BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KECEMERLANGAN KEMENTERIAN PELAJARAN MALAYSIA CHEMISTRY TRIAL-EXAM SPM 0 MARKING SCHEME PAPER PAPER PAPER 3
SKEMA KERTAS PENTAKSIRAN DIAGNOSTIK AKADEMIK SBP 0 CHEMISTRY 454/ A 6 C A 7 D 3 C 8 C 4 A 9 D 5 A 30 B 6 D 3 B 7 B 3 C 8 A 33 A 9 B 34 C 0 A 35 C B 36 D A 37 A 3 D 38 B 4 C 39 B 5 D 40 D 6 A 4 D 7 B 4 D 8 D 43 A 9 C 44 B 0 B 45 C C 46 B D 47 C 3 D 48 B 4 B 49 D 5 A 50 C
MARKING SCHEME CHEMISTRY PAPER SECTION A (454/) Question scheme Sub (a) A: Detergents B: Soap r : sodium salt magnesium ion // or calcium ion (b)(i) r : Mg +, Ca + (b)(ii) Soaps are biodegradable Total (c) (d)(i) (d)(ii) Analgesics Psychotherapeutic r: wrong spelling Sugar Aspartame To add / restore the colour of food // To enhance its visual appeal / appearance// more attractive 3 TOTAL 9 Question (a)(i) Magnesium r:mg scheme (a)(ii) Has 3 shell occupied with electrons (b). Proton number of chlorine is higher than magnesium // the number of proton of chlorine atom higher than magnesium atom // total positive charged in nucleus higher than magnesium. The attractive force between the nucleus and the electrons in chlorine atom stronger than magnesium atom // nuclei attraction towards electrons stronger. Sub Total (c) Light bulb // in welding process (d) Argon (e)(i) 3 Cl (g) + Fe (s) FeCl 3(s) (e)(ii) Mass iron (III) chloride = 0.05 x 6 // 8.05 g 3
(e)(ii) Chlorine gas Gas klorin Heat Panaskan Hot iron wool Wul besi panas Total 9 Question scheme Sub 3(a)(i) 6 3(a)(ii) To estimate the age of fossils and artifacts 3(a)(iii) C- // C-3 Total 3 C 6 Or C 6 3(a)(iv) 7 / 6 4 3(b)(i) A: covalent r: covalent bond B: ionic r: ionic bond 3(b)(ii).8.8. 3(b)(iii) High melting point and boiling point // conduct electricity in molten or aqueous solution // soluble in water // insoluble in organic solvent. [Any one] 3(c)(iv) K + Cl KCl. Formula of reactants and products correct. Balance the chemical equation TOTAL 6 0 4
Question 4(a)(i) 4(a)(ii) 4(a)(iii) 4(b)(i) 4(b)(ii) scheme Pipette r: wrong spelling Phenolphthalein // methyl orange r: wrong spelling Phenolphthalein : pink to colourless // methyl orange : yellow to orange KOH + H SO 4 K SO 4 + H O Formula of reactants and products correct Balance the chemical equation. Mol of KOH = (.0)(5) / 000 = 0.05 mol. Molarity of H SO 4 = (0.05)(000) / 0 =.5 mol dm -3 // Molarity of H SO 4 = 0.05 /0.0 =.5 mol dm -3 Or M av M V b a b a ; Molarity of H SO 4 = b r: wrong unit or without unit (5) x 0 =.5 mol dm -3 Sub Total 3 4(c)(i) 4(c)(ii) Hydrogen ion r: H + (symbol ion). no water. contain of molecule // no hydrogen ion, H + 3 TOTAL 0 5
Question scheme Sub Total 5(a) Chemical formula that shows the simplest whole number ratio of atom of each element in the compound. 5(b)(i) Mass of Mg =.4 g Mass of O =.6 g 5(b)(ii) Mole of Mg =.4 / 4 = 0. Mol of O =.6 / 6 = 0. Mg : O = : 5(b)(iii) MgO 5(b)(iv) to allow oxygen enter the crucible 5(b)(v) Repeat the process heating, cooling and weighing until a 6 constant mass is obtained. 5(c) crucible Magnesium ribbon Heat Apparatus set-up correct and functional Labeled : magnesium ribbon, heat 5(d)(i) Zinc oxide // aluminium oxide r: formula 5(d)(ii) Silver is less reactive / not reactive metal TOTAL 6
Question scheme Sub 6(a) Hydrogenation 6(b) C 4 H 8 + H C 4 H 0 6(c)(i) Mole of butene =.8 /56 = 0.05 mol Total 6(c)(ii) 6(d) 6(e)(i) 6(e)(ii) P. mol of butene burns in oxygen gas produce 4 mol of carbon dioxide. Therefore 0.05 mol of butene burns in oxygen gas produce 0. mol of carbon dioxide // C 4 H 8 : CO : 4 0.05 : 0. P. volume of CO = 0. x 4 = 4.8 dm 3 P: cm 3 of but--ene and cm 3 of butane gas are filled in two different test tubes. P : -3 drops of acidified potassium manganate (VII) solution is added to both test tubes. P3: But--ene decolourises purple acidified KMnO 4 Purple acidified KMnO 4 in butane remains unchange. Or P: cm 3 of but--ene and cm 3 of butane gas are filled in two different test tubes. P : -3 drops of bromine water is added to both test tubes. P3: But--ene decolourises brown bromine water. Brown bromine water in butane remains unchange. A: unvulcanised rubber B: vulcanised rubber 3 Rubber type B/ vulcanised rubber is more elastic than rubber type A/ unvulcanised rubber 3 TOTAL Or 5 7
