ESSAY QUESTION ANSWERS (2003-2008) Answer To Score Chemistry Essay Question Answers FORM 4 CHAPTER 2 THE STRUCTURE OF THE ATOM & CHAPTER 3 CHEMICAL FORMULAE AND EQUATION ANSWER 1 SPM 2004/P2/Q1(SECTION B) a) i) The electron configuration of atom X: 2.8.1 The electron configuration of atom Y: 2.8.7 ii) The number of neutron in Z = 6 Isotope of 13Z or 14 Z 6 6 b) 1. Atom X and atom Y will form ionic bond. The electron configuration of atom X is 2.8.1 and the electron configuration of atom Y is 2.8.7. 2. To attain the stable electron configuration with 8 electrons in the valence electron shell, atom X donates one electron to form a positive ion. X X+ + e 3. Atom Y will receive the electron to form Y- ions, and attain the stable electron configuration with 8 electrons in the valence electron shell. Y + e Y+ 4. The X ion will attract Y ion with strong electrostatic force and form an ionic compound with the formula XY. 6. Element Y and Z will form covalent compund. To attain stable electron configuration with 8 electrons in the valence electron shell, atom Z shares electrons with atom Y. 7. One atom of Z contributes 4 electrons and each atom of Y contributes one electron. 8. Atom Z shares electrons with four atoms Y to form a covalent compound with the formula ZY4. 2 Answer To Score Chemistry c) Essay Question Answers The apparatus was set-up as shown in the diagrams below: Figure (a) For molten compound: 1. The crucible was filled with XY powder until 2/3 full. 2. The crucible with its content was then heated strongly untill all the XY powder melts. 3. After that, two carbon electrodes, were dipped into the molten XY and the switch was switched on. 4. The ammeter shows a reading when XY powder is melted. Figure (b) For aqueous solution: 1. Water was filled into the beaker. 2. The XY powder was added into the beaker and dissolved in the water. 3. After that, two carbon rods, acting as electrodes, were immersed into the solution of XY and the switch was switched on. 4. The bulb lights up. SPM 2005/P2/Q10 SECTION C (a) i) Iodine-131 for cure cancer of thyroid glands ii) Carbon-14 is used to determine the age of a fossil or ancient artifacts b) The electronic arrangement of P is 2.4 whereas the electronic arrangement of Q is 2.6 Q is located in Group 16 because it has 6 valence electrons Q is located in Period 2 because it has 2 electrons shells filled with electrons c) Refer Answer Form4 Topic 5 Chemical Bond 2 3 11p + 12n (d) Na 4 SPM 2007/P2 / Q8 SECTION B a 1. The nucleus consists of 11 protons which are positively charged and 12 neutrons are neutral. electrons move around the nucleus 4. Nucleus is at the centre 2. element X is in liquid state 2. One shell is filled with electron b(i) Proton number Number of electron Chemical properties Number of neutron Nucleon number Physical properties (ii) c) 2 1 Atom of Diagram 4 1 1 Similar 1 2 Different Another atom 1 1 Similar 2 3 Different H State any ten of following information: At time t0 – t1 : 1. the electron is negatively charged 5. the particles are closed to each other 3. nucleus atom contains 1 proton and 1 neutron 3.Answer To Score Chemistry 3 SPM 2006/ P2/ Q9 SECTION C (a) Subatomic particles Relative mass Proton 1 Electron 1/1837 Neutron 1 Essay Question Answers Relative charge +1 -1 0 Row 1 Row 2 Row 3 Row 4 Note: state row 1 and any two from row 2/ 3/ 4 (b) 23 11 Na or 39 19 K or 86 37 Rb (c) The atom consists of two parts: the centre part called nucleus and the outer part called electron cloud. [ if answer in (b) is Na] The electron cloud consists of 11 electrons which are negatively charged and move around nucleus in orbits. the particles arrangement is not orderly 4. the kinetic energy increases 4 . There is an electroststic force between nucleus and electrons. Sample answer an atom of sodium. 1 3. Chlorine gas is poisonous.2.2 has four electron shell 2. the particle arrangement is not orderly 8. element X is is in liquid ad gaseous state 6. 2Fe(s) + 3Cl2(g) 2FeCl3(s) Part H The excess chlorine gas will flow into sodium hydroxide solution to produce sodium chloride. Therefore atom in Diagram 8. the kinetic energy is increases CHAPTER 4: PERIODIC TABLE OF THE ELEMENTS 1 SPM 2006/P2/Q8(SECTION B) a) i) The electron configuration is 2.1 is more reactive compared to the atom in Diagram 8. The attractive forces between the nucleus and the valence electron becomes weaker. the kinetic energy is constant At time t2 – t3 : 9. the particles are far away 11.2.1 is more electronegative compare to atom in Diagram 8. 2.2 is greater than atom in Diagram 8. element X is in gaseous state 10. The atom in Diagram 8. ii) Part G Chlorine gas will react with iron wool to produce iron (III) chloride solid. ii) Cl2(g) + 2NaOH(aq) b) NaCl(aq) + NaOCl(aq) + H2O(l) 1. The atom in Diagram 8. The distance between the nucleus and the valence electrons of atom in Diagram 8. sodium chlorate (I) and water. Less reactive i) 1. 5. Make sure that the apparatus are connected tightly to prevent leakage of chlorine gas.2.Answer To Score Chemistry Essay Question Answers At time t1 – t2 : 5. 6.8. 