jawapan module perfect score chemistry spm 2013

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@Hak cipta BPSBPSK/SBP/2013 1 Perfect Score & X A Plus Module/mark scheme 2013 BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER JAWAPAN MODUL PERFECT SCORE & X A-PLUS 2013 CHEMISTRY Set 1 Set 2 Set 3 Set 4 Set 5 http://cikguadura.wordpress.com/

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Jawapan Module Perfect Score Chemistry SPM 2013, untuk rujukan guru dan pelajar.

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  • [email protected] cipta BPSBPSK/SBP/2013BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER http://cikguadura.wordpress.com/JAWAPAN MODUL PERFECT SCORE & X A-PLUS 2013CHEMISTRY 1Set Set Set Set SetPerfect Score & X A Plus Module/mark scheme 20131 2 3 4 5

2. @Hak cipta BPSBPSK/SBP/2013MODULE 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 Melting1(ii) Molecule (b)Mark1The heat energy absorbed by the particles is used to overcome the forces of attraction1between the naphthalene molecules / particles. (c)The particles move faster(d)(i)1X : electron1Y : nucleus(ii) Electron (i)(e)1 1W and X(ii) W and X atom have different number of neutrons but same number of protons1+1Atom// Element W and X has different nucleon number but same proton number 10Question 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)Mark2.111(ii) eXXX3p 4nXeX e(d)(i)W and Y1(ii)Atom W and Y have the same number of valence electrons1(iii) To estimate the age of fossils /artefacts.1 102Perfect Score & X A Plus Module/mark scheme 2013 3. @Hak cipta BPSBPSK/SBP/2013Question No. 3Mark SchemeMarks(i)Total number of protons and neutrons in the nucleus of an atom1(ii)35 18 = 171(iii)(a)shows nucleus and three shells occupied with electron1 +1Label 12 proton, 12 neutron(iv) (b)Number of electrons = 2(i)1 ...5 1Liquid(ii)1+1QR ...3(c)Temperature/oC90671+1Time/s 1st mark - Label X and Y axis with correct unit 2 nd mark - Correct shape of curve 103Perfect Score & X A Plus Module/mark scheme 2013 4. @Hak cipta BPSBPSK/SBP/2013a)b)(i)F1(ii)4Atom F has achieve stable/octet electron arrangement // has 8 valence electron12D + 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+11Covalent bond(ii)1 1xE Yx xX Yx xE Yx(iii)1Show coloured ion//formed complex ion//has various oxidation number//act as catalyst(d)Cannot conduct electricity at any state/ low melting and boiling point/....1 115(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+1Chlorine more reactive than bromine Size of chlorine atom is smaller than bromine atom Chlorine atom is easier to receive one electron Al3+ Ionic compound1+1..4 (c)(d) (e)(i) (ii)1 11+1114Perfect Score & X A Plus Module/mark scheme 2013 5. @Hak cipta BPSBPSK/SBP/20136(a) (b)P : liquid (i)Q : solidR : gas1. 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 and1 +1+1 1 1 1 1 1move faster as the temperature increase3. The particles have enough energy to overcome the forces between them and gas is formed (iii)(c)(i)(ii) (iii)(iv)51. 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 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 SupercoolingPerfect Score & X A Plus Module/mark scheme 20131 1 1 1 1 1 1 1 1 1 1 1 20 6. @Hak cipta BPSBPSK/SBP/2013Question 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 // diagramMark 1 +1 1 1 1 1 1 1 1 1RRPRR(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+ + e1 1 1Atom R receives an electron to form ion R-//equation and achieve a stable octet electron arrangement. R+e R-1 1Ion Q+ and ion R- are attracted together by the strong electrostatic forces to form a compound with the formula QR// diagram1--+ Q6Perfect Score & X A Plus Module/mark scheme 2013R 7. @Hak cipta BPSBPSK/SBP/2013Question No 8 (a)(b)Mark schemeMark12 represents the nucleon number. 6 represents the proton number.1 1Able 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--e11p1 12n 2e-e- 3 e-ee-e-oree-11p + 12n e-ee--ee-eee-7Perfect Score & X A Plus Module/mark scheme 2013e-1 11 1 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 // diagram1 1 1 1 1YYWYY(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 11 11YThe 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 only1 1 1 1 1 or 1 1 1 1 1 208Perfect Score & X A Plus Module/mark scheme 2013 9. @Hak cipta BPSBPSK/SBP/20139(a)(i)Q1. 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 electrons1 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 1Melting/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 11 1 1 1Solubility6 7Ionic 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. 