Jawapan Module Perfect Score Chemistry SPM 2013

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Jawapan Module Perfect Score Chemistry SPM 2013

  1. 1. @Hak cipta BPSBPSK/SBP/2013 BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER http://cikguadura.wordpress.com/ JAWAPAN MODUL PERFECT SCORE & X A-PLUS 2013 CHEMISTRY      1 Set Set Set Set Set Perfect Score & X A –Plus Module/mark scheme 2013 1 2 3 4 5
  2. 2. @Hak cipta BPSBPSK/SBP/2013 MODULE PERFECT SCORE & X A-PLUS 2013 http://cikguadura.wordpress.com/ SET 1 :THE STRUCTURE OF ATOM, PERIODIC TABLE OF ELEMENTS AND CHEMICAL BONDS Question No 1 (a) (i) Mark schemes Melting 1 (ii) Molecule (b) Mark 1 The heat energy absorbed by the particles is used to overcome the forces of attraction 1 between the naphthalene molecules / particles. (c) The particles move faster (d) (i) 1 X : electron 1 Y : nucleus (ii) Electron (i) (e) 1 1 W and X (ii) W and X atom have different number of neutrons but same number of protons 1+1 Atom// Element W and X has different nucleon number but same proton number Σ 10 Question No 2 (a) Mark schemes No of electrons = 18, No of neutrons = 22 (b) 1+1 (i) The total number of protons and neutrons in the nucleus of an atom (ii) 40 (i) (c) Mark 2.1 1 1 (ii) e X XX 3p 4n Xe X e (d) (i) W and Y 1 (ii) Atom W and Y have the same number of valence electrons 1 (iii) To estimate the age of fossils /artefacts. 1 Σ 10 2 Perfect Score & X A –Plus Module/mark scheme 2013
  3. 3. @Hak cipta BPSBPSK/SBP/2013 Question No. 3 Mark Scheme Marks (i) Total number of protons and neutrons in the nucleus of an atom 1 (ii) 35 – 18 = 17 1 (iii) (a) shows nucleus and three shells occupied with electron 1 +1 Label 12 proton, 12 neutron (iv) (b) Number of electrons = 2 (i) 1 ...5 1 Liquid (ii) 1+1 Q R ...3 (c) Temperature/oC 90 67 1+1 Time/s 1st mark - Label X and Y axis with correct unit 2 nd mark - Correct shape of curve 10 3 Perfect Score & X A –Plus Module/mark scheme 2013
  4. 4. @Hak cipta BPSBPSK/SBP/2013 a) b) (i) F 1 (ii) 4 Atom F has achieve stable/octet electron arrangement // has 8 valence electron 1 2D + 2H2O  2DOH + H2  Correct reactant & correct product  Balance equation The nuclei attraction towards the valence electrons is weaker in atom G. More easier for atom G to lose / release an electron to form a positively charged ion. 1 (i) (ii) c) (i) 1 1+1 1 Covalent bond (ii) 1 1 x E Y x x X Y x x E Y x (iii) 1 Show coloured ion//formed complex ion//has various oxidation number//act as catalyst (d) Cannot conduct electricity at any state/ low melting and boiling point/.... 1 11 5 (a) 1 (i) Na/sodium, Mg/magnesium .... 1 (ii) Atomic size decreases across the period // Period 3. 1 (iii) (b) Increasing of proton number. 1. Number of protons in atom increases when across the period. 2. Force of attraction between nucleus and electrons in the shell is stronger. 1+1 Chlorine more reactive than bromine Size of chlorine atom is smaller than bromine atom Chlorine atom is easier to receive one electron Al3+ Ionic compound 1+1 ..4 (c) (d) (e) (i) (ii) 1 1 1+1 11 4 Perfect Score & X A –Plus Module/mark scheme 2013
  5. 5. @Hak cipta BPSBPSK/SBP/2013 6 (a) (b) P : liquid (i) Q : solid R : gas 1. P can be change to Q through freezing process. 2. When the liquid cooled, the particles in liquid lose energy and move slower. 3. As temperature drops, the liquid particles attract tone another and change into solid (ii) 1. P can change to R through boiling. 2. When liquid is heated, the particles of the liquid gain kinetic energy and 1 +1+1 1 1 1 1 1 move faster as the temperature increase 3. The particles have enough energy to overcome the forces between them and gas is formed (iii) (c) (i) (ii) (iii) (iv) 5 1. R can be change to P through condensation process. 2. When the gas cooled, the particles in gas lose energy and move slower. 3. Particles attract one another and change into liquid 1. Uniform scale for X-axis and Y-axis and labelled/size of graph plotted ¾ of graph paper. 2. Tranfer of point 3. Smooth curve 1. Dotted line on the graph from the horizontal line to Y-axis at 80oC. 2. Arrow mark freezing point at 80oC 1. Heat released to sorrounding 2. Is balanced when particles comes together to form a solid Supercooling Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1 1 1 1 1 1 1 1 1 1 20
  6. 6. @Hak cipta BPSBPSK/SBP/2013 Question No. (a) (i) 7 (ii) (b) (i) Mark Scheme Atom R is located in Group 17, Period 3 Electron arrangement of atom R is 2.8.7. Group 17 because it has seven valence electron. Period 3 because it has three shells filled with electron Atoms P and R form covalent bond. To achieve the stable electron arrangement, atom P needs 4 electrons while atom R needs one electron. Thus, atom P shares 4 pairs of electrons with 4 atoms of R, forming a molecule with the formula PR4 // diagram Mark 1 +1 1 1 1 1 1 1 1 1 R R P R R (ii) Atom Q and atom R form ionic bond. Electron arrangement for atom Q is 2.8.1 and electron arrangement for atom R is 2.8.7// Atom Q has 1 valence electron while atow R has 7 valence electron To achieve a stable (octet ) electron arrangement, atom Q donates 1 electron to form a positive ion// equation Q Q+ + e 1 1 1 Atom R receives an electron to form ion R-//equation and achieve a stable octet electron arrangement. R+e R- 1 1 Ion Q+ and ion R- are attracted together by the strong electrostatic forces to form a compound with the formula QR// diagram 1 -- + Q 6 Perfect Score & X A –Plus Module/mark scheme 2013 R
  7. 7. @Hak cipta BPSBPSK/SBP/2013 Question No 8 (a) (b) Mark scheme Mark 12 represents the nucleon number. 6 represents the proton number. 1 1 Able to draw the structure of an atom elements X. The diagram should be able to show the following informations: 1. correct number and position of proton in the nucleus/ at the centre of the atom. 2. correct number and position of neutron in the nucleus/ at the centre of the atom. 3. correct number and position of electron circulating the nucleus 4. correct number of valence electrons Sample answer: √4 ee- ee- - e 11p√1 12n √2 e- e- √3 e- ee- e- or ee- 11p + 12n e- ee- - e e- eee- 7 Perfect Score & X A –Plus Module/mark scheme 2013 e- 1 1 1 1
  8. 8. @Hak cipta BPSBPSK/SBP/2013 (c) (i) Atoms W and Y form covalent bond. To achieve the stable electron arrangement, atom W contributes 4 electrons while atom Y contributes one electron for sharing. Thus, atom W shares 4 pairs of electrons with 4 atoms of Y, forming a molecule with the formula WY4 // diagram 1 1 1 1 1 Y Y W Y Y (ii) Atom X and atom Y form ionic bond. Electron arrangement for atom X is 2.8.1 and electron arrangement for atom Y is 2.8.7 To achieve a stable (octet )electron arrangement, atom X donates 1 electron to form a positive ion // equation X X+ + e Atom Y receives an electron to form ion Y-//equation and achieve a stable octet electron arrangement. Y+e Y+ Ion X and ion Y are attracted together by the strong electrostatic forces to form a compound with the formula XY// diagram -- + X (d) 1 1 1 1 1 1 Y The melting point of the ionic compound/ (b)(ii) is higher than that of the covalent compound/ (b)(i) . This is because in ionic compounds oppositely ions are held by strong electrostatic forces. High energy is needed to overcome these forces. In covalent compounds, molecules are held by weak intermolecular forces. Only a little energy is required to overcome the attractive forces. OR The ionic compound/(b)(ii) conducts electricity in the molten or aqueous state whereas the covalent compound/(b)(i) does not conduct electricity. This is because in the molten or aqueous state, ionic compounds consist of freely moving ions carry electrical charges. Covalent compounds are made up of molecules only 1 1 1 1 1 or 1 1 1 1 1 20 8 Perfect Score & X A –Plus Module/mark scheme 2013
  9. 9. @Hak cipta BPSBPSK/SBP/2013 9 (a) (i) Q 1. Correct number of shells and valence electrons 2. Black dot or label Q at the center of the atom (ii) (b) 1. 2. 3. 4. (i) 1. Floats and moves fast on the water 2. ‘Hiss’ sound occurs 3. Gas liberates / bubble (ii) (c) (i) (ii) Group 14 There are 4 valence electrons Period 2 Atom consists of 2 shells occupied with electrons 1 1 1 1 1 1 1 1 [any two] 2Q + 2H2O  2QOH + H2 1. Correct reactant and product 2. Balanced equation Compound X Sharing electron between atom B and A Choose any one ionic compound and any one covalent compound. 1 1 1 1 Melting/boiling point Ionic compound 1. 2. High force of attraction between oppositely charged ions are strong. 3. more heat energy needs to overcome the forces. Electrical conductivity 4. 5. Ionic compound Conduct in molten state or aqueous solution. The free moving ions are able to carry electrical charges. Covalent compound low force of attraction between molecules are weak. less heat energy needs to overcome the forces. Covalent compound Not conduct electricity. Neutral molecules are not able to carry electrical charges. 1 1 1 1 1 1 1 1 Solubility 6 7 Ionic compound Soluble in water. Water molecule is polar solvent. Covalent compound soluble in benzene/ toluene / any organic solvents. The attraction forces between molecules in solute and solvent are the same. 20 9 Perfect Score & X A –Plus Module/mark scheme 2013
  10. 10. @Hak cipta BPSBPSK/SBP/2013 10 (i) Compound formed between X and Y Ionic bond is formed because X atom donates electrons and Y atom receives electrons to achieve stable octet electron arrangement/involve transfer electron High because a lot of heat energy needed to overcome the strong electrostatic forces between ions Types of chemical bonds Boiling point and melting point Molecule formed between Z and Y Covalent bond is formed because Z and Y atoms share the electrons to achieve stable electron arrangement // Inovelve sharing of electron Low because less heat energy is needed to overcome the weak forces of attraction between molecules 2 2 1.Correct electron arrangement of 2 ions 2.Correct charges and nuclei are shown 2+ XX X XX X XX X X X (b) X X X X X X X X X X X XX X 2- XX X 2+ X X X X X 1 1 X X XX X X Y2- 3. X atom with an electron arrangement of 2.8.2 donates 2 valence electrons to 1 achieve the stable octet electron arrangement, 2.8. X2+ ion is formed // X X2+ + 2e- 1 4. Y atom with an electron arrangement of 2.6 accept 2 electrons to achieve the 1 stable octet electron arrangement, 2.8. Y2- ion is formed // Y + 2eY2- 5. The oppositely-charged ions, X2+ and Y2- are attracted to each other by a strong electrostatic force. 6. An ionic compound XY is formed 10 Perfect Score & X A –Plus Module/mark scheme 2013 1 1
  11. 11. @Hak cipta BPSBPSK/SBP/2013 (c) 1. A crucible is filled with solid P until it is half full. 2. Two carbon electrodes are dipped in the solid P and connected to the batteries 3. 4. 5. 6. 7. using connecting wire. Switch is turned on and observation is recorded. The solid P is then heated until it melts completely. The switch is turned on again and observation is recorded. Steps 1 to 5 are repeated using solid Q to replace solid P. Observations: P does not light up the bulb in both solid and molten states. Q lights up the bulb in molten state only. P: naphthalene // any suitable answer Q: lead(II) bromide // any suitable answer 1 1 1 1 1 1 1 1 1 1 1 1 1 20 (a) (i) Z : 2.8.7 X : 2.4 (ii) 11 Z atom has 7 valence electrons needs one electron X atom has 4 valence electrons ,hence it needs 4 more electron each atom achieves stable octet electron arrangement share electrons between them four Z atoms , each contributes 1 electron // [ diagram one X atom contributes 4 electrons //[diagram] - four single covalent bonds are formed - the molecular formula is XZ4 - diagram [ no. of electrons in all the occupied shells in the X and Z atoms - correct] [ sharing of 4 pairs of single covalent bonds between 1 X atom and 4 Z atoms ] (iii) Colourless liquid b) [Procedures of the experiment] eg. 1. Add a quarter of spatula of YZ solid and add into a test tube. 2. Pour 2-5 cm3 of distilled water into the test tube containing theYZ2 3. Stopper the test tube and shake well. 4. Repeat Steps 1 to 3 using [ named organic solvent eg ether ] 5. Observe the changes and record them in a table . [Results] Eg Solvent Distilled water [named organic solvent] e.g ether Observation Colourless solution obtained Solid crystals insoluble in liquid [Conclusion] eg ZY is insoluble in organic solvent/[named organic solvent] but soluble in water. 11 Perfect Score & X A –Plus Module/mark scheme 2013 1 1 ..2 1 1 1 1 1 1 1 1 1 1 ..10 1 1 1 1 1 1 1 1 ..7
