1

BAHAGIAN PENGURUSAN

SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER

JAWAPAN

MODUL PERFECT SCORE

2012

CHEMISTRY

[KIMIA]

Set 1 Set 2 Set 3 Set 4 Set 5

2

JAWAPAN SET 1

PAPER 2 : STRUCTURED QUESTION

SECTION A

No Explanation Sub Mark Total Mark

1(a)(i) The number of proton in the nuclues of an atom 1 1

(ii) 18 1 1

(b) M and N,

because both atoms have same proton number but

different nucleon number/ the same the number of

proton but different the number of neutron

1

1

2

(c)(i) 2.8.6 1 1

(ii) Group 16 , Period 3 1 1

(iii) 6 valence electron ,

three shells fill / occupied with electron

1

1

2

(d) M-

1 1

Total 9

No Explanation Sub Mark Total Mark

2(a) 2.8.3 1 1

(b)(i) Z 1 1

(ii) Z atom has achieve a stable octet electron arrangement.

Atom does not donate or receive / share electron with

other atom.

1

1

2

(c)(i) VW

1 1

(ii) Draw

+ -

1. Correct number of shell and eletron for ion V and ion W 2. Correct charge of ion V and ion W

1

1

2

(d)(i) W / Y 1 1

(ii) U 1 1

(iii) T 1 1

Total 10

W V

3

No Explanation Sub Mark Total Mark

3(a)(i) 2.8.6 1 1

(ii) 16 1 1

(b)(i) Atomic size decrease / become smaller 1 1

(ii) The number of proton increase/ positive charge increase

Force of attraction between neucleus and valence

electron increase

1

1

2

(c) Has achieve a stable octet electron arrangement

Atom does not donate, receive or share electron with

other atom.

1

1

2

(d) Al / Aluminium 1 1

(e) 2Na + 2H2O 2NaOH + H2

Correct the formula reactant and product

Balance equation

1

1

2

Total 10

No Explanation Sub Mark Total Mark

4(a) K

J/ L / M

1

1

2

(b)(i) J/L/M 1 1

(ii) Atom of J/L/M is not stable// has 1/6/ 7 valence

electron

2 atoms of J/L/M share a pair of electron

To achieve duplet/octet electron arrangement

1

1

1

3

(c)(i) 17 1 1

(ii) 17, atom has 7 valence electron 1 1

(d) M,L, K 1 1

(e) K+ 1 1

(f) KM 1 1

Total 11

No Explanation Sub Mark Total Mark

5(a)(i) Magnesium / Mg 1 1

(ii) Mg2+ 1 1

(iii) Chlorine / Cl2 1 1

(iv) gas 1 1

(v) has free moving ions 1 1

(b)(i) covalent compound 1 1

(ii) T2 oC 1 1

(iii) becomes slower 1 1

(iv)

1 1

Total 9

4

No Explanation Sub Mark Total Mark

6(a)(i) 0.125 mol 0.125x 6.02 x 10

23 // 1.7525 x 10

23 molecules

1 1

3

(ii) 0.125 x 44 = 5.5 g 1 (b)(i) Chemical formula that shows simplest ratio of atoms of

elements/each element in a compound 1 7

(ii) Mg = 2.4g , O = 1.6g 1+1 (iii) 1 : 1 1 (iv) MgO 1 (v) To allow oxygen enter the crucible 1 (vi) Copper is less reactive towards oxygen than hydrogen 1

Total 10

No Explanation Sub Mark

Total Mark

7(a) Chemical formula that shows actual number of atoms of elements/each element in a molecule

1 1

(b) Number of mole 1 1 (c)(i) Method 1 1 3 (ii) Magnesium is reactive //Magnesium reacts easily with

oxygen 1

(iii) To allow oxygen/air to enter the crucible//To ensure magnesium reacts completely