4.7. The atom in Diagram 8. The element is chlorine. the particle arrangement are not orderly 12. Concentrated acid is corrosive and the experiment must be conducted in a fume chamber. Cl2(g) + 2NaOH(aq) NaCl(aq) + NaOCl(aq) + H2O(l) c) d) 5 . some particles are closed to each other and some are far apart 7.1 has a stronger attraction towards electron compared to the atom in Diagram 8. Dry the oil on the surface of the lithium with filter paper. distilled water. size of alkali metal c) Statement of the hypothesis : When going down Group 1. Place the lithium slowly onto the water surface in a water trough. 3. When the reactions stop. b) All the variables : Manipulated variable : Different type of alkali metal Responding variable : Reactivity of metals Fixed variable : Water. filter paper. test the solution produced with red litmus paper. alkali metals become more reactive in their reaction with water.Answer To Score Chemistry 1 SPM 2008/P3/Q2 a) Aim of the experiment : Essay Question Answers To investigate the reactivity of lithium. Repeat steps 1-5 using sodium and potassium to replace lithium one by one. sodium and potassium. 2. 6. Record the observation on the table. Cut a small piece of lithium using a knife and forceps. small knife. 5. 4. sodium and potassium with water. f) Tabulation of data: Alkali Metals Lithium Sodium Potassium Observation 6 . forceps e) Procedure of the experiment : 1. red litmus paper Apparatus : Water troughs. d) List of substances and apparatus : Substances : Small pieces of lithium. To attain stable electron configuration with 8 electrons in the valence electron shell.8.8.ions. Element Y and Z will form covalent compund. 7. The X ion will attract Y ion with strong electrostatic force and form an ionic compound with the formula XY. X X+ + e 3. The electron configuration of atom X is 2. Atom Z shares electrons with four atoms Y to form a covalent compound with the formula ZY4. 8. Y + e Y+ 4. atom Z shares electrons with atom Y.1 The electron configuration of atom Y: 2. To attain the stable electron configuration with 8 electrons in the valence electron shell. One atom of Z contributes 4 electrons and each atom of Y contributes one electron. Atom Y will receive the electron to form Y. 6. atom X donates one electron to form a positive ion.7. 7 .Answer To Score Chemistry CHAPTER 5 : CHEMICAL BOND 1 SPM 2004/P2/Q1(SECTION B) a) Essay Question Answers b) i) The electron configuration of atom X: 2. 2.8.7 ii) The number of neutron in Z = 6 Isotope of 13Z or 14 Z 6 6 1.8. and attain the stable electron configuration with 8 electrons in the valence electron shell.1 and the electron configuration of atom Y is 2. Atom X and atom Y will form ionic bond. Water was filled into the beaker. After that. The crucible with its content was then heated strongly untill all the XY powder melts. The bulb lights up. 6. Figure (b) For aqueous solution: 1. were dipped into the molten XY and the switch was switched on. After that. 4.Answer To Score Chemistry c) Essay Question Answers The apparatus was set-up as shown in the diagrams below: Figure (a) For molten compound: 5. 7. were immersed into the solution of XY and the switch was switched on. The crucible was filled with XY powder until 2/3 full. 8. The XY powder was added into the beaker and dissolved in the water. acting as electrodes. The ammeter shows a reading when XY powder is melted. two carbon rods. 2. 8 . 3. two carbon electrodes. 3. Atom Y donates 2 electron to two atoms of X to form Y2+ ion to attain a stable octet electron arrangement.4 whereas the electron arrangement of Q is 2. X + e X2+ 7. Each atom W contributes four valence electrons whereas each atom X contributes one valence electron. Y Y2+ + 2e 5. 6. Four atoms of X share electrons with one atom W. 1. Y and X ions are attracted by strong electrostatic forces to produce an ionic compound YX2. 2. There are two types of chemical bonds. The number of valence electrons of X and Y are 7 and 2 respectively. 10. The electron arrangement of P is 2. 9. 8. W and X tend to share valence electrons to attain a stable octet electron arrangement.6. 11. 9 . Q is located in Group 16 because it has 6 valence electrons 3. Q is located in Period 2 because it has 2 electron shells filled with electrons.Answer To Score Chemistry 2 SPM 2005/P2/Q10(SECTION C) a) b) Essay Question Answers c) Iodine-131 is used to cure cancer of thyroid glands Carbon-14 is used to determine the age of a fossil or ancient artifacts 1. The number of valence electrons of atom W and atom X are 4 and 7 respectively. 4. ionic bond and covalent bond. 2.ions to attain a stable octet electron arrangement. Atom X receives 1 electron from atom Y to form X. A covalent compound WX4 is formed. H+ . silver nitrate solution Procedures of the experiment: Key is made the cathode. OHZn → Zn2+ + 2e Cu2+ + 2e → Cu Zinc dissolves. Half equations Observation Brown solid forms. Silver is made the anode. SO42-. Electrolyte used is silver nitrate solution in a beaker. SO42-. Ions in electrolyte Cu2+. H+ . The silver anode and the key are immersed into the electroyte and connected to a battery. OHAt anode: Cu → Cu2+ + 2e At cathode: Cu2+ + 2e → Cu At anode: Copper dissolves. (c) To electroplate a key with silver: Chemicals required: silver. At cathode: Brown solid forms.Answer To Score Chemistry Essay Question Answers F4 Chapter 6 Electrochemistry F5 Chapter 