Chemistry perfect score module 2010 answer scheme

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Chemistry perfect score module 2010 answer scheme

  1. 1. SET 1 PAPER 2: STRUCTURE QUESTION 1. (a) bromine and phenol 1 (b) liquid 1 (c) Nickel 1 (d) 1 (e) ion 1 (f) (i) T1 1 (ii) Heat absorbed by the particles/molecules is used to overcome the attraction forces between the particles/molecules in solid naphthalene. 1 (iii) Become faster 1 ……8 o 2 (a) T C 1 (b) t2 1 (c) Heat energy released to the surroundings is balanced by the heat release as the particles attract one another to form a solid 1 (d) (i) molecules 1 (ii) P R 1+1 (e) (i) liquid and solid 1 (ii) Solid 1 (f) Sublimation 1 …….9 3 (a) Functional apparatus 1 Label 1 (b) 1 1
  2. 2. (c) 1. Label and unit of axes X and Y 1 2. Scale and size of graph 1 3. Transfer point 1 4. Correct and smooth curve 1 O (d) (i) Melting point is shown on the graph at 80 C 1 (ii) Melting point is the temperature at which solid change to liquid 1 (e) Heat absorbed by the particles/molecules is used to overcome the attraction forces between the particles/molecules in solid naphthalene. 1 …… 10 4 (a) A representation of a chemical substance using letters for atoms and subscripts for each type of atoms present in the substance. 1 (b) (i) Hydrochloric acid // nitric acid // sulphuric acid 1 Magnesium // zinc (ii) Correct formula of reactant and product 1 Balance equation 1 Example: Mg + HCl → MgCl2 + H2 (c) Flow the hydrogen gas into the combustion tube for a few minutes to remove air 1 before heating metal oxide (d) (i) Number of mole of copper = 1.62 // 0.025 mol 64 1 (ii) Number of mole of oxygen = 0.40 // 0.025 mol 1 16 (iii) Number of mole of copper : Number of mole of oxygen 0.025 : 0.025 // 1 1 : 1 CuO 1 (e) Iron oxide // Tin oxide // Lead(II) oxide // Silver oxide 1 (f) Burn metal M in oxygen 1 ...11 5 (a) (i) Magnesium // zinc 1 Hydrochloric acid // nitric acid // sulphuric acid 1 (ii) Correct formula of reactant and product 1 Balance equation 1 Example: Zn + 2HCl  ZnCl2 + H2 2
  3. 3. (b) (i) 1 1 1 Empirical formula is MO2 1 (ii) Correct formula of reactant and product 1 Balance equation 1 Example: MO2 + 2H2  M + 2H2O (c) Flow the hydrogen gas into the combustion tube for a few minutes to remove air 1 before heating metal oxide // Allow flow of hydrogen gas into the combustion tube during cooling (d) No. 1 Magnesium is more reactive than hydrogen. 1 …13 6 (a) The chemical formula that shows the simplest ratio of the number of atoms of each type of elements in the compound 1 (b) (i) Mass of magnesium = (26.4-24.0)g =2.4 g Mass of oxygen = (28.0 – 26.4) g = 1.6 g 1 (ii) moles of magnesium atoms : moles of oxygen atoms 1.6 : 2.4 // 24 16 0.1 : 0.1 1 (iii) MgO. 1 (iv) Correct formula of reactant and product 1 Balance equation 1 Example: 2Mg + O2  2MgO (c) To allow oxygen to enter the crucible for complete combustion to occur. 1 (d) Functional apparatus 1 Label 1 X oxide Dry hydrogen gas → (e) Collect the gas in a test tube.Place a burning wooden splinter at the mouth of the test tube. No pop sound. 1 Heat …..10 3
  4. 4. 7 (a) (i) 2.6 1 (ii) Period : 2 1 Group : 16 1 (b) (i) 9 1 (ii) 10 1 (iii) 19 P 1 9 (iv) 1. Element Q is more reactive than element R. 1 2. The size of atom Q is smaller tha atom R 1 3. The attraction forces between nucleus and valence electron of atom Q is stronger 1 than atom R 4. it is easier for atom Q to receive valence electron compare to atom R. 1 (c) (i) Ionic bond 1 (i) Has high melting and boiling point // Conduct electricity in aqueous solution and molten state // Soluble in water // insoluble in organic solvent 1 (d) 1+1 P C P ….14 8 (a) 2.8.2 1 (b) (i) Ionic bond 1 2+ (ii) 1. Atom X releases 2 electrons to form X ion 1 2. achieve octet electron arrangement 1 (iii) Each ion drawn correctly 2+ 2- X Y R Number of electrons for ion X and ion Y are correct 1 Charge of ions are correct 1 (iv) Has high melting and boiling point// Conduct electricity in aqueous solution and 1 molten state // Soluble in water // insoluble in organic solvent (c) (i) ZY2 1 (ii) 12 + 2(16) // 44 1 ……9 9 (a) Metal : P/Q/R/S/T 1 Non-metal : V/U/W/X 1 (b) 2.8.5 1 (c) R 1 (d) Electronegativity increases from Q to U 1 1. The atomic size decreases from Q to U 1 2. The nuclei attraction forces between electrons and nucleus increases from Q to U 1 + (e) Q 1 4
  5. 5. (f) X 1 Atom X has achieved octet electron arrangement 1 (g) Correct formula of reactant and product 1 Balance equation 1 Example: 2R + U2 → 2RU (h) 1. form coloured ions 1 2. has more than one oxidation number 3. as catalyst 4. form complex ions [ any one] …..13 10 (a) (i) seven / 7 1 (ii) 2.8.7 1 + (b) (i) P 1 (ii) Q 1 (iii) PQ2 1 (c) Element U is not reactive // do not cause explosion 1     (d)       Q                              Q T Q                         Q        1 st mark: - showing the sharing of electrons and correct number of 1 electrons in each shell 2 nd mark – correct label and correct number of electron pairs being 1 shared (e) low melting point & boiling point // does not conduct electricity 1 TOTAL 9 5
