essay question answers

ESSAY QUESTIONANSWERS

(2003-2008)

Answer To Score Chemistry Essay Question Answers

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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: 2.8.1The electron configuration of atom Y: 2.8.7

ii) The number of neutron in Z = 6

Isotope of 136 Z or 14

6 Z

b) 1. Atom X and atom Y will form ionic bond. The electron configuration of atomX 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 valenceelectron shell, atom X donates one electron to form a positive ion.

X X+ + e3. 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 ioniccompound with the formula XY.

6. Element Y and Z will form covalent compund. To attain stable electronconfiguration with 8 electrons in the valence electron shell, atom Z shareselectrons with atom Y.

7. One atom of Z contributes 4 electrons and each atom of Y contributes oneelectron.

8. Atom Z shares electrons with four atoms Y to form a covalent compoundwith the formula ZY4.


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c) 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 thesolution

of XY and the switch was switched on.4. The bulb lights up.

SPM 2005/P2/Q10 SECTION C2 (a) i) Iodine-131 for cure cancer of thyroid glands

ii) Carbon-14 is used to determine the age of a fossil or ancient artifactsb) The electronic arrangement of P is 2.4 whereas the electronic arrangement of Q is

2.6Q is located in Group 16 because it has 6 valence electronsQ is located in Period 2 because it has 2 electrons shells filled with electrons

c) Refer Answer Form4 Topic 5 Chemical Bond


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SPM 2006/ P2/ Q9 SECTION C3 (a)

Subatomic particles Relativemass

Relative charge Row 1

Proton 1 +1 Row 2Electron 1/1837 -1 Row 3Neutron 1 0 Row 4

Note: state row 1 and any two from row 2/ 3/ 4

(b) 23 39 8611 or 19 or 37

(c) The atom consists of two parts: the centre part called nucleus and the outer partcalled electron cloud.The nucleus consists of 11 protons which are positively charged and 12 neutronsare neutral. [ if answer in (b) is Na]The electron cloud consists of 11 electrons which are negatively charged andmove around nucleus in orbits.There is an electroststic force between nucleus and electrons.

(d) Sample answer an atom of sodium.

11p + 12n

Na

SPM 2007/P2 / Q8 SECTION B4 a 1. Nucleus is at the centre

2. nucleus atom contains 1 proton and 1 neutron3. electrons move around the nucleus4. the electron is negatively charged5. One shell is filled with electron

b(i) Atom of Diagram 4 Another atomProton number 1 1Number of electron 1 1Chemical properties Similar SimilarNumber of neutron 1 2Nucleon number 2 3Physical properties Different Different

(ii) 21 H

c) State any ten of following information:At time t0 – t1 :

1. element X is in liquid state2. the particles are closed to each other3. the particles arrangement is not orderly4. the kinetic energy increases

Na K Rb


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At time t1 – t2 :5. element X is is in liquid ad gaseous state6. some particles are closed to each other and some are far apart7. the particle arrangement is not orderly8. the kinetic energy is constant

At time t2 – t3 :9. element X is in gaseous state10. the particles are far away11. the particle arrangement are not orderly12. the kinetic energy is increases

CHAPTER 4: PERIODIC TABLE OF THE ELEMENTS


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 shell2. The distance between the nucleus and the valence electrons of atom in

Diagram 8.2 is greater than atom in Diagram 8.13. 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 conductedin a fume chamber.

2. Make sure that the apparatus are connected tightly to prevent leakageof chlorine gas. Chlorine gas is poisonous.

ii) Part GChlorine gas will react with iron wool to produce iron (III) chloride solid.2Fe(s) + 3Cl2(g) 2FeCl3(s)

Part HThe excess chlorine gas will flow into sodium hydroxide solution to producesodium chloride, sodium chlorate (I) and water.Cl2(g) + 2NaOH(aq) NaCl(aq) + NaOCl(aq) + H2O(l)


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1 SPM 2008/P3/Q2a) 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 metalResponding variable : Reactivity of metalsFixed variable : Water, size of alkali metal

c) Statement of the hypothesis :

When going down Group 1, alkali metals become more reactive in their reactionwith 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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CHAPTER 5 : CHEMICAL BOND


a) i) The electron configuration of atom X: 2.8.1The electron configuration of atom Y: 2.8.7

ii) The number of neutron in Z = 6

Isotope of 136 Z or 14

6 Z

b) 1. Atom X and atom Y will form ionic bond. The electron configuration of atomX 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 valenceelectron shell, atom X donates one electron to form a positive ion.