3 Oxidation & Reduction 1 SPM 2003/P2/Section C/Q3 (a) Hydrogen gas. 2H+ + 2e → H2 (b) Cell P Type of cell Electolytic cell Energy change electrical → chemical Name of electrode Both electrodes are copper Cell Q Chemical/voltaic cell chemical → electrical Copper = positive electrode zinc = negative elctrode Cu2+. Diagram showing the set-up of the apparatus 10 . Answer To Score Chemistry Chemical equation involved in the reaction: At anode: Ag → Ag+ + e At cathode: Ag+ + e → Ag Essay Question Answers Observation: At anode: Silver dissolves.ions. Electrons flow from iron to P because iron is more electropositive than P. Electrons flow from Q to Fe because Q is more electropositive than Fe. Blue solution shows the presence of Fe2+ ions. Iron is oxidised to Fe2+ ions. E. iron. The pink spots show the presence of OH. At cathode: Shiny solid forms. No change in the colour of the solution.g: Mg + Cu2+ → Mg2+ + Cu (b) (i) In Experiment I. In Experiment II.ions. (ii) Q. Q is oxidised. 2 SPM 2004/P2/Section C/Q4 (a) Redox reaction is a reaction in which oxidation and reduction occur simultaneously. P (c) Salt Solution W X Y Z Metal W √ √ √ X X √ √ Y X X √ Z X X X √ = Metal deposited X = No deposit 11 . Water and oxygen receive electrons to form OH. Z 3 SPM 2005/P2/Q9 (a) The iron key can be electroplated with nickel by electrolysis. Clean the metal strips with sandpaper. Put metal W in to every test tube. Nickel is made the anode. H+ . OH. No metal deposition occur when metal Y is immersed into salt solutions of W and X. X. W is the most electropositive. Na+ and H+ ions are attracted to the cathode. (b) Ions present in sodium chloride solution are Na+ . Cl. Hydrogen gas is produced at he cathode. X and Y.and OH. 12 . OH. Therefore. Therefore. H+ ions are selected to be discharged because it is lower than Na+ ions in the electrochemical series. Nickel(II) sulphate is used as the electrolyte.. Therefore. At the anode. Oxygen gas is produced at the anode.ions. For metal Z: No metal deposition when metal Z is immersed into salt solutions of W. Y. Observation: For metal W: Metal deposition occur when metal W is immersed into salt solutions of X.ions are attracted to the anode. X. Z is the least electropositive. Y and Z. Cl. Descending order: W. Y and Z. The iron key is made the cathode. X is more electropositive than Y and Z. Leave for a few minutes Repeat the steps above using metlas X.ions are selected to be discharged because it is llower than Cl. For metal X: There is metal deposition when metal X is immersed into salt solutions of Y and Z. Y and Z. At the cathode. Therefore.ions in the electrochemical series.Answer To Score Chemistry Essay Question Answers Fill 4 test tubes with salt solutions of metal W. For metal Y: There is metal deposition when metal Y is immersed into salt solution of Z. Y is more electropositive than Z. There is no metal deposition when metal X is immersed into salt solution of W. MnO4. slowly.Answer To Score Chemistry (c) Essay Question Answers Dilute sulphuric acid is filled into a U-tube.→ Cl2 + 2e // At cathode: Na+ + e → Na 13 . Therefore the roman number is used in naming copper(II) oxide. (iii) 5 SPM 2006/P2/Q10 (a) (b) sodium chloride At anode: 2Cl. 4 SPM 2006/P2/Q7 (a) (i) Oxidation number of Al = +3 Oxidation number of Cu = +2 (ii) Al2O3 = aluminium oxide Cu2O = copper(II) oxide Copper exhibits more than one oxidation number. The purple colour of potassium manganate(VII) solution fades. This is not required to name aluminium oxide because aluminium only exhibits one oxidation number.+ 8H+ + 5e → Mn2+ + 4H2O At positive terminal. (iii) (b) (i) (ii) acidified potassium manganate(VII) solution At positive terminal: MnO4. The wire is connected to complete the circuit. The green colour of iron(II) sulphate solution to brown. Aluminium and zic plates are immersed respectively into aluminium sulphate and zinc sulphate solution. Fe2+ ions donate electrons and are oxidised to Fe3+ ions. Aluminium sulphate solution is added into one arm of the U-tube and zinc sulphate solution is added into the other arm of the U-tube. At the negative terminal.ions receive electrons and are reduced to Mn2+ ions. M. Na+ are discharged to form sodium. Na+ + e → Na 6 SPM 2008/P2/Q8 (a) At electrode P/ cathode: the position of ions in electrochemical series At electrode Q/ anode: concentration of ions P/cathode Na+ . OH-Chloride ion Concentration of Cl. L can displace silver from its solution II M is more electropositive than silver.ions are discharged to form chlorine gas. 2Cl. At the anode.→ Cl2 + 2e (b) Electrode Ions attracted Ions selectively discharged Reason Half equation (c) (i) Experiment I L is more electropositive Explanation than silver.ions is higher than OH. M cannot displace L from its solution. Order of the three metals: silver. chloride ions are attracted to the anode whereas sodium ions are attracted to the cathode. Cl. M can displace silver from its solution III L is more electropositive than M.