  6. 6. 11 (a) (i) 2.8.1 1 (ii) Period : 2 1 Group : 16 1 (b) (i) Ionic 1 (ii) + 2- + P Q P R Number of electrons for ion X and ion Y are correct 1 Charge of ions and ratio of ion X to ion Y are correct 1           S (c)                              S R S                         S        Correct number of atom R and S 1 Correct number electrons and shells 1 (d) 1. Compound in (b) can conduct electricity at molten or aqueous state. 1 2. Compound in (c) cannot conduct electricity at any state 1 3. Compound in (b) consist of freely moving ions 1 1 4. Compound in (c) consist of neutral molecules / no freely moving ion OR 1. The melting point of compound (b) is high 2. The melting point of compound (c) is low 3. Ions in compound (b) are attracted by strong electrostatic forces 4. molecules in compound (c) are attracted by weak van der Waals forces ….12 6
  7. 7. 12 (a) Q 1 (b)(i) Ion 1 (b)(ii) solid state : Ions are not freely moving// ions are in a fixed position. 1 molten state : Ion can move freely 1 (c)(i) R : Gas 1 T : Liquid 1 (c)(ii) 1 (c)(iii) - Van der waal/intermolecular forces between molecules are weak 1 - Small amount of heat is required to overcome it 1 .......9 PAPER 2 :ESSAY QUESTION: 13 (a) (i) The number of neutrons : 18 and 20 respectively 2 (ii) Similarities Differences 1. having the same proton number// 1. different in the number of neutrons // Same number of electrons different in the nucleon number 2. having the same valence electron// 2. different in physical properties having the same chemical properties 4 (b) (i) 1. Nucleus contains 6 proton and 6 neutron 1 2. Electrons move around the nucleus 1 3. Two shells are filled with electrons 1 4. There are 6 valence electron// electron arrangement is 2.6 1 …….4 (ii) Comparison Diagram P Q Proton number 6 6 Number of valence electron 4 4 Chemical properties similar similar Number of neutron//nucleon number 6//12 7//13 Physical properties different different Standard representation of element different different Any four 4 (c) - Correct curve 1 - Mark 71 on the y axes which is at the same level with the 1 - X and Y axis with correct title and unit 1……3 Temperature/ ◦ C 100 71 60 Time/s 7
  8. 8. (ii) 1 Substance X in both solid and liquid state 1 2. heat energy is released 1 3. kinetic energy of particles decreases 1 4. They are closer to each other // Attraction force between the particles become stronger 1……4 20 14 (a) 1. Number of mole in 16 g of oxygen = 16/32 // 0.05 mole 1 -3 -3 2. Volume occupied by 16 g of oxygen = 0.05 mole x 24 dm // 12 dm 1 3. Number of mole in 22 g of CO2 = 22/44 // 0.05 mole 1 -3 -3 4. Volume occupied by 22g of CO2 = 0.05 moles x 24 dm // 12 dm 1…..4 (b) Element C H N O Mass /g 0.48 0.05 0.28 0.16 1 Number of 0.48/12 0.05/1 0.28/14 0.16/16 mole //0.04 //0.05 //0.02 //0.01 1 The simplest 0.04/0.01 0.05/0.01 0.02/0.01 0.01/0.01 1 ratio // 4 // 5 // 2 // 1 Empirical formula = C4H5N2O 1 [C4H5N2O ]n = 194 1 [ 97 ]n = 194 1 n = 194/97 // 2 1 Molecular formula = C8H10N4O2 1…..8 (c) 1 molar mass of ammonium sulphate = 132g/mol 1 2 percentage of nitrogen in ammonium sulphate = 28/132 x 100% // 21.2% 1 3 molar mass of urea = 60 g/mol 1 4 percentage of nitrogen in urea = 28/ 60 x 100% // 46.7% 1 5 molar mass of hydrazine = 32g/mol 1 6 percentage of nitrogen in hydrazine = 28/132 x 100% // 87.5% 1 7 Hydrazine has the richest source of nitrogen compares with other fertilizers. 1 8 The farmer should choose hydrazine 1…8 20 15 (a) 1. The proton number is 11 // Number of proton is 11 1 2. Nucleon number is 23 // Atomic mass is 23 1 3. Number of neutron = 23-11 = 12 1 4. Nucleus contains 11p and 12n 1 5. Position of electron circulating the nucleus 1 6. Correct number shell consists of electron 1.....6 7. Symbol of sodium is Na any 6 (b) (i) Formula that show simplest ratio number of atoms of each element in compound 1 8
  9. 9. (ii) 1. Relative molecular mass for n(CH2O) = 180 // 1 12n + 2n + 16n = 180 2. n = 6 1 3. C6H12O6 1 ....4 (c) (i) Element Fe Cl 1. Mass/g 2.80 5.32 1 2. No. of moles 2.80/56 = 0.05 5.32/35.5 = 0.15 1 3. Ratio of moles/ 0.05/0.05 = 1 0.15/0.05 = 3 1 Simplest ratio 4. Empirical formula = FeCl3 1...4 (ii) 1. Formula of the reactants 2. Formula of products 1 3. Balance equation 1 1 2Fe + 3Cl2 → 2FeCl3 4. No. of moles Fe = 2.80/56 = 0.05 mol 5. No. of moles Cl2 = (0.05 x 3)/2 = 0.075 mol 1 3 3 1 6. Volume of Cl2 = 0.075 x 24 = 1.8 dm / 1,800 cm 1 ...6 20 16 (a) Formula that shows the simplest ratio of the number of atoms for each element in the compound. 1…1 (b) Element C H Mass (%) 92.3 7.7 1 Number of 92.3 7 .7 = 7.7 = 7.7 1 moles 12 1 Ratio of moles 1 1 1 Empirical formula : CH RMM of (CH)n = 78 [ 12 + 1]n = 78 1 13 n = 78 n = 6 1…5 Molecular formula : C6H6 (c) Procedure: 1. Clean magnesium 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 the lid a 1 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 mass 1 is obtained Record all the mass 9
