.4/124 = 0.1 mol Mol of CuCl2 = 0.1 x 135g Mass = 13.5g 3 10 No. Answer Mark5 (a) Mg + 2HCl → MgCl2 + H2 1+1 (b) (i) 0.4/24 = 0.0167 mol 1 (ii The number of mole of HCl = MV/1000 = 1x 50/1000 = 0.05 mol 1 ) (c) From the chemical equation 1 mol of magnesium produce 1 mol hydrogen If 0.0167 mol produce 0.0167 mol hydrogen 1 Volume of hydrogen = 0.0167 x 24 dm3= 0.4 dm3/ 400 cm3 1 (d) I 400 /100 =4 cm3s-1 1 II 400 /60 = 6.67 cm3s-1 1 (e) As catalyst 1 (f) The temperature of hydrochloric acid 1 The concentration of hydrochloric acid 1 TOTAL 11 14
  15. 15. No. Answer Mark6 (a) The heat released when 1 one mole of copper is displaced from copper (II) sulphate 1 solution by zinc. (b) Cu2+ + Zn → Cu + Zn2+ 1 (c) The blue colour of the solution become colourless//Brown deposit is formed// 1 The polystyrene cup become hot//The reading of the thermometer increase 1 (d) (i) Heat release = 50 x 4.2 x 10 1 = 2100 J (ii) The number of moles = 50 x 0.5 = 0.025 mol 1000 1 (iii) Heat of displacement = 2100 = -84000 J 0.025 H = 84.0 kJ/mol 1 (e) To ensure all the copper(II) sulphate solution reacted completely 1 (f) Energy Zn + Cu2+ H= - 84.0 kJ/mol Zn2+ + Cu 1+1 TOTAL 10 No. Answer Mark7 (a) Graph : Axes labeled with units 1 All points plotted correctly 1 & Shape of graph correct 1 (i) 50 cm3 ( marked on the graph) 1 (ii) NaOH + HCl NaCl + H2O Mol of NaOH = 50 x 1 = 0.05 1 1000 From the equation : 1 mol NaOH : 1 mol HCl 0.05 mol NaOH : 0.05 mol HCl 1+1 Concentration HCl = 0.05 x 1000 = 1 moldm-3 50 (c) To ensure uniform temperature of mixture in the polystyrene cup 1 (d) All the sodium hydroxide has reacted completely 1 (e) (i) 0.1 mole of NaOH when reacted releases 5.6 kJ 1 Therefore for 1 mole of NaOH reacted, 5.6/0.1 = 56 kJ heat energy released 15
  16. 16. (ii) Energy NaOH + HCl H= - 56.0 kJ/mol 1+1 H2O + NaCl (i) Less than 5.6 kJ 1 (ii) - Hydrochlolric acid is strong acid dissociates completely in water ; ethanoic 1 acid is a weak acid dissociates in partially water - Part of the heat released during neutralisation is absorbed to ionise further 1 ethanoic acid molecules, therefore heat released will be less than 5.6 kJ TOTAL 14PAPER 2 : ESSAY QUESTIONSection B No. Answer Mark8 (a) (i) Label axes with units 1 All points are transferred correctly 1 Shape of the graph is smooth and correct 1 (ii) 2.5 cm3 1 (iii) moles of Pb2+ ions = 2.5 x 1.0 / / 0.0025 1 1000 moles of I- ions = 5 x 1.0 // 0.005 1000 1 Pb2+ : I- 0.0025 : 0.0005 1 1 : 2 1 (b) Test tube 1: 1. Ion exist : K+, I- and NO3- 2. All lead(II) nitrate reacts completely 1 3. Excess of potassium iodide 1 4. Solution contains soluble salt of potassium iodide and potassium nitrate 1 1 Test tube 5: 5. Ion exist : K+ and NO3- 6. All lead(II) nitrate reacts completely and all potassium iodide reacts 1 completely 1 7. Solution contains soluble salt of potassium nitrate 1 Test tube 7: 8. Ion exist : K+, Pb2+ and NO3- 1 9. All potassium iodide reacts completely 1 10. Excess of lead(II) nitrate 1 11. Solution contains soluble salt of lead(II) nitrate and potassium nitrate 1 Max 10 16
  17. 17. No. Answer Mark9 (a) (i) Size of the reactant/the total surface area of the reactant 1 Concentration of the reactant 1 Temperature of the reactant 1 Catalyst 1 (ii) Temperature : 450-550oC 1 Catalyst : iron 1 Pressure : 200 atm 1 (b) (i)  The axes are labeled together with its unit 1  The scale is correct 1  The points are transferred correctly 1 1  The curve is smooth (ii) Average rate of reaction for experiment I = 26.0 1 210 = 0.12 cm3 s-1 1 Average rate of reaction for experiment II = 26.0 150 1 = 0.17 cm3 s-1 [correct unit] 1 (iii) 1. The rate of reaction for Experiment II is higher than in Experiment I 1 2. The concentration of HCl in Experiment II is more/higher than in 1 Experiment I 3. The