209Perfect Score & X A Plus Module/mark scheme 2013 10. @Hak cipta BPSBPSK/SBP/201310(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 ionsTypes of chemical bondsBoiling point and melting pointMolecule 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 molecules221.Correct electron arrangement of 2 ions 2.Correct charges and nuclei are shown2+XXXXXXXXX X X(b)XX X X XX X XXXXXX X2-XXX2+XX X X X1 1XXXX XXY2-3. X atom with an electron arrangement of 2.8.2 donates 2 valence electrons to1achieve the stable octet electron arrangement, 2.8. X2+ ion is formed // X X2+ + 2e-14. Y atom with an electron arrangement of 2.6 accept 2 electrons to achieve the1stable 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 formed10Perfect Score & X A Plus Module/mark scheme 20131 1 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 answer1 1 1 1 1 1 1 1 1 1 1 1 1 20(a)(i)Z : 2.8.7 X : 2.4(ii)11Z 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 etherObservation Colourless solution obtained Solid crystals insoluble in liquid[Conclusion] eg ZY is insoluble in organic solvent/[named organic solvent] but soluble in water.11Perfect Score & X A Plus Module/mark scheme 20131 1 ..2 1 1 1 1 1 1 1 1 1 1 ..10 11 1 1 1 111..7 12. @Hak cipta BPSBPSK/SBP/2013NoExplanation12 (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 equationNo13. (a)(b)(c)ExplanationGlucose // 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/galvanometer12Perfect Score & X A Plus Module/mark scheme 2013Sub 1 1 1 1 1 1 1 1 1 1 1Total5321 1 1 1 1 1 1 Max 5 11 1 Total`1 1 1 1 1 11 1 1 1 1120Total47 13. @Hak cipta BPSBPSK/SBP/2013Procedure 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 Diagram1 1 1 1 111 19Functional diagram Labeled TOTAL20SET 1:CHEMICAL FORMULAE AND EQUATIONSQuestion No 1Mark 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 N212+2(16) = 44H2S2(1)+ 32 = 34H2O2(1)+161Molar mass / gmol-1 14x2 = 28CO2(b)Mark4= 18 1Mole of water= 0.9/ 18 = 0.05Number of molecules(c)= 0.05 x 6.02 x 1023 = 0.3 x 1023 // 3 x 1022Mole of carbon dioxide = 2.2 / 44 = 0.0513Perfect Score & X A Plus Module/mark scheme 20131 1 1 14. @Hak cipta BPSBPSK/SBP/2013= 0.05 x 6.02 x 1023Number of molecules1= 0.3 x 1023 // 3 x 1022Number of molecule is simmilarMass of CO2Number of molecules = 0.1 mol x 6.02 x 10231(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-11 1= 4.4 gNumber of atomstimes until a(ii) CompoundAnhydrous CoCl2 (34.10-31.50)g = 2.60 g 2.60/130 = 0.020.12/0.02 = 61Number of moles(36.26-34.10)g = 2.16 g 2.16/18 = 0.120.02/0.02 = 1Mass/gH2O6Ratio of moles Simplest ratio of moles1 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)111Percentage of water 1 =6(18) x 100% 59 2(35.5) 6(18)= 108 x 100% 238=45.4%1Total 10314concentrated sulphuric acid1zink 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 16Perfect Score & X A Plus Module/mark scheme 20131 15. @Hak cipta BPSBPSK/SBP/2013Mole of copper = 45.15 - 40.35 641= 4.8 = 0.075 64 Empirical formula = CuO1(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/air1(iii)Repeat heating, cooling and weighing processes until a constant mass obtained.1(ii)(iii) (c)1Total4(a) (b)(i) (ii) (i)Pb(NO3)2 AgCl Pb2+ + 2 Cl- PbCl2111 1 1+1Correct 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)12 Pb(NO3)2 2 PbO + 4NO2 + O2CompoundColour of the residue when hot Brown1Colour of the residue when coldPbO1Yellow1GasesColour of the gas releasedNO2Brown1O2Colourless1 Total15Perfect Score & X A Plus Module/mark scheme 201310 16. @Hak cipta BPSBPSK/SBP/2013NoExplanation (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)CuCO31 1 11.Green solid turn Black 2. Lime water becomes cloudy(ii)(c)Mark 1+ 1 1+14+(CH3COO)2Ca + H2O + CO22 1 1CuO + CO21+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 state1(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 g1 1Mass of oxygen is 0.8g Simplest mol ratio : Cu : O = 3.2/64 : 0.8/16 = 1 : 11 1(v)1120Mark(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 C10H81H2SO416Empirical formula C5H4 H2SO41Perfect Score & X A Plus Module/mark scheme 2013 17. @Hak cipta BPSBPSK/SBP/2013(b) Element Percentage (%) Mass/ g MoleSimplest mole ratioCarbon 62.07 62.07 62.07/12 = 5.17 5.17/1.72 =3Hydrogen 10.34 10.34 10.34/1 = 10.34Oxygen 27.59 27.59 27.59/16 = 1.7210.34/1.72 = 611.72/1.72 =11Empirical formula = C3H6O 1n [C3H6O ] = 116 [ 3(12) + 6(1) + 16 ] n = 116 58 n = 116 n= 21 1Molecular 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 =17Magnesium b-a b-a/ 24 xMass/ g a b c Oxygen c-b c-b / 16 y11 1 1MgxOyPerfect Score & X A Plus Module/mark scheme 2013Max 11 Total 20 18. @Hak cipta BPSBPSK/SBP/2013No 7. (a)Sub1. 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 C2H42. 3.ElementCarbonHydrogen40.006.66Ratio of moles40 12 3.336.66 11 6.66 253.33 16 3.331Empirical formula is CH2O n(CH2O) = 180 12n + 2n + 16n = 180 30n = 180 n=6 molecular formula = C6H12O6(ii) (iii)31111 1Magnesium is more reactive than hydrogen//Position of magnesium is above hydrogen in the reactivity series Lead(II) oxide / Stanum oxide / iron oxide / copper(II) oxide1. 2. 3. 4. 5. 6. 7. 8. 9.153.33Number of moles(ii)1 1OxygenPercentage(b)(i)(c)(i)T51Clean [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. DescriptionMass(g)10Crucible + lid Crucible + lid + Mg / Zn / Al Crucible + lid + MgO / ZnO / Al2O3Total18Perfect Score & X A Plus Module/mark scheme 201320 19. @Hak cipta BPSBPSK/SBP/2013SET 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 schemeMarkElectrical 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 Total1 1 1 1 1Chloride 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 + 2eHydrogen gas Gas hidrogen(ii)-(iii)-191 1 1 1 1 111 1+1 1Oxygen gas Gas oksigenSodium sulphate solution Larutan natrium sulfat Functional 1 Label - 11Carbon electrodes Elektrod karbon Aplace 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 TotalPerfect Score & X A Plus Module/mark scheme 20131 1 1 11 111 20. @Hak cipta BPSBPSK/SBP/2013Question 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 airMark 1 1 11 1 1 1 11 1 111Question 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 / PlumbumMark 1 1 1(ii)Pb2+ + 2e Pb1(iii)Brown gas / Gas berwarna perang1(c)(i)hydroxide ion / ion hidroksida1(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)11 920Perfect Score & X A Plus Module/mark scheme 2013 21. @Hak cipta BPSBPSK/SBP/2013Rubric (i) Q, R, S , Cu5(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 toAnode 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 H21+1 1+1Ions selectively discharged ReasonHalf equation2Cl- Cl2 + 2eTotal 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 and21Perfect Score & X A Plus Module/mark scheme 20131+11+1 . 