  10. 10. 10. Results: Mass/g 1 Crucible + lid x Crucible + lid + magnesium y 1 Crucible + lid + magnesium oxide z 1 Calculations: 1 Empirical formula: MgaOb / MgO 1 Element Mg O Mass (g) y-x z-y Max yx zy 12 Number of moles 24 16 Simplest ratio of moles a b (d) - Cannot separate copper from magnesium oxide 1 - Cannot weigh copper 1 20 17 (a) (i) 1.The electron arrangement of atom Q : 2.7 1+1 2. The electron arrangement of atom R : 2.4 2 (ii) 1.The number of neutrons in atom Q is 10 1+1 2. Number of electron in atom Q is 9 2 (b) 1.Q and R form covalent bond. 1 2.Atom Q has an electron arrangement of 2.7 1 3.Atom R has an electron arrangement of 2.4 1 4.To achieve the stable electron arrangement, atom R 1 shares electrons with atom Q. 5.One atom R contributes 4 electrons. 6. Each atom R contributes one electron. 1 7. Atom R shares four of its valence electrons each with 4 1 atoms of Q 1 8. molecule with the formula RQ4. 1 Q Q Q R Q Notes : points 4, 7 and 8 can be obtained from the diagram 8 10
  11. 11. (c) For group 1 elements, 1. Going down the group ,atomic size increases 2. the valence electron becomes further away from the nucleus 3. Forces of attraction between nucleus and the valence electron 1 becomes weaker. 1 4. It is easier for the atom to donate / release the valence electrons. 5. The reactivity increases down the group 1 For Group 17 elements , 6. Atomic size increases when descending the group 1 7. the valence electrons become further away from nucleus. 8. Forces of attraction between the protons / nucleus and the valence electrons become weaker 1 9. It is more difficult for the atom to accept /gain/receive electrons. 1 8.The reactivity decreases down the Group 1 1 1 max 8 Total 20 18 (a) (i) Atom Y : 2.8.7 1 Atom Z : 2.8.8.1 1 Group 17 1 Because atom Y has 7 valence electron 1 Period 3 1 Because atom Y has three shells occupied with electrons 1….6 (ii) 2Na + Y2 → 2NaY Correct formula of reactants and product correct 1 Balance equation 1 Y2 + 2NaOH → NaY + NaOY + H2O Correct formula of reactants and products correct 1 Balance equation 1…4 (b) Ionic compound/bond 1- The electron arrangement of atom P = 2, 8, 1 / 2. 8. 1 The electron arrangement of atom Q = 2, 8, 7 / 2. 8. 7 1 2- to achieve the stable electron arrangement 1 + 3- atom of P donates / gives one electron to form P // half equation 1 - 4 - atom of Q receives / accepts one electron to form Q // half equation 1 + - 5 -the P ion and Q ions are attracted by a strong electrostatic force to form ionic bond . 1 6- formula PQ / correct diagram 1 Covalent compound / bond 7-The electron arrangement of atom R = 2, 6 / 2. 6 1 8-to achieve the stable electron arrangement 11
  12. 12. 9-atom R and atom Q share electrons 1 10 - atom R contributes 4 electrons and atom Q contributes one electron 1 11- one atom R and 4 atom Q share 4 pairs of electrons 12- to form covalent compound with the formula RQ4 / diagram 1Max 10 PAPER 3 STRUCTURE Explanation Score 19 (a) Suggested Answer 3 (a) 0 s = 95.0 ° C, 30 s = 85.0 ° C, 60 s = 82.0 ° C, 90 s = 80.0 ° C, 120 s = 80.0 ° C,150 s = 80.0 ° C, 180 s = 78.0 ° C, 210 s = 70.0 ° C. (b) Suggested Answer 3 Time (second) Temperature (° C) 0.0 95.0 30.0 85.0 60.0 82.0 90.0 80.0 120.0 80.0 150.0 80.0 180.0 78.0 210.0 70.0 (c) (i) & (ii) [ Score 3 & 3 ] Temperature / ° C 95 x x 90 x 80.0 x x x Freezing x point 70 x 60 30 60 90 120 150 180 210 Time / second 12
  13. 13. d(i) Suggested Answer The constant temperature at which liquid becomes a solid 3 d(ii) Suggested Answer The heat released when the particles in the liquid arrange 3 to form solid balanced by the heat loss to the surroundings. e Suggested Answer The air trapped in the conical flask is a poor conductor 3 of heat. The air helped to minimize heat loss to the surroundings. / to ensure uniform cooling. f Suggested Answer 3 Element Compound P, R Q, S 20 (a) Observation Inference (i) White fume is released (i) Magnesium oxide is formed (ii) White solid is formed/The mass of crucible (ii) Magnesium reacts with oxygen and its content increases. (b) The mass of crucible and lid = 25.35 g. The mass of crucible, lid and magnesium ribbon =27.75 g. The mass of crucible, lid and magnesium oxide when cooled = 29.35g (c) (i) The mass of Mg= (27.75 -25.35)g =2.40g (ii) The mass of O2=(29.35-27.75)=1.60 g (iii) The number of moles Mg=0.1 mole The number of moles O = 0.1 mole The ratio of Mg : O = 1 : 1 The empirical formula is MgO. (d) 0.1 mole of Mg reacts with 0.1 mole of O/1 mole of Mg reacts with 1 mole of O 21 (a) Able to predict the manipulated variable, the Able to state how to control the manipulated responding variable and the constant variable variables correctly completely. Manipulated variable : metals of Group 1 elements // sodium, lithium, Repeat the experiment by using the metals of potassium. sodium, lithium and potassium Responding variable: Able to state correctly the way to control the the reactivity of the reaction with water // the manipulated variable speed of movement on the water surface To observe how fast the metals move on the surface of water. Constant variable: size/mass of metals. Volume of water Able to use the metal granules with the same size Use the metal granules with the same size. 21 (b) Able to state the relationship between the manipulated variable and the responding variable correctly.. Suggested answer: The reactivity of Group 1 elements increases going down the group. 