1

(d)(i) Mass of Pb = 49.68 g 1 5 (ii) No.of mole of Pb = 0.24 mol 1 (iii) Mass of oxygen = 3.84 g 1 (iv) No. of mole of O = 0.24 mol 1 (v) Empirical formula = PbO 1

Total 10

No Explanation Sub Mark

Total Mark

8(a)(i) Chemical formula that shows the simplest ratio of atoms of each element in the compound

1 2

(ii) To dry hydrogen gas//To absorb water/moisture 1 (b)(i) Copper = 8.00g

Oxygen = 2.00g 1 1

4

(ii) 8/64 : 2/16 = 1:1 1 (iii) CuO 1 (c)(i) To avoid the oxidation of copper//To avoid the formation of

copper oxide 1 2

(ii) Repeat the process of heating, cooling and weighing until the mass of copper is constant/no change

1

(d)(i) Magnesium is more reactive towards oxygen than hydrogen//Magnesium is placed above hydrogen in the

reactivity series

1 2

(ii) Iron oxide/Tin oxide/Lead oxide/Silver oxide 1

Total 10

5

SECTION B

No Rubric Sub Mark Marks

9(a) 1.Number of proton for both isotopes is 6

2.Number of electron for both isotopes is 6

3.Nucleon number for first isotope is 12

4.Nucleon number for second isotope is 14

5.Both isotope has different physical properties

6.Both isotope has same chemical properties

1

1

1

1

1

1

6

(b) (i) Naphthalene / acetamide 1

(ii) 1. Water 2. Melting point of X below than boiling point of

water / uniform heating

1

1

2

(iii) From 0 to t1 minutes

1. In solid state 2. Particles are closely packed and in orderly

arrangement

3. Kinetic energy of particles very low 4. Attraction force between particles very strong

From t1 to t2 minutes

1. In solid and liquid state 2. Some particle are closely packed and in orderly

manner and some particle are slightly lose packed

and in orderly manner

3. Kinetic energy of particles constant

From t2 to t3 minutes

1. In liquid state 2. Particles are closely packed and in orderly manner 3. Attraction forces between particles become weaker

1

1

1

1

1

1

1

1

1

1

Max 9

(v) 1. correct curve

2. Label freezing point ( 80 o C )

Sample answer

1

1

2

TOTAL 20

Temperature (oC)

Time (min)

80 oC

6

No Explanation Sub Mark Total Mark

10 (a) Electron arrangement of helium atom is 2 // Helium atom has two valence electrons. Achieved a duplet electron arrangement. Helium atom will not gain, lose or share electrons with other

atoms.

Electron arrangement of chlorine atom is 2.8.7 // Chlorine

atom has seven valence electrons. Chlorine atom needs one electron to achieve the octet

electron arrangement. Thus, two chlorine atoms share one pair of electrons.

1 1 1

1

1

1

6

(b)(i)

Covalent compound

1

1 + 1

3

(b)(ii) Electron arrangement of magnesium atom is 2.8.2 Magnesium atom donate 2 electrons to achieve the stable

octet electron arrangement A positive ion / magnesium ion / Mg

2+ ion is formed //

Mg Mg2+ + 2e // [Diagram] Electron arrangement of chlorine atom 2.8.7 chlorine atom accepts 1 electron to achieve the stable octet

electron arrangement A negative ion / chloride ion / Cl

ion is formed //

Cl2 + 2e 2Cl

// [Diagram]

Mg2+

and Cl ions are attracted to each other by a strong

electrostatic/ionic force.

1

1

1

1

1

1

1

7

(c) Naphthalene cannot conduct electricity/non-electrolyte. Naphthalene does not has ions // exist as molecule. Sodium chloride solution conduct electricity/ an electrolyte. In sodium chloride solution, sodium ions/Na

+ and chloride

ions/ Cl are free to move.