→ Cl2 + 2e At the cathode.Answer To Score Chemistry (c) Essay Question Answers (d) During electrolysis of molten sodium chloride. L (ii) copper(II) nitrate 14 . H+ Hydrogen ion H+ ions is lower in the electrochemical series 2H+ + 2e → H2 Q/anode Cl-.ions 2Cl. copper(II) sulphate solution (ii) (iii) (iv) 15 . [Alternate Answer: any other metal less electropositive than iron. Copper is less electropositive than iron. 8 SPM 2007/P3/Q2 (i) Statement of the problem: How does the distance between two metals in the electrochemical series affect the voltage produced in a cell? Manipulated variable: Pair of metals Responding variable: Voltage of cell Constant variable: Type of electrolyte Hypothesis: The further apart the distance between two metals in the electrochemical series. Filter the mixture. Fe2+ Fe3+ Halogen is bromine. A green precipitate is formed. Add sodium hydroxide solution. Therefore. zinc. Materials: iron. Oxidation: Na Na + e 2Reduction: 02 + 4e 20 (b) X is copper. the higher the voltage produced. magnesium prevents iron from rusting.Answer To Score Chemistry 7 SPM 2008/P2/Q9 (a) M is sodium. + Y is magnesium. magnesium. [Alternate Answer: aluminium or zinc] Magnesium is more electropositive than iron. Add magnesium to a solution containing Fe3+. Therfore. copper and aluminium strips. Add sodium hydroxide solution. iron rusts. Add bromine water to a solution containing Fe2+. Shake the mixture. Brown precipitate is formed. Heat the mixture. [Alternate Answer: any other Group 1 metals] Essay Question Answers Sodium burns with a yellow flame to produce a white solid. (c) Fe3+ Fe2+ Reducing agent is magnesium. 4 are repeated with zinc. CO32- heat White precipitate dissolve Confirmed the presence of lead(II) ions. 3. iron and aluminium. 4.Cu Zn .Cu Fe .Answer To Score Chemistry (v) Essay Question Answers Apparatus: voltmeter. 2. The copper and magnesium strips are cleaned with sand paper. Steps 1 . A beaker is filled with copper(II) sulphate solution.fertilizers or organic fertilizers or compost [calcium phosphate or polyphosphate fertilizers or superphosphate fertilizers] (b) Salt X Cation test Anion test Salt X + HNO3 Salt X + HNO3 heat A colourless solution is formed Gas evolved is flowed into lime water Add hydrochloric acid into the colourless solution Lime water turns chalky / milky A white precipitate is formed Confirmed the presence of carbonate ion.Cu (vi) FORM 4 CHAPTER 7 ACIDS AND BASES 2003/P2/Q4/SECTION C 1 (a) By adding .Cu Al .Pb2+ 16 . The reading of the voltmeter is recorded. sand paper and connecting wires Procedure: 1. Tabulation of data: Metal pair Voltage/ V Mg .quick lime and slaked lime or calcium oxide and calcium hydroxide to the soil . The strips are immersed into the solution and connected by wires to a voltmeter. 5. beaker. spatula and wire gauze Measure required volume (50 -100 cm3) of molarity(0. distilled water Apparatus: 100 cm3 measuring cylinder. filter funnel. beakers.Answer To Score Chemistry (c) Essay Question Answers To prepare a dry magnesium chloride salt:Materials : magnesium sulphate solution. potassium carbonate solution.1.magnesium carbonate.0 mol dm-3) potassium carbonate solution by using a measuring cylinder and poured into another beaker Mix the two solutions and a white precipitate .0 mol dm-3 )of hydrochloric acid solution by using a measuring cylinder and poured into a beaker Carefully warm the acid Add magnesium carbonate powder bit by bit by using a spatula and stir using a glass rod until some of it no longer dissolves MgCO3 + 2HCl → MgCl2 + H2O + CO2 Remove the unreacted magnesium carbonate (MgCO3) by filtration Pour the filtrate into an evaporating dish. filter papers. MgCl2 . Gently heat the salt to produce a saturated solution Cool the saturated solution until crystals are formed Filter out the magnesium chloride. dilute hydrochloric acid.5 – 1.0 mol dm-3) magnesium sulphate solution by using a measuring cylinder and poured into a beaker Measure required volume (50 -100 cm3)of molarity(0. crystals Wash or rinse the crystals with distilled water Press the crystals with a few pieces of filter paper to dry them [refer to the Chemistry Practical Book on Page 114 – 115. 119] 17 . glass rod. Bunsen burner.5 . evaporating dish.5 – 1.(MgCO3) is formed K2CO3 + MgCl2 → 2KCl + MgCO3 Filter out magnesium carbonate (MgCO3) to remove potassium sulphate or impurities - - Magnesium carbonate (MgCO3) is washed with a little cold distilled water Measure required volume (50 -100 cm3) of molarity(0. retort stand and clamp. K2SO4 Lead(II) sulphate.in HCl 1.025 x 124 = 0. Add Nessler’s reagent 3. black in colour Carbon dioxide . confirmed the presence of chloride ions (c) (i) (ii) (iii) n = 0. Add dilute sodium hydroxide solution and warm it 3. confirmed the presence of chloride ions Test for NH4+ ions in ammonium chloride solution 1. the pH value is low .Answer To Score Chemistry 2 Essay Question Answers 2005/P2/Q8/SECTION B (a) . Add nitric acid solution . ZnSO4 (ii) lead(II) nitrate. then add silver nitrate solution 3.5 x 50/1000 Mass of CuCl2 = 0. A white precipitate is formed. A colourless gas given off that change moist red litmus paper to blue Test for Cl.Hydrochloric acid(strong acid) ionises completely in water to produce a high concentration of hydrogen ions . PbSO4 Zinc sulphate. Pour 2 cm3 of ammonium chloride solution into a test tube 2. A white precipitate is formed.Ethanoic acid (weak acid) ionises partially in water to produce low concentration of hydrogen