Essay Question Answers

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 136Z or 146 Z

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.

2

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

3

Answer To Score Chemistry 3

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SPM 2006/ P2/ Q9 SECTION C (a) Subatomic particles Relative mass Proton 1 Electron 1/1837 Neutron 1

Relative charge

Row 1

+1 -1 0

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. The nucleus consists of 11 protons which are positively charged and 12 neutrons are neutral. [ if answer in (b) is Na] The electron cloud consists of 11 electrons which are negatively charged and move around nucleus in orbits. There is an electroststic force between nucleus and electrons.

(d)

Sample answer an atom of sodium. 11p + 12n

Na

4

SPM 2007/P2 / Q8 SECTION B a 1. Nucleus is at the centre 2. nucleus atom contains 1 proton and 1 neutron 3. electrons move around the nucleus 4. the electron is negatively charged 5. 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. element X is in liquid state 2. the particles are closed to each other 3. the particles arrangement is not orderly 4. the kinetic energy increases

4

Answer To Score Chemistry

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At time t1 – t2 : 5. element X is is in liquid ad gaseous state 6. some particles are closed to each other and some are far apart 7. the particle arrangement is not orderly 8. the kinetic energy is constant At time t2 – t3 : 9. element X is in gaseous state 10. the particles are far away 11. the particle arrangement are not orderly 12. 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 2.8.7. The element is chlorine. ii) Cl2(g) + 2NaOH(aq)

NaCl(aq) + NaOCl(aq) + H2O(l)

b)

1. The atom in Diagram 8.2 has four electron shell 2. The distance between the nucleus and the valence electrons of atom in Diagram 8.2 is greater than atom in Diagram 8.1 3. The attractive forces between the nucleus and the valence electron becomes weaker. 4. The atom in Diagram 8.1 has a stronger attraction towards electron compared to the atom in Diagram 8.2. 5. The atom in Diagram 8.1 is more electronegative compare to atom in Diagram 8.2. 6. Therefore atom in Diagram 8.1 is more reactive compared to the atom in Diagram 8.2.

c)

Less reactive

d)

i) 1. Concentrated acid is corrosive and the experiment must be conducted in a fume chamber. 2. Make sure that the apparatus are connected tightly to prevent leakage of chlorine gas. Chlorine gas is poisonous. ii) Part G Chlorine gas will react with iron wool to produce iron (III) chloride solid. 2Fe(s) + 3Cl2(g) 2FeCl3(s) Part H The excess chlorine gas will flow into sodium hydroxide solution to produce sodium chloride, sodium chlorate (I) and water. Cl2(g) + 2NaOH(aq) NaCl(aq) + NaOCl(aq) + H2O(l)

5

Answer To Score Chemistry 1

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SPM 2008/P3/Q2 a) Aim of the experiment : To investigate the reactivity of lithium, sodium and potassium with water. b)

All the variables : Manipulated variable : Different type of alkali metal Responding variable : Reactivity of metals Fixed variable : Water, size of alkali metal

c)

Statement of the hypothesis : When going down Group 1, alkali metals become more reactive in their reaction with water.

d)

List of substances and apparatus : Substances : Small pieces of lithium, sodium and potassium, filter paper, distilled water, red litmus paper Apparatus : Water troughs, small knife, forceps

e)

Procedure of the experiment : 1. Cut a small piece of lithium using a knife and forceps. 2. Dry the oil on the surface of the lithium with filter paper. 3. Place the lithium slowly onto the water surface in a water trough. 4. When the reactions stop, test the solution produced with red litmus paper. 5. Record the observation on the table. 6. Repeat steps 1-5 using sodium and potassium to replace lithium one by one.

f)

Tabulation of data:

Alkali Metals

Observation

Lithium Sodium Potassium

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Answer To Score Chemistry

Essay Question Answers CHAPTER 5 : CHEMICAL BOND

1

SPM 2004/P2/Q1(SECTION B) a)

b)

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 136Z or 146 Z 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.

7

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. The crucible was filled with XY powder until 2/3 full. 6. The crucible with its content was then heated strongly untill all the XY powder melts. 7. After that, two carbon electrodes, were dipped into the molten XY and the switch was switched on. 8. 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.