1 1 1

1

4

Total 20

C Cl Cl

Cl

Cl

7

No Explanation Sub Mark Total Mark

11(a)(i) Al3+ , Pb4+ 1+ 1 4 (ii) Aluminium oxide

Lead(IV) oxide 1 + 1

(b)(i) (CH2O)n = 60 12n + 2n + 16n = 60 n = 2 Molecular formula = C2H4O2//CH3COOH

1

1 1

3

(ii) CaCO3 + 2CH3COOH (CH3COO)2Ca + H2O + CO2

2 2

(c)(i) 1.Green solid turn Black 2. Lime water becomes cloudy

1 1

2

(ii) CaCO3 CaO + CO2 1 + 1 2 (iii) 1. mol of calcium carbonate decomposed into 1 mol of

calcium oxide and 1 mol of carbon dioxide 2. Calcium carbonate is in solid state, calcium oxide is in

solid state and carbon dioxide is in gaseous state

1

1

2

(iv) 1. No. of mole for CuCO3 = 12.4 / 124 = 0.1 mol

2. 1 mol of CuCO3 produces 1 mol of CuO

Therefor No. of mole for CuO = 0.1 mol

3. Mass of CuO = 0.1 X 80 g = 8 g

1 1

1

3

(v) Mass of oxygen is 0.8g

Simplest mol ratio : Cu : O = 3.2/64 : 0.8/16 = 1 : 1

1 1

2

Total 20

8

No Explanation Sub Mark Total Mark

12(a)(i) Cation Anion

1 1

5

(ii) Iron(III) ion and chloride ion 1 + 1 (iii) K2SO4 1 (b)(i) Lead(II) sulphate

Pb(NO3)2 + K2SO4 PbSO4 + 2KNO3 1

1 + 1 8

(ii) 1 mol of lead(II) nitrate reacts with 1mol of potassium sulphate producing 1 mol of lead(II) sulphate and 2 mol of potassium

nitrate.

1

(iii) 1.No.of mol of K2SO4 = 10 x 0.5 = 0.005 mol 1000 2. 1 mol of K2SO4 producing 1 mol of PbSO4 No of mol PbSO4 = 0.005 mol 3. Mass of PbSO4 = 0.005 x 303= 1.52 g

1

1

1 + 1

(c)(i) 1.No.of mole:- C = 52.2/12 H = 13/1 O = 34.8/16 = 4.35 = 13 = 2.175 2. Mol Ratio:- C : H : O 4.35 / 2.175 : 13.0 / 2.175 : 2.175 / 2.175 Simplest mol ratio:- C : H : O 2 : 6 : 1 3. Empirical formula: C2H6O 4. (C2H6O)n = 46 (2 x12)n + (6 x1)n + (1x16)n = 46 5. n = 1 6. Molecular formula : C2H6O 7. Structural formula : H H H C C OH H H

1

1 1 1

1 1

1

7

Total 20

9

SECTION C

No Explanation Sub Mark

Total

Mark

13 (a)(i) Y more reactive 1 5 Atomic size of Y bigger than X // The number of shell

occupied with electron atom Y more than X. 1

The single valence electron becomes further away from the

nucleus. 1

the valence electron becomes weakly pulled by the nucleus. 1 The valence electron can be released more easily. 1

(ii) Name : Sodium 4Na + O2 2Na2O Chemical formulae Balance equation

1

1 1

3

(b) Put group1 metal into bottle that contain paraffin oil To avoid react with air

1 1

2

(c) Name : Sodium/any group 1 element Material : group 1 elements, water, Apparatus : forceps , knife, filter paper, basin, litmus

paper.