  12. 12. @Hak cipta BPSBPSK/SBP/2013 No Explanation 12 (a)(i) Y more reactive Atomic size of Y bigger than X // The number of shell occupied with electron atom Y more than X. The single valence electron becomes further away from the nucleus. the valence electron becomes weakly pulled by the nucleus. The valence electron can be released more easily. Name : Sodium 4Na + O2  2Na2O Chemical formulae Balance equation Put group1 metal into bottle that contain paraffin oil Group 1 metal readily reacts with air/moisture in atmosphere/ water Name : Sodium/any group 1 element Material : group 1 elements, water, Apparatus : forceps , knife, filter paper, basin, litmus paper. [procedure] 3. Pour some water into the basin 4. Group 1 metal is take out from paraffin oil using forceps 5. A small piece of group 1 element is cut using a small knife 6. Oil on group 1 element is dried using a filter paper 7. The group 1 element is placed in the basin contain water. 8. Dip a red litmus paper into water (ii) (b) (c) [observation] 9. Color of red litmus paper turn to blue [chemical equation ] Sample answer 2 Na + 2 H2O  2NaOH + H2 Chemical formulae Balance equation No 13. (a) (b) (c) Explanation Glucose // naphthalene // any solid covalent compound covalent Intermolecular forces are weak Small amount of heat energy needed to overcomes the forces X = 2.1 X = 2.2 Y = 2.7 // Y = 2.6 // 1. Suitable electron aranggement 2. Ionic bond 3. to achieve octet electron arrangement + 4. One atom of X donates 1 electron to form ion X 5. One atom of Y receives an electron to form ion Y + 6. Ion X and ion Y are attracted together by the strong electrostatic forces material and apparatus; compound XY, Carbon electrode, cell, wire, crucible, bulb/ammeter/galvanometer 12 Perfect Score & X A –Plus Module/mark scheme 2013 Sub 1 1 1 1 1 1 1 1 1 1 1 Total 5 3 2 1 1 1 1 1 1 1 Max 5 1 1 1 Total ` 1 1 1 1 1 1 1 1 1 1 1 1 20 Total 4 7
  13. 13. @Hak cipta BPSBPSK/SBP/2013 Procedure A crucible is half fill with solid XY powder Dipped two carbon electrode Connect the electrodes with connecting wire to the battery and bulb Observed whether bulb glow Heated the solid XY in the crucible Observed whether bulb glow Observation Solid XY - bulb does not glow Molten XY - bulb glow Diagram 1 1 1 1 1 1 1 1 9 Functional diagram Labeled TOTAL 20 SET 1:CHEMICAL FORMULAE AND EQUATIONS Question No 1 Mark scheme (a) Molar mass is the mass of a substance that contains one mole of the substance. Example : Molar mass of one mole of magnesium is 24gmol -1 . Substance N2 12+2(16) = 44 H2S 2(1)+ 32 = 34 H2O 2(1)+16 1 Molar mass / gmol-1 14x2 = 28 CO2 (b) Mark 4 = 18 1 Mole of water = 0.9/ 18 = 0.05 Number of molecules (c) = 0.05 x 6.02 x 1023 = 0.3 x 1023 // 3 x 1022 Mole of carbon dioxide = 2.2 / 44 = 0.05 13 Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1
  14. 14. @Hak cipta BPSBPSK/SBP/2013 = 0.05 x 6.02 x 1023 Number of molecules 1 = 0.3 x 1023 // 3 x 1022 Number of molecule is simmilar Mass of CO2 Number of molecules = 0.1 mol x 6.02 x 1023 1 (iv) (b) Volume CO2 = 0.1 mol x 24dm3mol-1 = 2.4 dm3 (iii) (a) (i) (ii) 2 = 6.02 x 1022 x 3 = 1.806 x 1023 Heating, cooling and weighing processes are repeated a few constant mass is obtained. 1+1 (i) = 0.1 mol x 44 gmol-1 1 1 = 4.4 g Number of atoms times until a (ii) Compound Anhydrous CoCl2 (34.10-31.50)g = 2.60 g 2.60/130 = 0.02 0.12/0.02 = 6 1 Number of moles (36.26-34.10)g = 2.16 g 2.16/18 = 0.12 0.02/0.02 = 1 Mass/g H2O 6 Ratio of moles Simplest ratio of moles 1 mole of CoCl2 combines with 6 moles of H2O Therefore, the molecular formula of hydrated cobalt(II) chloride crystal is CoCl2.6H2O. Hence, the value of x in CoCl2.xH2O is 6. (iii) 1 1 1 Percentage of water 1 = 6(18) x 100% 59  2(35.5)  6(18) = 108 x 100% 238 = 45.4% 1 Total 10 3 14 concentrated sulphuric acid 1 zink and hydrochloric acid[ any suitable metal and acid ] 1 (iii) (b) (i) (ii) (a) Zn + 2HCl  ZnCl2 + H2 (i) Mole of oxygen = 46.35 - 45.15 16 = 1.2 = 0.075 16 Perfect Score & X A –Plus Module/mark scheme 2013 1
  15. 15. @Hak cipta BPSBPSK/SBP/2013 Mole of copper = 45.15 - 40.35 64 1 = 4.8 = 0.075 64 Empirical formula = CuO 1 (i) Collect the hydrogen gas in a test tube Put a burning wooden splinter at the mouth of the test tube ‘No pop sound ‘ produced. 1 1 1 (ii) To avoid the hot copper react with oxygen/air 1 (iii) Repeat heating, cooling and weighing processes until a constant mass obtained. 1 (ii) (iii) (c) 1 Total 4 (a) (b) (i) (ii) (i) Pb(NO3)2 AgCl Pb2+ + 2 Cl-  PbCl2 11 1 1 1+1 Correct formula for reactants and product Balance ionic equation (ii) Qualitative aspect : Lead(II) nitrate and sodium chloride are the reactants and lead (II) chloride and sodium nitrate are the products // Lead(II) nitrate solution reacts with sodium chloride solution to form lead(II) chloride precipitate and sodium nitrate solution. Quantitative aspect : One mole of lead(II) nitrate reacts with 2 mole sodium chloride to produce 1 mole of lead(II) chloride and 2 mole of sodium nitrate. (c) (i) 1 2 Pb(NO3)2  2 PbO + 4NO2 + O2 Compound Colour of the residue when hot Brown 1 Colour of the residue when cold PbO 1 Yellow 1 Gases Colour of the gas released NO2 Brown 1 O2 Colourless 1 Total 15 Perfect Score & X A –Plus Module/mark scheme 2013 10
  16. 16. @Hak cipta BPSBPSK/SBP/2013 No Explanation (a) (i) (ii) (b) 5 (i) 3+ Al , Pb Aluminium oxide Lead(IV) oxide (CH2O)n = 60 12n + 2n + 16n = 60 n= 2 Molecular formula = C2H4O2//CH3COOH (ii) CaCO3 + 2CH3COOH (i) CuCO3 1 1 1 1.Green solid turn Black 2. Lime water becomes cloudy (ii) (c) Mark 1+ 1 1+1 4+ (CH3COO)2Ca + H2O + CO2 2 1 1 CuO + CO2 1+1 (iii) 1. 1 mol of copper(II) carbonate decomposed into 1 mol of copper(II) oxide and 1 mol of carbon dioxide 2. copper(II) carbonate is in solid state, copper(II) oxide is in solid state and carbon dioxide is in gaseous state 1 (iv) 1. No. of mole for CuCO3 = 12.4 / 124 = 0.1 mol 2. 1 mol of CuCO3 produces 1 mol of CuO Therefor No. of mole for CuO = 0.1 mol 3. Mass of CuO = 0.1 mol X 80 g mol-1 = 8 g 1 1 Mass of oxygen is 0.8g Simplest mol ratio : Cu : O = 3.2/64 : 0.8/16 = 1 : 1 1 1 (v) 1 1 20 Mark (a) (i) 6 (ii) Empirical formula of a compound is a formula that shows the simplest whole number ratio of each atoms of each element in a compound. 1 (ii) Substance C10H8 1 H2SO4 16 Empirical formula C5H4 H2SO4 1 Perfect Score & X A –Plus Module/mark scheme 2013
  17. 17. @Hak cipta BPSBPSK/SBP/2013 (b) Element Percentage (%) Mass/ g Mole Simplest mole ratio Carbon 62.07 62.07 62.07/12 = 5.17 5.17/1.72 =3 Hydrogen 10.34 10.34 10.34/1 = 10.34 Oxygen 27.59 27.59 27.59/16 = 1.72 10.34/1.72 = 6 1 1.72/1.72 =1 1 Empirical formula = C3H6O 1 n [C3H6O ] = 116 [ 3(12) + 6(1) + 16 ] n = 116 58 n = 116 n= 2 1 1 Molecular formula = C6H12O2 (c) Procedure : 1. Clean magnesium ribbon with sand paper. 2.Weigh crucible and its lid. 3. Put magnesium ribbon into the crucible and weigh the crucible with its lid. 4. Heat strongly the crucible without its lid. 5. Cover the crucible when the magnesium starts to burn and lift/raise the lid a little at intervals. 6. Remove the lid when the magnesium burnt completely. 7.Heat strongly the crucible for a few minutes. 8.Cool and weigh the crucible with its lid and the content. 9. Repeat the processes of heating, cooling and weighing until a constant mass is obtained. 10.Record all the mass. 10 Tabulation of result : Description Crucible + lid Crucible + lid + magnesium Crucible + lid + magnesium oxide Element Mass / g Mole Simplest ratio of mole Empirical formula = 17 Magnesium b-a b-a/ 24 x Mass/ g a b c Oxygen c-b c-b / 16 y 1 1 1 1 MgxOy Perfect Score & X A –Plus Module/mark scheme 2013 Max 11 Total 20
  18. 18. @Hak cipta BPSBPSK/SBP/2013 No 7. (a) Sub 1. Empirical formula is the chemical formula that shows the simplest ratio of atoms of each element in the compound. Molecular formula is the formula that shows the actual number of atoms of each element in the compound. Example : empirical formula of ethene is CH2 and the molecular formula is C2H4 2. 3. Element Carbon Hydrogen 40.00 6.66 Ratio of moles 40 12  3.33 6.66 1 1  6.66 2 53.33 16  3.33 1 Empirical formula is CH2O n(CH2O) = 180 12n + 2n + 16n = 180 30n = 180 n=6 molecular formula = C6H12O6 (ii) (iii) 3 1 1 1 1 1 Magnesium is more reactive than hydrogen//Position of magnesium is above hydrogen in the reactivity series Lead(II) oxide / Stanum oxide / iron oxide / copper(II) oxide 1. 2. 3. 4. 5. 6. 7. 8. 9. 1 53.33 Number of moles (ii) 1 1 Oxygen Percentage (b)(i) (c)(i) T 5 1 Clean [5 – 15] cm magnesium ribbon with sandpaper and coil it. Weigh an empty crucible with its lid. Place the magnesium in the crucible and weigh again. Record the reading. Heat the crucible very strongly. Open and close the lid very quickly. When burning is complete stop the heating Let the crucible cool and then weigh it again The heating, cooling and weighing process is repeated until a constant mass is recorded. 10. Description Mass(g) 10 Crucible + lid Crucible + lid + Mg / Zn / Al Crucible + lid + MgO / ZnO / Al2O3 Total 18 Perfect Score & X A –Plus Module/mark scheme 2013 20
  19. 19. @Hak cipta BPSBPSK/SBP/2013 SET 2 :ELECTROCHEMISTRY http://cikguadura.wordpress.com/ Question No 1(a) (b) (c) (d)(i) (ii) (e) (f) (g) 2(a)(i) (ii) (iii) (b)(i) Mark scheme Mark Electrical to chemical energy / Tenaga elektrik kepada tenaga kimia Pure copper / Kuprum tulen Cu2+ and H+ Become thicker / brown solid formed Bertambah tebal / pepejal perang terbentuk Cu2+ + 2e  Cu Blue solution remain unchanged // the intensity of blue solution is the same. Larutan biru tidak berubah // keamatan warna biru larutan adalah sama. (i) the concentration of Cu2+ ions remains the same. kepekatan ion kuprum(II) tidak berubah (ii) the rate of ionized copper at the anode same as the rate of discharged copper(II) ion at the cathode . kadar pengionan kuprum di anode sama dengan kadar ion kuprum(II) dinyahcaskan di katod Oxidation / pengoksidaan Copper atom released electron to form copper(II) ion. Atom kuprum menderMarkan / membebaskan elektron menghasilkan ion kuprum(II). Electroplating of metal // extraction of metal Penyaduran logam // pengekstrakan logam Total 1 1 1 1 1 Chloride ion / Cl-, hydroxide ion / OH-, sodium ion / Na+ and hydrogen ion / H+ Ion klorida / Cl-, ion hidroksida /OH-, ion natrium , Na+ dan ion hidrogen / H+ Cl-. The concentration of chloride ion is higher than hydroxide ion. Cl-. Kepekatan ion klorida lebih tinggi daripada ion hidroksida 2Cl-  Cl2 + 2e Hydrogen gas Gas hidrogen (ii) - (iii) - 19 1 1 1 1 1 11 1 1+1 1 Oxygen gas Gas oksigen Sodium sulphate solution Larutan natrium sulfat Functional – 1 Label - 1 1 Carbon electrodes Elektrod karbon A place lighted splinter at the mouth of the test tube containing hydrogen gas “pop” sound produced Letakkan kayu uji menyala ke dalam tabung uji berisi gas hydrogen Bunyi “pop” terhasil Sodium ion and hydrogen ions move to the cathode, hydrogen ion is selectively discharged hydrogen ion is lower than sodium ion in the Electrochemical Series. Ion natrium dan ion hydrogen bergerak / tertarik ke katod, ion hidrogen terpilih untuk nyahcas / discas Ion hidrogen terletak di bawah ion natrium dalam Siri Elektrokimia Total Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1 1 1 1 11