ions . Filter the solution to remove impurities and pour the filtrate into an evaporating dish 18 . Add nitric acid solution .ions in ammonium chloride solution 1. Pour 2 cm3 of hydrochloric acid into a test tube 2. lime water turns chalky 3 2007/P2/Q7/SECTION B (a) (i) Soluble salt Insoluble salt Potassium sulphate.025 = 3. confirmed the presence of ammonium ions OR 1. flow the gas into lime water. Pour 2 cm3 of hydrochloric acid into a test tube 2.1 g Solid X = Copper(II) oxide . the pH value is high (b) Test for Cl.If the concentration of H+ ions is high. Pour 2 cm3 of ammonium chloride solution into a test tube 2. then add silver nitrate solution 3. Pb(NO3)2 or lead(II) ethanoate(CH3COO)2Pb and sodium sulphate or potassium sulphate or all soluble sulphate salt (b) The crystallisation method for preparing a soluble salt from its aqueous solution. A brown precipitate is formed.if the concentration of H+ ions is high. Pour 2 cm3 salt X solution into a test tube 2. A white precipitate is formed insoluble in excess of NaOH. shows the present of lead(II) ions or aluminium ions 4.Magnesium ions. filter paper. filter funnel . If no precipitate is formed. glass rod Materials : lead(II) nitrate solution (1.0 – 2. wash bottle 19 . Add potassium sulphate solution or potassium chloride solution or potassium iodide solution or sulphuric acid or hydrochloric acid into the test tube 3. aluminium ions and lead(II) ions (any two ions) 1. A white precipitate is formed soluble in excess of NaOH. Add 2 cm3 of iron(II) sulphate solution and shake to mix well 4. Pour 2 cm3 salt X solution into a test tube 2. measuring cylinder.0 mol dm-3). Do not shake the test tube.0 mol dm-3) or (0. Slant the test tube and carefully add concentrated sulphuric acid down the side of the test tube.Answer To Score Chemistry Essay Question Answers Gently heat the solution to obtain a saturated solution Cool the hot saturated solution to allow it to crystallise Filter and wash or rinse the crystals using distilled water Press the crystals with a few pieces of filter papers to dry them [refer to the Chemistry Practical Book on Page 114] (c) (i) Brown gas is nitrogen dioxide gas and the anion is nitrate Test for nitrate ions 1. A white or yellow precipitate is formed. indicates the presence of magnesium ions or aluminium ions 4 2008/P2/Q10/SECTION C (a) Acid A Hydrochloric acid or sulphuric acid or nitric acid Strong acid Ionises completely in water HCl → H+ + ClConcentration of H+ is high (b) To prepare lead(II) sulphate in the laboratory Acid B ethanoic acid or phosphoric acid weak acid Ionises partially in water CH3COOH ↔ CH3COOConcentration of H+ is low + H+ Apparatus: beakers. sodium sulphate solution or any soluble sulphates (0.5 – 1. Pour 2 cm3 salt X solution into a test tube 2. 5. A brown ring is formed [refer to the Chemistry Practical Book on Page 133] (ii) The cations :. shows the present of lead(II) ions 4.0 mol dm-3) sulphuric acid. shows the present of magnesium ions OR 1. Acidify the solution with about 2 cm3 of dilute sulphuric acid 3. Add sodium hydroxide solution into the test tube until in excess 3.5 – 1. retort stand. measure (20 .does not produce too much heat . Quality of air decreases. 3. K2SO4 . Acid rain reduces the pH of the soil (or toxic waste poison the soil). 4. A dent is formed on the cooper block and its diameter is measured and recorded. Process: Sulphur dioxide dissolves in the rain water to form acid rain. metre rule.solution by using a measuring cylinder and pour into a beaker Then. solution by using a measuring cylinder and pour into another beaker Pour the lead(II) nitrate solution into the sodium sulphate solution or sulphuric acid and stir the mixture A white precipitate. 2. A steel ball is stuck onto a copper block by cellophane tape. steel ball.100 cm3) of potassium sulphate. retort stand and clamp Materials: copper block. bronze block. thread. Plants. Consequently the nutrients in the soils are destroyed.100 cm3) of lead(II) nitrate . page 155 for a more detailed answer] Source: Sulphur dioxide gas (or toxic waste) is produced in the factories (or due to the burning of sulphur in the factories). The weight is dropped from 100 cm. cellophane tape. 5. (c) Procedure 1.→ PbSO4 Filter the solution mixture Wash / rinse the residue/solid/salt with distilled water Press the crystals with a few pieces of filter papers to dry them (c) Substance that can be used – vinegar . Step 1 to step 4 is repeated by replacing the copper block with the bronze block. 20 . roots of tree and fishes in the rivers die. Effect: 1. 3. 4. 