8

Answer To Score Chemistry 2

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SPM 2005/P2/Q10(SECTION C) a) b)

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 electron arrangement of P is 2.4 whereas the electron arrangement of Q is 2.6. 2. 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. 1. There are two types of chemical bonds, ionic bond and covalent bond. 2. The number of valence electrons of X and Y are 7 and 2 respectively. 3. Atom Y donates 2 electron to two atoms of X to form Y2+ ion to attain a stable octet electron arrangement. 4. Y Y2+ + 2e 5. Atom X receives 1 electron from atom Y to form X- ions to attain a stable octet electron arrangement. 6. X + e X2+ 7. Y and X ions are attracted by strong electrostatic forces to produce an ionic compound YX2.

8. The number of valence electrons of atom W and atom X are 4 and 7 respectively. 9. W and X tend to share valence electrons to attain a stable octet electron arrangement. 10. Each atom W contributes four valence electrons whereas each atom X contributes one valence electron. 11. Four atoms of X share electrons with one atom W. A covalent compound WX4 is formed.

9

Answer To Score Chemistry

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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

Ions in electrolyte

Cu2+, SO42-, H+ , OH-

Cell Q Chemical/voltaic cell chemical → electrical Copper = positive electrode zinc = negative elctrode Cu2+, SO42-, H+ , OH-

Half equations

At anode: Cu → Cu2+ + 2e

Zn → Zn2+ + 2e

Observation

At cathode: Cu2+ + 2e → Cu At anode: Copper dissolves. At cathode: Brown solid forms.

(c)

Cu2+ + 2e → Cu Zinc dissolves.

Brown solid forms.

To electroplate a key with silver: Chemicals required: silver, silver nitrate solution Procedures of the experiment: Key is made the cathode. Silver is made the anode. Electrolyte used is silver nitrate solution in a beaker. The silver anode and the key are immersed into the electroyte and connected to a battery. Diagram showing the set-up of the apparatus

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Answer To Score Chemistry

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Chemical equation involved in the reaction: At anode: Ag → Ag+ + e At cathode: Ag+ + e → Ag

Observation: At anode: Silver dissolves. At cathode: Shiny solid forms. No change in the colour of the solution.

2

SPM 2004/P2/Section C/Q4 (a) Redox reaction is a reaction in which oxidation and reduction occur simultaneously. E.g: Mg + Cu2+ → Mg2+ + Cu

(b)

(i)

In Experiment I, Iron is oxidised to Fe2+ ions. Electrons flow from iron to P because iron is more electropositive than P. Blue solution shows the presence of Fe2+ ions. In Experiment II, Q is oxidised. Electrons flow from Q to Fe because Q is more electropositive than Fe. Water and oxygen receive electrons to form OH- ions. The pink spots show the presence of OH- ions.

(ii)

Q, iron, 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

Answer To Score Chemistry

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Fill 4 test tubes with salt solutions of metal W, X, Y and Z. Clean the metal strips with sandpaper. Put metal W in to every test tube. Leave for a few minutes Repeat the steps above using metlas X, Y and Z. Observation: For metal W: Metal deposition occur when metal W is immersed into salt solutions of X, Y and Z. Therefore, W is the most electropositive. For metal Z: No metal deposition when metal Z is immersed into salt solutions of W, X and Y. Therefore, Z is the least electropositive. For metal X: There is metal deposition when metal X is immersed into salt solutions of Y and Z. There is no metal deposition when metal X is immersed into salt solution of W. Therefore, X is more electropositive than Y and Z. For metal Y: There is metal deposition when metal Y is immersed into salt solution of Z. No metal deposition occur when metal Y is immersed into salt solutions of W and X. Therefore, Y is more electropositive than Z. Descending order: W, X, Y, Z

3

SPM 2005/P2/Q9 (a) The iron key can be electroplated with nickel by electrolysis. The iron key is made the cathode. Nickel is made the anode. Nickel(II) sulphate is used as the electrolyte. (b)

Ions present in sodium chloride solution are Na+ , Cl- , H+ , OH- ions. Na+ and H+ ions are attracted to the cathode. Cl- and OH- ions are attracted to the anode. At the cathode, H+ ions are selected to be discharged because it is lower than Na+ ions in the electrochemical series. Hydrogen gas is produced at he cathode. At the anode, OH- ions are selected to be discharged because it is llower than Cl- ions in the electrochemical series. Oxygen gas is produced at the anode.