  20. 20. @Hak cipta BPSBPSK/SBP/2013 Question Mark scheme No 3(a) Cu2+ , H+ (b) Carbon electrode which connect to copper electrode in cell A. Because oxidation takes place Elektrod karbon yang disambung kepada elektrod kuprum dalam sell A Kerana proses pengoksidaan berlaku (c)(i) X – silver electrode / elektrod argentum Y – impure silver electrode / elektrod argentum tak tulen (ii) Ag+ + e  Ag (d)(i) - The electrode become thinner - Silver atom ionized / silver atom oxidized to form silver ion - elektrod seMarkin nipis - atom argentum mengion / atom argentum dioksidakan membentuk argentum ion. (ii) Y : Ag  Ag+ + e Z : Ag+ + e  Ag (e) The waste chemicals emitted contain poisonous heavy metal ions and cyanide ions / alter the pH of water. Bahan buangan kimia dibebaskan mengandungi logam berat yang beracun dan sianid / mengubah nilai pH air Mark 1 1 1 1 1 1 1 1 1 1 1 11 Question Mark scheme No 4(a)(i) Lead(II) ion// Pb2+, bromide ion// BrIon plumbum(II)// Pb2+, ion bromida// Br(ii) Sodium ion // Na+, hydrogen ion// H+, sulphate ion// SO42-, hydroxide ion//OHion natrium // Na+, ion hidrogen// H+, ion sulfat // SO42-, ion hidroksida //OH(b)(i) Lead / Plumbum Mark 1 1 1 (ii) Pb2+ + 2e  Pb 1 (iii) Brown gas / Gas berwarna perang 1 (c)(i) hydroxide ion / ion hidroksida 1 (ii) Anode : Oxygen gas anod : Gas oksigen Cathode : hydrogen gas Katod : gas hidrogen Sodium nitrate solution // sulphuric acid Larutan natrium nitrat // asid sulfurik (Any suitable electrolyte) 1 (iii) 1 1 9 20 Perfect Score & X A –Plus Module/mark scheme 2013
  21. 21. @Hak cipta BPSBPSK/SBP/2013 Rubric (i) Q, R, S , Cu 5(a) Mark 1 …. 1 1 1 1 ..... 3 1 1 1 (ii) positive terminal : Cu Potential difference : 0.7 V S is higher than Cu in the Electrochemical Series (b) (i) positive terminal : copper / Cu Negative terminal : Metal P (ii) metal P : Zinc / Zn // Magnesium/Mg (any suitable metal) Solution Q : Zinc sulphate // magnesium sulphate (any suitable electrolyte) (c) 1 ..... 4 (i) anode : greenish yellow gas cathode : colourless gas (bubbles) 1 1 ….. 2 1 1 ….. 2 (ii) gas X : hydrogen gas Y : chlorine (iii) Ions move to / ion attracted to Anode Hydroxide ion/OHChloride ion/Cl- Cathode Hydrogen ion/H+ , Potassium ion/K+ Cl- H+ Concentration Cl- higher than OH- Position of hydrogen ion/H+ is lower than potassium ion/K+ in the Electrochemical Series. 2H+ + 2e  H2 1+1 1+1 Ions selectively discharged Reason Half equation 2Cl-  Cl2 + 2e Total Question Mark scheme No 6(a) (i) Substance R : Glucose / ethanol (any suitable covalent compound) Substance S : Sodium chloride solution ( any salt solution / acid / alkali) (ii) 1. S conducts electricity but R does not 2. S has free moving ions // ions free to move 3. R consists of molecules / no free moving ions (b) (i) negative terminal : zinc positive terminal : copper (ii) 1. zinc electrode become thinner 2. Zn  Zn2+ + 2e (iii) 1. the potential difference decreases 2. iron is lower than zinc in the Electrochemical Series // iron is less electropositive than zinc // distance between iron and 21 Perfect Score & X A –Plus Module/mark scheme 2013 1+1 1+1 …. 8 20 Mark 1 1 ….. 2 1 1 1 ….. 3 1 1 ….. 2 1 1 ….. 2 1 1
  22. 22. @Hak cipta BPSBPSK/SBP/2013 (c) copper is shorter than distance between zinc and copper in the Electrochemical Series (i) Sample answer Lead(II) bromide / lead(II) iodide /sodium chloride/sodium iodide (any suitable ionic compound) r : substance that decompose when heated. Example : lead(II) nitrate, lead(II) carbonate ….. 2 1 (ii) PbI2 // PbBr2 // NaCl Diagram: Functional Label Observation: Anode : brown gas Cathode: grey solid Carbon electrodes Elektrod karbon Heat Panaskan 1 1 Note : Observations and half-equations are based on the substance suggested. 1 1 Half equation: Anode : 2Br-  Br2 + 2e Cathode : Pb2+ + 2e  Pb 1 1 Product: Anode : lead Cathode : bromine gas Total Question No 7(a) Sample answer Silver nitrate solution Mark scheme 1 1 ….. 8 20 Mark 1 Silver Iron spoon Silver nitrate solution Functional – 1 Label - 1 Anode : Ag  Ag+ + e Cathode : Ag+ + e  Ag (b) 22 1 1 1 1 ….. 5 1. metal X is more electropositive than copper // X is higher than copper in the Electrochemical Series Perfect Score & X A –Plus Module/mark scheme 2013 1
  23. 23. @Hak cipta BPSBPSK/SBP/2013 1 1 1 1 ….. 5 Apparatus Test tube, test tube rack, sand paper 1 Procedure 1. Clean the metal strips with sand paper 2. Pour 5 cm3 of P nitrate solution , R nitrate solution , S nitrate solution into different test tubes. 3. Place a strip of metal P into each test tube 4. Record the observation after 5 minutes 5. Repeat steps 2 to 4 using strip of metal Q, R and S to replace metal P. (c) 2. atom X oxidises to X ion // atom X releases electron 3. copper(II) ion accepts electron to form copper 4. the concentration of copper(II) ion decreases 5. metal Y is less electropositive than copper // Y is lower than copper in the Electrochemical Series Material 0.5 mol dm-3 of P nitrate, Q nitrate, R nitrate, S nitrate solutions, metal P, Q, R and S 1 Observation Metal P Q R S Metal ion P / / / Metal ion Q X / / Metal ion R X X Metal ion S X X X / Conclusion The electropositivity of metals increases in the order of P,Q,R,S 1 1 1 1 1 1 1 1 …..10 TOTAL 20 SET 2 :OXIDATION AND REDUCTION Question No 1 Mark scheme (a) (b) ( c) To allow the flow / movement / transfer of ions through it chemical energy to electrical energy mark at electrodes Cell 1 Cell 2 Positive Negative Positive Negative electrode electrode electrode electrode Q P R S (d)(i) magnesium more electropositive than copper // above copper in the Electrochemical Series (ii) blue becomes paler / colourless Concentration / number of Cu2+ ion decreases (iii) Mg→ Mg2+ + 2e (iv) Oxidation (e)(i) copper become thicker // brown solid deposited (ii) zinc (iii) zinc undergoes oxidation // zinc atom release electron to form zinc ion 23 Perfect Score & X A –Plus Module/mark scheme 2013 Mark 1 1 1 1 1 1 1 1 1 1 11
  24. 24. @Hak cipta BPSBPSK/SBP/2013 Question No 2(a) (b) (c) (d) Mark scheme Mark A reaction which involves oxidation and reduction occur at the same time (i) green to yellow/brown (ii) oxidation (iii) Fe2+ → Fe3+ + e (iv) 0 (i) magnesium (ii) Mg +Fe2+ → Mg2+ + Fe (iii) +2 to 0 1. label for iron, water and oxygen 2. ionization of iron in the water droplet (at anode) 3. flow of electron in the iron to the edge of water droplet Water droplet 1 1 1 1 1 1 1 1 1 1 1 O2 e e 2+ Fe  Fe +2e Iron 11 3 (a) 1.....4 (i) Oxidation number of copper in compound P is + 2 Oxidation number of copper in compound Q is + 1 1 1.....2 (ii) Compound P : Copper(II) oxide Compound Q : Copper(I) oxide Oxidation number of copper in compound P is +2 Oxidation number of copper in compound P is +1 1 1 1 1.....4 (iii)     1 1 1 1.....4 (i) X, Z, Y 1 Y : Copper Z : Lead X : Magnesium 24 1 Reaction B: Oxidation number of magnesium changes/increases from 0 to +2 // Oxidation number of zinc changes/decreases from +2 to 0 (c) 1 1 Reaction A: No change in oxidation number (b) Reaction A : not a redox reaction Reaction B : a redox reaction 1 1 1.....3 Substance that is oxidised Substance that is reduced Oxidizing agent Reducing agent : H2 : CuO : CuO : H2 Perfect Score & X A –Plus Module/mark scheme 2013
  25. 25. @Hak cipta BPSBPSK/SBP/2013 2Mg + O2 → 2MgO // 2X + O2 → 2XO 1 1.....2 [Correct formulae of reactants and product] [Balanced equation] TOTAL 4 (a) (b) (i) 20 Iron(II) ion releases / loses one electron and is oxidised to iron(III) ion// Oxidation number of iron in iron(II) ion increases from +2 to +3. Iron(II) ion undergoes oxidation, Iron(II) ion acts as a reducing agent (ii) Iron(II) ion receives/ gain one electron and is reduced to iron.// Oxidization number of iron in iron(II) iron decreases from +2 to 0. iron(II) ion undergoes reduction, Iron(II) ion acts as an oxidising agent 1 Mg  Mg 2  2e Oxidation number of magnesium increases from 0 to +2 magnesium undergoes oxidation Cu 2   2e  Cu oxidation number of copper in copper(II) ion decreases from +2 to 0 copper(II) ion undergoes reduction 1 1 1 (c) 1 1 1 1 1 1 At the negative terminal: Iron(II) ion release / lose one electron and is oxidised to iron(III) ion. Fe2+  Fe3+ + e The green coloured solution of iron(II) sulphate turns brown. Fe2+ act as a reducing agent. 1 1 1 1 1 At the positive terminal: Bromine molecules accepts electrons and is reduced to bromide ions, BrBr2 + 2e  2BrThe brown colour of bromine water turns colourless. Bromine acts as an oxidising agent 1 1 1 1 1 20 Question No 5 (a) Mark scheme 1. Mg/Al/Fe/Pb/Zn Magnesium undergoes oxidation as oxidation number of magnesium increases from 0 to +2 and 3. Copper (II) oxide undergoes reduction as oxidation number of copper in copper(II) oxide decreases from +2 to 0 4. Oxidation and reduction occur at the same time. 2. (b) 1 1 1 1 Experiment I Fe2+ ion present Metal X lower than iron in the Electrochemical Series // Metal X is less electropositive than iron 3. Iron atoms releases electrons to form iron(II) ions 1. 2. 25 Mark Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1
  26. 26. @Hak cipta BPSBPSK/SBP/2013 Experiment II 1. OH ion present 2. Metal Y higher than iron in the Electrochemical Series // Metal Y is more electropositive than iron n+ 3. Atom Y releases electrons to form Y ions 4. Water and oxygen gain electron to form OH ion // 2H2O + O2 + 4e → 4OH 1 1 1 1 Max 3 (c) Procedure 1. One spatula of copper(II)oxide powder and one spatula of carbon powder is placed into a crucible 2. The crucible and its content are heated strongly 3. The reaction and the changes that occur are observed 4. Steps 1 to 3 are repeated by replacing copper(II)oxide powder with zinc oxide powder and magnesium oxide powder. 1 1 1 1 Observation Mixture Carbon and copper(II)oxide Carbon and zinc oxide Carbon and magnesium oxide Observation The mixture burns brightly. The black powder turns brown The mixture glows dimly. The white powder turns grey. No Changes 1+1 Explanation Carbon can react with copper(II)oxide and zinc oxide Carbon more reactive than copper and zinc / carbon is above copper and zinc in the Reactivity Series Carbon cannot react with magnesium oxide Carbon less reactive than magnesium / carbon is below magnesium in the Reactivity Series 1 1 1 1 20 6 Sample answer (a) Magnesium/Aluminium/zinc/iron/lead Magnesium dissolve//The blue colour of copper(II)sulphate solution become paler // brown solid deposited Mg→Mg2+ + 2e Cu2+ + 2e→ Cu Oxidising agent- Cu2+ ion / copper(II) sulphate Reducing agent- Mg 1 1 1 1 1 1..6 (b) sample answer Pb(NO3)2 Oxidation number: 26 +2 +5 + -2  2KI +1 -1 Perfect Score & X A –Plus Module/mark scheme 2013 Pbl2 +2 -1 + 2KNO3 +1 +5 -2 1 1
  27. 27. @Hak cipta BPSBPSK/SBP/2013 no changes of oxidation number of all elements in the compounds of reactants and products. 1 Neutralization 1...4 (c ) sample answer [Material : Any suitable oxidizing agent (example : acidified potassium manganate(VII) solution, acidified potassium dichromate(VI) solution, chlorine water, bromine water), any suitable reducing agent (example : potassium iodide solution, iron(II) sulphate solution) and any suitable electrolyte] 1 [ Apparatus : U-tube , carbon electrodes , connecting wires and galvanometer] 1 Diagram Functional Labelled 1 1 Procedure 1 Sulphuric acid is added into a U-tube until 1/3 full 2 Bromine water is added into one end of the U-tube while potassium iodide solution is added into the other end of the U-tube 3 carefully 4 Two carbon electrodes connected by connecting wires to a galvanometer are dipped into the two solution at the two ends of the U-tube. Observation The colour of bromine water change from brown to colourless// The colour of potassium iodide solution change from colourless to yellow/brown// The needle of the galvanometer is deflected Oxidation reaction : Br2 + 2e→ 2BrReduction reaction: 2I- → I2 + 2e 27 Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1 1 1 1 1 Max : 10 20