21(c) Able to arrange correctly the reactivity series of the metals according to descending order. Answer: potassium, sodium, lithium 21 (d) Able to classify the ions correctly. [to name or write all the formula of the ions correctly at the cations and anions group.] Answer: positive ion/ cation : sodium ion/ Na+, hydrogen ion/ H+ Negative ion/anion : hydroxide ion/ OH- Paper 3: Essay 13
  14. 14. 22(a) [Able to state the aim of experiment accurately] 3 To compare the reactivity of lithium, sodium and potassium based on the reaction with water and describing the effect of the solution towards the red litmus paper. 22(b) Hypothesis 3 Metals of lithium, sodium and potassium show different rate of reactivity with water and the solution formed turns red litmus paper to blue. 22(c) Variables a) Manipulated variable :type of metals 3 b) Responding variable : reactivity of reaction c) Constant variable : water and temperature 22(d) [Able to list the correct and complete substances and apparatus.] 3 Substances and Apparatus Lithium, sodium and potassium metals with water, basin, knife, forceps, blue litmus paper and white tile. 22(e) [Able to give all the procedures correctly, steps 1 - 7] 3 1. Lithium metal is cut into a small piece 2. The paraffin oil on the surface of the metal is wiped with the filter. 3. A basin is filled with water. 4. Lithium metal is put on the surface of the water with a pair of forceps. 5. Reactivity of the reaction is observed and recorded. 6. The experiment is repeated with sodium and potassium metals. 22(f) [Able to show the accurate tabulation of data with correct title.] 2 Metals Observations Lithium Sodium Potassium 14
  15. 15. SET 2 1 (a) Electrolysis is a process whereby an electrolyte is decomposed to its constituent elements when electric current passes through it. 1 2+ - (b) Pb / lead(II) ions , Br / bromide ions 1 (c ) electrical to chemical 1 (d) In solid : no free moving ions 1 In molten : Ions free to move 1 (e) (i) brown gas are released 1 - (ii) 2Br → Br2 + 2e 1 (f) (i) Lead 1 (ii) reduction 1 9 2 (a) (i) Copper ion and hydrogen ion 1 2+ + r: Cu and H (ii) [Able to mark the positive and the negative electrode correctly] 1 (b) (i) The blue colour turns colourless/becomes paler 1 2+ (ii) The concentration/ number of Cu decreases 1 2+ (c) (i) Mg / magnesium ion 1 2+ (ii) Mg → Mg + 2e 1 (d) (i) Reduction 1 (ii) The voltmeter reading increases 1 The distances between Mg and Cu is greater than the distance between Mg and Zn in electrochemical series 1 9 3 (a) To allow the movement of ions. 1 (b) e G e e e Electrode P Electrode Q Potassium iodide Chlorine water solution Dilute sulphuric acid 1 (c) (i) Colourless change to brown 1 (ii) Put a few drops of starch solution. 1 15
  16. 16. A blue precipitate is formed. 1 (d) Iodide ion // potassium iodide 1 Loss electron//increase in oxidation number 1 Cl2 + 2e  2Cl - (e) 1 (f) Bromine water // acidified KMnO4 solution // acidified K2Cr2O7 solution 1 (g) 0 to -1 1 10 4 (a) (i) Iron(III) nitrate // Iron (III) ions 1 (ii) Oxidising agent 1 (iii) Add / pour NaOH /NH3 solution 1 Brown precipitate is formed 1 2+ 2+ (b ) (i) Zn + Cu → Zn + Cu 1 (ii) 0 to +2 1 (iii) Oxidation 1 (c) (i) Magnesium // Aluminium 1 Metal M higher than zinc in the reactivity series// Metal M reactive than zinc 1 (ii) M , Zn , N 1 10 5 (a) (i) Mg /Zn / [ any suitable metal ] 1 MgSO4 / ZnSO4 / [ any suitable solution ] 1 (ii) Positive terminal : Cu 1 Negative terminal : X/Mg/Zn 1 2+ (iii) Positive terminal : Cu + 2e → Cu 1 2+ Negative terminal : X → X + 2e // 2+ Mg → Mg + 2e 1…. 6 (b) (i) Z Y X W 1 (ii) X 1 X below Z in the electrochemical series// X less electropositive than Z 1 0.6 V 1…. 4 (c ) Observation on the electrolyte Experiment I 2+ The concentration of Cu ions decrease 1 Copper ions discharge to form copper atom at cathode 1 2+ Cu + 2e → Cu 1 Experiment II 2+ The concentration of Cu ions unchanged 1 2+ The rate of Cu ions discharge at cathode is the same the rate of copper ions formed at anode. 1 16