12

Answer To Score Chemistry

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(c)

Dilute sulphuric acid is filled into a U-tube. 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, slowly. Aluminium and zic plates are immersed respectively into aluminium sulphate and zinc sulphate solution. The wire is connected to complete the circuit.

4

SPM 2006/P2/Q7 (a)

(i)

Oxidation number of Al = +3 Oxidation number of Cu = +2

(b)

5

(ii)

Al2O3 = aluminium oxide Cu2O = copper(II) oxide

(iii)

Copper exhibits more than one oxidation number. Therefore the roman number is used in naming copper(II) oxide. This is not required to name aluminium oxide because aluminium only exhibits one oxidation number.

(i)

acidified potassium manganate(VII) solution

(ii)

At positive terminal: MnO4- + 8H+ + 5e → Mn2+ + 4H2O

(iii)

At positive terminal, MnO4- ions receive electrons and are reduced to Mn2+ ions. At the negative terminal, Fe2+ ions donate electrons and are oxidised to Fe3+ ions. The purple colour of potassium manganate(VII) solution fades. The green colour of iron(II) sulphate solution to brown.

SPM 2006/P2/Q10 (a)

sodium chloride

(b)

At anode: 2Cl- → Cl2 + 2e // At cathode: Na+ + e → Na

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Answer To Score Chemistry

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(c)

(d)

6

During electrolysis of molten sodium chloride, chloride ions are attracted to the anode whereas sodium ions are attracted to the cathode. At the anode, Cl- ions are discharged to form chlorine gas. 2Cl- → Cl2 + 2e At the cathode, Na+ are discharged to form sodium. Na+ + e → Na

SPM 2008/P2/Q8 (a)

At electrode P/ cathode: the position of ions in electrochemical series At electrode Q/ anode:

concentration of ions

(b) Electrode Ions attracted Ions selectively discharged Reason

Half equation

(c)

(i)

P/cathode Na+ , H+ Hydrogen ion H+ ions is lower in the electrochemical series 2H+ + 2e → H2

Experiment

Q/anode Cl-, OH-Chloride ion Concentration of Cl- ions is higher than OH- ions 2Cl- → Cl2 + 2e

I L is more electropositive Explanation than silver.

II M is more electropositive than silver.

III L is more electropositive than M.

L can displace silver from its solution

M can displace silver from its solution

M cannot displace L from its solution.

Order of the three metals: silver, M, L (ii)

copper(II) nitrate

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Answer To Score Chemistry 7

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SPM 2008/P2/Q9 (a) M is sodium. [Alternate Answer: any other Group 1 metals] Sodium burns with a yellow flame to produce a white solid. +

Oxidation: Na  Na + e 2Reduction: 02 + 4e  20 (b) X is copper. [Alternate Answer: any other metal less electropositive than iron. Copper is less electropositive than iron. Therfore, iron rusts.

Y is magnesium. [Alternate Answer: aluminium or zinc] Magnesium is more electropositive than iron. Therefore, magnesium prevents iron from rusting. (c)

Fe3+  Fe2+ Reducing agent is magnesium. Add magnesium to a solution containing Fe3+. Heat the mixture. Filter the mixture. Add sodium hydroxide solution. A green precipitate is formed. Fe2+  Fe3+ Halogen is bromine. Add bromine water to a solution containing Fe2+. Shake the mixture. Add sodium hydroxide solution. Brown precipitate is formed.