  28. 28. @Hak cipta BPSBPSK/SBP/2013 SET 3 :ACIDS, BASES AND SALTS http://cikguadura.wordpress.com/ Question No 1 (a)(i) (ii) (b)(i) Mark scheme Propanone / Methylbenzene / [any suitable organic solvent] Water Molecule Mark 1 1 1 1 (ii) Ion (c) 1. Beaker A : No observable change Beaker B : Gas bubbles released 2. H+ ion does not present in beaker A but H+ ion present in beaker B // Hydrogen chloride in beaker A does not show acidic properties but hydrogen chloride in beaker B shows acidic properties 1 1. Correct formula of reactants and products 2. Balanced equation 1 1 (d)(i) Mg + 2HCl → MgCl2 (ii) + H2 1. Mole of HCl 2. Mole ratio 3. Answer with correct unit Mole HCl = 1 1 1 1 // 0.005 2 mol HCl reacts with 1 mol Mg 0.005 moles HCl reacts with 0.0025 moles Mg Mass Mg = 0.0025 x 24 // 0.06 g TOTAL Question Mark scheme No 2 (a)(i) Substance that ionize / dissociate in water to produce H + ion 10 Mark 1 (ii) 3 1 (iii) 1. Concentration of acid / H+ ion in Set II is lower than Set I 2. The lower the concentration of H+ ion the higher the pH value 1 1 (iv) 1. Ethanoic acid is weak acid while hydrochloric acid is strong acid 2. Ethanoic acid ionises partially in water to produce low concentration of H+ ion 1 1 while 3. hydrochloric acid ionises completely in water to produce high concentration of H + ion 1 (b)(i) 28 1. The pH value of sodium hydroxide in volumetric flask B is lower than A 2. Concentration of sodium hydroxide / OH- ion in volumetric flask B is lower than A Perfect Score & X A –Plus Module/mark scheme 2013 1 1
  29. 29. @Hak cipta BPSBPSK/SBP/2013 (ii) 1 1 1. Mole of NaOH 2. Mass of NaOH with correct unit Mole NaOH = // 0.005 Mass NaOH = 0.005 x 40 g // 0.2 g (iii) 0.01 x V = 0.002 x 100 // 20 cm3 TOTAL Question No 3 (a) Pink to colourless (b) (c)(i) (ii) Mark scheme KOH 1 → KNO3 + H2O 1 1 1 1 1. Mole of HNO3 // Substitution 2. Mole ratio 3. Concentration of KOH with Mole HNO3 = Mark 1 Potassium nitrate HNO3 + 10 // 0.01 0.01 mole HNO3 reacts with 0.01 mole KOH Molarity KOH = (d)(i) (ii) mol dm-3 // 0.4 mol dm-3 10 cm3 1 1. Sulphuric acid is diprotic acid but nitric acid is monoprotic acid // 1 mole of 1 + sulphuric acid produce 2 moles of H ion but 1 mole of nitric acid produce 1 mole of H+ ion 2. Concentration of H+ ion in sulphuric acid is double compare to nitric acid 3. Volume of sulphuric acid needed is half TOTAL Question Mark scheme No 4 (a) Ionic compound formed when H+ ion from an acid is replaced by a metal ion or ammonium ion 1 1 10 Mark 1 (b) Pb(NO3)2 1 (c) To ensure all the nitric acid reacts completely 1 (d)(i) 1. Correct formula of reactants and products 2. Balanced equation 1 1 2H+ + PbO → Pb2+ + H2O 29 Perfect Score & X A –Plus Module/mark scheme 2013
  30. 30. @Hak cipta BPSBPSK/SBP/2013 (ii) 1. Mole of acid 2. Mole ratio 3. Answer with correct unit 1 1 1 // 0.05 Mole HNO3 = 0.05 moles HNO3 produce 0.025 moles salt G Mass of salt G = 0.025 x 331 g // 8.275 g (e) 1. Add 2 cm3 dilute sulphuric acid followed by 2 cm3 of Iron(II) sulphate solution Slowly add concentrated sulphuric acid by slanted the test tube. Then turn it upright. 2. Brown ring is formed. TOTAL Question No 5 (a)(i) Salt W : Copper(II) carbonate Solid X : Copper(II) oxide Mark scheme 1 1 Mark 1 1 (ii) (iii) Neutralisation (iv) 1 1 1. Flow gas into lime water 2. Lime water turns cloudy / chalky 3. 1. Correct formula of reactants and products 2. Balanced equation CuO + 2HCl → CuCl2 (b) (c)(i) (ii) 1 1 + H2O 2+ Cation : Cu ion // copper(II) ion Anion : Cl- ion // chloride ion Ag+ + Cl- → 1 1 AgCl 1 Double decomposition reaction 1 TOTAL Question No 6 (a)(i) Green (ii) (b)(i) Mark scheme Mark 1 Double decomposition reaction 1 Carbon dioxide 1 (ii) CuCO3 → CuO + CO2 1 (iii) 1. Functional apparatus 2. Label Copper(II) carbonate 1 1 Heat (c)(i) 30 Sulphuric acid // H2SO4 Perfect Score & X A –Plus Module/mark scheme 2013 Lime water 1
  31. 31. @Hak cipta BPSBPSK/SBP/2013 (ii) 1. Mole of CuCO3 2. Mole ratio 3. Answer with correct unit 1 1 1 // 0.1 Mole CuCO3 = 0.1 moles CuCO3 produces 0.1 mole CuO Mass CuO = 0.1 x 80 g // 8 g TOTAL 7 (a) (b) (c) 1. Vinegar 2. Wasp sting is alkali 3. Vinegar can neutralize wasp sting 1 1 1 1. 2. 3. 4. 5. 1 1 1 1 1 6. 1. 2. 3. 4. (d)(i) Water is present in test tube X but in test tube Y there is no water. Water helps ammonia to ionise // ammonia ionise in water OH- ion present OH- ion causes ammonia to show its alkaline properties Without water ammonia exist as molecule // without water OH- ion does not present When OH- ion does not present, ammonia cannot show its alkaline properties Sulphuric acid is a diprotic acid but nitric acid is a monoprotic acid 1 mole of sulphuric acid ionize in water to produce two moles of H+ ion but 1 mole of nitric acid ionize in water to produce one mole of H+ ion The concentration of H+ ion in sulphuric acid is double / higher The higher the concentration of H+ ion the lower the pH value 1. Mole of KOH 2. Molarity of KOH and correct unit Mole KOH = 1. 2. 3. 4. 5. mol dm-3 Mole KOH = 1 1 1 1 // 1 mol dm-3 1 1 1 1 1 Correct formula of reactants Correct formula of products Mole of KOH // Substitution Mole ratio Answer with correct unit HCl + KOH 1 1 // 0.25 Molarity = (ii) 1 → KCl + H2O // 0.025 0.025 mole KOH produce 0.025 mole KCl Mass KCl = 0.025 x 74.5 g // 1.86 g TOTAL 31 Perfect Score & X A –Plus Module/mark scheme 2013 20
  32. 32. @Hak cipta BPSBPSK/SBP/2013 Question No 8 (a)(i) (ii) (b)(i) (ii) (c)(i) Mark scheme Mark 1 1 1. PbCl2 2. Double decomposition reaction Copper (II) chloride : Copper(II) oxide / copper(II) carbonate , Hydrochloric acid Lead (II) chloride : Lead (II) nitrate solution , sodium chloride solution ( any solution that contains Cl- ion) 1+1 1+1 1 1 1 1 1. 2. 3. 4. S = zinc nitrate T = zinc oxide U = nitrogen dioxide W = oxygen 2Zn(NO3)2  2ZnO + 4NO2 + O2 1+1 1 1 1 1. Both axes are label and have correct unit 2. Scale and size of graph is more than half of graph paper 3. All points are transferred correctly (ii) 1 5 (iii) Mole Ba2+ ion = 1 // 0.0025 Mole SO4 2- ion = // 0.0025 1 Ba2+ ion : SO4 2- ion 0.0025 : 0.0025 // 1 : 1 (iv) Ba 2+ + SO42- 1 1 → BaSO4 TOTAL Question Mark scheme No 9 (a) 1. HCl // HNO3 2. 1 mole acid ionises in water to produce 1 mole of H+ ion 3. H2SO4 4. 1 mole acid ionises in water to produce 2 moles of H+ ion (b) 1. Sodium hydroxide is a strong alkali 2. Ammonia is a weak alkali 3. Sodium hydroxide ionises completely in water to produce high concentration of OH ion 4. Ammonia ionises partially in water to produce low concentration of OH - ion 5. Concentration of OH- ion in sodium hydroxide is higher than in ammonia 6. The higher the concentration of OH- ion the higher the pH value 32 Perfect Score & X A –Plus Module/mark scheme 2013 20 Mark 1 1 1 1 1 1 1 1 1 1
  33. 33. @Hak cipta BPSBPSK/SBP/2013 Volumetric flask used is 250 cm3 Mass of potassium hydroxide needed = 0.25 X 56 = 14 g Weigh 14 g of KOH in a beaker Add water Stir until all KOH dissolve Pour the solution into volumetric flask Rinse beaker, glass rod and filter funnel. Add water when near the graduation mark, add water drop by drop until meniscus reaches the graduation mark 10. stopper the volumetric flask and shake the solution TOTAL (c) 1. 2. 3. 4. 5. 6. 7. 8. 9. Question Mark scheme No 10 (a)(i) Substance C : Glacial ethanoic acid Solvent D : Propanone [ or any organic solvent] (ii) Solution E 1. Ethanoic acid ionises in water 2. Can conduct electricity because presence of freely moving ions 3. blue litmus paper turns to red because of H+ ions is present Solution F 4. Ethanoic acid exist as molecules 5. Cannot conduct electricity because no freely moving ion 6. Cannot change the colour of blue litmus paper because no H+ ion 1 1 1 1 1 1 1 1 1 1 20 Mark 1 1 1 1 1 1 1 1 1. Measure and pour [20-100 cm3] of [0.1-2.0 mol dm-3]zinc nitrate solution into a 1 beaker 2. Add [20-100 cm3] of [0.1-2.0 mol dm-3]sodium carbonate solution 3. Stir the mixture and filter 4. Rinse the residue with distilled water 5. Zn(NO3)2 + Na2CO3ZnCO3 + 2NaNO3 6. Measure and pour [20-100cm3]of [0.1-1.0mol dm-3]sulphuric acid into a beaker 7. Add the residue/ zinc carbonate into the acid until in excess 8. Stir the mixture and filter 9. Heat the filtrate until saturated / 1/3 of original volume 10. Cool the solution and filter 11. Dry the crystal by pressing between two filter papers 12. ZnCO3 + H2SO4 ZnSO4 + H2O + CO2 (b) 1 1 1 1 1 1 1 1 1 1 1 TOTAL 33 Perfect Score & X A –Plus Module/mark scheme 2013 20
  34. 34. @Hak cipta BPSBPSK/SBP/2013 SET 3 :RATE OF REACTION http://cikguadura.wordpress.com/ Question No 1(a)(i) (ii) (b)(i) Mark scheme Set II 1 Able to draw the graph with these criterion: 1 Labelled axis with correct unit 2. Uniform scale for X and Y axis & size of the graph is at least half of the graph paper 3. All points are transferred correctly 4. Curve is smooth. Set I : 1.Tangen shown in graph correctly 2.Rate of reaction = 0.19 cm3s-1 ( +- 0.05) Set II : 1.Tangen shown in graph correctly 2.Rate of reaction = 0.23 cm3s-1 (+- 0.05) (ii) Question No 2 (a) Mark Add catalyst Increase the temperature Use smaller size/ metal powder Increases the concentration of acid// Double the concentration of acid but half volume [Any two] Mark scheme 1 1 1 1 1 1 1 1 1 1 Mark 1 1 CaCO3+ 2HNO3 → Ca(NO3) 2+ CO2 + H2O Functional diagram Label (b) 1. Correct formulae of reactants and product 2. Balanced equation 1 1 Water Nitric acid Calcium carbonate (c) 1. Mole of nitric acid 2. Mole ratio 3. Answer with correct unit Number of moles of HNO 3 = 0.2 X 50 = 0.01 mol // 1000 2 mol of HNO3 produce 1 mol of CO2 0.01 mol of HNO3 produce 0.005 mol of CO2 34 Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1
  35. 35. @Hak cipta BPSBPSK/SBP/2013 Maximum volume of CO2 = 0.005 x 24 = 0.12 dm3 // 120 cm3 (d) (e)(i) (ii) Experiment I = 0.12 X 1000 // 0.2 cm3 s-1 // 10 X 60 //0.12 //0.012 dm3 min-1 10 Experiment II = 0.12 X 1000 // 0.4 cm3 s-1 // 5 X 60 // 0.12 // 0.024 dm3 min-1 5 Rate of reaction in Experiment II is higher than I - The size of calcium carbonate in Experiment II is smaller than Experiment I // calcium carbonate powder in Experiment II has a larger total surface area exposed to collision than Experiment I. - The frequency of collision between between calcium carbonate and hydrogen ion in Experiment II is higher than Experiment I. - The frequency of effective collision s in Experiment II is higher than Experiment I 1 1 1 1 1 1 Question No 3 (a) (b)(i) (ii) Mark scheme -Total surface area of smaller pieces wood is larger/bigger/ greater than the bigger pieces of wood - More surface area exposed to air for burning 1. Experiment II 2. Present of catalyst /manganase(IV) oxide in Experiment I 1 1 1.Correct formulae of reactants and product 2.Balanced equation 1 1 2H2O2 → (iii) Mark 1 1 2H2O + O2 Energy Ea Ea’ 2H2O2 2 H2O + O2 1. Arrow upward with energy label ,two levels and position of reactant and products are correct 2. Curve of Experiment I and experiment II are correct and label 3. Activation energy of experiment I and experiment II are shown and labelled (c)(i) 1.Correct formulae of reactants and product 2.Balanced equation 1 1 1 1 1 Zn + 2HCl  ZnCl2 + H2 (ii) 35 No. of mol HCl = 50 X 0.5 // 0.025 1000 Perfect Score & X A –Plus Module/mark scheme 2013 1
  36. 36. @Hak cipta BPSBPSK/SBP/2013 2 mol HCl 0.025 mol HCl : 1 mol H2 : 0.0125 mol H2 1 1 Volume of H2 = 0.0125 x 24 // 0.3dm3 // 300 cm3 1. Add excess zinc powder with 12.5 cm3 of 1 mol dm-3hydrochloric acid . 2. At the same temperature (iii) 1 1 OR 1. Add excess zinc powder with 25 cm3 of 0.5 mol dm-3hydrochloric acid 2. At the higher temperature //present of catalyst (iv) 1. 2. 3. 4. 1 Rate of reaction using sulphuric acid is higher The concentration of H+ ion in sulphuric acid is higher Maximum volume of gas collected is double The number of mole of H+ ion in sulphuric acid is double 1 1 1 1 1 20 Question No 4 (a) Mark scheme 1. Temperature in refrigerator is lower than in cabinet 2. The activity of microorganisme (bacteria) in refrigerator is lower than in Mark 1 1 refrigerator 3. The amount of toxin produced in the refrigerator is less then in the kitchen (b)(i) 1. 2. 3. 4. Zn + 1 cabinet. Correct formula of reactants and products Mol of sulphuric acid Mole ratio Volume and ratio H2SO4 ------- ZnSO4 + H2 No. Of mol H2SO4 = 1 X 50/1000 // 0.05 1 1 mol of H2SO4 0.05 mol of H2SO4 1 : : 1 mol of H2 0.05 mol of H2 Volume of H2 = 0.05 x 24 dm3 //1.2 dm3 //0.05 x 24000//1200 cm3 (ii) (iii) 36 = 1200 // 15 cm3 s-1 80 Experiment II = 1200 // 7.5 cm3 s-1 160 Experiment III = 600 // 2.5 cm3 s-1 240 Exp I and II 1.Rate of reaction of Expt I is higher 2.The size of zinc in Expt I is smaller 3.Total surface area of zinc in Expt I is bigger/larger 4.The frequency of collision between zinc atom and hydrogen ion/H+ in Expt I is higher 5. The frequency of effective collision in Exp I is higher Experiment I Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1 1 1 1 1 1 1 1