number of hydrogen ion/ H+ per unit volume of the solution in 1 1 Experiment II is more than in Experiment I 4. The frequency of collisions between hydrogen ion and calcium carbonate 1 in Experiment II is more than in Experiment I 5. The frequency of effective collisions hydrogen ion and calcium carbonate in Experiment II is more than in Experiment I TOTAL 20PAPER 2 : ESSAY QUESTION Section C No. Answer Mark10 (a) (i) Experiment I – hydrochloric acid or 1 Experiment II – sulphuric acid Mg + 2HCl → MgCl2 + H2 1+1 (ii) The number of mole of HCl = MV/1000 = 1.0 x 50 = 0.05 mol 1000 or 1 The number of mole of H2SO4 = MV/1000 = 1.0 x 50 = 0.05 mol 1000 (iii) The rate of reaction is the change of volume of hydrogen gas per unit time 1 17
  18. 18. (b) (i) Volume of hydrogen/ cm3 Experiment II Experiment I 1 1 Time/s 1. Curve with label 2. Axis with title and correct unit (ii) 1. Sulphuric acid in experiment II is diprotic acid, hydrochloric acid in experiment I is monoprotic acid//Concentration of hydrogen ion, H+ in experiment II is higher than experiment I 2. The number of hydrogen ion per unit volume in experiment II is higher than experiment I 3. Frequency of collisions between hydrogen ions and magnesium atoms in experiment II is higher than experiment I 4. Frequency of effective collisions between hydrogen ions and magnesium atoms in experiment II is higher than experiment I 5. Rate of reaction in experiment II is higher than experiment I …5 (c) Diagram : Functional apparatus set-up 1 Label correctly 1…..2 Procedure : 1. A burette is filled with water and inverted over a basin containing water. The burette is clamped vertically to the retort stand. 1 2. The water level in the burette is adjusted and the initial burette reading is recorded. 1 3. 50 cm3 of 0.2 moldm-3 hydrocloric acid / sulphuric acid is measured and poured into a conical flask 4. 4. 5 cm of magnesium ribbon are added into the conical flask 1 5. 5. close conical flask immediately with the stopper fitted with 1 delivery tube. 1 6. At the same time the stopwatch is started shake the conical flask. 1 7. The burette readings are recorded at 30 second intervals for 5 1 minutes 7 max 5 Time/s 0 30 60 90 120 150 180 ……1 Volume of gas / cm3 TOTAL 20 No. Answer Mark11 (a) Water on the wet shirt evaporated 1 Evaporation absorbs heat energy from body 1 (b) (i) C2H5OH + 3 O2  2 CO2 + 3 H2O H = - 1,376 kJ / mol 1+ 1 1. Heat of combustion for propanol is higher than ethanol 1 2. No. of carbon and hydrogen atoms per molecule propanol is higher 1 18
  19. 19. than ethanol 3. No. of mole of CO2 and H2O produced during combustion of 1 propanol is more than ethanol 1..4 4. Formation of CO2 and H2O releases heat energy (ii) Diagram – labelled and functional 1 1….2 Material : Water , ethanol Apparatus : spirit lamp. weighing balance, copper can, clay-pipe 1 triangle, thermometer, wind shield 1…..2 Procedure : 1. Measure (100 – 250) cm3 of water and pour into the copper can and initial temperature is recorded after 5 minutes 1 2. Weigh the spirit lamp filled with ethanol 3. Light the spirit lamp to heat the water in the can and stir 1 4. Extinguish the spirit lamp when the temperature increase reaches 1 30˚C, record the maximum temperature of water reached 5. Weigh the spirit lamp with its remain. 