8

SPM 2007/P3/Q2 (i)

(ii)

(iii)

(iv)

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, the higher the voltage produced. Materials: iron, zinc, magnesium, copper and aluminium strips; copper(II) sulphate solution

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Answer To Score Chemistry (v)

(vi)

Essay Question Answers

Apparatus: voltmeter, beaker, sand paper and connecting wires Procedure: 1. A beaker is filled with copper(II) sulphate solution. 2. The copper and magnesium strips are cleaned with sand paper. 3. The strips are immersed into the solution and connected by wires to a voltmeter. 4. The reading of the voltmeter is recorded. 5. Steps 1 - 4 are repeated with zinc, iron and aluminium. Tabulation of data: Metal pair Voltage/ V Mg - Cu Zn - Cu Fe - Cu Al - Cu

FORM 4 CHAPTER 7 ACIDS AND BASES 2003/P2/Q4/SECTION C 1 (a) By adding - quick lime and slaked lime or calcium oxide and calcium hydroxide to the soil - fertilizers or organic fertilizers or compost [calcium phosphate or polyphosphate fertilizers or superphosphate fertilizers] (b) Salt X

Cation test

Salt X + HNO3

Anion test

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, CO32-

heat White precipitate dissolve

Confirmed the presence of lead(II) ions,Pb2+

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Answer To Score Chemistry (c)

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To prepare a dry magnesium chloride salt:Materials : magnesium sulphate solution, dilute hydrochloric acid, potassium carbonate solution, filter papers, distilled water Apparatus: 100 cm3 measuring cylinder, filter funnel, evaporating dish, Bunsen burner, retort stand and clamp, beakers, glass rod,spatula and wire gauze -

Measure required volume (50 -100 cm3) of molarity(0.5 - 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(0.5 – 1.0 mol dm-3) potassium carbonate solution by using a measuring cylinder and poured into another beaker Mix the two solutions and a white precipitate ,magnesium carbonate,(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(0.5 – 1.0 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. Gently heat the salt to produce a saturated solution Cool the saturated solution until crystals are formed Filter out the magnesium chloride, MgCl2 , 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]

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Answer To Score Chemistry 2

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2005/P2/Q8/SECTION B (a) - Hydrochloric acid(strong acid) ionises completely in water to produce a high concentration of hydrogen ions - if the concentration of H+ ions is high, the pH value is low - Ethanoic acid (weak acid) ionises partially in water to produce low concentration of hydrogen ions - If the concentration of H+ ions is high, the pH value is high (b)

Test for Cl- in HCl 1. Pour 2 cm3 of hydrochloric acid into a test tube 2. Add nitric acid solution , then add silver nitrate solution 3. A white precipitate is formed, confirmed the presence of chloride ions Test for NH4+ ions in ammonium chloride solution 1. Pour 2 cm3 of ammonium chloride solution into a test tube 2. Add Nessler’s reagent 3. A brown precipitate is formed, confirmed the presence of ammonium ions OR 1. Pour 2 cm3 of ammonium chloride solution into a test tube 2. Add dilute sodium hydroxide solution and warm it 3. A colourless gas given off that change moist red litmus paper to blue Test for Cl- ions in ammonium chloride solution 1. Pour 2 cm3 of hydrochloric acid into a test tube 2. Add nitric acid solution , then add silver nitrate solution 3. A white precipitate is formed, confirmed the presence of chloride ions

(c)

(i) (ii) (iii)

3

n = 0.5 x 50/1000 Mass of CuCl2 = 0.025 x 124 = 0.025 = 3.1 g Solid X = Copper(II) oxide , black in colour Carbon dioxide , flow the gas into lime water, lime water turns chalky

2007/P2/Q7/SECTION B (a) (i) Soluble salt Insoluble salt Potassium sulphate, K2SO4 Lead(II) sulphate, PbSO4 Zinc sulphate, ZnSO4 (ii)  lead(II) nitrate, 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; Filter the solution to remove impurities and pour the filtrate into an evaporating dish

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Answer To Score Chemistry    

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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. Pour 2 cm3 salt X solution into a test tube 2. Acidify the solution with about 2 cm3 of dilute sulphuric acid 3. Add 2 cm3 of iron(II) sulphate solution and shake to mix well 4. Slant the test tube and carefully add concentrated sulphuric acid down the side of the test tube. Do not shake the test tube. 5. A brown ring is formed [refer to the Chemistry Practical Book on Page 133]