  37. 37. @Hak cipta BPSBPSK/SBP/2013 Exp II and III 1. Rate of reaction in Expt II is higher 2.The concentration of sulphuric acid/ H+ ion in Exp II is higher 3. The no. of H+ per unit volume in Expt II is higher/greater in Expt II// 4. The frequency of collision between zinc atom and H + in Expt II is higher 5. The frequency of effective collision in Expt II is higher 1 1 1 1 1 20 Question No 5.(a) (i) (ii) Mark scheme Mark N2 + 3H2 ------- 2NH3 1+1 Temperature : 450 – 550 ˚ C Pressure : 200 – 300 atm Catalyst : Powdered iron// Iron filling 1 1 [ Any two] (b)(i) Example of acid Sample answer : Hydrochloric acid / HCl// Sulphuric acid // Nitric acid 1 1 (ii) (iii) Correct formula of reactant and product Balance Sample answer 2HCl + Mg → MgCl2 + H2 1. Experiment I : 20 cm3 / 60 s // 0.33 cm3s-1 2. Experiment II : 20 cm3 / 50 s // 0.4 cm3s-1 1 1 1 (Catalyst) Experiment 1: 1.Pour /measure (50-100) cm3 of (0.1-2 mol dm-3 ) hydrochloric acid . 2.Add excess zinc powder/granules 3.Add a (2-5 cm3 ) of copper(II) sulphate solution 4.At the same temperature Experiment II : 1. Pour /measure (50-100) cm3 of (0.1-2 mol dm-3 ) hydrochloric acid . 2. Add excess zinc powder/granule 3. At the same temperature (Temperature) Experiment 1: 1. Pour /measure (50-100) cm3 of (0.1-2 mol dm-3 ) hydrochloric acid 2. Heat acid to (30-80OC) 3. Add excess zinc powder/granule Experiment II : 1 1 1 1 1 1 3 -3 Pour /measure (50-100) cm of (0.1-2 mol dm ) hydrochloric acid . Without heating Add excess zinc powder/granules OR (Concentration) Experiment 1: 1.Pour /measure (50-100) cm3 of (0.2-2 mol dm-3 ) hydrochloric acid . 2. Add excess zinc powder/granules 3.At the same temperature 37 1 1 OR 1. 2. 3. 1 Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1 1
  38. 38. @Hak cipta BPSBPSK/SBP/2013 Experiment II : 1 1. Pour /measure (50-100) cm3 of (0.1-1 mol dm-3 ) hydrochloric acid . 2. Add excess zinc powder/granules 3. At the same temperature 1 OR (Size) Experiment 1: 1.Pour /measure (50-100) cm3 of (0.1-2 mol dm-3 ) hydrochloric acid . 2. Add excess zinc powder 3.At the same temperature 1 Experiment II : 1 3 1 -3 1. Pour /measure (50-100) cm of (0.1-2 mol dm ) hydrochloric acid . 2. Add excess zinc granule 3. At the same temperature (iv) 1 (Catalyst) 1.Catalyst/copper(II) sulphate is used in Experiment I 2. Catalyst/(copper(II) sulphate) lower activation energy (and provide an alternative path) 3. More colliding particles / ions are able to achieve that lower activation energy. 4.The frequency of effective collision between magnesium atoms and hydrogen ion increases. 5. The rate of reaction of Experiment I is higher. (Any 4) (Temperature) 1. Rate of reaction in Experiment I is higher. 2. The temperature of reaction in Experiment I is higher 3. The kinetic energy of particles increases in Experiment I // The particles move faster 4. Frequency of collision between magnesium atom and H+ ion in Experiment I is higher 5. Frequency of effective collision in Experiment I is higher (Any 4) (Concentration) 1. Rate of reaction in Experiment II is higher 2. The concentration of acid in Experiment I is higher 3. The number of hydrogen ion per unit volume in Experiment II is higher 4. Frequency of collision between magnesium atom and H+ ion in Experiment I is higher 5. Frequency of effective collision in Experiment II is higher (Any 4) 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 (Size) 1.Rate of reaction in Experiment I is higher 2.The size of magnesium in Experiment I is smaller 3.Total surface area of magnesium in Experiment I is bigger/larger 4.The frequency of collision between magnesium atoms and hydrogen ions in Experiment I higher 5.The frequency of effective collision between in Experiment I is higher (Any 4) (v) 38 1 The number of mol are same // The concentration and volume of acid are same 1 Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1 1 1
  39. 39. @Hak cipta BPSBPSK/SBP/2013 Question No 6.(a) (i) Mark scheme Mark 1. First minute = 24/60 =0.4 cm3 s-1 // 24 cm3 min-1 2. 2 nd minute = 34-24/60 =0.167 cm3 s-1 // 10 cm3 min-1 1 3. rate in 1 st minute higher than 2 nd minute (vice versa) 4. concentration of sulphuric acid / mass of zinc decreases 1 (iii) All hydrogen ion from acid was completely reacts 1 (iv) A catalyst lower activation energy provide an alternative path More colliding particles /zinc atoms and hydrogen ions are able to overcome the lower activation energy. The frequency of effective collisions between zinc atom and hydrogen ion in is higher. (any 2 ) - hydrogen and oxygen molecules collide - with correct orientation -total energy of particles higher or equal to activation /minimum energy (Temperature) 1 (ii) (b) Materials: 0.2 mol dm-3 sodium thiosulphate, 1.0 mol dm-3 sulphuric acid, a piece of white paper marked ‘X’ at the centre. Apparatus: 150 cm3 conical flask, stopwatch, 50 cm3 measuring cylinder, 10 cm3 measuring cylinder, thermometer, Bunsen burner, wire gauze. Procedure: 1 1 1 1 1 1 1 1 1 3 -3 1.Using a measuring cylinder, 50 cm of 0.2 mol dm sodium thiosulphate solution is measured and poured into a conical flask. 1 2.The conical flask is placed on top of a piece of white paper marked ‘X’ at the centre. 3.5 cm3 of 1.0 mol dm-3 sulphuric acid is measured using another measuring cylinder. 1 4.The sulphuric acid is poured immediately and carefully into the conical flask. At the same time, the stop watch is started 1 5.The mixture in a conical flask is swirled. 1 6.The ‘X’ mark is observed vertically from the top of the conical flask through the solution. 1 7.The stopwatch is stopped once the ‘X’ mark disappears from view. 1 8.Step 1 – 7 are repeated using 50 cm3 of 0.2 mol dm-3 sodium thiosulphate solution at 40oC, 50oC, 60 oC by heating the solution before 5 cm3 of sulphuric acid is added in. (Max 7) Conclusion When the temperature of sodium thiosulphate solution is higher , the rate of reaction is higher 39 Perfect Score & X A –Plus Module/mark scheme 2013 1 1
  40. 40. @Hak cipta BPSBPSK/SBP/2013 (Temperature) 1 Materials: 0.2 mol dm-3 sodium thiosulphate, 1.0 mol dm-3 sulphuric acid, water, a piece of white paper marked ‘X’ at the centre. 1 Apparatus: 150 cm3 conical flask, stopwatch, 50 cm3 measuring cylinder, 10 cm3 measuring cylinder, wire gauze. 1 Procedure: 1.Using a measuring cylinder, 50 cm3 of 0.2 mol dm-3 sodium thiosulphate solution is measured and poured into a conical flask. 1 2.The conical flask is placed on top of a piece of white paper marked ‘X’ at the centre. 1 3.5 cm3 of 1.0 mol dm-3 sulphuric acid is measured using another measuring cylinder. 1 4.The sulphuric acid is poured immediately and carefully into the conical flask. At the same time, the stop watch is atarted 1 5.The mixture in a conical flask is swirled. 6.The ‘X’ mark is observed vertically from the top of the conical flask through the solution. 1 7.The stopwatch is stopped once the ‘X’ mark disappears from view. 1 8.Step 1 – 7 are repeated by adding 5 cm3, 10 cm3, 15 cm3, 20 cm3 and 40 cm3 of distilled water .(at the same time) maintaining the total volume of solution at 50 cm3 after dilution//table of dilution (Max 7) Conclusion When the temperature of sodium thiosulphate solution is higher , the rate of reaction is higher 1 1 SET 3 :THERMOCHEMISTRY Question No 1 (a) Heat change /released when 1 mol copper is displaced from copper (II) sulphate solution by zinc (b) (c) Mark scheme Blue to colourless 50 X 4.2 X 6 J // 1260 J (ii) (1.0 )(50) 1000 (iii) 40 (i) 1260 -1 0.05 J // 25200 J mol Mark 1 1 1 // 0.05 Perfect Score & X A –Plus Module/mark scheme 2013 1 1
  41. 41. @Hak cipta BPSBPSK/SBP/2013 = - 25.2 kJ mol-1 1. Correct reactant and product 2. Correct two energy level for exothermic reaction 3. Correct value heat of displacement and unit (d) 1 1 1 1 Sample answer Energy Zn + CuSO4 //Zn + Cu2+ ∆H = - 25.2 kJmol-1 ZnSO4 + Cu //Zn2+ (e) (i) 3°C (ii) + Cu Number of mole copper displaced is half Heat released is half / 1260 J // 630 J 2 1 1 1 TOTAL Question No 2 (a) (b) (c) (d) (i) (ii) (iii) (iv) (e) Mark scheme Heat of precipitation is the heat change when one mole of a precipitate is formed from its solution. To reduce heat loss to the surrounding. Reject : prevent Ag+ + Cl- → AgCl The heat released =(50 + 50) x 4.2 x 3.5 =1470 J Number of moles of Ag+ = (50 x 0.5) = 0.025 mol 1000 Number of moles of Cl= (50 x 0.5) = 0.025 mol 1000 0.025 mole of Ag+ reacts with 0.025 mole of Cl- to form 0.025 mole of AgCl Number of moles of AgCl = 0.025 mol = x 1470 J =58 800 J Heat of precipitation of AgCl = -58.8 kJ mol-1 Ag+ + Cl-→AgCl ∆H = -58.8kJmol-1 // AgNO3 + NaCl →AgCl + NaNO3 ∆H = -58.8kJmol-1 (i) 41 12 Perfect Score & X A –Plus Module/mark scheme 2013 Mark 1 1 1 1 1 1 1 1 1 1
  42. 42. @Hak cipta BPSBPSK/SBP/2013 Energy Ag+ + Cl- ∆H = -58.8kJmol-1 (ii) AgCl 1. Label axes 2. Energy levels of reactants and products correct with formula of reactants and products 3. Heat of precipitation written Total Question No 3. (a) Mark scheme 1 1 1 Mark (i) Ethanol 1 (ii) 1260 kJ of heat energy is released when one mole of ethanol is burnt completely in excess oxygen 1 (i) No of moles of alcohol = 0.23 / 46 = 0.005 mol 1 mol of alcohol burnt released 1260 kJ Thus, 0.005 mol of alcohol burnt released 6.3 kJ 1 (b) (ii) ( c) (d) mc = 6.3 kJ mc = 6.3 x 1000 = 6300/ 200 x 4.2 = 7.5 0 C 1 1 1 Heat is lost to the surrounding // Heat is absorbed by the apparatus or containers // Incomplete combustion of alcohol 1 (i) Energy C2 H5 O H + 3 O2 ∆ H = - 1260 kJmol-1 2 CO2 + H2 O 1. Label axes 2. Energy levels of reactants and products correct with formula of reactants and products 3. Heat of combustion written 42 Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1
  43. 43. @Hak cipta BPSBPSK/SBP/2013 (ii) 1 1 1. Label 2. Functional (e) (i) (ii) - 2656 kJmol-1 // 2500-2700 kJmol-1 1 1. The molecular size/number of carbon atom per molecule propanol is bigger/higher methanol 2. Combustion of propanol produce more carbon dioxide and water molecules 3. More heat is released during formation of carbon dioxide and water molecules 1 1 1 Total marks Question No 4 (a) Mark scheme (i) Characteristic Change in temperature Type of chemical reaction Energy content of reactants and products (ii) 43 Mark Diagram 4.1 Increase Diagram 4.2 Decrease Exothermic reaction Endothermic reaction The total energy content of the reactants more than the energy content of the products The total energy content of the reactants less than the energy content of the products Amount of Amount of heat absorbed Amount of heat absorbed for heat absorbed for the breaking of bond in the breaking of bond in the reactant is more than heat /realeased the reactant is less than heat released during released during formation of during breaking of formation of bond in the bond in the products products bonds Number of moles of FeSO4 = MV 1000 = (0.2)(50) = 0.01 mol 1000 Heat change = 0.01 x 200 kJ = 2 kJ // 2000 J Heat change = mcθ θ = 2000 (50)(4.2) θ = 9.5 oC Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1+1 1+1 1 1 1