1 Result : 7. The initial mass of the spirit lamp + ethanol = a g The final mass of the spirit lamp + ethanol = b g 8. The mass of ethanol burnt = (a-b) g 1 9. The initial temperature of water = t1˚C The maximum temperature of water = t2˚C 1 10. Increase in temperature of the water = (t2 – t1) t˚C Calculation : 1 RMM of ethanol C2H5OH = 46 11. The no. of mol of ethanol burnt = ab = y mol 60 12. The released heat = mc  = 100 x 4.2 x t 1 = xJ x 1 13. The heat of combustion of propanol = - J mol-1 or -Z y 1 kJ mol-1 13 max 8 TOTAL 20 No. Answer Mark12 (a) Exothermic reaction is a reaction that releases heat to the surrounding 1 The total energy content of the products is lower than the total energy content of the 1 reactants Endothermic reaction is a reaction that absorbs heat from the surrounding 1 The total energy content of the products is higher than the total energy content of the 1 reactants (b) A reacts with B to form C and D 1 A and B are the reactants while C and D are the products 1 Heat energy is absorbed from surrounding //It is an endothermic reaction 1 Total energy content of C and D/ product is higher than total energy content of A and B/ 1 reactants When reaction occurs, the temperature of mixture of solutions increases / becomes hot 1 (any 4 of the above) (c) 1. 1 mole of silver nitrate solution produces 1 mole of Ag+ ion 1 2. 1 mole of sodium chloride solution produces 1 mole of Cl- ion 1 19
  20. 20. 3. One e mole of potassium chloride produces 1 mole of Cl- ion 1 4. The heat of precipitation of silver chloride is heat that released when 1 mole of AgCl 1 is formed from Ag+ ion and Cl- ion // Ag+ + Cl-  AgCl 5. Number of mole of AgCl produced in bothe reactions are the same, heat released are 1 the same. Max 4 (d) Materials : calcium nitrate solution, sodium carbonate solution 1 Procedures : - measure 50 cm3 of 1.0 mol/ dm3 Ca(NO3)2 solution and 50 cm3 of 1.0 mol / dm3 1 Na2CO3 solution separately and poured into a plastic cup - measure and record the initial temperature of both solutions after 5 minutes 1 - pour quickly and carefully Ca(NO3)2 solution into the plastic cup that contains 1 Na2CO3 solution and stir continuously - measure and record the lowest temperature reached 1 Tabulation of data : Initial temperature of Ca(NO3)2 / oC Ө1 Initial temperature of Na2CO3 / oC Ө2 Average initial temperature / oC (Ө1 + Ө2)/2 Ө3 Lowest temperature of the mixture / oC Ө4 1 Change in temperature / oC Ө3- Ө4 Calculation : No. of moles of CaCO3 = No. of moles of Ca(NO3)2 1 = mv/1000 = 1.0(50)/1000 = 0.05 heat change mc(Ө4 – Ө3) 1 = x kJ heat of reaction = + x kJmol-1 0.05 = + y kJmol-1 TOTAL 20 No. Answer Mark13 a (i) 1. Zinc nitrate, zinc sulphate 1 2. Zinc carbonate 1 (ii) I :Sodium carbonate solution/ potassium carbonate solution / ammonium 1 carbonate solution 1 II : Sulphuric acid (iii) 1. 50 cm3 of 1 mol dm-3 magnesium nitrate solution is measured and 1 poured into a beaker 2. 50 cm3 of 1 mol dm-3 Sodium carbonate solution/ potassium carbonate solution / ammonium carbonate solution solution is measured and 1 poured into the beaker. 1 3. The mixture is stirred with a glass rod and a white solid, magnesium 1 carbonate is formed. 1 4. The mixture is filtered 5. and the residue is rinsed with distilled water 1 6. The white precipitate is dried by pressing it between filter papers. 1…6 20
  21. 21. c (i) 1. nitrate ion / NO3- ion 1 2. Add dilute sulphuric acid followed by iron(II) sulphate solution into test tube 1 containing salt X solution 3. Add a few drops of concentrated sulphuric acid through the wall of test tube 1 4. A brown ring is formed. 1 (ii) 1. Zn2+ , Pb2+ , Al3+ 1 2. Add ammonia solution into test tube containing salt X solution until excess 1 3. White precipitate dissolves in excess ammonia solution showing the 1 presence of Zn2+ ions 4. White precipitate insoluble in excess ammonia solution showing the 1 presence of Pb2+ and Al3+ ions. 1 5. Add potassium iodide solution into test tube containing salt X solution 1 Yellow precipitate formed showing the presence of Pb2+ ions // 6. No change showing the presence of Al3+ ions. 20 21