(ii)

4

The cations :- Magnesium ions, aluminium ions and lead(II) ions (any two ions) 1. Pour 2 cm3 salt X solution into a test tube 2. Add sodium hydroxide solution into the test tube until in excess 3. A white precipitate is formed soluble in excess of NaOH, shows the present of lead(II) ions or aluminium ions 4. A white precipitate is formed insoluble in excess of NaOH, shows the present of magnesium ions OR 1. Pour 2 cm3 salt X solution into a test tube 2. Add potassium sulphate solution or potassium chloride solution or potassium iodide solution or sulphuric acid or hydrochloric acid into the test tube 3. A white or yellow precipitate is formed, shows the present of lead(II) ions 4. If no precipitate is formed, indicates the presence of magnesium ions or aluminium ions

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)

Acid B ethanoic acid or phosphoric acid weak acid Ionises partially in water CH3COOH ↔ CH3COOConcentration of H+ is low

+ H+

To prepare lead(II) sulphate in the laboratory Apparatus: beakers, filter funnel , retort stand, measuring cylinder, glass rod Materials : lead(II) nitrate solution (1.0 – 2.0 mol dm-3), sodium sulphate solution or any soluble sulphates (0.5 – 1.0 mol dm-3) or (0.5 – 1.0 mol dm-3) sulphuric acid, filter paper, wash bottle

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Answer To Score Chemistry

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

Measure (20 - 100 cm3) of lead(II) nitrate , Pb(NO3) ,solution by using a measuring cylinder and pour into a beaker  Then, measure (20 - 100 cm3) of potassium sulphate, K2SO4 , 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, PbSO4 , is formed Pb(NO3)2 + K2SO4 → PbSO4 + 2KNO3 or Pb2+ + SO42- → 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 - Vinegar is a weak acid - vinegar will neutralise the sting of the jelly-fish which is alkali - does not produce too much heat - 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)

Stage II : 2SO2 + O2  2SO2 Stage IV : H2S2O7 + H2O  2H2SO4

(b)

[Please refer to the pictorial diagram in the text book. 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). Process:  Sulphur dioxide dissolves in the rain water to form acid rain.  Rain water (or toxic waste) from the factories were channeled into rivers. Effect: 1. Acid rain reduces the pH of the soil (or toxic waste poison the soil). 2. Consequently the nutrients in the soils are destroyed. 3. Plants, roots of tree and fishes in the rivers die. 4. Acid rain and acidic river causes corrosion of buildings and bridges. 5. Quality of air decreases.

(c)

Apparatus: one kilogram weight, metre rule, retort stand and clamp Materials: copper block, bronze block, cellophane tape, thread, steel ball.

Procedure 1. A steel ball is stuck onto a copper block by cellophane tape. 2. A weight is hung at a height of 100 cm above the steel ball. 3. The weight is dropped from 100 cm. 4. A dent is formed on the cooper block and its diameter is measured and recorded. 5. Step 1 to step 4 is repeated by replacing the copper block with the bronze block.

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Answer To Score Chemistry

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Observation : It is found that the diameter of the dent in the copper block is larger than that of bronze. [Explaining the difference in hardness in terms of atomic arrangement.] 1. The arrangement of atoms in the copper block is uniform (or orderly)

2. The orderly arrangement of atoms in pure metals enables the layers of atoms to slide on one another when a force is applied. 3. Tin atom which is of different size disturb the orderly arrangement. 4. Hence this reduces sliding between layers of copper atoms. Question 2 2003/P3/Q3 Answer on laboratory experiment based on the property hardness (i)

Problem statement Is steel harder than iron?

(ii)

Hypothesis Steel is harder than iron. (or Dent formed in steel has a smaller diameter while dent formed on iron has a bigger diameter.)

(iii)

List of substances and apparatus Iron block, steel block, cellophane tape, thread, one kilogram weight, steel ball bearing, meter rule, retort stand and clamp.

(iv)

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.