8 20 Mark 1 1 .. 2 1 1 1 .. 3 1 1 .. 2 1 1 .. 2 1 1 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 // NaClDiagram: Functional Label Observation: Anode : brown gas Cathode: grey solidCarbon electrodes Elektrod karbonHeat Panaskan 1 1Note : Observations and half-equations are based on the substance suggested.1 1Half equation: Anode : 2Br- Br2 + 2e Cathode : Pb2+ + 2e Pb1 1Product: Anode : lead Cathode : bromine gas Total Question No 7(a) Sample answer Silver nitrate solutionMark scheme1 1 .. 8 20 Mark1Silver Iron spoonSilver nitrate solutionFunctional 1 Label - 1 Anode : Ag Ag+ + e Cathode : Ag+ + e Ag (b)221 1 1 1 .. 51. metal X is more electropositive than copper // X is higher than copper in the Electrochemical Series Perfect Score & X A Plus Module/mark scheme 20131 23. @Hak cipta BPSBPSK/SBP/20131 1 1 1 .. 5Apparatus Test tube, test tube rack, sand paper1Procedure 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 S1Observation Metal P Q R SMetal ion P / / /Metal ion Q X / /Metal ion R X XMetal ion S X X X/Conclusion The electropositivity of metals increases in the order of P,Q,R,S1 1 1 1 11 11 ..10 TOTAL 20SET 2 :OXIDATION AND REDUCTIONQuestion No 1Mark 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 ion23Perfect Score & X A Plus Module/mark scheme 2013Mark 1 1 11 1 1 1 1 1 1 11 24. @Hak cipta BPSBPSK/SBP/2013Question No 2(a) (b)(c)(d)Mark schemeMarkA 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 droplet1 1 11 1 1 1 1 1 1 1O2e e 2+ Fe Fe +2e Iron11 3(a)1.....4(i)Oxidation number of copper in compound P is + 2 Oxidation number of copper in compound Q is + 11 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 +11 1 1 1.....4(iii) 1 1 1 1.....4(i)X, Z, Y1Y : Copper Z : Lead X : Magnesium241Reaction B: Oxidation number of magnesium changes/increases from 0 to +2 // Oxidation number of zinc changes/decreases from +2 to 0(c)1 1Reaction A: No change in oxidation number(b)Reaction A : not a redox reaction Reaction B : a redox reaction1 1 1.....3Substance that is oxidised Substance that is reduced Oxidizing agent Reducing agent: H2 : CuO : CuO : H2Perfect Score & X A Plus Module/mark scheme 2013 25. @Hak cipta BPSBPSK/SBP/20132Mg + O2 2MgO // 2X + O2 2XO 1 1.....2[Correct formulae of reactants and product] [Balanced equation] TOTAL4(a)(b)(i)20Iron(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 agent1Mg 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 reduction1 1 1(c)1 1 11 1 1At 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 1At 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 agent1 1 1 1 1 20Question No 5(a)Mark scheme 1.Mg/Al/Fe/Pb/ZnMagnesium 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 1Experiment 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) ions1. 2.25MarkPerfect Score & X A Plus Module/mark scheme 20131 1 1 26. @Hak cipta BPSBPSK/SBP/2013Experiment 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 4OH1 1 1 1 Max 3(c)Procedure1. 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 1Observation Mixture Carbon and copper(II)oxide Carbon and zinc oxide Carbon and magnesium oxideObservation The mixture burns brightly. The black powder turns brown The mixture glows dimly. The white powder turns grey. No Changes1+1Explanation 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 Series1 1 1 120 6 Sample answer (a)Magnesium/Aluminium/zinc/iron/lead Magnesium dissolve//The blue colour of copper(II)sulphate solution become paler // brown solid deposited MgMg2+ + 2e Cu2+ + 2e Cu Oxidising agent- Cu2+ ion / copper(II) sulphate Reducing agent- Mg1 1 1 1 1 1..6(b) sample answer Pb(NO3)2Oxidation number:26+2 +5+-22KI+1-1Perfect Score & X A Plus Module/mark scheme 2013Pbl2+2 -1+2KNO3+1 +5 -211 27. @Hak cipta BPSBPSK/SBP/2013no changes of oxidation number of all elements in the compounds of reactants and products.1Neutralization1...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 Labelled1 1Procedure 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 + 2e27Perfect Score & X A Plus Module/mark scheme 20131 1 1 11 1 1 Max : 10 20 28. @Hak cipta BPSBPSK/SBP/2013SET 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 MoleculeMark1 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 properties11. Correct formula of reactants and products 2. Balanced equation1 1(d)(i)Mg + 2HCl MgCl2 (ii)+ H21. Mole of HCl 2. Mole ratio 3. Answer with correct unit Mole HCl =11 1 1// 0.0052 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 + ion10 Mark1(ii)31(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 value1 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+ ion1 1while 3. hydrochloric acid ionises completely in water to produce high concentration of H + ion1(b)(i)281. 