  22. 22. JAWAPAN SET 4PAPER 2 : STRUCTURED QUESTIONSection A No Answer Mark1 (a) Compound that contains only carbon and hydrogen 1 Has double bonds between carbon – carbon atoms 1 (b) Alkene 1 (c) Propene 1 (d) (i) Hydrogenation / Addition reaction 1 (ii) 1 (e) (i) C3H6 + 9/2 O2 → 3CO2 + 3H2O or 2 2C3H6 + 9O2 → 6CO2 + 6H2O (ii) 2 .1 1 No. of mole of C3H6 = 42 = 0.05 Volume of gas CO2 = 0.05 x 3 x 24 = 3.6 dm3 1 TOTAL 10 No Answer Mark2 (a) Ethanol 1 (b) Hydroxyl group 1 (c) (i) Oxidation 1 (ii) Orange colour of potassium dichromate (VI) solution turns to green 1 (iii) 1 H O H CC O H H (d) (i) Esterification 1 (ii) Ethyl ethanoate 1 (iii) Pleasant smell 1 (iv) CH3COOH + C2H5OH → CH3COOC2H5 + H2O 2 TOTAL 10 22
  23. 23. No. Explanation Mark3 (a) (i) Haber process 1 (ii) N2 + 3H2 2NH3 Correct formula 1 Balanced 1 (iii) 450 oC --- 550oC 1 Vanadium(V) oxide 1 (iv) As a fertiliser 1 (b) (i) Polyvinyl chloride // polychloroethene 1 (ii) 1 (c) Correct arrangement 1 Tin atom Correct label 1 Copper atom TOTAL 10 No. Explanation Mark4 (a) (i) glycerol 1 (ii) saponification / alkaline hydrolysis 1 (iii) to cause precipitation of soap 1 (b) (i) 1 X: detergent 1 Y :soap (ii) magnesium stearate or calcium stearate 1 (iii) Mg2+ and Ca2+ 1+1 (iv) causes water pollution / non-biodegradable 1 TOTAL 9 23
  24. 24. PAPER 2 : ESSAY QUESTIONSection B No Answer Mark5 (a) (i) 14.3 % 1 (ii) Element C H Mass/ % 85.7 14.3 1 No. of moles 85.7 14.3 = 7.14 = 14.3 12 1 2 Ratio of moles/ 7.14 14.3 Simplest ratio =1 =2 7.14 7.14 3 Empirical formula = CH2 RMM of (CH2)n = 56 .............1 [(12 + 1(2)]n = 56 14n = 56 56 6 max 5 n = 14 = 4 ………..1 Molecular formula : C4H8 ………………..1 (iii) 1+1 1+1 But-1-ene But-2-ene Max 4 2-methylpropene (iv) Compound M (Butene, C4H8) has a higher percentage of carbon atom in their molecule than butane, C4H10 …………….1 4(12) % of C in C4H8 = x 100% 4(12)  8 48 = x 100% 56 = 85.7% …………1 4(12) % of C in C4H10 = x 100% 4(12)  10 48 = x 100% 58 = 82.7% ………..1 .....3 (b) (i) Starch 1 Protein 1 (ii) H H CH3 H I I I I C = C– C = C I I 1 H H 1..2 2-methylbut-1,3-diene or isoprene (c) (i) Rubber that has been treated with sulphur 1 24
  25. 25. In vulcanised rubber sulphur atoms form cross-links between the rubber 1 (ii) molecules These prevent rubber molecules from sliding too much when stretched 1 TOTAL 20 No. Explanation Mark6 (a) Examples of food preservatives and their functions:  Sodium nitrite – slow down the growth of microorganisms in meat 1+1  Vinegar – provide an acidic condition that inhibits the growth of microorganisms in pickled foods 1+1 (b) (i) Paracetamol 1 Codeine 1 (ii) To follow the instructions given by the doctor concerning the dosage and 1 method of taking the medicine To visit the doctor immediately if there are symtoms of allergy or other side 1 effects of thye medicine (iii) If the correct dosage is not given by the doctor, it will cause abuse of the 1 medicine. For instance, if the child is given a overdose of codeine, it may lead to addition. If the child is given paracetamol on a regular basis for a long time, it may cause skin rashes, blood disorders and acute inflammation of the pancreas. 1 (c) Type of food Examples Function additives Preservatives Sugar, salt To slow down the growth 2 of microorganisms Flavourings Monosodium To improve and enhance glutamate, spice, the taste of food 2 garlic Antioxidants Ascorbic acid To prevent oxidation of 2 food Dyes/ Colourings Tartrazine To add or restore the Turmeric colour in food 2 Disadvantages of any two food additives: Sugar – eating too much can cause obesity, tooth decay and diabetes 1 Salt – may cause high blood pressure, heart attack and stroke. 1 Tartrazine – can worsen the condition of asthma patients - May cause children to be hyperactive MSG – can cause difficult in breathing, headaches and vomiting. TOTAL 20PAPER 2 : ESSAY QUESTIONSection C Mar Questions Marking criteria ks7 (a) (i) 1. Sulphur is burnt in air to produce sulphur dioxide // 1 2. Burning of metal sulphides/zinc sulphide / lead sulphide produce sulphur 1 dioxide 1 1 3. Sulphur dioxide is oxidised to sulphur trioxide in excess oxygen 1 4. Sulphur trioxide is dissolved in concentrated sulphuric acidto form oleum. 5. The oleum is diluted with water to produce concentrated sulphuric acid 25