  44. 44. @Hak cipta BPSBPSK/SBP/2013 (b) (c) 1. Number of mole of Ag+ ion in both experiment = 25 x 0.5 // 0.0125 mol 1000 2. Number of mole of Cl- ion in both experiment = 25 x 0.5 // 0.0125 mol 1000 3. Number of mole of silver chloride formed is the same 4. Na+ ion and K+ ion not involved in the reaction // Ag+ ion and Cl- involved in the reaction Heat change = mcθ = (100)(4.2)(42.2 – 30.2) = 5040 J / 5.04 kJ (i) 1 1 1 1 1 Number of moles of HCl / H + ion = 0.1 mol = (50)(2 1000 Number of moles of NaOH / OH - ion = (50)(2) = 0.1 mol 1000 The heat of neutralization = 5.04 0.1 ΔH = - 50.4 kJ mol-1 1 1 1 Temperature change is 12.0 oC // same Number of moles of sodium hydroxide reacted when hydrochloric acid or sulphuric acid is used is the same // 0.01 mol Number of mole of water formed when hydrochloric acid or sulphuric acid used is the same // 0.01 mol H+ ion in excess when sulphuric acid is used 1 Total marks (ii) 20 Question No 5 (a) Mark scheme 1 1 1 Mark (i) Neutralisation//Exothermic reaction 1 (ii) Total energy content of reactant is higher than total energy content in product 1. The heat of neutralization of Experiment 1 is higher than Experiment 2 2. HCl is strong acid while ethanoic acid is weak acid 3. HCl ionises completely in water to produce high concentration of H+ ion 4. CH3COOH ionizes partially in water to produce low concentration of H + ion and most of ethanoic acid exist as molecules 5. In Expt 2,Some of heat given out during neutralization reaction is used to dissociate the ethanoic acid molecules completely in water//part of heat that is released is used to break the bonds in the molecules of ethanoic acid that has not been ionised No of mol acid/alkali= 50 X 1 /1000= 0.05 Q = ∆ H X no of mol = 57.3 X 0.05 = 2.865 kJ // 2865 J 1 (iii) (b) (i) (ii) 44 2865 = 100 X 4.2 X 0 θ = 2865 ÷ 420 Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1 1 1 1 1 1 1 1
  45. 45. @Hak cipta BPSBPSK/SBP/2013 = 6.8 oC ( correct unit) (iii) (c ) 1 1. Some of heat is lost to the sorrounding 2. Heat is absorbed by polystyrene cup A The reaction is exothermic// Heat is released to the surrounding during the reaction Heat released is x kJ when 1 mol product is formed The total energy content in reactant is higher than total energy content in product The temperature increases during the reaction Heat released during the formation of bond in product is higher than heat absorbed during the breaking of bond in reactant 1 1 B The reaction is endothermic// Heat is absorbed from the surrounding during the reaction Heat absorbed is y kJ when 1 mol product is formed. The total energy content in reactant is lower than total energy content in product The temperature decreases during the reaaction Heat absorbed during the breaking of bond in reactant is higher than heat released during the formation of bond in product 1 1 1 1 1 TOTAL 6 (a) (i) 20 energy Zn + CuSO4 ∆H = -152 kJmol-1 ZnSO4 + Cu 1. Y-axes : energy 2. Two different level of energy (ii) (b) (c) 45 1. reactants have more energy // products have less energy 2.energy is released during the experiment // this is exothermic reaction No. of mol of H+ ion/OH- = 1x50/1000// 0.05 Heat change = 100x 4.2 x7//2940 Joule//2.94 kJ Heat of neutralization= -2940/0.05 = -58800 J mol -1//-58.8 kJ mol-1 1. Heat of combustion of propane is higher 2. The molecular size/number of carbon atom per molecule propane is bigger/higher 3. Produce more carbon dioxide and water molecules//released more heat energy 1. Methanol/ethanol/ propanol, Diagram: 2. -labelled diagram 3. -arrangement of apparatus is functional Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1 1 1 1 1 1 1 1 1 1 1 1..3
  46. 46. @Hak cipta BPSBPSK/SBP/2013 1. (100-250 cm3 )of water is measured and poured into a copper can and the copper can is placed on a tripod stand 2. the initial temperature of the water is measured and recorded 3. a spirit lamp with ethanol is weighed and its mass is recorded 4. the lamp is then placed under the copper can and the wick of the lamp is lighted up immediately 5. the water in the can is stirred continuously until the temperature of the water increases by about 30oC. 6. the flame is put off and the highest temperature reached by the water is recorded 7. The lamp and its content is weighed and the mass is recorded …. 8 max 4 Data The highest temperature of water The initial temperature of water Increase in temperature,  = = = t2 t1 t2 Mass of lamp after burning Mass of lamp before burning Mass of lamp ethanol burnt, m = = = m2 m1 m1 - m2 = m - ..4 t1 =  …..1 Calculation : Number of mole of ethanol, C2H5OH, n = m 46 ……1 The heat energy given out during combustion by ethanol = the heat energy absorbed by water= 100x x c x  J Heat of combustion of ethanol = m c  KJ mol-1 n = -p kJ/mol …1 Total marks 46 Perfect Score & X A –Plus Module/mark scheme 2013 ..3 20
  47. 47. @Hak cipta BPSBPSK/SBP/2013 7 Question No (a) (i) Mark scheme Heat change = mc = (25+25)(4.2)(33-29) = 445 J Mark 1 Heat of precipitation of AgCl = - 445 / 0.0125 = -35600 J mol-1 // 35.6 kJ mol-1 1 Energy AgNO3 + NaCl H = -35.6 kJ mol-1 AgCl + NaNO3* * Accept ionic equation 1. The position and name /formulae of reactants and products are correct. 2. Label for the energy axis and arrow for two levels are shown. (b) (i) (ii) 1. HCl is a strong acid // CH3COOH is a weak acid. 2. HCl ionised completely in water to produce higher concentration of H + ion. // 3. CH3COOH ionised partially in water to produce lower concentration of H+ ion. 4. during neutralisation reaction, some of the heat released are absorbed by CH3COOH molecules to dissociate further in the molecules. 1. H2SO4 is a diprotic acid// HCl is a monoprotic acid. 2. H2SO4 produced two moles of hydrogen ion/H+ when one mole of the acid ionised in water // 3. HCl produced one mole of hydrogen ion/ H+ when one mole of the acid ionised in water. 4. When one mole of OH- reacts with two moles of H+ will produce one mole of water, the heat of neutralisation is still the same as Experiment I because the definition of heat of neutralisation is based on the formation of one mole of water. (c) 4Max 3 4Max 3 - apparatus and material : 2 marks - procedures : 5 marks - Table : 1 mark - Calculation : 2 marks Sample answer: Apparatus : Polystyrene cup, thermometer, measuring cylinder. Materials : Copper (II) sulphate, CuSO4 solution, zinc powder. Procedures : 1. Measure 25 cm3 of 0.2 mol dm-3 copper (II) sulphate, CuSO4 solution and pour it into a polystyrene cup. 2. Put the thermometer in the polystyrene cup and record the initial temperature of the solution. 3. Add half a spatula of zinc powder quickly and carefully into the polystyrene cup. 4. Stir the reaction mixture with the thermometer to mix the reactants. 5. Record the highest temperature reached. 47 1 1 Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1 1 1 1 1
  48. 48. @Hak cipta BPSBPSK/SBP/2013 Tabulation of data: 1 2 2 - 1 ....1 Initial temperature of CuSO4 solution (oC) Highest temperature of the reaction mixture (oC) Temperature change (oC) Calculation : Number of mole of CuSO 4 = MV/1000 = (0.2)(25)/1000 = 0.005 mol ……1 Heat change = mc(2 - 1) = x J Heat of displacement = x / 0.005 kJ mol-1 = y kJ mol-1 …….1 TOTAL 20 SET 4 :CARBON COMPOUNDS http://cikguadura.wordpress.com/ 1 Question No (a) Mark scheme Mark 1 Or C3H7OH + 9/2O2  3CO2 + 4H2O 1 (i) Sweet/ pleasant smell /// fruity smell 1 (ii) Methanoic acid 1 (b) (c) (iii) O H H HC  OC C C H (d) 1+1 H H H H (i) Oxidation 1 (ii) Orange colour of acidified potassium dichromate (VI) solution turns green 1 (iii) C3H7OH + 2[O]  C2H5COOH + H2O 1 (e) C3H7OH (ii)  C3H6 + H2O propanol propene 1+1 48 Perfect Score & X A –Plus Module/mark scheme 2013
  49. 49. @Hak cipta BPSBPSK/SBP/2013 2 Question No (a) (i) (ii) Mark scheme Fermentation Ethanol Mark 1 1 H H (iii) H C C H 1 H OH (b) (c) (i) (ii) C2H5OH + 3O2 → 2CO2 + 3H2O Ethene H ৷ C ৷ H H ৷ -C ৷ H 1+1 1 n Purple to colourless 1 1 (i) Ethyl ethanoate 1 (ii) CH3COOH + C2H5OH  CH3COOC2H5 + H2O (d) (e) Mark scheme Question No 3 1+1 Mark (a) Characteristics Same general formula Explanation CnH2n + 1OH 1+1 successive member is different from each other by – CH2 Relative atomic mass is different by 14 1+1 Gradual change in physical properties // Melting / boiling point increase Number of carbon atom per molecules increase // size of molecule increase 1+1 Similar chemical properties // oxidation produce carboxylic acid Have same chemical/similar functional group 1+1 Can be prepared by similar method // can be prepared by hydration of alkene Have same chemical properties // have same functional group 1+1 (b) (i) (CH2O)n = 60 (12 + 2 + 16)n = 60 n=2 C2H4O2 (ii) Carboxylic acid React with carbonate to produce carbon dioxide 49 Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1 1
  50. 50. @Hak cipta BPSBPSK/SBP/2013 (iii) 2 CH3COOH + CaCO3 → (CH3COO)2Ca + H2O + CO2 Correct formula of reactants and products Balanced equation 1 1 (c) Compound The number of carbon atom P 2 Q 2 1 1 The number of hydrogen atom 4 6 number of hydrogen atom Q is higher Type of covalent bond between // carbon/ Type of hydrocarbon Type of homologous series // // Name of compound Double bond / / Unsaturated Single bond/ / Saturated 1 Alkene// Ethene // Alkane // Ethane 1 General formula// Molecular formula of the compound CnH2n // C2H4 CnH2n+2 // C2H6 1 Max 4 20 Question No (a) (i) 4 (ii) Mark scheme Mark 1 14.3 % Element Mass/ % No. of moles Ratio of moles/ Simplest ratio C 85.7 85.7 = 7.14 12 H 14.3 14.3 = 14.3 1 7.14 = 1 7.14 1 1 14.3 = 2 7.14 1 Empirical formula = CH2 RMM of (CH2)n [(12 + 1(2)]n 14n = 56 .............1 = 56 = 56 n = 56 14 = 4 ………..1 Molecular formula : C4H8 ………………..1 6 max 5 (iii) 1+1 1+1 But-2-ene 2-methylpropene [any 2] 50 Perfect Score & X A –Plus Module/mark scheme 2013 But-1-ene Max 4
  51. 51. @Hak cipta BPSBPSK/SBP/2013 (iv) Compound M (Butene, C4H8) has a higher percentage of carbon atom in their molecule than butane, C4H10 …………….1 % of C in C4H8 = 4(12) x 100% 4(12)  8 = 48 x 100% 56 = 85.7% …………1 4(12) x 100% % of C in C4H10 = 4(12)  10 = 48 x 100% 58 = 82.7% ………..1 (b) (i) (ii) (c) (i) (ii) .....3 1 1 Starch Protein / natural silk H H CH3 H I I I I C = C– C = C I I H H 1 1..2 2-methylbut-1,3-diene or isoprene Rubber that has been treated with sulphur In vulcanised rubber sulphur atoms form cross-links between the rubber molecules These prevent rubber molecules from sliding too much when stretched TOTAL Question No (a) (i) 5 Mark scheme 1 1 1 20 Mark Hydrocarbon General formula covalent alkane CnH2n+2 3 B (iii) Homologous series A (ii) Type of bond covalent alkene CnH2n 3 Carbon dioxide 2C4H10 + 13O2 → 8CO2 + 10H2O [Chemical formulae of reactants and products] [Balanced] 1 Hydrocarbon B. Hydrocarbon B is an unsaturated hydrocarbon which react with bromine. Hydrocarbon A is a saturated hydrocarbon which do not react with bromine. 1 1 1 1 1 51 Perfect Score & X A –Plus Module/mark scheme 2013
  52. 52. @Hak cipta BPSBPSK/SBP/2013 (iv) 1 1 Hydrocarbon B more sootiness. B has higher percentage of carbon by mass. % of carbon by mass ; Hydrocarbon A : Hydrocarbon B : (b) 4(12) 4(12) + 10(1) 4(12) 4(12) + 8(1) × 100 × 100 // 82.76 % 1 // 85.71 % 1 Carboxylic acid X : 1 Propanoic acid 1 Alcohol Y: 1 Ethanol 1 TOTAL 6 Question No (a) (i) 20 1. 2. 3. 4. 5. (iii) Ammonia aqueous solution contains hydroxide ions Hydroxide ions neutralise hydrogen ions (acid) produced by activities of bacteria 1 1 1 1 1 5 max 4 1 (i) (ii) Alcohol Burns in oxygen to form carbon dioxide and water Oxidised by oxidising agent (acidified potassium dichromate (VI) solution) to form carboxylic acid (iii) 52 Mark 1 1 (ii) (b) Mark scheme X - any acid – methanoic acid Y - any alkali – ammonia aqueous solution Procedure: 1. Place glass wool in a boiling tube 3 2. Soak the glass wool with 2 cm of ethanol 3. Place pieces of porous pot chips in the boiling tube 4. Heat the porous pot chips strongly 5. Heat glass wool gently Methanoic acid contains hydrogen ions Hydrogen ions neutralise the negative charges of protein membrane Rubber particles collide, Protein membrane breaks Rubber polymers combine together Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1 1