5. Pb(NO3) . Rain water (or toxic waste) from the factories were channeled into rivers.is less corrosive or do not harmful to the skin FORM 4 CHAPTER 9 MANUFACTURED SUBSTANCES IN INDUSTRY Question 1 2004/P2/Section C Q3 (a) (b) Stage II : 2SO2 + O2 2SO2 Stage IV : H2S2O7 + H2O 2H2SO4 [Please refer to the pictorial diagram in the text book. 2. Apparatus: one kilogram weight.Vinegar is a weak acid .Answer To Score Chemistry Essay Question Answers Measure (20 . Acid rain and acidic river causes corrosion of buildings and bridges. is formed Pb(NO3)2 + K2SO4 → PbSO4 + 2KNO3 or Pb2+ + SO42. PbSO4 .vinegar will neutralise the sting of the jelly-fish which is alkali . A weight is hung at a height of 100 cm above the steel ball. cellophane tape. retort stand and clamp. steel nail. Procedure Please refer to Form 4 Chemistry Practical Book page 148 for the procedure and diagram Note: A two dimensional diagram should be drawn (not a 3-D diagram) Replace the words ‘copper’ and ‘bronze’ (in the practical text book) with the words ‘iron’ and ‘steel’ respectively. The orderly arrangement of atoms in pure metals enables the layers of atoms to slide on one another when a force is applied. one kilogram weight. steel ball bearing. (or Iron is less resistance to rust while steel has a higher resistance to rust. (ii) (iii) 21 . steel block. sandpaper.) List of substances and apparatus Iron nail. Question 2 2003/P3/Q3 Answer on laboratory experiment based on the property hardness (i) Problem statement Is steel harder than iron? Hypothesis Steel is harder than iron. thread. 3. meter rule. potassium hexacynoferrate(III) solution. jelly solution. Tabulation of data Please refer to Form 4 Chemistry Practical Book page 149 for the data table (ii) (iii) (iv) (v) Answer on laboratory experiment based on the property rust resistant (i) Problem statement Does iron rusts more easily than steel? Hypothesis Iron has a higher rate of rusting compared to that of steel. [Explaining the difference in hardness in terms of atomic arrangement. 4. (or Dent formed in steel has a smaller diameter while dent formed on iron has a bigger diameter.) List of substances and apparatus Iron block.] 1. The arrangement of atoms in the copper block is uniform (or orderly) 2. Hence this reduces sliding between layers of copper atoms.Answer To Score Chemistry Essay Question Answers Observation : It is found that the diameter of the dent in the copper block is larger than that of bronze. Tin atom which is of different size disturb the orderly arrangement. test tubes. Tabulation of data Please refer to Form 4 Chemistry Practical Book page 151 for the data table (v) Question 3 2005/P3/Q3 [Answer for Task 1] (i) Statement of the problem Are alloys harder than its pure metal? (Is alloying of metal increases its hardness?) All the variables Manipulated variable: Metal and its metal alloy (or copper and bronze. retort stand and clamp. the rate of reaction is higher (i) Mg + 2HCl → MgCl2 + H2 Mole of Mg = 0. Step 1 to step 5 is repeated by replacing the copper block with the bronze block. Tabulation of data Metal Copper Bronze Diameter of dent (cm) Average diameter (cm) CHAPTER 1 : RATE OF REACTION 1. modify the text book procedure by ignoring the use of stainless nail. A dent is formed on the copper block and its diameter is measured and recorded. Hence to answer this question. (v) A steel ball is stuck onto a copper block by cellophane tape. The experiment is repeated to obtain more dents and its diameter reading. thread. 2.2 cm3 (b) 22 . 3. Small size has bigger total surface area 2.0083 x 24000 = 199. The weight is dropped from 100 cm. (or height of the weight. 4. metre rule. Procedure (ii) (iv) 1. bronze block.Answer To Score Chemistry (iv) Essay Question Answers Procedure Please refer to Form 4 Chemistry Practical Book page 151 for the procedure and diagram Note: In the practical text book. In this question. steel ball bearing.2 = 0.0083 mol 24 1 mol Mg produce 1 mol H2 Volume of H2 = 0. three types of iron nails were suggested to be used in the experiment. 5. SPM 2003/P2/Q1(SECTION B) (a) 1. we need only to use iron nail and steel nail. cellophane tape. A weight is hung at a height of 100 cm above the steel ball. one kilogram weight. type of ball bearing) (iii) Lists of substances and apparatus Copper block. or iron and steel) Responding variable: Diameter of dent Fixed variable : Mass of weight. 6. 2 15 = 3. 3. 9. H+ and calcium carbonate collide with one another more rapidly. High temperature causes the reactants particles to have more kinetic energy.2 20 Experiment III = 199.984 or 4 cm3 s-1 = 9. 7. 2.96 or 10 cm3 s-1 = 13. 8.2 Experiment II Experiment I Time/s (iii) Average rate of reaction : Experiment I = 199. (ii) 23 .28 or 13 cm3 s-1 (iv) 1.Answer To Score Chemistry (ii) Essay Question Answers 199. 3.2 50 Experiment II = 199. The temperature for Experiment II is higher than Experiment I. The rate of reaction for Experiment III is higher than Experiment II.909 cm3s-1 55 1. 5. 6. rate of reaction in experiment II is higher than experiment I the temperature of experiment II is higher than experiment I the kinetic energy of particles increase the collision between H+ ion and magnesium occur frequency of effective collision increase rate of reaction in experiment III is higher than experiment II CuSO4 is used as a catalyst in experiment III The presence of catalyst lower the activation energy frequency of effective collision increase 2. Hydrogen ions. The frequency of effective collision between hydrogen ions and calcium carbonate increases. 6. The rate of reaction for Experiment II is higher than Experiment I. SPM 2005/P2/Q7 (SECTION B) (a) Refrigerator Low temperature Low bacterial activity Less toxin produced by bacteria Rate of food spoilage decreases Kitchen Cabinet High temperature High bacterial activity More toxin produced by bacteria Rate of food spoilage is high (b) (i) Average rate of reaction for Experiment I = 50 = 0. 