  26. 26. (ii) H2SO4 + 2NH3→ (NH4)2SO4 Formula for reactants and product correct 1 Balanced 1 (b) 1. Bronze is harder than copper 1 2. Atoms of pure copper are same size and arrange in layers 1 3. when force applied the layers will slide. 1 1 4. In bronze tin atom has different size compare to pure copper 1 5. and interrupt the orderly arrangement of pure copper. max4 (c) Procedure: 1. Iron nail and steel nail are cleaned using sandpaper. 1 2. Iron nail is placed into test tube A and steel nail is placed into test tube B. 1 3. Pour the agar-agar solution mixed with potassium hexacyanoferrate(III) 1+ 1 solution into test tubes A and B until it covers the nails. 4. Leave for 1 day. 1 5. Both test tubes are observed to determine whether there is any blue spots formed or if there are any changes on the nails. 1 6. The observations are recorded 1 Results: Test tube The intensity of blue spots 1 A High 1 B Low 1 Conclusion: Iron rust faster than steel. TOTAL 20 No Answer Mark8 (a) (i) X - any acid – methanoic acid 1 Y - any alkali – ammonia aqueous solution 1 (ii) 1. Methanoic acid contains hydrogen ions 1 2. Hydrogen ions neutralise the negative charges of protein membrane 1 3. Rubber particles collide, 1 4. Protein membrane breaks 1 5. Rubber polymers combine together 1 5 max 4 (iii) Ammonia aqueous solution contains hydroxide ions 1 Hydroxide ions neutralise hydrogen ions (acid) produced by activities of 1 bacteria (b) (i) Alcohol 1 (ii) Burns in oxygen to form carbon dioxide and water 1 Oxidised by oxidising agent (acidified potassium dichromate (VI) solution) to 1 form carboxylic acid (iii) Procedure: 1. Place glass wool in a boiling tube 3 2. Pour 2 cm of ethanol into the boiling tube 3. Place pieces of porous pot chips in the boiling tube 4. Heat the porous pot chips strongly 5. Heat ethanol gently 6 max 5 6. Using test tube collect the gas given off Diagram: 26
  27. 27. Porous pot chips Glass wool soaked with ethanol Heat Heat Water[Functional diagram] ….1 ...2[Labeled – porous pot, water, named alcohol, heat] ….1Test:Put a few drops of bromine water .....1Brown colour of bromine water decolourised .....1 ...2 Total 20 27
  28. 28. JAWAPAN SET 5 PAPER 3 SET 1 EXPLANATION SCORE1. (a) [Able to record all reading accurately with unit](i) Sample answer Experiment Metal X Metal Y 3 I 1.70 cm 1.40 cm II 1.75 cm 1.45 cm III 1.75 cm 1.45 cm [Able to record all reading correctly without unit] 2 [able to record three to five reading correctly 1 No response or wrong response 0 EXPLANATION SCORE1(a) [Able to construct a table to record the diameter of the dents and average(ii) diameters for material X and Y that contain: 1. correct title 2. Reading and unit Sample answer: Material Diameter of the dents(cm) Average 1 2 3 diameter,(cm) X 1.70 1.75 1.75 1.73 3 Y 1.40 1.45 1.45 1.43 [Able to construct a table to record the diameter of the dents and average 2 diameters for material X and Y that contain 1. title 2. Reading [Able to construct a table with at least one title / reading 1 No response or wrong response 0 EXPLANATION SCORE1.(b) [Able to state correct observation] 3 Sample answer: The diameter of dents made on material Y is smaller than material X// The diameter of dents made on material X is bigger than material Y [Able to state correct observation, incompletely] 2 Sample answer: The diameter of dents made on material Y is smaller// The diameter of dents made on material X is bigger [Able to state an idea of the observation] 1 Sample answer: The diameter of dents for Y is small// The diameter of dents for X is big No response or wrong response 0 28