1

1

max10

[procedure] 3. Pour some water into the basin 4. Group 1 elements is take out from paraffin oil using

forceps 5. A small piece of group 1 element is cut using a small

knife 6. Oil on group 1 element is dried using a filter paper 7. The group 1 element is placed in the basin contain water. 8. Put litmus paper into water

1 1 1 1 1 1

[observation] 9. Color of red litmus paper turn to blue

1

[chemical equation ] Sample answer 2 Na + 2 H2O 2NaOH + H2 Chemical formulae Balance equation

1 1

Total 20

10

No Explanation Sub Mark

Total

Mark

14. (a) CO2 // any covalent compound covalent

Intermolecular forces are weak

Small amount of heat energy needed to overcomes the

forces

1

1

1

1

4

(b) X = 2.1 X = 2.2

Y = 2.7 // Y = 2.6 // suitable

electron aranggement

Ionic bond

to achieve octet electron arrangement

One atom of X donates 1 electron to form ion X+

One atom of Y receives an electron to form ion Y-

Ion X+ and ion Y

- are attracted together by the strong

electrostatic forces

1

1

1

1

1

1

1

7

(c) material and apparatus;

compound XY, Carbon electrode, cell, wire, crucible,

bulb/ammeter/galvanometer

Procedure

A crucible is filled with solid XY

Dipped two carbon electrode

Connect two electrode with connecting wire with bulb

Observed whether bulb glow

Heated the solid XY in the crucible

Observed whether bulb glow

Observation

Solid XY - bulb does not glow

Molten XY - bulb glow

Diagram

Functional diagram

labeled

1

1

1

1

1

1

1

1

1

9

TOTAL 20

11

No Explanation Sub Mark

Total Mark

15 (a) By burning metal P in the air/oxygen Because metal P is a reactive metal/ react actively with

oxygen/place higher than hydrogen in the reactivity series

1

1

2

(b)(i) Element Y O

1. Mass of element (g) 2.07g 0.32g

2. Number of moles 2.07

207

0.32

16

Ratio of moles 0.01 0.02

3. Simplest ratio of moles 1 2

Empirical formula of copper oxide : YO2

1

1

1

1

4

(ii) 1. Collect the gas into a test tube 2. Put/place a lighted wooden splinter at the mouth of the test tube 3. no pop sound,

1 1 1

3

1. Weigh A crucible and its lid 2. Clean the magnesium ribbon with sandpaper. 3. Weigh the crucible with its lid and content 4. Heat the crucible without its lid with a strong flame 5. When the magnesium ribbon starts to burns cover the crucible

with its lid. 6. Using a pair of tongs lift the lid a little at intervals (open once in

a while) (to allow oxygen from the air to enter for the

combustion of magnesium). 7. When the burning is complete, remove the lid and heat the

crucible strongly. 8. Allow the crucible to cool to room temperature with its lid still

on 9. Weigh the crucible with its lid and content again 10. Repeat the process of heating, cooling and weighing until a

constant mass is obtained. Record the constant mass obtained. 11.Result: Description Mass Mass of crucible + lid (a) = a g Mass of crucible + lid + magnesium ribbon (b) = b g Mass of crucible + lid + magnesium oxide(c) = c g 12.

Element Mg O

Mass of element (g)

(b-a) (c-b)

. Number of moles (b-a) 24

(c-b) 16

Ratio of moles x y

Or: Simplest ratio of

mole

1

1

13. Empirical formula of magnesium oxide: MgxOy / MgO

1 1 1 1 1

1

1

1

1 1

1

1

1

Max;

11

Total 20

12

No Explanation Sub Mark

Total Mark

16 (a) 1. Name of the c ompound is sulphur trioxide 2. Sulphur trioxide is made up of two elements, which is sulphur and oxygen 3.One sulphur atom combine with three oxygen atoms

1 1

1

3

(b)(i) Hydrochloric acid Calcium chloride n = 2

1 1 1

3

(ii) 1.No. of moles of HCl = 0.5 x 30 = 0.015 1000 2. 2 mol of HCl produces 1 mol of CO2 No. of moles of CO2 = 0.015/2 = 0.0075 mol 3. Volume of CO2 = 0.0075 x 24 = 0.18 dm

3

1

1

1 + 1`

4

(c) 1. Weigh a combustion tube with the porcelain dish in it and record the reading.