(v)

Tabulation of data Please refer to Form 4 Chemistry Practical Book page 149 for the data table

Answer on laboratory experiment based on the property rust resistant (i)

Problem statement Does iron rusts more easily than steel?

(ii)

Hypothesis Iron has a higher rate of rusting compared to that of steel. (or Iron is less resistance to rust while steel has a higher resistance to rust.)

(iii)

List of substances and apparatus Iron nail, steel nail, jelly solution, potassium hexacynoferrate(III) solution, sandpaper, test tubes.

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Answer To Score Chemistry

Essay Question Answers

(iv)

Procedure Please refer to Form 4 Chemistry Practical Book page 151 for the procedure and diagram Note:  In the practical text book, three types of iron nails were suggested to be used in the experiment. In this question, we need only to use iron nail and steel nail. Hence to answer this question, modify the text book procedure by ignoring the use of stainless nail.

(v)

Tabulation of data Please refer to Form 4 Chemistry Practical Book page 151 for the data table

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?)

(ii)

All the variables Manipulated variable: Metal and its metal alloy (or copper and bronze; or iron and steel) Responding variable: Diameter of dent Fixed variable : Mass of weight, (or height of the weight, type of ball bearing)

(iii)

(iv)

Lists of substances and apparatus Copper block, bronze block, cellophane tape, thread, one kilogram weight, steel ball bearing, metre rule, retort stand and clamp. Procedure

1. 2. 3. 4. 5. 6. (v)

A steel ball is stuck onto a copper block by cellophane tape. A weight is hung at a height of 100 cm above the steel ball. The weight is dropped from 100 cm. A dent is formed on the copper block and its diameter is measured and recorded. The experiment is repeated to obtain more dents and its diameter reading. 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.

SPM 2003/P2/Q1(SECTION B) 1. Small size has bigger total surface area 2. the rate of reaction is higher

(a)

(b)

(i)

Mg + 2HCl → MgCl2 + H2 Mole of Mg = 0.2 = 0.0083 mol 24 1 mol Mg produce 1 mol H2 Volume of H2 = 0.0083 x 24000 = 199.2 cm3

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(ii)

199.2 Experiment II Experiment I

Time/s

(iii)

(iv)

2.

Average rate of reaction : Experiment I = 199.2 50

= 3.984 or 4 cm3 s-1

Experiment II = 199.2 20

= 9.96 or 10 cm3 s-1

Experiment III = 199.2 15

= 13.28 or 13 cm3 s-1

1. 2. 3. 4. 5. 6. 7. 8. 9.

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

SPM 2005/P2/Q7 (SECTION B) (a) Refrigerator Low temperature Low bacterial activity Less toxin produced by bacteria Rate of food spoilage decreases

(b)

(i)

(ii)

Kitchen Cabinet High temperature High bacterial activity More toxin produced by bacteria Rate of food spoilage is high

Average rate of reaction for Experiment I = 50 = 0.909 cm3s-1 55 1. The rate of reaction for Experiment II is higher than Experiment I. 2. The temperature for Experiment II is higher than Experiment I. 3. High temperature causes the reactants particles to have more kinetic energy. 4. Hydrogen ions, H+ and calcium carbonate collide with one another more rapidly. 5. The frequency of effective collision between hydrogen ions and calcium carbonate increases. 6. The rate of reaction for Experiment III is higher than Experiment II.

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7.

The calcium carbonate in Experiment III have a bigger total surface area. 8. The frequency of collision between hydrogen ions and calcium carbonate increases. 9. Therefore, hydrogen ions and calcium carbonate can collide with each other more rapidly. 10. The frequency of effective collision between hydrogen ions and calcium carbonate increases. (iii)

Number of moles HCl = 0.5 x 30 1000 = 0.015 mol 2 moles of HCl produces 1 mole of CO2 Therefore, the number of moles CO2 = 0.015 2

= 0.0075 mol

Volume of CO2 = 0.0075 x 24 = 0.18 dm3 3.

SPM 2007/P2/Q10 (SECTION C) (a)

(i)

Experiment I : P is hydrochloric acid. Zn + 2HCl → ZnCl2 + H2

(ii)

1. label of energy on vertical axis 2. the position of the energy level of the reactants is higher than the energy of the product. 3. correct position for ∆H 4. correct position for Ea 5. correct position for Ea’ 1. 2. 3. 4. 5. 6. 7.