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 APerfect Score & X A Plus Module/mark scheme 20131 1 29. @Hak cipta BPSBPSK/SBP/2013(ii)1 11. Mole of NaOH 2. Mass of NaOH with correct unit Mole NaOH =// 0.005Mass NaOH = 0.005 x 40 g // 0.2 g (iii)0.01 x V = 0.002 x 100 //20 cm3 TOTALQuestion No 3 (a) Pink to colourless (b) (c)(i) (ii)Mark schemeKOH1 KNO3 + H2O1 1 1 11. Mole of HNO3 // Substitution 2. Mole ratio 3. Concentration of KOH with Mole HNO3 =Mark 1Potassium nitrate HNO3 +10// 0.010.01 mole HNO3 reacts with 0.01 mole KOH Molarity KOH = (d)(i) (ii)mol dm-3 // 0.4 mol dm-310 cm311. Sulphuric acid is diprotic acid but nitric acid is monoprotic acid // 1 mole of1+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 ion1 110 Mark 1(b)Pb(NO3)21(c)To ensure all the nitric acid reacts completely1(d)(i)1. Correct formula of reactants and products 2. Balanced equation1 12H+ + PbO Pb2+ + H2O29Perfect Score & X A Plus Module/mark scheme 2013 30. @Hak cipta BPSBPSK/SBP/2013(ii)1. Mole of acid 2. Mole ratio 3. Answer with correct unit1 1 1// 0.05Mole 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. TOTALQuestion No 5 (a)(i) Salt W : Copper(II) carbonate Solid X : Copper(II) oxideMark scheme1 1Mark 1 1(ii)(iii)Neutralisation(iv)1 11. 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+ H2O2+Cation : Cu ion // copper(II) ion Anion : Cl- ion // chloride ion Ag++ Cl-1 1AgCl1Double decomposition reaction1 TOTALQuestion No 6 (a)(i) Green (ii) (b)(i)Mark schemeMark 1Double decomposition reaction1Carbon dioxide1(ii)CuCO3 CuO + CO21(iii)1. Functional apparatus 2. Label Copper(II) carbonate1 1Heat (c)(i)30Sulphuric acid // H2SO4 Perfect Score & X A Plus Module/mark scheme 2013Lime water 1 31. @Hak cipta BPSBPSK/SBP/2013(ii)1. Mole of CuCO3 2. Mole ratio 3. Answer with correct unit1 1 1// 0.1Mole 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 sting1 1 11. 2. 3. 4. 5.1 1 1 1 16. 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 value1. Mole of KOH 2. Molarity of KOH and correct unit Mole KOH =1. 2. 3. 4. 5.mol dm-3Mole KOH =1 1 1 1// 1 mol dm-3 1 1 1 1 1Correct formula of reactants Correct formula of products Mole of KOH // Substitution Mole ratio Answer with correct unitHCl + KOH1 1// 0.25Molarity = (ii)1KCl+H2O// 0.0250.025 mole KOH produce 0.025 mole KCl Mass KCl = 0.025 x 74.5 g // 1.86 g TOTAL31Perfect Score & X A Plus Module/mark scheme 201320 32. @Hak cipta BPSBPSK/SBP/2013Question No 8 (a)(i) (ii)(b)(i)(ii) (c)(i)Mark schemeMark 1 11. 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 11. 2. 3. 4.S = zinc nitrate T = zinc oxide U = nitrogen dioxide W = oxygen 2Zn(NO3)2 2ZnO + 4NO2 + O21+1 1 1 11. 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)15 (iii)Mole Ba2+ ion =1// 0.0025Mole SO4 2- ion =// 0.00251Ba2+ ion : SO4 2- ion 0.0025 : 0.0025 // 1 : 1 (iv)Ba2++SO42-1 1 BaSO4 TOTALQuestion 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 value32Perfect Score & X A Plus Module/mark scheme 201320 Mark 1 1 1 1 1 1 1 1 1 1 33. @Hak cipta BPSBPSK/SBP/2013Volumetric 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+ ion1 1 1 1 1 1 1 1 1 120 Mark 1 1 1 1 1 1 1 11. Measure and pour [20-100 cm3] of [0.1-2.0 mol dm-3]zinc nitrate solution into a1beaker 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 + Na2CO3ZnCO3 + 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 1TOTAL33Perfect Score & X A Plus Module/mark scheme 201320 34. @Hak cipta BPSBPSK/SBP/2013SET 3 :RATE OF REACTION http://cikguadura.wordpress.com/Question No 1(a)(i) (ii)(b)(i)Mark scheme Set II1Able 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)MarkAdd 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 scheme1 1 1 1 1 11 1 1 1Mark 1 1CaCO3+ 2HNO3 Ca(NO3) 2+ CO2 + H2O Functional diagram Label(b)1. Correct formulae of reactants and product 2. Balanced equation1 1WaterNitric acidCalcium 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 CO234Perfect Score & X A Plus Module/mark scheme 20131 1 1 35. @Hak cipta BPSBPSK/SBP/2013Maximum 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 10Experiment 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 I111 111 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 I1 11.Correct formulae of reactants and product 2.Balanced equation1 12H2O2 (iii)Mark 1 12H2O + O2EnergyEa Ea2H2O22 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 equation1 1 11 1Zn + 2HCl ZnCl2 + H2 (ii)35No. of mol HCl =50 X 0.5 // 0.025 1000Perfect Score & X A Plus Module/mark scheme 20131 36. @Hak cipta BPSBPSK/SBP/20132 mol HCl 0.025 mol HCl: 1 mol H2 : 0.0125 mol H21 1Volume of H2 = 0.0125 x 24 // 0.3dm3 // 300 cm31. Add excess zinc powder with 12.5 cm3 of 1 mol dm-3hydrochloric acid . 2. At the same temperature(iii)1 1OR1. 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. 4.1Rate 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 double1 1 1 1120 Question No 4 (a)Mark scheme1. Temperature in refrigerator is lower than in cabinet 2. The activity of microorganisme (bacteria) in refrigerator is lower than inMark 1 1refrigerator3. The amount of toxin produced in the refrigerator is less then in the kitchen (b)(i)1. 