  53. 53. @Hak cipta BPSBPSK/SBP/2013 6. Using test tube collect the gas given off 6 max 5 Diagram: Glass wool soaked with ethanol Heat Porcelain chips Heat Water [Functional diagram] [Labeled – porcelain chips, water, named alcohol, heat] Test: Add a few drops of bromine water Brown colour of bromine water decolourised Total Question No (a) 7 1 1 20 Mark scheme Carbon dioxide/ CO2 and water/ H2O Any one correct chemical equation Example 2C4H10 + 13O2 → 8CO2 + Chemical formula of reactants Balanced 1 1 Mark 1 10H2O 1 1 (b) Compound B & Compound D Same molecular formula / C4H8 Different structural formula 1 1 1 (c) Pour compound A/B into a test tube Add bromine water to the test tube and shake Test tube contain compound A unchanged Test tube contain compound B brown colour turn colourless or Pour compound A/B into a test tube Add acidified Potassium manganate(VII) solution to the test tube and shake Test tube contain compound A unchanged Test tube contain compound B purple colour turn colourless 1 1 1 1 (d) (i) Any members of carboxylic acid and correct ester Example [Methanoic acid] [Propylmethanoate] 1 1 1 1 53 Perfect Score & X A –Plus Module/mark scheme 2013
  54. 54. @Hak cipta BPSBPSK/SBP/2013 (d) (ii) Pour 2 cm3 of [methanoic acid] into a boiling tube Add 2 cm3 of propanol/compound E into the boiling tube Slowly/carefully/drop 1 cm3 of concentrated sulphuric acid Heat the mixture gently Pour the mixture in a beaker that contain water Observation : Colorless liquid with fruity smell is formed / Colorless liquid float on water surface TOTAL 20 Mark scheme Mark Question No 8(a) H H H C H H C C C H C H But-2-ene H C C C H H H 2-methylpropene H 1+1 H H H 1 1 1 1 1 1 1+1 H (b) (i) Propanoic acid Ethanol (ii) Chemical properties for propanoic acid: 1. React with reactive metal to produce salt and hydrogen gas 2. React with bases/alkali to produce salt and water 3. React with carbonates metal to produce salt, carbon dioxide gas and water 4. React with alcohol to produce ester 1 1 1 1 1 1 1 [any three] Chemical properties for ethanol: 1. Undergo combustion to produce carbon dioxide and water 2. Burnt in excess oxygen to produce CO2 and H2O 3. Undergo oxidation to produce carboxylic acid / ethanoic acid 4. React with acidified K2Cr2O7 /KMnO4 to produce carboxylic acid / ethanoic acid 5. Undergo dehydration to produce alkene / ethene. [Any three answers] (c) (i) P : Hexane Q : Hexene // Hex-1-ene (ii) Reaction with bromine // acidified potassium manganate(VII) solution Procedure: 54 Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1 1 1 1 1 1 1
  55. 55. @Hak cipta BPSBPSK/SBP/2013 1. Pour about [2 -5 cm3] of P into a test tube. 2. Add 4-5 drops of bromine water / acidified potassium manganate(VII) solution 1 1 1 and shake. 3. Observe and record any changes. 4. Repeat steps 1 to 3 by replacing P with Q 1 1 Observation: P : Brown/ Purple colour remains unchanged. Q : Brown/ Purple colours decolourise / turn colourless. Max 6 20 SET 4 :MANUFACTURED SUBSTANCE IN INDUSTRY http://cikguadura.wordpress.com/ 1 Question No (a) (i) Mark scheme Mark 1 Contact process (ii) 1 (iii) Vanadium(V) oxide, 450 oC - 500oC 1 (iv) Ammonium sulphate 1 (v) (i) (b) Ammonia 2NH3 + H2SO4  (NH4)2SO4 Composite material 1+1 1 (ii) Tin atom Correct arrangement Correct label 1 1 Copper atom (iii) nC2H3Cl  --( C2H3Cl )n 1 (iv) It has low thermal expansion coefficient // resistant to thermal shock 1 TOTAL Question No 2 (a) (i) (ii) 55 11 Mark scheme SO2 + H2O  H2SO3     Corrodes buildings Corrodes metal structures pH of the soil decreases Lakes and rivers become acidic [Able to state any three items correctly] Perfect Score & X A –Plus Module/mark scheme 2013 Mark 1 3 4
  56. 56. @Hak cipta BPSBPSK/SBP/2013 (b) (i)     1 1 1     Pure metal are made up of same type of atoms and are of the same size. The atoms are arranged in an orderly manner. The layer of atoms can slide over each other. Thus, pure copper are ductile. 1    (ii) (iii) Oleum 2SO2 + O2  2SO3 Moles of sulphur = 48 / 32 =1.5 Moles of SO2 = moles of sulphur = 1.5 Volume of SO2 = 1.5  24 dm3 = 36 dm3 There are empty spaces in between the atoms. When a pure copper is knocked, atoms slide. Thus, pure copper are malleable.    Zinc. Zinc atoms are of different size, The presence of zinc atoms distrupt the orderly arrangement of copper atoms. This reduce the layer of atoms from sliding.  (c) (i) (ii)  1 1 1 6 1 1 1 1 1 1 Max:5 1 1 1 1 Zinc atom Copper atom 1 Arrangement of atoms – 1; Label - 1 1 Max: 5 Total 20 Question No 3 (a) Mark scheme    (b) Haber process Iron N2 + 3H2 Pure copper Mark 1 1 1+1 2NH3 1 Bronze Tin atom 1+1 Copper atom Bronze is harder than pure copper   56 Tin atoms are of different size The presence of tin atoms distrupt the orderly arrangement of copper Perfect Score & X A –Plus Module/mark scheme 2013 1 1
  57. 57. @Hak cipta BPSBPSK/SBP/2013  atoms. This reduce the layer of atoms from sliding. 1 1 MAX 6 Procedure: 1. Iron nail and steel nail are cleaned using sandpaper. 2. Iron nail is placed into test tube A and steel nail is placed into test tube B. 3. Pour the agar-agar solution mixed with potassium hexacyanoferrate(III) solution into test tubes A and B until it covers the nails. 4. Leave for 1 day. 5. Both test tubes are observed to determine whether there is any blue spots formed or if there are any changes on the nails. 6. The observations are recorded Results: Test tube A B 1 1+ 1 1 1 1 1 1 1 The intensity of blue spots High Low Conclusion: Iron rust faster than steel. TOTAL 20 SET 4 :CHEMICALS FOR CONSUMERS Question No 1 (a) (i) Mark scheme Mark 1 To improve the colour of food (ii) (c) 1 (iii) (b) Absorbs water /inhibits the growth of microorganisms 1. Preservative 2. Flavouring Analgesic To relieve pain Saponification // alkaline hydrolysis 1 1 1 1 1 (i) (ii) (i) (ii) 1+1 Hydrophobic (iii) hydrophilic Soap form scum/insoluble salts in hard water. 1 TOTAL 57 Perfect Score & X A –Plus Module/mark scheme 2013 10
  58. 58. @Hak cipta BPSBPSK/SBP/2013 Question No 2 (a) (b) (i) (ii) Mark scheme Examples of food preservatives and their functions:  Sodium nitrite – slow down the growth of microorganisms in meat  Vinegar – provide an acidic condition that inhibits the growth of microorganisms in pickled foods No // cannot Because aspirin can cause brain and liver damage if given to children with flu or chicken pox. // It causes internal bleeding and ulceration Paracetamol Codeine (iii) 1. If the child is given a overdose of codeine, it may lead to addition. 2. If the child is given paracetamol on a regular basis for a long time, it may cause skin rashes/ blood disorders /acute inflammation of the pancreas. Mark 1+1 1+1 1 1 1 1 1 1 (c) Type of food additives Preservatives Examples Function Sugar, salt 2 Flavourings Monosodium glutamate, spice, garlic Ascorbic acid To slow down the growth of microorganisms To improve and enhance the taste of food To prevent oxidation of food To add or restore the colour in food 2 Antioxidants Dyes/ Colourings Tartrazine Turmeric Disadvantages of any two food additives: Sugar – eating too much can cause obesity, tooth decay and diabetes Salt – may cause high blood pressure, heart attack and stroke. Tartrazine – can worsen the condition of asthma patients - May cause children to be hyperactive MSG – can cause difficult in breathing, headaches and vomiting. 2 1 1 TOTAL Question No 3 (a) (i) 2 Mark scheme  Traditional medicines are derived from plants or animals.  Modern medicines are made by scientists in laboratory and based on substances found in nature. 20 Mark 1 1 (ii) Type Analgesics Antibiotics Psychotherapeutic (iii) 58 Modern medicine Aspirin Paracetamol Codein Penicillin Chloropromazin Caffeina Penicillin Cause allergic reaction, diarrhoea, difficulty breathing and easily bruising Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1 1 1 1 MAX 5 1
  59. 59. @Hak cipta BPSBPSK/SBP/2013 Codeine Cause addiction, drowsiness, trouble sleeping, irregular heartbeat and hallucinations. 1 1 Aspirin Cause brain and liver damage if given to children with flu or chicken pox. Cause internal bleeding and ulceration (b) Hard water contains calcium ions and magnesium ions. Example : sea water 1 1 Procedure 1. 20cm3 of hard water (magnesium sulphate solution) is poured into two separate beakers X and Y. 2. 50 cm3 of soap and detergent solutions are added separately in beaker X and beaker Y. 3. A small piece of cloth with oily stains is dipped into each beaker. 4. Each cloth is washed. 5. The cleansing action of the soap and detergent is observed. Results Beaker X Y Observation The cloth is still dirty. The cloth becomes clean. Conclusion The cleansing action of detergent is more effective than soap in hard water 59 Perfect Score & X A –Plus Module/mark scheme 2013 1 1 1 1 1 1 1 1
  60. 60. @Hak cipta BPSBPSK/SBP/2013 SET 5 :PAPER 3 SET 1 http://cikguadura.wordpress.com/ Rubric 1(a)(i) Score Able to give correct observation 3 Sample answer: Colourless solution formed//Aluminium oxide powder dissolved in nitric acid/sodium hydroxide solution. Rubric 1(a)(ii) Able to give the correct inference. Sample answer Aluminium oxide is soluble in nitric acid/sodium solution//Aluminium oxide shows basic/acidic properties 1(a) (iii) Score 3 hydroxide Rubric Able to give the correct property of aluminium oxide. Score 3 Answer: amphoteric Rubric Able to state the hypothesis correctly. 1(b) Sample answer: When aluminium oxide dissolves in nitric acid, it shows basic properties, when aluminium oxide dissolves in sodium hydroxide solution, shows acidic properties. Rubric Able to state all the variables correctly. 1(c) Answer: Manipulated variable: type of solutions // nitric acid and sodium hydroxide solution Responding variable: solubility of aluminium oxide in acid and alkali//property of aluminium oxide Fixed variable: aluminium oxide Rubric Able to state the operational definition correctly. 1(d) Score 3 Score 3 Score 3 Sample answer. When aluminium oxide solid is added into sodium hydroxide solution, the solid dissolved. 60 Perfect Score & X A –Plus Module/mark scheme 2013
  61. 61. @Hak cipta BPSBPSK/SBP/2013 1(e)(i) 1(e)(ii) Rubric Able to give the correct observations for both experiments. Red litmus paper turns blue Blue litmus paper turns red Score Rubric Able to classify all the oxides correctly. Acidic oxide Basic axide Carbon dioxide Magnesium oxide Phosphorous pentoxide Calcium oxide Score Rubric 2(a) 3 3 Score Able to state the observation Sample Answer: 1. Iron glowed brightly 2. Iron ignited rapidly with bright flame. 3. Iron glowed dimly Rubric Able to state the observation and the way on how to control variable 2(b) Sample Answer : 1. change bromine with chlorine and iodine 2. Ignition or glowing of halogen 3. Use the same quantity of iron wool in each experiment. Rubric Able to state the correct hypothesis by relating the manipulated variable and responding variable 2(c) 3 Score 3 Score Sample Answer : 1. The higher the position of halogen in group 17 the higher the reactivity towards iron. 2. The higher the position of halogen in group 17 the greater the ignition or glowing reaction with iron. Rubric Able to state the inference correctly. 2(d) Score 3 Sample answer: The solid of Iron(lll) bromide formed//Bromine combined with iron //Iron is oxidized by bromine//Bromine is reduced by iron Rubric Able to arrange the three position of halogen based on the reactivity toward iron in ascending order Answer : Iodine. Bromine, Chlorine, 3(a) Score Rubric Able to give the correct arrangement of the metals 2(e) Score 3 Answer: Magnesium, Y, copper 61 Perfect Score & X A –Plus Module/mark scheme 2013 3