  29. 29. EXPLANATION SCORE1.(c) [Able to state the inference correctly] 3 Sample answer: Material Y is harder than material X// Material X softer than material Y [Able to state the inference correctly/ 2 Sample answer: Material Y is harder // Material X softer [Able to state an idea of inference. 1 Sample answer: Material Y is hard// Material X is soft No response or wrong response 0 EXPLANATION SCORE1.(d) [Able to state the correct operational definition for alloy] 3 1. what should be done and 2. what should be observe correctly Sample answer: When the weight of 1 kilogram is dropped at height of 50 cm to hit the ball bearing which is taped onto the alloy block using cellophane tape a smaller dent is formed. [Able to state the meaning of alloy, incompletely] 2 Sample answer: Material that form small dent is hard [Able to state an idea of alloy] 1 Sample answer: Alloy form dent//alloy is hard No response or wrong response 0KK0508 EXPLANATION SCORE1.(e) [able to give all three explanations correctly] 3 Sample answer: 1. atoms in material X are in orderly arrangement 2. atoms in material Y are not in orderly arrangement 3. layer of atoms in material Y difficult to slide on each other [able to give any two explanations ] 2 [able to give any one explanations ] 1 No response given / wrong response 0 29
  30. 30. EXPLANATION SCORE 1.(f) [Able to state any alloy for material Y and its major pure metal for materials X correctly] 3 Sample answer: Material X: copper // iron// any suitable metal Material Y: bronze/ brass//stainless steel// any suitable alloy for pure metal given. [Able to state any alloy for material Y and its major pure metal for materials X correctly] 2 Sample answer: Material X: tin/ zinc// chromium / nickel // any suitable metal Material Y: bronze// brass//stainless steel// any suitable alloy for pure metal given. [Able to state any alloy for material Y and its major pure metal for materials X correctly] 1 Sample answer: Material X: magnesium // aluminium//zinc // any metal Material Y: pewter // bronze // stainless steel //any alloy No response given / wrong response 0 EXPLANATION SCORE1.(g) [Able to state the relationship correctly between the manipulated variable and responding 3 variable with direction] Sample answer: The harder/ softer the material, the smaller / bigger the diameter of the dent. [Able to state the relationship correctly between the manipulated variable and responding 2 variable with direction] Sample answer: Alloy/ pure metal will form smaller/ bigger dent than pure / alloy // The smaller / bigger the diameter of the dent, the harder/softer the material [able to state the idea of hypothesis] 1 Sample answer: Y is harder // X is softer // alloy is harder No response given / wrong response 01.(h)EXPLANATION SCORE[Able to state all the three variables and all the three actions correctly] 3Sample answer: Names of variables Action to be taken (i) manipulated : (i) the way to manipulate variable: Type of materials / material X and Y Change pure metal/ alloy with alloy /pure metal (ii) responding: (ii) what to observe in the responding variable: Diameter of dent The diameter of the dent formed on material X and Y. (iii) controlled: (iii) the way to maintain the controlled Mass of the weight // height of the weight variable: // size of steel ball bearing. Uses same mass of weight // same height of weight // same size of steel ball bearing[able to state any two variables and any two actions correctly] 2[able to state any one variablesand any two action correctly] 1No response given / wrong response 0 30

×