4. 4. 2. 5. 9. 24 . The frequency of collision between hydrogen ions and calcium carbonate increases. 6.0075 x 24 = 0. the position of the energy level of the reactants is higher than the energy of the product. Activation energy must be overcome in order for the reaction to take place. Essay Question Answers The calcium carbonate in Experiment III have a bigger total surface area. correct position for Ea’ 1. hydrogen ions and calcium carbonate can collide with each other more rapidly. the number of moles CO2 = 0.Answer To Score Chemistry 7. correct position for ∆H 4. 3. The reaction is exothermic reaction. 3.18 dm3 3. The use of catalyst reduces the activation energy. ∆H is the energy difference in the reactants and in the products. 8. Heat given out during bond formation is greater than heat absorb during bond breaking.5 x 30 1000 = 0. The reactants contains more energy than the products. 4. label of energy on vertical axis 2. 7. The frequency of effective collision between hydrogen ions and calcium carbonate increases. correct position for Ea 5. SPM 2007/P2/Q10 (SECTION C) (a) (i) Experiment I : P is hydrochloric acid.0075 mol Volume of CO2 = 0. Zn + 2HCl → ZnCl2 + H2 (ii) 1. 2.015 mol 2 moles of HCl produces 1 mole of CO2 Therefore. (iii) Number of moles HCl = 0. The use of a catalyst increases the frequency of effective collisions. 10. Therefore.015 2 = 0. 5. SPM 2005/P3/Q3 Statements of problem : (i) Does the increase in concentration of acid will increase the rate of reaction? (ii) All the variables : Manipulated variable : concentration of acid Responding variable : rate of reaction Controlled variables : volume of acid. The time required for all the metal dissolved is recorded.0 mol dm-3 of hydrochloric acid. The time taken to collect 50cm3 of hydrogen gas is recorded. 5. 4. 2. 50 cm3 of 0. Using a measuring cylinder. 5. Step 1 to 5 is repeated by replacing 1. The frequency of effective collisions in experiment II is higher than in experiment I. Lists of substances and apparatus : Materials : hydrochloric acid with concentration 0. Weigh 2 g of zinc granule and drop into the conical flask.Answer To Score Chemistry (b) (i) Experiment I Rate = 960 cm3 240 s = 4 cm3 s-1 Experiment II Rate = 960 cm3 160 s = 6 cm3 s-1 (ii) Essay Question Answers 1. The frequency of collision between H+ ion and zinc in experiment II is higher than in experiment I.0 mol dm-3.5 mol dm-3 of hydrochloric and pour into a conical flask. 6. 4. (iv) Procedure : 1. zinc granule. 4. Immediately close the conical flask with the stopper connected to a inverted burette filled with water. Diprotic acid has higher concentration of H+ ion. 10 cm3 measuring cylinder. 3. temperature. This is because the H2SO4 is a diprotic acid whereas HCl is monoprotic acid. (v) Tabulation of data : Concentration of acid/ mol dm-3 Time taken to collect 50 cm3 of the hydrogen gas (iii) 25 . stopwatch. 3. Apparatus : 100 cm3 measuring cylinder. 7. At the same time start the stopwatch. 2.5 mol dm-3 and 1. The rate of reaction for experiment II is higher than experiment I. 6. 1. 3. Propane is a saturated hydrocarbon with carbon – carbon single bond.Answer To Score Chemistry Essay Question Answers FORM 5 CHAPTER 2 CARBON COMPOUNDS Question 1 2004/P2/Section B/Q2 (a) Propan-1-ol (b) (i) Propan-2-ol 85. 4.083 14. of mol of carbon atom No. 3. latex coagulates immediately.3 1 2 (integer) (integer) Empirical formula = CH2 Molecular formula of alkene Y : CnH2n Given relative molecular mass = 42 Hence 12n + 2n = 42 14n = 42 n=3 molecular formula of alkene Y is C3H6 (ii) (iii) (iv) (c) (i) Propene CnH2n 1. Hence when latex is left under natural conditions.083 14. Hence when propanoic acid is added. it coagulates slowly. No reaction occurs when bromine is added. Propanoic acid contains H+ ions 2. brown colour is decolourised because addition reaction (or bromination ) occurs. The growth and spread of bacteria produce lactic acid slowly. 2. of mol of carbon atom = 12 = 7. Alkene Y is an unsaturated hydrocarbon with a carbon-carbon double bond.1 No. 4. 5. The H+ ions immediately neutralized the negative charge on the protein membrane.3 No.7 No. of mol of hydrogen atom 7. Bacteria from the air enter the latex. (ii) Question 2 2007/P2/Q9 (a) (i) Ethene (ii) [ the other accepted answer is propene] Compound F is ethanol Compound G is ethene 26 . of mol of hydrogen atom = 1 = 14. When bromine water is added to alkene Y. Substances Methanol. test tubes. Ethanol burns completely in air (oxygen) to produce carbon dioxide and water. Using a measuring cylinder. 5. butanoic acid. Ethene undergoes polymerization to form polyethane [Choose any three. Ethene undergoes hydrogenation to produce ethane 3. [Choose any three. ethanol. 4. dropper. Ethanol react with acidified potassium dichromate(VI) to produce ethanoic acid. Bunsen burner.] Chemical properties of compound G (ethene) 1. 3. 2. For other accepted answers please refer to text book page 46] (b) Homologous series Alkene General formula CnH2n Functional group Carbon – carbon double bonds (or C = C) Hydroxyl group (or –OH ) Carboxyl group (or COOH) Member Ethene Alcohol Carboxylic acid CnH2n+1OH CnH2n+1COOH Ethanol Ethanoic acid Question 3 2004/P3/Q3 (a) Aim To prepare two different types of ester using the same carboxylic acid with different alcohols and describe their scents. condenser Liebig (d) Procedure 1. concentrated sulphuric acid. The mixture is then stirred. 