  17. 17. Observation at anode Experiment I - OH ions discharge to form oxygen gas 1 - 2- OH ion lower than SO 4 ion in electrochemical series. 1 - 4OH → 2H2O + O2 + 4e 1 Experiment II 2+ Copper ionise to form Cu ions 1 2+ Cu→ Cu + 2e 1…10 20 6 (a) Reaction II 1 Oxidation number of magnesium changes from 0 to +2, Oxidation number of zinc changes from +2 to 0 1 No change in oxidation number for each elements in 1 reaction I 1…4 (b) Test tube P: The solution changes colour from pale green to yellow 1 2+ 3+ - 2Fe + Cl2  2Fe +2Cl Correct formulae of reactants and products 1 Balance equation 1 Test tube Q: The solution changes colour from colourless to yellow/brown. 1 - - 2I + Cl2  I2 + 2Cl Correct formulae of reactants and products Balance equation 1 1…6 (c ) Experiment I Reaction between carbon and oxide of metal P occurs 1 Carbon is more reactive than metal P 1 Experiment II Reaction between carbon and oxide of metal Q does not occur 1 Metal Q is more reactive than carbon 1 Experiment III Reaction between carbon and oxide of metal R occurs. 1 Carbon is more reactive than metal R 1 Reaction between carbon and oxide of metal P produces flame whereas reaction between carbon and oxide of metal R produces glow. 1 Metal P is less reactive than metal R. 1 Reactivity of metals in descending order is Q, carbon, R, P 1 Q is Aluminium // Magnesium 1..10 Total marks [20] 17
  18. 18. + 7 (a) 2H + 2e  H2 Correct formulae of reactants and product 1 2 Balanced 1 (b) Properties Cell A Cell B 1. Type of cell Voltaic cell Electrolytic cell 2. Energy change Chemical  electrical Electrical  chemical 1 3. Electrodes Positive terminal: Anode: Copper 1 Copper Cathode: Copper Negative terminal: 1 Magnesium 2+ 2- + - 2+ 2- + - 4. Ions in electrolyte Cu , SO4 , H and OH Cu , SO4 , H and OH ions ions 5. Half equation Positive terminal: Anode: 1 2+ 2+ Cu + 2e  Cu Cu  Cu + 2e Negative terminal Cathode: 1 2+ 2+ Mg  Mg + 2e Cu + 2e  Cu 6. Observation Positive terminal: Anode: 1 Copper plate becomes Copper thicker dissolves//become 1 thinner Magnesium becomes Cathode: thinner/dissolve Copper becomes thicker 1 8 (c) Chemicals silver plate , Silver nitrate solution 1 Procedure: 1. Iron ring is connected to the negative plate on the battery while the silver plate is connected to the positive terminal of the battery//Iron ring is made as 1 cathode while silver plate is made as anode 1 2. Both plates are immersed into the silver nitrate solution. 1 3. The circuit is completed Diagram Functional apparatus set-up Label correctly: 1 Chemical Equation 1 + Cathode: Ag + e  Ag + Anode : Ag  Ag + e 1 1 Observation: Cathode :Grey /silvery solid is deposited Anode/silver become thinner//dissolve 1 1 10 Total 20 18
  19. 19. 8 (a) (i) Sampel answer. 2+ Mg → Mg + 2e // any suitable equation 1 Metal / magnesium atom lose electron // metal / magnesium is oxidized 1…2 (ii) Experiment I 2+ Iron nail is oxidized to form Fe ions 1 Metal P speeds up the process of rusting 1 Because iron is more electropositive than P 1 2+ Dark blue precipitate indicates the presence of Fe ions 1 Experiment II Metal Q is oxidized to form Q ions 1 Because metal Q is more electropositive than iron 1 - Water and oxygen accept electron to become OH ions // - - 4OH → 2H2O + OH + 4e 1 - Pink colour of solution indicate the presence of OH ions 1 Arrangement : metal Q , iron , metal P. 1 Any 8….8 (b) Sample answer Zinc as a reducing agent 1 Add zinc to iron(III) chloride solution 1 Heat the solution 1 Filter the solution / mixture 1 2+ Add sodium hydroxide solution to the solution produced/ Fe 1 Green precipitate is formed 1 Chlorine as an oxidising agent 1 Add chlorine water to iron(II) nitrate solution 1 Stir/ shake the solution 1 Add sodium hydroxide solution 1 Brown precipitate 1 Max …10 20 Questio Rubric Score n [Able to state the relationship correctly between the manipulated variable 9 (a) and the responding variable ] Example: 3 When a more/less electropositive metal in contact with iron, the metal inhibits/speeds up rusting. [Able to state three variables correctly] 9 (b) Example: Manipulated variable: metals in contact with 3 iron//magnesium,copper,zinc Responding variable: The intensity of blue colouration // rusting of ion Constant variable: Iron nails//temperature of solution. 19
  20. 20. [Able to state the inference based on the observation correctly] 9(c) Example: Test tube W X Y Z Inferences The iron The iron The iron The iron 3 nail does nail does nail rust nail rust a not rust not rust quickly little 9(d) [Able to write the balanced half equations correctly ] Example: Oxidation: Fe Fe 2+ + 2e 3 - Reduction: O2 + 2H2O + 4e 4OH [Able to state the operational definition for rusting correctly ] 9(e) Example: 3 Rusting of iron is the formation of blue colouration when iron is in contact with less electropositive metals or without contact with other metals. 9(f) [Able to classify all the three metals correctly] Metals that can provide Metals that cannot provide 3 sacrificial protection sacrificial protection Magnesium Copper Zinc Question Rubric Score Able to state the relationship between the manipulated variable 10(a) and the responding variable correctly and with direction Sample answer: The higher the metal in reactivity series, the brighter the flame / glow produced // 3 The higher the metal in reactivity series, the reactivity of the metal increases. Able to state three variables and the way to control them correctly: 10(b) Sample answer: (i) Manipulated variable Type of metals (ii) Responding variable Reactivity of metals / Brightness of flame or glow 3 (iii) Controlled variable Mass of metal powder / quantity of potassium manganate(VII) 10(c) Able to give the operational definition accurately Sample answer: Metal that burns brightly when reacts with oxygen is the most reactive metal // 3 Metal that glows faintly when reacts with oxygen is the least reactive metal 20