  62. 62. @Hak cipta BPSBPSK/SBP/2013 Rubric Able to give the name of metal Y correctly. 3(b) Score 3 Answer: Zinc//Iron//Lead Rubric Able to give the three observations correctly. 3 (c) Answer: 1. Brown solid deposited 2. Blue solution turns light blue 3. Zinc strip becomes pale blue. Rubric Able to give the problem statement correctly. 4(a) Score 3 Score 3 Sample answer: How is the effect of other metals on the rusting of iron when the metals are in contact with iron. Rubric Able to state the three variables correctly. 4(b) Answer: Manipulated variable: Type of metals//Zinc and copper Responding variable: Rusting of iron Fixed variable: iron nail Rubric Able to state the hypothesis correctly. 4(c) Score 3 Score Sample answer: When iron is in contact with a more electropositive metal/zinc, rusting will not occur, when iron is in contact with less electropositive metal/copper, rusting will occur. Rubric Able to list the apparatus and materials needed for the experiment. Apparatus: two test tubes, test-tube rack, Materials: hot agar-agar solution added with phenolphthalein and potassium hexacyanoferrate(III) solution, iron nails, zinc strip, copper strip, sand paper. 4(e) Score Rubric Able to give the procedures correctly 4(d) 3 Score Sample answer: 1. Clean 2 pieces of iron nails, zinc strip and copper strip with sand paper. 2. Coil the iron nails with zinc strip and copper strip each. 3. Put the iron nails into two different test tubes 4. Pour hot agar into each test tube until the iron nail is immersed. 5. Leave the apparatus for about 1 day and record the observations. 62 Perfect Score & X A –Plus Module/mark scheme 2013 3 3
  63. 63. @Hak cipta BPSBPSK/SBP/2013 Rubric Able to tabulate the data correctly 4(f) Answer: Experiment Iron nail coiled with zinc Iron nail coiled with copper Score 2 Observation PAPER 3 SET 2 http://cikguadura.wordpress.com/ Rubric Able to construct the table correctly with the following aspects: 1(a) Experiment I II III Ammeter reading/A 0.0 0.5 0.0 3 Rubric 1(b) Score Score Able to state the inference correctly. 3 Sample answer: Lead(II) bromide can conduct electricity in molten state//Naphthalene/Glucose cannot conduct electricity in molten state Rubric Able to state the type of compound correctly 1(c) Score 3 Answer: ionic compound Rubric Able to state all the three variables correctly: 1(d) Answer: Manipulated variable: type of compound Responding variable: ammeter reading//conductivity of electricity Fixed variable: state of compound//ammeter Rubric Able to state the hypothesis correctly. 1(e) Score 3 Score 3 Sample answer: Molten ionic compound can conduct electricity but molten covalent compound cannot conduct electricity. Rubric Able to state the operational definition correctly. Sample answer: When carbon electrodes are dipped into molten lead(II) bromide, ammeter shows a reading/ammeter needle deflects 1(f) 63 Perfect Score & X A –Plus Module/mark scheme 2013 Score 3
  64. 64. @Hak cipta BPSBPSK/SBP/2013 Rubric Able to explain the difference in conductivity of electricity in Experiment I and II. Sample answer: In Experiment II, molten lead(II) bromide consists of free moving ions that carry the electrical current, In Experiment I molten naphthalene consists of neutral molecules. 1(g) Rubric Able to classify the substances correctly. Answer: Substance can conduct electricity Substance cannot conduct electricity Carbon rod Glacial ethanoic acid Copper(II) sulphate solution Molten polyvinyl chloride 1(h) Rubric Able to give the correct value of the reading. 2(a) Score 3 Score 3 Score 3 Answer: Final burette reading = 40.20 cm3 Initial burette reading = 47.20 cm3 X = 5.0 cm3 Rubric Able to draw the correct graph with the following aspects. 2(b) Score 3 1. X –axis and y-axis with label and unit 2. Correct scale 3. Correct shape of graph Rubric Able to determine the correct mole ratio. 2(c) Answer: 3 Ag+ : Cl1.0 x 5 : 1.0 x 5 1000 1000 0.005 : 0.005 1 : 1 Rubric Able to write the ionic equation correctly. 2(d) Score Score 3 Answer: Ag+ + Cl- → AgCl Rubric Rubric 2(e) Score 3 Score Able to sketch the correct curve: Graph constant at V = 10 cm3 64 Perfect Score & X A –Plus Module/mark scheme 2013
  65. 65. @Hak cipta BPSBPSK/SBP/2013 2(f) Able to classify the salts correctly. Soluble salt Potassium chloride Nickel nitrate Ammonium carbonate Insoluble salt Barium sulphate Rubric Able to state the problem statement correctly. 3. (a) 3 Score 3 Sample answer: What is the effect of size of zinc on the rate of reaction with sulphuric acid? Rubric Able to state the hypothesis correctly 3(b) Score 3 Sample answer: When size of zinc is smaller, the rate of reaction is higher. Rubric Able to state the all the variables correctly 3(c) Score 3 Answer: Manipulated variable: big sized granulated zinc and small sized granulated zinc Responding variable: rate of reaction Fixed variable: volume and concentration of sulphuric acid Rubric Able to list the necessary materials and apparatus needed. 3(d) Sample answer: Materials: big sized granulated zinc, small sized granulated zinc, 0.1 mol dm-3 sulphuric acid, water. Apparatus: burette, conical flask, delivery tube with stopper, basin, retort, basin, weighing balance, stop watch, measuring cylinder. Rubric Able to list procedures for the experiment 3(e) Sample answer. 1. [5-10] g of big sized granulated zinc is weighed and put into the conical flask. 2. Half filled a basin with water. 3. Fill burette with water and invert into the basin and record the initial reading. 4. Measure 50 cm3 of sulphuric acid and pour into the conical flask. 5. Stopper the conical flask and immediately start the stop watch. 6. Record the burette reading every 30 s intervals for 5 minutes. 7. Repeat the experiment by replacing the big sized granulated zinc with small sized granulated zinc. 65 Perfect Score & X A –Plus Module/mark scheme 2013 Score 3 Score 3
  66. 66. @Hak cipta BPSBPSK/SBP/2013 Rubric Able to tabulate the data with the following aspects: 3(f) Score 2 Time/s Burette reading/cm3 Volume of gas/cm3 0 30 60 90 120 150 180 210 PAPER 3 SET 3 1(a) RUBRIC Able to record all the temperature accurately SCORE 3 Sample answer : Experiment 1 Initial temperature = 28.0 Highest temperature = 40.0 Change of temperature = 12.0 Experiment II Initial temperature = 28.0 Highest temperature = 38.0 Change of temperature = 10.0 1(b) RUBRIC Able to construct table accurately with correct title and unit SCORE 3 Sample answer : Temperature Initial temperature of mixture, oC Highest temperature of mixture, oC Change of temperature, oC 1(c) Experiment I 28.0 40.0 12.0 Experiment II 28.0 38.0 10.0 RUBRIC Able to state the relationship between manipulated variable and responding variable with direction correctly Sample answer : Manipulated variable : type of acid Responding variable : heat of neutralisation Direction : ? The reaction between a strong acid and strong alkali produce a greater heat of 66 Perfect Score & X A –Plus Module/mark scheme 2013 SCORE 3
  67. 67. @Hak cipta BPSBPSK/SBP/2013 neutralization than the reaction between a weak acid and strong alkali.// The reaction between hydrochloric acid and sodium hydroxide produce a greater heat of neutralization than the reaction between ethanoic acid and sodium hydroxide// The heat of neutralization between a strong acid and a strong alkali is greater than the heat of neutralization between a weak acid and a strong alkali 1(d) RUBRIC Able to explain with two correct reasons SCORE 3 Sample answer :   This is to enable the change in temperature to be measured. The change of temperature is needed to calculate the heat of neutralization RUBRIC 1(e) Able to state the formula accurately SCORE 3 Sample answer : Change in temperature = Highest temperature of mixture - initial temperature of mixture 1(f) RUBRIC Able to state three observation correctly SCORE 3 Sample answer : 1. A colourless mixture of solution is obtained 2. The vinegar smell of ethanoic acid disappears 3. The polystyrene cup becomes warmer 1(g) RUBRIC Able to state three constant variables correctly SCORE 3 Sample answer : 1. 2. 3. 1(h) The volumes and concentration of the acid and the alkali The type of cup used in the experiment The type of alkali RUBRIC Able to calculate the heat of neutralisation for experiment I and II correctly Sample answer : Experiment I Heat released = mcƟ = 50 x 4.2 x 12 = 2520 J 67 Perfect Score & X A –Plus Module/mark scheme 2013 SCORE 3
  68. 68. @Hak cipta BPSBPSK/SBP/2013 Number of mole of sodium hydroxide = MV = 2.0 x 25/1000 = 0.05 mol 0.05 mole of sodium hydroxide releases 2520 J heat energy 1.0 mole of sodium hydroxide releases = heat released / number of mole = 2520 / 0.05 = 50400 J Heat of neutralisation = - 50.40 kJ/mol Experiment II Heat released = mcƟ = 50 x 4.2 x 10 = 2100 J Number of mole of sodium hydroxide = MV = 2.0 x 25/1000 = 0.05 mol 0.05 mole of sodium hydroxide releases 2100 J heat energy 1.0 mole of sodium hydroxide releases = heat released / number of mole = 2100 / 0.05 = 42000 J Heat of neutralisation = - 42.0 kJ/mol RUBRIC 1(i) Able to write the operational definition for the heat of neutralisation correctly. Able to describe the following criteria (i) (ii) SCORE 3 What should be done What should be observed Sample answer : The heat of neutralization is defined as the temperature rises when one mole of water is produced from reaction between acid and alkali 1(j) RUBRIC SCORE Able to state the relationship between type of acid and value of heat of neutralization and 3 explain the difference correctly. Sample answer : 1. The heat of neutralization of a weak acid by a strong alkali is less than the heat of neutralization of a strong acid by a strong alkali. Explanation : 68 Perfect Score & X A –Plus Module/mark scheme 2013
  69. 69. @Hak cipta BPSBPSK/SBP/2013 2. Experiment I uses a strong acid whereas Experiment II uses a weak acid. 3. During neutralization of a weak acid such as ethanoic acid, small portion of the heat released in experiment II is absorbed to help the dissociation of the ethanoic acid molecules 1(k) RUBRIC Able to predict the temperature change accurately SCORE 3 Sample answer : Lower than 10oC 1(l) RUBRIC Able to classify the acids as strong acid or weak acid. SCORE 3 Sample answer : Heat of neutralization /kJmol-1 Type of acid Ethanoic acid - 50.3 Weak acid Hydrochloric acid - 57.2 Strong acid Methanoic acid - 50.5 Weak acid Name of acid 2(a) RUBRIC Able to record all the temperature accurately one decimal places. SCORE 3 Time 55.0 s at 30oC Time 48.0 s at 35oC Time 42.0 s at 40oC Time 37.0 s at 45oC Time 33.0 s at 50oC 2(b) RUBRIC Able to construct table accurately with correct title and unit SCORE 3 Sample answer : Temperature/oC Time/s 1/time / s-1 30 55.0 0.018 35 48.0 0.021 40 42.0 0.024 RUBRIC 2(c)(i) Able to draw the graph of temperature against 1/time correctly i) Axis x : temperature / 0C and axis y : 1/time /1/s ii) Consistent scale and the graph half of graph paper iii) All the points are transferred correctly iv) Correct curve 69 Perfect Score & X A –Plus Module/mark scheme 2013 45 37.0 0.027 50 33.0 0.030 SCORE 3
  70. 70. @Hak cipta BPSBPSK/SBP/2013 RUBRIC 2(c)(ii) state the relationship between the rate of reaction and temperature correctly SCORE 3 The rate of reaction increases with the increase in temperature RUBRIC 2(d ) Able to predict the time taken SCORE 3 From the graph, when temperature = 55oC, 1/time = 0.033 s-1 Time = 1/0.033 = 30.3 s RUBRIC 2(e)(i) Able to state all variables correctly SCORE 3 Manipulated variable : Temperature of sodium thiosulphate solution Responding variable : Rate of reaction between sodium thiosulphate and hydrochloric acid//time taken for the sign X disappear Constant variable : Concentration and volume of sodium thiosulphate solution and hydrochloric acid RUBRIC 2(e)(ii) Able to state how to manipulate one variable while keeping the other variables constant. Temperature is the manipulated variable. Heating sodium thiosulphate with several different temperatures by remaining the 70 Perfect Score & X A –Plus Module/mark scheme 2013 SCORE 3

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