2. Add a spatulaful of metal oxide on to the porcelain dish and weigh the combustion tube again

3. Allow the dry hydrogen gas to flow .into the combustion tube for 10 minutes to expell ( remove) all the air in the tube.

4. Collect a sample of gas from the small hole of the tube. Introducing a burning splinter into the test tube. If there is no

pop sound, then all the air has been removed. Burn the excess hydrogen gas.

5. Heat metal oxide strongly while the hydrogen gas is flowing. 6. Stop the heating when metal oxide turns completely from

green to grey. 7. Continue the flow of hydrogen gas until the set of apparatus

cools down to room temperature 8. Weigh the mass of the combustion tube with its content . 9. Repeat the heating, cooling and weighing process until a

constant mass is obtained. 10 Result: Mass of combustion tube + porcelain dish = a g Mass of combustion tube + porcelain dish + Metal oxide (b)

= b g Mass of combustion tube + porcelain dish + Metal M

= c g 11.

Element M O S1: Mass of M (c-a) (b-c) S3: No. of moles c-a

56 b-c 16

S3: Ratio of moles x

y

S4: Simplest ratio of

moles

1 1

12.Therefore: Empirical formula of copper oxide : MxOy/MO

1

1

1

1

1 1

1

1 1

1

1

1

Max:

11

Total 20

13

SET 2

SECTION A

No Explanation Sub Mark Total Mark

1(a) Chemical to electrical energy 1 1 (b) Cu2+ , H+ , OH- and SO4

2- 1 1 (c) (i) zinc 1 1

(ii) Zinc more electropositive than copper// zinc above copper in the Electrochemical series.

1 1

(d)(i) Brown solid deposited // Copper becomes thicker. 1 1 (ii) Cu2+ + 2e Cu 1 1 (e) Blue to colourless // Blue become fade /paler // The intensity

of blue colour decreases The number of copper(II)ions decreases // The concentration

of copper(II)ions decreases

1

1

2

(f)(i) Cu , R , Q ,P 1 1 (ii) 1.5V 1 1

Total 10

No Explanation Sub Mark Total Mark

2(a)(i) Redox reaction // oxidation and reduction 1 1

(ii) Zn + 2Fe3+ Zn2+ + 2Fe2+

[Correct chemical formulae of reactants and products] [Balance the equation correctly.]

1 1

2

(iii) Sodium hydroxide solution is added slowly into the solution until in excess. Green precipitate that is insoluble in excess sodium

hydroxide solution is produced//White precipitate that is

soluble in excess sodium hydroxide to produce colourless

solution.

1

1

2

(b)(i) Pink colouration / spot is observed 1 1

(ii) Blue colouration / spot is observed 1 1

(c) When iron is in contact with zinc, iron does not rust. When iron is in contact with copper, iron rusts.

1 1

2

(d) Apply grease on the surface//apply paint on the surface // galvanising // tin plating

1 1

Total 10

14

SECTION B

No Explanation Sub Mark

Total

Mark

3(a)(i) Cathode : hydrogen Anode : Chlorine

Electrode Cathode Anode

Ions that are

attracted

Mg2+

and H+ Cl

- and OH

-

Ions that are

selectively

discharged

H+ Cl

-

Reason H+ is lower in the

electrochemical

series.

Concentration of Cl-

higher than OH-

Half

equation

2H+ + 2eH2 2Cl

- Cl2 + 2e

1

1

1+1

1+1

1+1

1+1

10

(ii) cathode : hydrogen anode : oxygen

1

1

2

(b) Aspect Cell X Cell Y

Types of cells Electrolytic cell Voltaic cell

Energy

changes

Electrical to chemical Chemical to electrical

Name of

electrodes

Anode : Copper

Cathode : Copper

Positive electrode :

Copper

Negative electrode:

Magnesium

Ions in the

electrolyte

Cu2+

,H+ ,SO

2-4 , OH

- Cu

2+ ,H

+ ,SO

2-4 , OH

-

Half equations Anode :

CuCu2+ + 2e Cathode :

Cu2+

+ 2e Cu

Negative electrode :

MgMg2+ + 2e Positive electrode:

Cu2+

+ 2e Cu

Observations Anode : Copper

becomes thinner

Cathode :brown solid

deposited//copper

becomes thicker

Positive electrode:

brown solid

deposited//copper

becomes thicker

Negative electrode :

Magnesium become

thinner.