  21. 21. 10(d) Able to state one observation accurately Sample answer: 3 Brown solid when hot, yellow solid when cold 10(e) Able to state an accurate inference for this experiment: Sample answer: Magnesium is the most reactive metal 3 10(f) Able to arrange the metals in ascending order of reactivity series of metals towards oxygen accurately Sample answer: Copper, lead, zinc, magnesium 3 Question Rubric Score number 10(g) Able to predict the position of iron in the reactivity series of metals accurately Answer: 3 Between zinc and lead 10(h) Able to explain the relationship between the time to light up and the reactivity of metal accurately Sample answer: 3 Magnesium is more reactive than zinc // Zinc is less reactive than magnesium 10(i) Able to make the classification of more reactive metals and less reactive metals when reacts with oxygen accurately Sample answer: 3 More reactive metals : Magnesium, zinc Less reactive metals : Lead, copper 10(j)(i) Able to record the masses accurately in two decimal places with unit Answer: 14.63g 3 17.03g 18.63g 10(j) (ii) Able to construct a table that contains: 1. Crucible + lid, crucible + lid + magnesium, crucible + lid + magnesium oxide and mass with correct unit. 3 2. Transfer all the readings from (j)(i) correctly. Answer: Description Mass (g) Crucible + lid 14.63 Crucible + lid + magnesium 17.03 Crucible + lid + magnesium oxide 18.63 21
  22. 22. 11. (a) - Statement of the problem Score Rubric [ Able to give the statement of problem correctly ] 3 Example : How is the effect on rusting of iron when iron is in contact with another metals? (b) - variables Score Rubric [ Able to state All variables correctly ] Suggested answer : 3 Manipulated variable : Different types of metals// Different metals Responding variable : Rate of rusting // Rusting of iron Constant variable : Iron nails/temperature (c) - hypothesis Score Rubric 3 [Able to give the hypothesis accurately] Suggested answer : When a more electropositive metal is in contact with iron, the metal inhibits rusting // When a less electropositive metal is in contact with iron, the metal speed up rusting // Iron rusts faster when in contact with metal less electropositive (d) - Apparatus and materials Score Rubric [ Able to give the list of the apparatus and substances correctly and completely] Suggested answer : Apparatus : Five test tubes, test tube rack 3 Materials : sand paper, five iron nails, magnesium strip, zinc strip, tin strip, copper strip, hot agar-agar/jelly solution mixed with potassium hexacyanoferrate(III) solution and phenolphthalein indicator (e) - Procedure of the experiment Score Rubric [ Able to state all procedures correctly ] Suggested answer : 1. Clean all the metal strips with sand paper 2. Coil the metal strip around the iron nails and then put in the each test tube 3. Pour the same volume of hot agar-agar/jelly solution has been mixed with 3 potassium hexacyanoferrate(III) and phenolphthalein indicator 4. Leave the test tubes aside for one day 5. Compare the intensity of the blue and pink colour in each test tube and recorded 3 (f) - Tabulation of data Score Rubric [ Able to exhibit the tabulation of data correctly ] Tabulation of data has 6 columns and 3 rows Example : 2 Test tube A B C D E Intensity of blue colour Intensity of pink colour 22
  23. 23. Question Rubric Skor No. 12(a) [Able to write the statement of the problem accurately.] Sample answer: 3 Does the difference in the position of two metals in the Electrochemical Series causes the difference in the voltage? 12(b) [Able to state the three variables correctly.] Sample answer: 3 Manipulated variable : Type of metal Responding variable : The voltage of the cell. Constant variable : The type and concentration of electrolyte, copper electrod/ positive terminal. 12(c) Able to give the hypothesis accurately. Sample answer: If the distance between two metals is further away in the 3 electrochemical series, the voltage produced will be higher/bigger. [Able to state complete materials and apparatus.] 12(d) Sample answer: 3 Materials: Zinc, Magnesium, Copper, Iron and Copper(II) sulphate solution/ any other electrolytes. Apparatus: Voltmeter, connecting wires, beaker. Able to state a complete procedure. Sample answer: 1. Fill a beaker with copper(II) sulphate solution until it is two-thirds full. 3 2. Dip the zinc strip and copper strip into copper(II) sulphate solution. 3. Connect the two metals to a voltmeter(by using connecting wires) 12(e) 4. Record the potential difference produced between the metals. Repeat steps 1 to 4 using iron strip, magnesium strip to replace zinc strip. Able to construct a labelled tabulation of data with suitable unit. Sample answer: 12(f) Experiment Pairs of metal. Reading of voltmeter /V 3 I II III 23