2. 3. 4. Zn+1cabinet. Correct formula of reactants and products Mol of sulphuric acid Mole ratio Volume and ratio H2SO4------- ZnSO4+ H2No. Of mol H2SO4 = 1 X 50/1000 // 0.0511 mol of H2SO4 0.05 mol of H2SO41: :1 mol of H2 0.05 mol of H2Volume 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 IPerfect Score & X A Plus Module/mark scheme 20131 11 1 11 1 1 1 1 37. @Hak cipta BPSBPSK/SBP/2013Exp 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 higher1 1 1 1 120 Question No 5.(a) (i) (ii)Mark schemeMarkN2 + 3H2 ------- 2NH31+1Temperature : 450 550 C Pressure : 200 300 atm Catalyst : Powdered iron// Iron filling1 1 [ Any two](b)(i)Example of acid Sample answer : Hydrochloric acid / HCl// Sulphuric acid // Nitric acid1 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-111 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 :11 11 1 13-3Pour /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 temperature3711OR1. 2. 3.1Perfect Score & X A Plus Module/mark scheme 20131 1 11 38. @Hak cipta BPSBPSK/SBP/2013Experiment II :11. Pour /measure (50-100) cm3 of (0.1-1 mol dm-3 ) hydrochloric acid . 2. Add excess zinc powder/granules 3. At the same temperature1OR (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 temperature1Experiment II :1 31-31. 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 11 1 1 111 1 1 1 11 11 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)381The number of mol are same // The concentration and volume of acid are same1Perfect Score & X A Plus Module/mark scheme 20131 1 1 1 1 39. @Hak cipta BPSBPSK/SBP/2013Question No 6.(a) (i)Mark schemeMark1. 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-113. rate in 1 st minute higher than 2 nd minute (vice versa) 4. concentration of sulphuric acid / mass of zinc decreases1(iii)All hydrogen ion from acid was completely reacts1(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:1111 1 1 111 3-31.Using a measuring cylinder, 50 cm of 0.2 mol dm sodium thiosulphate solution is measured and poured into a conical flask.12.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.14.The sulphuric acid is poured immediately and carefully into the conical flask. At the same time, the stop watch is started15.The mixture in a conical flask is swirled.16.The X mark is observed vertically from the top of the conical flask through the solution.17.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 higher39Perfect Score & X A Plus Module/mark scheme 201311 40. @Hak cipta BPSBPSK/SBP/2013(Temperature) 1Materials: 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.1Apparatus: 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.12.The conical flask is placed on top of a piece of white paper marked X at the centre.13.5 cm3 of 1.0 mol dm-3 sulphuric acid is measured using another measuring cylinder.14.The sulphuric acid is poured immediately and carefully into the conical flask. At the same time, the stop watch is atarted15.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.17.The stopwatch is stopped once the X mark disappears from view.18.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 higher11SET 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 schemeBlue 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 molMark1 1 1// 0.05Perfect Score & X A Plus Module/mark scheme 20131 1 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 1Sample answer Energy Zn + CuSO4 //Zn + Cu2+ H = - 25.2 kJmol-1 ZnSO4 + Cu //Zn2+(e)(i)3C(ii)+ CuNumber of mole copper displaced is half Heat released is half / 1260 J // 630 J 21 1 1 TOTALQuestion 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)4112Perfect Score & X A Plus Module/mark scheme 2013Mark 1 1 1 111 1 1 1 1 42. @Hak cipta BPSBPSK/SBP/2013Energy 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 scheme1 1 1Mark(i)Ethanol1(ii)1260 kJ of heat energy is released when one mole of ethanol is burnt completely in excess oxygen1(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 kJ1(b)(ii)( c)(d)mc = 6.3 kJ mc = 6.3 x 1000 = 6300/ 200 x 4.2 = 7.5 0 C11 1Heat is lost to the surrounding // Heat is absorbed by the apparatus or containers // Incomplete combustion of alcohol1(i)EnergyC2 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 written42Perfect Score & X A Plus Module/mark scheme 20131 1 1 43. @Hak cipta BPSBPSK/SBP/2013(ii)1 11. Label 2. Functional (e)(i) (ii)- 2656 kJmol-1 // 2500-2700 kJmol-111. 