X X+ + e3. 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 ioniccompound with the formula XY.

6. Element Y and Z will form covalent compund. To attain stable electronconfiguration with 8 electrons in the valence electron shell, atom Z shareselectrons with atom Y.

7. One atom of Z contributes 4 electrons and each atom of Y contributes oneelectron.

8. Atom Z shares electrons with four atoms Y to form a covalent compoundwith

the formula ZY4.


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c) 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 thesolution

of XY and the switch was switched on.4. The bulb lights up.


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2 SPM 2005/P2/Q10(SECTION C)

a) Iodine-131 is used to cure cancer of thyroid glandsCarbon-14 is used to determine the age of a fossil or ancient artifacts

b) 1. The electron arrangement of P is 2.4 whereas the electron arrangement ofQ is 2.6.

2. Q is located in Group 16 because it has 6 valence electrons3. Q is located in Period 2 because it has 2 electron shells filled with electrons.

c) 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+ + 2e5. Atom X receives 1 electron from atom Y to form X- ions to attain a

stable octet electron arrangement.6. X + e X-

7. Y2+ and X- ions are attracted by strong electrostatic forces to produce anionic compound YX2.

8. The number of valence electrons of atom W and atom X are 4 and 7respectively.

9. W and X tend to share valence electrons to attain a stable octet electronarrangement.

10. Each atom W contributes four valence electrons whereas each atom Xcontributes one valence electron.

11. Four atoms of X share electrons with one atom W. A covalent compoundWX4 is formed.


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F4 Chapter 6 ElectrochemistryF5 Chapter 3 Oxidation & Reduction

1 SPM 2003/P2/Section C/Q3(a) Hydrogen gas.

2H+ + 2e → H2

(b) Cell P Cell QType of cell Electolytic cell Chemical/voltaic cellEnergy change electrical → chemical chemical → electricalName of electrode Both electrodes are copper Copper = positive

electrodezinc = negative

elctrodeIons in electrolyte Cu2+, SO4

2-,H+ , OH-

Cu2+, SO42-,

H+ , OH-

Half equations At anode:Cu → Cu2+ + 2e

At cathode:Cu2+ + 2e → Cu

Zn → Zn2+ + 2e

Cu2+ + 2e → CuObservation At anode:

Copper dissolves.

At cathode:Brown solid forms.

Zinc dissolves.

Brown solid forms.

(c) To electroplate a key with silver:Chemicals required:

silver, silver nitrate solutionProcedures 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 andconnected to a battery.

Diagram showing the set-up of the apparatus


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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 SolutionMetal

W X Y Z

W √ √ √X X √ √Y X X √Z X X X

√ = Metal deposited X = No deposit


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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 minutesRepeat 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 andZ.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 andZ.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 Wand 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 thanNa+ 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 thanCl- ions in the electrochemical series.Oxygen gas is produced at the anode.


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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 zincsulphate solution is added into the other arm of the U-tube, slowly.Aluminium and zic plates are immersed respectively into aluminium sulphateand 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

(ii) Al2O3 = aluminium oxideCu2O = copper(II) oxide

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

(b) (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 toFe3+ ions.The purple colour of potassium manganate(VII) solution fades.The green colour of iron(II) sulphate solution to brown.

5 SPM 2006/P2/Q10

(a) sodium chloride

(b) At anode: 2Cl- → Cl2 + 2e //

At cathode: Na+ + e → Na


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(c)

(d) During electrolysis of molten sodium chloride,chloride ions are attracted to the anode whereas sodium ions are attracted tothe cathode.At the anode, Cl- ions are discharged to form chlorine gas.2Cl- → Cl2 + 2eAt the cathode, 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(b)

Electrode P/cathode Q/anodeIons attracted Na+ , H+ Cl-, OH--

Ions selectively discharged Hydrogen ion Chloride ionReason H+ ions is lower in the

electrochemicalseries

Concentration ofCl- ions is higherthan OH- ions

Half equation 2H+ + 2e → H2 2Cl- → Cl2 + 2e

(c) (i) Experiment I II III

Explanation

L is moreelectropositivethan silver.

L can displacesilverfrom itssolution

M is moreelectropositivethan silver.

M can displacesilverfrom itssolution

L is moreelectropositivethan M.

M cannot displace Lfrom its solution.

Order of the three metals: silver, M, L

(ii) copper(II) nitrate


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7 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+ + eReduction: 02 + 4e 202-

(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) Statement of the problem:How does the distance between two metals in the electrochemical seriesaffect the voltage produced in a cell?