  24. 24. Set 3 1 a. Certain volume of acid completely neutralises a given volume of alkali// Certain volume of acid at which change of colour of indicator when a drop of acid is added to alkali 1 b. Pink solution to colourless solution 1 c. Experiment 1 = 22.40 2=22.20, 3=22.00 1 d. H2SO4 + 2NaOH  Na2SO4 + 2H2O 1 3 e. (i) Average volume = (22.40 + 22.30 + 22.00)÷3 = 22.30 cm 1 (ii) Mol of sulphuric acid, H2SO4 = (22.30 X 1 )/1000 = 0.0223 mol 1 f. (i) Functional set-up: Conical flask, Burette 1 (ii) Label: (H2SO4, KOH and Phenolphthalein) 1 g. (i) Add drops of acid a little at a time - towards the end point 1 (ii) Conical flask with content - shaken during experiment 1 10 2 a. (i) Red 1 (ii) Yellow 1 (iii) Orange 1 3 b. 15.00 cm 1 c (i) H2SO4 + 2KOH  K2SO4 + 2H2O 1 + - (ii) H + OH  H2O 1 -3 d. 0.1 x 20 = 2  Ma = 0.067 mol dm Ma x 15.00 1 1 e. (i) Yellow 1 (ii) Red 1 3 f. 30 cm 1 10 3 a. Ba(OH)2 + H2SO4  2H2O + BaSO4 reactants & product correct – 1, balanced -1 2 b. Neutralisation / double decomposition 1 c. - Methyl orange indicator changes from yellow to orange 1 - White precipitate formed 1 d. (i) The titration has achieved the end-point / there is no more free moving ions present in the beaker 1 (ii) Ba(OH)2 + H2SO4  2H2O + BaSO4 Mb= 0.2M Ma=1.0M 3 3 Vb= 50 cm Va= x cm MaVa = 1  1.0M(x) = 1 1 MbVb 1 0.2M(50) 1 x = 10 cm3 1 Ammeter e. (i) reading/ A 0 5 Volume of sulphuric acid added/ Correct shape - 1 cm3 Correct volume indicated and ammeter reading not reaching zero 1 2 (ii) - Ammeter does not show a zero reading 1 + 2- because at end-point there are free moving Na and SO4 ions 1 OR 24
  25. 25. - The volume of sulphuric acid required is halved 1 because sodium hydroxide is monobasic alkali whereas - barium hydroxide is dibasic alkali / the concentration of OH ions in NaOH is half of Ba(OH)2 1 2 4 (a) Neutralisation 1 (b) To ensure all nitric acid is completely reacted 1 (c) ZnO + 2HNO3  Zn(NO3)2 + H2O 1 (d) zinc oxide [functional apparatus] 1 [label] 1 (e) Number mole of nitric acid = 2(50) = 0.1 mol 1 1000 Mass of salt = 0.1 x 189 = 9.45 g 1 1 (f) Zinc carbonate, 1 Zinc 1 10 5 (a) Silver chloride 1 (b) White 1 3 (c) 3.0 cm 1 (d) (i) 0.5(6.0) = 0.003 mol 1 1000 (ii) Number of mole of sodium chloride = 3.0(1) = 0.003 mol 1 1000 0.003 mol of sodium chloride reacts with 0.003 mol of silver nitrate 1 Therefore 1 mol of sodium chloride reacts with 1 mol of silver nitrate. 1 + - (e) Ag + Cl  AgCl 1 (f) 1.5 cm 1 Height of precipitate (cm) 1.5__________ Ι Ι Ι Ι 3 1.5 Volume of sodium chloride (cm ) [labeled axis + correct shape of graph -1] 1 [height of precipitate and volume of NaCl required – correct & indicated – 1] 1 10 6 (a) (i) Copper(II) sulphate 1 (ii) Heat the solution until saturated then cool 1 Filter and then rinse the salt with distilled water 1 Dry the salt using filter paper 1 (b) (i) Blue precipitate 1 Cannot dissolve in excess NaOH 1 (ii) Cu(OH)2 1 25
  26. 26. (c) Ammonia 1 (d) (i) White 1 2+ 2- (ii) Pb + SO4  PbSO4 1 (iii) Filter the mixture 1 10 7(a) 1. x-axis and y-axis labelled with units 1 2. plots transferred correctly 1 3. smooth curve 1……3 (b) 1. Rate of reaction of Experiment 4 is higher than Experiment 2 1 2. Temperature in Experiment 4 is higher. 1 3. The kinetic energy of thiosulphate ions is higher 1 + 2- 4. The frequency of collisions between the H ions and the S2O3 ions 1 increases 5. The frequency of effective collisions increases. 1…..5 (c) Na2S2O3 + H2SO4 → Na2SO4 + S + H2O + SO2 1 (d) Sulphur 1 (e) Concentration of sodium thiosulphate solution 1 13 8(a) 1. Functional apparatus set-up 1 2. Complete labels 1…..2 (b) 25 = 3 -1 1 0.78 cm s 32 + 2+ (c) Zn + 2H → Zn + H2 1 (d) 1. Rate of reaction in Experiment II is higher than in Experiment I 1 2. In Experiment II copper(II) sulphate is a catalyst. 1 3. Catalyst will lower the activation energy. 1 + 4. The frequency of effective collisions between zinc atoms and the H ions increases. 1……4 (e) 1. Rate of reaction of Experiment III is higher than in Experiment I 1 2. Sulphuric acid in Experiment III is a diprotic acid (hydrchloric acid is a monoprotic acid) 1 + 3. Concentration of H ions per unit volume in Experiment III is 2 times higher than in Experiment I. 1 + 4. The frequency of effective collisions between zinc atoms and the H ions increases 1…..4 (f) (i) Hydrogen 1 (ii) Bring a lighted wooden splinter to the mouth of the test tube 1 Pop sound 1….3 13 9 (a) Heat released when 1 mole of silver is displaced from its salt solution by copper metal. 1 (b) Silvery solid formed// Colourless solution of silver nitrate becomes blue// Amount of copper powder decreases 1 + (c) (i) No. of moles of Ag reacted = No. of moles of AgNO3 used = mv/1000 = 0.5(50)/1000 = 0.025 mol 1 (ii) Heat released = No. of moles of Ag x ΔH = 0.025 x 105 kJ = 2.625 kJ = 2625 J 1 (iii) Heat change = mc   2625 J = 50 (4.2)    = 12.5 C o 1 (d) Assumptions - no heat loss to surrounding 1 - specific heat capacity of solution = specific heat capacity of water 1 - density of solution = density of water ( any two) 1 26