10 drops of concentrated sulphuric acid is added and the apparatus is set up for reflux. The mixture is then heated under reflux. 2. test tube holder. Ethanol react with carboxylic acid to produce ester. 4. 2. Ester is collected in a conical flask. smelled and its scent recorded. beakers. Step 1 to step 5 is repeated by replacing methanol with ethanol while butanoic acid is used in both experiments. Ethene reacts with water to form ethanol. 6. (c) Apparatus Measuring cylinder. Ethanol undergoes dehydration to form ethene. 25 cm3 of methanol and 50 cm3 of butanoic acid is separately measured and poured into a round bottom flask. 27 . round bottom flask. 4. Ethene burns completely in air to produce carbon dioxide and water. retort stand. (b) Hypothesis Different alcohol produces different ester. Using a dropper. 3.Answer To Score Chemistry (iii) Essay Question Answers Chemical properties of compound F (ethanol) 1. ] Tabulation of data Initial length / cm Vulcanized rubber Unvulcanized rubber Length with weight /cm Length after removal of weight / cm (c) (d) (f) CHAPTER 4 : THERMOCHEMISTRY No essay question in this chapter 28 . unvulcanized rubber strip Apparatus Retort stand and clamps. metre rule. (b) All the variables Manipulated variable: Vulcanized rubber and unvulcanized rubber Responding variable: Change in length of rubber strip Fixed variable : Length (size) of rubber strip. Substances Vulcanized rubber strip. mass of weight Statement of the hypothesis Vulcanized rubber is more elastic than vulcanized rubber.Answer To Score Chemistry (e) Tabulation of data Essay Question Answers Alcohol Carboxylic acid Scent Question 4 2006/P3/Q2 (a) Aim of experiment To compare the elasticity of vulcanized and unvulcanized rubber. Bulldog clips. 50 g weight (e) Procedure [Please refer to Practical chemistry book page 63 for complete diagram and procedure. Some of the common examples of antibiotics are penicillin and streptomycin. Citric acid is used as an antioxidant 4. Some of the common examples are aspirin. and sulphur dioxide Freezing or deep freezing Drying or drain out water from food Canned or sterilization or pasteurisation or heating or vacuum b) The cleansing action of soap: The socks are dipped in a soap solution Soap reduces the surface tension of water Soap increases the wetting ability of water on the surface of the socks The hydrophobic part of the soap dissolves in the oily stains The hydrophilic part is attracted to the water molecules 2 Mechanical agitation during scrubbing helps pull the oily stains free and break the oily stains into small droplets The droplets do not coagulate and redeposit on the surface of the socks due to the repulsion between the negative charges on the surface The droplets are suspended in water forming an emulsion Rinsing washes away these droplets and leaves the surface clean SPM 2008/P2/Q7 a) 1. paracetamol and codeine. Sodium benzoate is used to slow down or prevent the growth of microorganism b) The medicine prescribed to Aida is an analgesic. It must be taken at the recommended dose. May Ling must take the full course of the antibiotic 29 . Antibiotics are used to kill or slow down the growth of bacteria. benzoic acid or sodium benzoate. Ethyl butanoate is used as a flavouring agent 2. The medicine prescribe to May Ling is antibiotic. An analgesic is a medicine used to relieve pain. Sucrose is used as a flavouring agent 3.Answer To Score Chemistry Essay Question Answers CHAPTER 5 : CHEMICAL FOR CONSUMERS 1 SPM 2003/P2/Q2 SECTION B a) Methods of food preservation Adding salt or sugar How the methods work Draws the water out of the cells of microorganisms Retards the growth of microorganisms Provides an acidic condition that inhibits the growth of microorganisms Slow down the growth of microorganisms Low temperature slows down the growth of bacteria or microorganisms Microorganisms cannot live without water Inhibit the growth of microorganisms Adding vinegar or spices Adding sodium nitrite or sodium nitrate. Paracetamol is prescribed to Aida. Gelatin is used to thicken food 5. 30 . 3. c) i) Experiment I and III Essay Question Answers 1. These ions react with the cleaning agent A to form an insoluble precipitate (scum). Its more effective in cleaning action than cleaning agent A. ii) Experiment II and IV 1. Hard water contains calcium and magnesium ions. 2. 4. It can perform its cleaning action in hard water. 3.Answer To Score Chemistry prescribed. Cleaning agent B is effective both in soft water and hard water. Both the cleaning agents A and B are effective in soft water. Cleaning agent B does not form precipitate (scum) in hard water. Both are dissolves in soft water. Soft water does not contains calcium and magnesium ions. They are able to lower the surface tension of water. Formation of scum greatly reduces the number of cleaning agent A molecules available for cleaning. 2. The water wets the surface of the cloth thoroughly. Cleaning agent A is not effective in hard water.