The reaction is exothermic reaction. The reactants contains more energy than the products. ∆H is the energy difference in the reactants and in the products. Heat given out during bond formation is greater than heat absorb during bond breaking. Activation energy must be overcome in order for the reaction to take place. The use of catalyst reduces the activation energy. The use of a catalyst increases the frequency of effective collisions.

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Answer To Score Chemistry (b)

(i)

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Experiment I Rate = 960 cm3 240 s = 4 cm3 s-1 Experiment II Rate = 960 cm3 160 s = 6 cm3 s-1

(ii)

4.

1. The rate of reaction for experiment II is higher than experiment I. 2. This is because the H2SO4 is a diprotic acid whereas HCl is monoprotic acid. 3. Diprotic acid has higher concentration of H+ ion. 4. The frequency of collision between H+ ion and zinc in experiment II is higher than in experiment I. 5. The frequency of effective collisions in experiment II is higher than in experiment I.

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, temperature.

(iii)

Lists of substances and apparatus : Materials : hydrochloric acid with concentration 0.5 mol dm-3 and 1.0 mol dm-3, zinc granule. Apparatus : 100 cm3 measuring cylinder, 10 cm3 measuring cylinder, stopwatch.

(iv)

Procedure : 1. Using a measuring cylinder, 50 cm3 of 0.5 mol dm-3 of hydrochloric and pour into a conical flask. 2. Weigh 2 g of zinc granule and drop into the conical flask. 3. Immediately close the conical flask with the stopper connected to a inverted burette filled with water. 4. At the same time start the stopwatch. 5. The time taken to collect 50cm3 of hydrogen gas is recorded. 6. The time required for all the metal dissolved is recorded. 7. Step 1 to 5 is repeated by replacing 1.0 mol dm-3 of hydrochloric acid.

(v)

Tabulation of data : Concentration of acid/ mol dm-3

Time taken to collect 50 cm3 of the hydrogen gas

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FORM 5 CHAPTER 2 CARBON COMPOUNDS Question 1 2004/P2/Section B/Q2 (a)

Propan-1-ol (b)

Propan-2-ol

(i) 85.7 No. of mol of carbon atom = 12 = 7.083 14.3 No. of mol of hydrogen atom = 1 = 14.1

No. of mol of carbon atom No. of mol of hydrogen atom



7.083 14.3



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)

(c)

(iii) (iv)

Propene CnH2n

(i)

1. Alkene Y is an unsaturated hydrocarbon with a carbon-carbon double bond. 2. When bromine water is added to alkene Y, brown colour is decolourised because addition reaction (or bromination ) occurs. 3. Propane is a saturated hydrocarbon with carbon – carbon single bond. 4. No reaction occurs when bromine is added.

(ii)

1. Propanoic acid contains H+ ions 2. The H+ ions immediately neutralized the negative charge on the protein membrane. 3. Hence when propanoic acid is added, latex coagulates immediately. 4. Bacteria from the air enter the latex. 5. The growth and spread of bacteria produce lactic acid slowly. 6. Hence when latex is left under natural conditions, it coagulates slowly.

Question 2 2007/P2/Q9 (a)

(i) Ethene (ii)

[ the other accepted answer is propene]

Compound F is ethanol Compound G is ethene

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Answer To Score Chemistry (iii)

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Chemical properties of compound F (ethanol) 1. Ethanol burns completely in air (oxygen) to produce carbon dioxide and water. 2. Ethanol react with acidified potassium dichromate(VI) to produce ethanoic acid. 3. Ethanol undergoes dehydration to form ethene. 4. Ethanol react with carboxylic acid to produce ester. [Choose any three.]