  27. 27. (e) Energy 2 Ag+ + Cu ΔH = -105kJ -1 mol 2 Ag + Cu2+ 1. The position and name/formulae for the reactants and products are correct 1 2 Label for the energy axis and arrow for the two levels are shown. 1 (f) Lower/smaller 1 The total surface area exposed to the air is larger 1 Heat is lost to the environment 1 (g) To ensure all the silver nitrate solution reacted completely 1 (h) Bigger / Higher because magnesium is more electropositive than copper. 1 15 10 (a) Strong acid – nitric acid/ hydrochloric acid / sulphuric acid 1 Strong alkali – sodium hydroxide / potassium hydroxide 1 (b) Heat released when 1 mole of water is formed from the reaction between an acid and an alkali 1 (c) - It is an exothermic reaction // heat energy is released to the surrounding 1 - The total energy of reactants is higher than the products 1 - 57 kJ of heat energy is released when 1 mole of water is formed ( any 2) 1 (d) (i) No. of moles alkali used = mv/1000 = 1(50)/1000 = 0.05 mol 1 (ii) Heat change = mc  = (50+50)x4.2x 6.5 = 2730 J 1 Heat of neutralization = - 2730/0.05 = - 54600 J/ mol = - 54.6kJ / mol 1 (e) (i) The heat of neutralization for Experiment I is higher than Experiment II 1 (ii) Ethanoic acid – weak acid, dissociates partially in water 1 Part of heat released in Experiment t II during neutralization is absorbed to dissociate further the molecules of ethanoic acid 1 (f) The number of moles of water produced doubled, hence amount of heat energy released is doubled but the total volume of solution used also doubled, therefore the temperature increase remain the same 1 13 ESSAY SECTION B Question Explanation Marks Number 10(a) Acid that ionises completely in water 1 to produce high concentration of hydrogen ions 1..2 (b) Number of sulphuric acid = 0.5(50)/1000 = 0.025 mol 1 Number of hydrochloric acid = 1.0(50)/1000 = 0.05 mol 1 Both are strong acids ionizes completely in water 1 + To produce the same concentration of H 1…..4 (c) (i) Test Aqueous HCl solution Solution of HCl in methylbenzene Universal Green to red No changes 1+1 indicator Add zinc Bubbles of colourless No changes 1+1 powder gas formed 1+1….6 Add Blue solution formed No changes copper(II) oxide powder 27
  28. 28. Question Explanation Marks Number (ii) Aqueous hydrogen chloride solution 1 In presence of water 1 hydrogen chloride ionizes 1 + to produce hydrogen ions/H which gives rise to acidic properties 1…..4 (iii) Aqueous HCl HCl in methylbenzene 1 Observation Ammeter needle Ammeter needle does 1 deflected not deflect 1 Explanation Consists of free Consists of molecules // 1…..4 moving ions no free moving ions 20 Question Explanation Marks Number 3 -3 11 (a) - 25 cm of 1.0 mol dm sodium hydroxide solution is measured 1 and poured into a conical flask. - Two drops of phenolphthalein are added to the solution. 1 - A burette is filled with nitric acid and the nitric acid is added slowly dropwise into the potassium hydroxide solution and shaken until the pink solution turned coloruless. 1 - The volume of acid added is calculated. 1 - The experiment is repeatedby using the same volume of sodium hydroxide and nitric acid but without the 1 phenolphthalein. - The salt solution formed is heated until saturated. 1 - The saturated solution is allowed to cool. 1 - The salt crystals are filtered and rinsed with distilled water. 1 - The salt crystals are dried by pressing them between sheets of 1 filter papers. - NaOH + HNO3  NaNO3 + H2O 1…..10 3 11(b) - 50 cm of 1.0M copper(II) sulphate solution is measured and 3 added into 50 cm of 1.0M sodium carbonate solution in a 1 beaker. - The mixture is stirred and filtered to obtain green precipitate, 1 copper(II) carbonate. - The copper(II) carbonate is rinsed with water. 1 - The copper(II) carbonate is added a little at a time into heated 3 50 cm of 1.0M dilute hydrochloric acid until some copper(II) 1 carbonate solid no longer dissolved anymore. - The mixture is filtered to remove excess copper(II) carbonate. 1 - The filtrate is copper(II) chloride solution. 1…..6 11(c)(i) Lead(II) oxide 1 Brown when hot, yellow when cold 1 Carbon dioxide 1 Bubble the gas through lime water and lime water turns chalky 1….4 20 28

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