SECTION B Question scheme Σ 7 (a) (i) Ascending order : Cu, Y,X, W (ii) P : Positive terminal : Cu P : Potential difference :.3V P3: Copper is less electropositive // X is more electropositive correct value and unit 4 7 ( b) (i) Experiment I : Pb +, Cl - Experiment II: H +. OH -, Cl r: lead(ii) ion, chloride ion Hydrogen ion, hydroxide ion, chloride ion Experiment Experiment II Product at anode: Chlorine gas Products at cathode: Oxygen gas + Reason: Reason: P: Cl - is discharged P:OH - is selected to be discharged + 7(b)(ii) P: the only anion presence and discharged at anode P:the position of OH - is lower than Cl - in electrochemical series +.Half equation:. Half equation: Cl - Cl + e 4OH - H O + O + 4e P:Correct formula of reactant and product : P: Balance equation P: Correct formula of reactant and product P: Balance equation + + 0 8
7(c) P: Positive terminal: R P: Negative terminal:cu P3: Suitable metal for R : Silver P4: Suitable solution for R : Silver nitate solution 4 Total 0 Question scheme Σ 8(a) P. Smaller size has larger total surface area. P. Absorb heat faster. P3. Bigger size has smaller total surface area. P4. Absorb heat slower 4 (b)(i) Copper(II) sulphate (b)(ii) (b)(iii). Experiment I Rate of reaction = 40/ = 0 cm 3 min -. Experiment II Rate of reaction = 60/ = 30 cm 3 min - P. Rate of reaction in Experiment II is higher than Experiment I. P. Substance X used in Experiment II is a catalyst. P3. Catalyst provided an alternative path with requires a lower activation energy. P4. More particles are able to achieve lower activation energy. P5. Frequency of effective collisions between zinc atoms and hydrogen ions are higher. 5 9
(b)(iv). Label of axes and unit. Correct curve and label Volume of gas / cm 3 Exp II Exp I Time / min (v). Correct formula of reactants and product. Balanced equation Zn + H + Zn + + H (vi). Rate of reaction using sulphuric acid is higher.. Volume of hydrogen gas released is doubled. 3. Sulphuric acid is a diprotic acid. 4. Concentration of hydrogen ions in sulphuric acid is double than that in hydrochloric acid. 4 Total 0 0
Question scheme Σ 9(a) P: Magnesium atom undergoes oxidation P: oxidation number increases from 0 to + P3: Copper (II) ion undergoes reduction P4: oxidation number decreases from + to 0 MAX P5: The reaction involving oxidation and reduction 4 (b) Experiment I L can reduce copper(ii) oxide// L can react with copper(ii) oxide L is more reactive than copper. Experiment II M can reduce copper(ii) oxide//m can react with copper(ii) oxide M is more reactive than copper. Experiment III M cannot reduce L oxide // M cannot react with L oxide. M is less reactive than L//L is more reactive than M. Max 6 5 The arrangement in order of increasing reactivity toward oxygen is Cu, M and L. (c) Procedure : P. Pour cm 3 of potassium bromide solution into a test tube. P. Add cm 3 of chlorine water to the test tube and shake the mixture. P3. Add cm 3 of,, trichloroethane to the test tube and shake the mixture. P4. Record the observation P5. Repeat steps -4 using another halogens and halide solution..
Result : Chlorine water Bromine water Iodine water Potassium chloride X X Potassium bromide X Potassium iodide Ionic equation:. Cl + Br - Cl - + Br. Cl + I - Cl - + I 3. Br + I - Br - + I 0 0
Question scheme Σ Precipitation / double decomposition reaction 0(a) Barium nitrate solution/barium chloride solution [Any sulphate solution] Example: sodium sulphate, potasium sulphate, sulphuric acid Reject : Lead(II) sulphate, calcium sulphate Ba + + SO 4 BaSO 4 4 0(b)(i) Cation : Iron(II) ion / Fe + Anion: Chloride ion / Cl Test for NO 3 0(b)(ii) P: Add cm 3 of dilute sulphuric acid into the test tube follow by cm 3 of iron(ii) sulphate solution. P: Add a few drops of concentrated sulphuric acid P3: carefully and slowly along the side of slanting test tube into the mixture. P: A brown ring is formed. 4 Procedure: 0(c) P. Add zinc nitrate solution to sodium carbonate solution in a beaker. P. Stir the mixture. P3. Filter the white precipitate/solid zinc carbonate formed. P4. Add zinc carbonate to sulphuric acid in a beaker until some zinc carbonate solid no longer dissolve. P5. Filter the mixture. P6. Transfer the filtrate to a evaporating dish. P7. Heat the filtrate(zinc sulphate solution) until saturated// Heat the filtrate to about one-third (/3) of its initial volume P8. Allow the saturated solution to cool at room temperature. P9. Filter the crystals formed. P0. Dry the crystals by pressing it between two sheets of filter papers. 0 Total 0 3