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 molecules1 1 1Total marksQuestion No 4(a)Mark scheme(i) Characteristic Change in temperature Type of chemical reaction Energy content of reactants and products(ii)43MarkDiagram 4.1 IncreaseDiagram 4.2 DecreaseExothermic reactionEndothermic reactionThe total energy content of the reactants more than the energy content of the productsThe total energy content of the reactants less than the energy content of the productsAmount 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 = mc = 2000 (50)(4.2) = 9.5 oCPerfect Score & X A Plus Module/mark scheme 20131 11+11+111 1 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 = mc = (100)(4.2)(42.2 30.2) = 5040 J / 5.04 kJ(i)111 11Number 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 H = - 50.4 kJ mol-111 1Temperature 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 used1Total marks(ii)20Question No 5(a)Mark scheme1 1 1Mark(i)Neutralisation//Exothermic reaction1(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 J1(iii)(b)(i)(ii)442865 = 100 X 4.2 X 0 = 2865 420Perfect Score & X A Plus Module/mark scheme 20131 1 1 11 1 1 1 1 1 45. @Hak cipta BPSBPSK/SBP/2013= 6.8 oC ( correct unit) (iii)(c )11. Some of heat is lost to the sorrounding 2. Heat is absorbed by polystyrene cupA 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 reactant1 1B 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 product11111 TOTAL 6(a)(i)20energyZn + CuSO4 H = -152 kJmol-1 ZnSO4 + Cu 1. Y-axes : energy 2. Two different level of energy (ii)(b)(c)451. reactants have more energy // products have less energy 2.energy is released during the experiment // this is exothermic reactionNo. 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 functionalPerfect Score & X A Plus Module/mark scheme 20131 1 1 1 1 1 1 1 1 1 1 1 1 1..3 46. @Hak cipta BPSBPSK/SBP/20131. (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 t2Mass of lamp after burning Mass of lamp before burning Mass of lamp ethanol burnt, m= = =m2 m1 m1 - m2 = m-..4t1 = ..1Calculation : 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 marks46Perfect Score & X A Plus Module/mark scheme 2013..320 47. @Hak cipta BPSBPSK/SBP/20137Question No (a) (i)Mark scheme Heat change = mc = (25+25)(4.2)(33-29) = 445 JMark 1Heat of precipitation of AgCl = - 445 / 0.0125 = -35600 J mol-1 // 35.6 kJ mol-11Energy AgNO3 + NaClH = -35.6 kJ mol-1 AgCl + NaNO3** Accept ionic equation1. 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 34Max 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.471 1Perfect Score & X A Plus Module/mark scheme 20131 11 1 1 1 1 48. @Hak cipta BPSBPSK/SBP/2013Tabulation of data: 1 2 2 - 1 ....1Initial 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 mol1Heat change = mc(2 - 1) = x J Heat of displacement = x / 0.005 kJ mol-1 = y kJ mol-1.1 TOTAL 20SET 4 :CARBON COMPOUNDS http://cikguadura.wordpress.com/1Question No (a)Mark schemeMark 1OrC3H7OH + 9/2O2 3CO2 + 4H2O1(i)Sweet/ pleasant smell /// fruity smell1(ii)Methanoic acid1(b) (c)(iii)OH HHC OCCCH (d)1+1HHHH(i)Oxidation1(ii)Orange colour of acidified potassium dichromate (VI) solution turns green1(iii)C3H7OH + 2[O] C2H5COOH + H2O1(e)C3H7OH (ii) C3H6 + H2O propanolpropene1+148Perfect Score & X A Plus Module/mark scheme 2013 49. @Hak cipta BPSBPSK/SBP/20132Question No (a) (i) (ii)Mark scheme Fermentation EthanolMark 1 1H H(iii)HC C H1H OH (b) (c)(i) (ii)C2H5OH + 3O2 2CO2 + 3H2O Ethene H C HH -C H1+1 1nPurple to colourless1 1(i)Ethyl ethanoate1(ii)CH3COOH + C2H5OH CH3COOC2H5 + H2O(d) (e)Mark schemeQuestion No 31+1 Mark(a) Characteristics Same general formulaExplanation CnH2n + 1OH1+1successive member is different from each other by CH2Relative atomic mass is different by 141+1Gradual change in physical properties // Melting / boiling point increaseNumber of carbon atom per molecules increase // size of molecule increase1+1Similar chemical properties // oxidation produce carboxylic acidHave same chemical/similar functional group1+1Can be prepared by similar method // can be prepared by hydration of alkeneHave same chemical properties // have same functional group1+1(b) (i) (CH2O)n = 60 (12 + 2 + 16)n = 60 n=2 C2H4O2 (ii) Carboxylic acid React with carbonate to produce carbon dioxide49Perfect Score & X A Plus Module/mark scheme 20131 1 1 1 50. @Hak cipta BPSBPSK/SBP/2013(iii) 2 CH3COOH + CaCO3 (CH3COO)2Ca + H2O+ CO2Correct formula of reactants and products Balanced equation1 1(c) Compound The number of carbon atomP 2Q 21 1The number of hydrogen atom4 6 number of hydrogen atom Q is higherType of covalent bond between // carbon/ Type of hydrocarbon Type of homologous series // // Name of compoundDouble bond / / UnsaturatedSingle bond/ / Saturated1Alkene// Ethene //Alkane // Ethane1General formula// Molecular formula of the compoundCnH2n // C2H4CnH2n+2 // C2H61 Max 420 Question No (a) (i) 4 (ii)Mark schemeMark 114.3 % Element Mass/ % No. of moles Ratio of moles/ Simplest ratioC 85.7 85.7 = 7.14 12H 14.3 14.3 = 14.3 17.14 = 1 7.141 114.3 = 2 7.141Empirical formula = CH2 RMM of (CH2)n [(12 + 1(2)]n 14n= 56 .............1 = 56 = 56 n = 56 14 = 4 ..1 Molecular formula : C4H8 ..16 max 5(iii)1+11+1 But-2-ene2-methylpropene[any 2]50Perfect Score & X A Plus Module/mark scheme 2013But-1-ene Max 4 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 1Starch Protein / natural silk H H CH3 H I I I I C = C C = C I I H H1 1..22-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 TOTALQuestion No (a) (i) 5Mark scheme1 1 1 20 MarkHydrocarbonGeneral formulacovalentalkaneCnH2n+23B(iii)Homologous seriesA(ii)Type of bondcovalentalkeneCnH2n3Carbon dioxide 2C4H10 + 13O2 8CO2 + 10H2O [Chemical formulae of reactants and products] [Balanced]1Hydrocarbon B. Hydrocarbon B is an unsaturated hydrocarbon which react with bromine. Hydrocarbon A is a saturated hydrocarbon which do not react with bromine.11 11 151Perfect Score & X A Plus Module/mark scheme 2013 52. @Hak cipta BPSBPSK/SBP/2013(iv)1 1Hydrocarbon 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 %1Carboxylic acid X : 1Propanoic acid1Alcohol Y:1 Ethanol1 TOTAL6Question No (a) (i)201. 