(ii) Manipulated variable: Pair of metalsResponding variable: Voltage of cellConstant variable: Type of electrolyte

(iii) Hypothesis:The further apart the distance between two metals in the electrochemicalseries, the higher the voltage produced.

(iv) Materials:iron, zinc, magnesium, copper and aluminium strips; copper(II) sulphatesolution


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Apparatus:voltmeter, beaker, sand paper and connecting wires

(v) 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.

(vi) Tabulation of data:

Metal pair Voltage/ VMg - CuZn - CuFe - CuAl - Cu

FORM 4 CHAPTER 7 ACIDS AND BASES2003/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 Anion test

Salt X + HNO3 Salt X + HNO3

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+

heat


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(c) To prepare a dry magnesium chloride salt:-Materials : magnesium sulphate solution, dilute hydrochloric acid, potassium

carbonate solution, filter papers, distilled waterApparatus: 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 intoanother 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 orimpurities

- 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 abeaker

- 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 dissolvesMgCO3 + 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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2005/P2/Q8/SECTION B2 (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 tube2. Add nitric acid solution , then add silver nitrate solution3. 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 tube2. Add Nessler’s reagent3. A brown precipitate is formed, confirmed the presence of ammonium ionsOR1. Pour 2 cm3 of ammonium chloride solution into a test tube2. Add dilute sodium hydroxide solution and warm it3. A colourless gas given off that change moist red litmus paper to blue

Test for Cl- ions in ammonium chloride solution1. Pour 2 cm3 of hydrochloric acid into a test tube2. Add nitric acid solution , then add silver nitrate solution3. A white precipitate is formed, confirmed the presence of chloride ions

(c) (i) n = 0.5 x 50/1000 Mass of CuCl2 = 0.025 x 124= 0.025 = 3.1 g

(ii) Solid X = Copper(II) oxide , black in colour(iii) Carbon dioxide , flow the gas into lime water, lime water turns chalky

2007/P2/Q7/SECTION B3 (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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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 nitrateTest for nitrate ions1. Pour 2 cm3 salt X solution into a test tube2. Acidify the solution with about 2 cm3 of dilute sulphuric acid3. Add 2 cm3 of iron(II) sulphate solution and shake to mix well4. 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) The cations :- Magnesium ions, aluminium ions and lead(II) ions (any two ions)1. Pour 2 cm3 salt X solution into a test tube2. Add sodium hydroxide solution into the test tube until in excess3. A white precipitate is formed soluble in excess of NaOH, shows the present

of lead(II) ions or aluminium ions4. A white precipitate is formed insoluble in excess of NaOH, shows the

present of magnesium ionsOR1. Pour 2 cm3 salt X solution into a test tube2. Add potassium sulphate solution or potassium chloride solution or

potassium iodide solution or sulphuric acid or hydrochloric acid into the testtube

3. A white or yellow precipitate is formed, shows the present of lead(II) ions4. If no precipitate is formed, indicates the presence of magnesium ions or

aluminium ions

2008/P2/Q10/SECTION C4 (a)

Acid A Acid BHydrochloric acid or sulphuric acid ornitric acid

ethanoic acid or phosphoric acid

Strong acid weak acidIonises completely in waterHCl → H+ + Cl-

Ionises partially in waterCH3COOH ↔ CH3COO- + H+

Concentration of H+ is high Concentration of H+ is low

(b) 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 orany 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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Measure (20 - 100 cm3) of lead(II) nitrate , Pb(NO3) ,solution by using ameasuring cylinder and pour into a beaker

Then, measure (20 - 100 cm3) of potassium sulphate, K2SO4 , solution byusing a measuring cylinder and pour into another beaker

Pour the lead(II) nitrate solution into the sodium sulphate solution orsulphuric acid and stir the mixture

A white precipitate, PbSO4 , is formedPb(NO3)2 + K2SO4 → PbSO4 + 2KNO3 or Pb2+ + SO4

2- → 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 detailedanswer]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 clampMaterials: copper block, bronze block, cellophane tape, thread, steel ball.

Procedure1. 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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Observation :It is found that the diameter of the dent in the copper block is larger than that ofbronze.[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/Q3Answer on laboratory experiment based on the property hardness

(i) Problem statementIs steel harder than iron?

(ii) HypothesisSteel is harder than iron. (or Dent formed in steel has a smaller diameter while dentformed on iron has a bigger diameter.)