Chemical properties of compound G (ethene) 1. Ethene burns completely in air to produce carbon dioxide and water. 2. Ethene undergoes hydrogenation to produce ethane 3. Ethene reacts with water to form ethanol. 4. Ethene undergoes polymerization to form polyethane [Choose any three. For other accepted answers please refer to text book page 46] (b)

Homologous series Alkene

CnH2n

Alcohol

CnH2n+1OH

Carboxylic acid

CnH2n+1COOH

General formula

Functional group Carbon – carbon double bonds (or C = C) Hydroxyl group (or –OH ) Carboxyl group (or COOH)

Member Ethene

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. (b)

Hypothesis Different alcohol produces different ester.

(c)

Substances Methanol, ethanol, butanoic acid, concentrated sulphuric acid.

Apparatus Measuring cylinder, test tubes, beakers, round bottom flask, Bunsen burner, dropper, retort stand, test tube holder, condenser Liebig (d)

Procedure 1. Using a measuring cylinder, 25 cm3 of methanol and 50 cm3 of butanoic acid is separately measured and poured into a round bottom flask. 2. The mixture is then stirred. 3. Using a dropper, 10 drops of concentrated sulphuric acid is added and the apparatus is set up for reflux. 4. The mixture is then heated under reflux. 5. Ester is collected in a conical flask, smelled and its scent recorded. 6. Step 1 to step 5 is repeated by replacing methanol with ethanol while butanoic acid is used in both experiments.

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Answer To Score Chemistry (e)

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Tabulation of data

Alcohol

Carboxylic acid

Scent

Question 4 2006/P3/Q2 (a) Aim of experiment To compare the elasticity of vulcanized and unvulcanized rubber. (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, mass of weight

(c)

Statement of the hypothesis Vulcanized rubber is more elastic than vulcanized rubber.

(d)

Substances Vulcanized rubber strip, unvulcanized rubber strip Apparatus Retort stand and clamps, Bulldog clips, metre rule, 50 g weight

(e)

Procedure [Please refer to Practical chemistry book page 63 for complete diagram and procedure.]

(f)

Tabulation of data Initial length / cm

Length with weight /cm

Length after removal of weight / cm

Vulcanized rubber Unvulcanized rubber

CHAPTER 4 : THERMOCHEMISTRY No essay question in this chapter

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CHAPTER 5 : CHEMICAL FOR CONSUMERS 1

SPM 2003/P2/Q2 SECTION B a)

Methods of food preservation Adding salt or sugar

Adding vinegar or spices Adding sodium nitrite or sodium nitrate, benzoic acid or sodium benzoate, and sulphur dioxide Freezing or deep freezing Drying or drain out water from food Canned or sterilization or pasteurisation or heating or vacuum b)

Low temperature slows down the growth of bacteria or microorganisms Microorganisms cannot live without water Inhibit the growth of microorganisms

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

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

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. Ethyl butanoate is used as a flavouring agent 2. Sucrose is used as a flavouring agent 3. Citric acid is used as an antioxidant 4. Gelatin is used to thicken food 5. Sodium benzoate is used to slow down or prevent the growth of microorganism b) The medicine prescribed to Aida is an analgesic. An analgesic is a medicine used to relieve pain. Some of the common examples are aspirin, paracetamol and codeine. Paracetamol is prescribed to Aida. It must be taken at the recommended dose. The medicine prescribe to May Ling is antibiotic. Antibiotics are used to kill or slow down the growth of bacteria. Some of the common examples of antibiotics are penicillin and streptomycin. May Ling must take the full course of the antibiotic

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Answer To Score Chemistry

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prescribed. c)

i) Experiment I and III 1. Both the cleaning agents A and B are effective in soft water. Soft water does not contains calcium and magnesium ions. 2. Both are dissolves in soft water. 3. They are able to lower the surface tension of water. The water wets the surface of the cloth thoroughly. ii) Experiment II and IV 1. Cleaning agent A is not effective in hard water. Hard water contains calcium and magnesium ions. 2. These ions react with the cleaning agent A to form an insoluble precipitate (scum). Formation of scum greatly reduces the number of cleaning agent A molecules available for cleaning. 3. Cleaning agent B is effective both in soft water and hard water. It can perform its cleaning action in hard water. 4. Cleaning agent B does not form precipitate (scum) in hard water. Its more effective in cleaning action than cleaning agent A.

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