2. 3. 4. 5.(iii)Ammonia aqueous solution contains hydroxide ions Hydroxide ions neutralise hydrogen ions (acid) produced by activities of bacteria1 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)52Mark 1 1(ii)(b)Mark scheme X - any acid methanoic acid Y - any alkali ammonia aqueous solutionProcedure: 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 gentlyMethanoic acid contains hydrogen ions Hydrogen ions neutralise the negative charges of protein membrane Rubber particles collide, Protein membrane breaks Rubber polymers combine togetherPerfect Score & X A Plus Module/mark scheme 20131 1 1 1 53. @Hak cipta BPSBPSK/SBP/2013 6.Using test tube collect the gas given off6 max 5Diagram: Glass wool soaked with ethanolHeatPorcelain chipsHeatWater[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) 71 1 20Mark scheme Carbon dioxide/ CO2 and water/ H2O Any one correct chemical equation Example 2C4H10 + 13O2 8CO2 + Chemical formula of reactants Balanced1 1Mark 110H2O 1 1(b)Compound B & Compound D Same molecular formula / C4H8 Different structural formula1 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 colourless1 1 1 1(d) (i)Any members of carboxylic acid and correct ester Example [Methanoic acid] [Propylmethanoate]1 1 1153Perfect Score & X A Plus Module/mark scheme 2013 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 TOTAL20Mark schemeMarkQuestion No 8(a)HHHCHH CCCHCHBut-2-eneH CCCHH H2-methylpropeneH1+1HH H1 1 1 1 1 11+1H (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 ester1 11 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:54Perfect Score & X A Plus Module/mark scheme 20131 1 1 1 1 1 1 11 55. @Hak cipta BPSBPSK/SBP/20131. Pour about [2 -5 cm3] of P into a test tube. 2. Add 4-5 drops of bromine water / acidified potassium manganate(VII) solution1 1 1and shake.3. Observe and record any changes. 4. Repeat steps 1 to 3 by replacing P with Q1 1Observation: P : Brown/ Purple colour remains unchanged. Q : Brown/ Purple colours decolourise / turn colourless.Max 6 20SET 4 :MANUFACTURED SUBSTANCE IN INDUSTRY http://cikguadura.wordpress.com/1Question No (a) (i)Mark schemeMark 1Contact process(ii)1(iii)Vanadium(V) oxide, 450 oC - 500oC1(iv)Ammonium sulphate1(v) (i)(b)Ammonia2NH3 + H2SO4 (NH4)2SO4 Composite material1+1 1(ii)Tin atomCorrect arrangement Correct label1 1Copper atom (iii)nC2H3Cl --( C2H3Cl )n1(iv)It has low thermal expansion coefficient // resistant to thermal shock1TOTAL Question No 2(a) (i) (ii)5511Mark 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 2013Mark 134 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 dm3There 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 1 161 11 1 1 1 Max:5 1 1 1 1Zinc atomCopper atom1 Arrangement of atoms 1; Label - 1 1 Max: 5 Total 20Question No 3 (a)Mark scheme (b)Haber process Iron N2 + 3H2 Pure copperMark 1 1 1+12NH31Bronze Tin atom1+1Copper atomBronze is harder than pure copper 56Tin atoms are of different size The presence of tin atoms distrupt the orderly arrangement of copperPerfect Score & X A Plus Module/mark scheme 20131 1 57. @Hak cipta BPSBPSK/SBP/2013atoms. This reduce the layer of atoms from sliding.1 1 MAX 6Procedure: 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 B1 1+ 1 1 1 11 1 1The intensity of blue spots High LowConclusion: Iron rust faster than steel. TOTAL20SET 4 :CHEMICALS FOR CONSUMERSQuestion No 1 (a) (i)Mark schemeMark 1To 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 hydrolysis1 1 1 1 1(i) (ii) (i) (ii)1+1Hydrophobic (iii)hydrophilicSoap form scum/insoluble salts in hard water.1 TOTAL57Perfect Score & X A Plus Module/mark scheme 201310 58. @Hak cipta BPSBPSK/SBP/2013Question 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 PreservativesExamplesFunctionSugar, salt2FlavouringsMonosodium glutamate, spice, garlic Ascorbic acidTo 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 food2Antioxidants Dyes/ ColouringsTartrazine 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.21 1 TOTALQuestion No 3 (a) (i)2Mark 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) TypeAnalgesics Antibiotics Psychotherapeutic(iii)58Modern medicine Aspirin Paracetamol Codein Penicillin Chloropromazin CaffeinaPenicillin Cause allergic reaction, diarrhoea, difficulty breathing and easily bruisingPerfect Score & X A Plus Module/mark scheme 20131 1 1 1 1 1 MAX 5 1 59. @Hak cipta BPSBPSK/SBP/2013Codeine Cause addiction, drowsiness, trouble sleeping, irregular heartbeat and hallucinations.11 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 water1 1Procedure 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 YObservation The cloth is still dirty. The cloth becomes clean.Conclusion The cleansing action of detergent is more effective than soap in hard water59Perfect Score & X A Plus Module/mark scheme 20131 1 1 1 11 11 60. @Hak cipta BPSBPSK/SBP/2013SET 5 :PAPER 3 SET 1 http://cikguadura.wordpress.com/Rubric 1(a)(i)ScoreAble 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 properties1(a) (iii)Score3 hydroxideRubric Able to give the correct property of aluminium oxide.Score 3Answer: amphotericRubric 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 