(iii) List of substances and apparatusIron block, steel block, cellophane tape, thread, one kilogram weight, steel ballbearing, meter rule, retort stand and clamp.

(iv) ProcedurePlease refer to Form 4 Chemistry Practical Book page 148 for the procedure and

diagramNote: 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 dataPlease 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 statementDoes iron rusts more easily than steel?

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

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


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(iv) ProcedurePlease refer to Form 4 Chemistry Practical Book page 151 for the procedure and

diagramNote: 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 toanswer this question, modify the text book procedure by ignoring the use ofstainless nail.

(v) Tabulation of dataPlease 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 variablesManipulated variable: Metal and its metal alloy (or copper and bronze; or iron and

steel)

Responding variable: Diameter of dentFixed variable : Mass of weight, (or height of the weight, type of ball bearing)

(iii) Lists of substances and apparatusCopper block, bronze block, cellophane tape, thread, one kilogram weight, steel ballbearing, metre rule, retort stand and clamp.

(iv) Procedure1. 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 copper block and its diameter is measured and recorded.5. The experiment is repeated to obtain more dents and its diameter reading.6. Step 1 to step 5 is repeated by replacing the copper block with the bronze block.

(v) Tabulation of data

CHAPTER 1 : RATE OF REACTION

1. SPM 2003/P2/Q1(SECTION B)

(a) 1. Small size has bigger total surface area2. the rate of reaction is higher

(b) (i) Mg + 2HCl → MgCl2 + H2

Mole of Mg = 0.2 = 0.0083 mol24

1 mol Mg produce 1 mol H2

Volume of H2 = 0.0083 x 24000= 199.2 cm3

Metal Diameter of dent (cm) Average diameter (cm)CopperBronze


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(ii)

(iii) Average rate of reaction :

Experiment I = 199.2 = 3.984 or 4 cm3 s-1

50

Experiment II = 199.2 = 9.96 or 10 cm3 s-1

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Experiment III = 199.2 = 13.28 or 13 cm3 s-1

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(iv) 1. rate of reaction in experiment II is higher than experiment I2. the temperature of experiment II is higher than experiment I3. the kinetic energy of particles increase4. the collision between H+ ion and magnesium occur5. frequency of effective collision increase6. rate of reaction in experiment III is higher than experiment II7. CuSO4 is used as a catalyst in experiment III8. The presence of catalyst lower the activation energy9. frequency of effective collision increase

2. SPM 2005/P2/Q7 (SECTION B)(a)

Refrigerator Kitchen CabinetLow temperature High temperatureLow bacterial activity High bacterial activityLess toxin produced by bacteria More toxin produced by bacteriaRate of food spoilage decreases Rate of food spoilage is high

(b) (i) Average rate of reaction for Experiment I= 50 = 0.909 cm3s-1

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(ii) 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.

Time/s

199.2

Experiment II

Experiment I


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7. The calcium carbonate in Experiment III have a bigger total surfacearea.

8. The frequency of collision between hydrogen ions and calciumcarbonate increases.

9. Therefore, hydrogen ions and calcium carbonate can collide with eachother more rapidly.

10. The frequency of effective collision between hydrogen ions andcalcium carbonate increases.

(iii) Number of moles HCl = 0.5 x 301000

= 0.015 mol

2 moles of HCl produces 1 mole of CO2

Therefore, the number of moles CO2

= 0.015 = 0.0075 mol2

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 axis2. the position of the energy level of the reactants is higher than the

energy of the product.3. correct position for ∆H4. correct position for Ea5. correct position for Ea’

1. The reaction is exothermic reaction.2. The reactants contains more energy than the products.3. ∆H is the energy difference in the reactants and in the products.4. Heat given out during bond formation is greater than heat absorb during

bond breaking.5. Activation energy must be overcome in order for the reaction to take

place.6. The use of catalyst reduces the activation energy.7. The use of a catalyst increases the frequency of effective collisions.


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(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) 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.

4. SPM 2005/P3/Q3(i) Statements of problem :

Does the increase in concentration of acid will increase the rate of reaction?

(ii) All the variables :Manipulated variable : concentration of acidResponding variable : rate of reactionControlled variables : volume of acid, temperature.