oxideRubric Able to state the operational definition correctly.1(d)Score3Score3Score 3Sample answer. When aluminium oxide solid is added into sodium hydroxide solution, the solid dissolved.60Perfect Score & X A Plus Module/mark scheme 2013 61. @Hak cipta BPSBPSK/SBP/20131(e)(i)1(e)(ii)Rubric Able to give the correct observations for both experiments. Red litmus paper turns blue Blue litmus paper turns redScoreRubric Able to classify all the oxides correctly. Acidic oxide Basic axide Carbon dioxide Magnesium oxide Phosphorous pentoxide Calcium oxideScoreRubric 2(a)33ScoreAble 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 variable2(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 variable2(c)3Score3ScoreSample 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 3Sample 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)ScoreRubric Able to give the correct arrangement of the metals2(e)Score 3Answer: Magnesium, Y, copper61Perfect Score & X A Plus Module/mark scheme 20133 62. @Hak cipta BPSBPSK/SBP/2013Rubric Able to give the name of metal Y correctly.3(b)Score 3Answer: 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)Score3Score 3Sample 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)Score3ScoreSample 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)ScoreRubric Able to give the procedures correctly4(d)3ScoreSample 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.62Perfect Score & X A Plus Module/mark scheme 201333 63. @Hak cipta BPSBPSK/SBP/2013Rubric Able to tabulate the data correctly4(f)Answer: Experiment Iron nail coiled with zinc Iron nail coiled with copperScore2 ObservationPAPER 3 SET 2 http://cikguadura.wordpress.com/Rubric Able to construct the table correctly with the following aspects:1(a)Experiment I II IIIAmmeter reading/A 0.0 0.5 0.03Rubric 1(b)ScoreScoreAble 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 correctly1(c)Score 3Answer: 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)Score3Score 3Sample 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 deflects1(f)63Perfect Score & X A Plus Module/mark scheme 2013Score 3 64. @Hak cipta BPSBPSK/SBP/2013Rubric 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 chloride1(h)Rubric Able to give the correct value of the reading.2(a)Score3Score3Score 3Answer: 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 31. 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:3Ag+ : 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)ScoreScore 3Answer: Ag+ + Cl- AgCl RubricRubric2(e)Score 3 ScoreAble to sketch the correct curve: Graph constant at V = 10 cm364Perfect Score & X A Plus Module/mark scheme 2013 65. @Hak cipta BPSBPSK/SBP/20132(f)Able to classify the salts correctly. Soluble salt Potassium chloride Nickel nitrate Ammonium carbonateInsoluble salt Barium sulphateRubric Able to state the problem statement correctly.3. (a)3Score 3Sample answer: What is the effect of size of zinc on the rate of reaction with sulphuric acid?Rubric Able to state the hypothesis correctly3(b)Score 3Sample answer: When size of zinc is smaller, the rate of reaction is higher. Rubric Able to state the all the variables correctly3(c)Score 3Answer: 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 experiment3(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.65Perfect Score & X A Plus Module/mark scheme 2013Score3Score3 66. @Hak cipta BPSBPSK/SBP/2013Rubric Able to tabulate the data with the following aspects:3(f)Score 2Time/s Burette reading/cm3 Volume of gas/cm30306090120150180210PAPER 3 SET 31(a)RUBRIC Able to record all the temperature accuratelySCORE 3Sample 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 = 10.01(b)RUBRIC Able to construct table accurately with correct title and unitSCORE 3Sample answer : Temperature Initial temperature of mixture, oC Highest temperature of mixture, oC Change of temperature, oC1(c)Experiment I 28.0 40.0 12.0Experiment II 28.0 38.0 10.0RUBRIC 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 of66Perfect Score & X A Plus Module/mark scheme 2013SCORE 3 67. @Hak cipta BPSBPSK/SBP/2013neutralization 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 alkali1(d)RUBRIC Able to explain with two correct reasonsSCORE 3Sample answer : This is to enable the change in temperature to be measured. The change of temperature is needed to calculate the heat of neutralizationRUBRIC 1(e)Able to state the formula accuratelySCORE 3Sample answer : Change in temperature = Highest temperature of mixture - initial temperature of mixture1(f)RUBRIC Able to state three observation correctlySCORE 3Sample answer : 1. A colourless mixture of solution is obtained 2. The vinegar smell of ethanoic acid disappears 3. The polystyrene cup becomes warmer1(g)RUBRIC Able to state three constant variables correctlySCORE 3Sample 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 alkaliRUBRIC Able to calculate the heat of neutralisation for experiment I and II correctly Sample answer : Experiment I Heat released = mc = 50 x 4.2 x 12 = 2520 J67Perfect Score & X A Plus Module/mark scheme 2013SCORE 3 68. @Hak cipta BPSBPSK/SBP/2013Number 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 = mc = 50 x 4.2 x 10 = 2100 JNumber of mole of sodium hydroxide = MV = 2.0 x 25/