(iii) Lists of substances and apparatus :Materials : hydrochloric acid with concentration 0.5 mol dm-3 and 1.0 moldm-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 hydrochloricand 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 Propan-2-ol

(b) (i)

No. of mol of carbon atom =85.712 = 7.083

No. of mol of hydrogen atom =14.3

1 = 14.1

No. of mol of carbon atom 7.083 1No. of mol of hydrogen atom 14.3 2 (integer) (integer)

Empirical formula = CH2

Molecular formula of alkene Y : CnH2nGiven relative molecular mass = 42Hence 12n + 2n = 4214n = 42n = 3 molecular formula of alkene Y is C3H6

(ii)

(iii) Propene(iv) CnH2n

(c) (i) 1. Alkene Y is an unsaturated hydrocarbon with a carbon-carbon doublebond.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+ ions2. 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 [ the other accepted answer is propene]

(ii) Compound F is ethanolCompound G is ethene


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(iii) 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 ethane3. 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)

Question 3 2004/P3/Q3(a) Aim

To prepare two different types of ester using the same carboxylic acid with differentalcohols and describe their scents.

(b) HypothesisDifferent alcohol produces different ester.

(c) SubstancesMethanol, ethanol, butanoic acid, concentrated sulphuric acid.ApparatusMeasuring cylinder, test tubes, beakers, round bottom flask, Bunsen burner, dropper,retort stand, test tube holder, condenser Liebig

(d) Procedure1. 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.

Homologousseries

General formula Functional group Member

Alkene CnH2n Carbon – carbondouble bonds(or C = C)

Ethene

Alcohol CnH2n+1OH Hydroxyl group(or –OH )

Ethanol

Carboxylicacid

CnH2n+1COOH Carboxyl group(or COOH)

Ethanoic acid


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(e) Tabulation of data

Question 4 2006/P3/Q2(a) Aim of experiment

To compare the elasticity of vulcanized and unvulcanized rubber.

(b) All the variablesManipulated variable: Vulcanized rubber and unvulcanized rubberResponding variable: Change in length of rubber stripFixed variable : Length (size) of rubber strip, mass of weight

(c) Statement of the hypothesisVulcanized rubber is more elastic than vulcanized rubber.

(d) SubstancesVulcanized rubber strip, unvulcanized rubber strip

ApparatusRetort stand and clamps, Bulldog clips, metre rule, 50 g weight

(e) Procedure[Please refer to Practical chemistry book page 63 for complete diagram andprocedure.]

(f) Tabulation of data

Initial length / cmLength with weight

/cmLength after removal

of weight / cmVulcanizedrubberUnvulcanizedrubber

CHAPTER 4 : THERMOCHEMISTRYNo essay question in this chapter

Alcohol Carboxylic acid Scent


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CHAPTER 5 : CHEMICAL FOR CONSUMERS

1 SPM 2003/P2/Q2 SECTION B

a) Methods of food preservation How the methods workAdding salt or sugar Draws the water out of the cells of

microorganismsRetards the growth of microorganisms

Adding vinegar or spices Provides an acidic condition thatinhibits the growth of microorganisms

Adding sodium nitrite or sodium nitrate,benzoic acid or sodium benzoate, andsulphur dioxide

Slow down the growth ofmicroorganisms

Freezing or deep freezing Low temperature slows down thegrowth of bacteria or microorganisms

Drying or drain out water from food Microorganisms cannot live withoutwater

Canned or sterilization orpasteurisation or heating or vacuum

Inhibit the growth of microorganisms

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

Mechanical agitation during scrubbing helps pull the oily stains free andbreak the oily stains into small droplets

The droplets do not coagulate and redeposit on the surface of the socksdue 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

2 SPM 2008/P2/Q7a) 1. Ethyl butanoate is used as a flavouring agent

2. Sucrose is used as a flavouring agent3. Citric acid is used as an antioxidant4. Gelatin is used to thicken food5. 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 usedto relieve pain. Some of the common examples are aspirin, paracetamol andcodeine. Paracetamol is prescribed to Aida. It must be taken at therecommended dose.The medicine prescribe to May Ling is antibiotic. Antibiotics are used to kill orslow down the growth of bacteria. Some of the common examples of antibiotics arepenicillin and streptomycin. May Ling must take the full course of the antibiotic


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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 waterdoes 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 containscalcium and magnesium ions.

2. These ions react with the cleaning agent A to form an insolubleprecipitate (scum). Formation of scum greatly reduces the number ofcleaning agent A molecules available for cleaning.

3. Cleaning agent B is effective both in soft water and hard water. It canperform its cleaning action in hard water.

4. Cleaning agent B does not form precipitate (scum) in hard water. Itsmore effective in cleaning action than cleaning agent A.

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