1

1

1

1

1

1

1

1

8

Total 20

15

No Explanation Sub Mark

Total

Mark

4(a)(i) Oxidation number of sodium is +1 Oxidation number of lead is + 4

1

1

2

(ii) Na2 O Sodium oxide PbO2 Lead(IV) oxide

1

1

2

(b)(i) Metal P = Copper Metal Q = Zinc

1

1

10

(ii) Experiment 1 Iron rust Iron more electropositive than copper // Iron above copper

in the electrochemical series Fe Fe2+ + 2e Experiment 2 Iron does not rust // The solution is alkaline Iron less electropositive than zinc // Iron below zinc in the

electrochemical series. O2 + 2H2O + 4e 4OH

-

1

1

1+1

1

1

1+1

(c) Sample answer Metal X : Magnesium Magnesium atom releases two electrons to form magnesium ion Magnesium is oxidized to magnesium ion/ undergoes oxidation Magnesium is a reducing agent Copper (II) ion accepts two electrons to form copper atom Copper (II) ion is reduced to copper atom/ undergoes reduction Copper (II) ion is an oxidizing agent

1

1

1

1

1

1

1

1

Max 5

Total 20

16

SECTION C

No Explanation Sub Mark Total Mark

5(a) Electrode X : Hydrogen Electrode Y : Oxygen

1

1

2

(b) Electrode X Y

Ions that are

attracted to Na

+ , H

+ SO42-

, OH -

Name of ions

that are

selectively

discharged

Hydrogen ion hydroxide ion

Reason -

Position of OH- is

lower than SO42-

in

the electrochemical

series

1+1

1+1

1

5

(c) [Sample answer : Sulphuric acid // any suitable solution ] 2H

+ + 2e H2

1

1+1

3

(d) [Sample answer : metal A = Copper ;A nitrate solution = Copper(II) nitrate solution or any suitable answer ]

1 10

1. [Functional set up of apparatus] 2. [Label : Zinc , copper , zinc nitrate solution , copper(II)

nitrate solution and porous pot]

3. [Mark of negative electrode zinc plate Mark of positive electrode copper plate]

4. [Mark of the electron flow from zinc to copper]

1

1

1

1

Put/fill/pour zinc nitrate solution a beaker. Put copper(II) nitrate solution into a porous pot.

Put the porous pot into the beaker.

Dip/put zinc plate into zinc nitrate solution.

Dip/put copper plate into copper(II) nitrate solution.

Connect the wire // complete the circuit.

1 1 1 1 1 1

Total 20

Copper plate Zinc plate

Copper(II) nitrate Zinc nitrate

Porous pot

17

No Explanation Sub Mark Total Mark

6(a) Reaction II is a redox reaction Oxidation number of magnesium changes from 0 to +2 Oxidation number of zinc changes from +2 to 0 No change in oxidation number of elements in reaction I

1

1

1

1

4

(b) Reactivity of metals in descending order is Q, carbon, R, P Experiment I Reaction between carbon and oxide of metal P occurs carbon

is more reactive than metal P Experiment II Reaction between carbon and oxide of metal Q does not

occur, metal Q is more reactive than carbon Experiment III Reaction between carbon and oxide of metal R occurs carbon is more reactive than metal R Reaction between carbon and oxide of metal P more

vigorous than reaction between carbon and oxide of metal R Metal P is less reactive than metal R.

1

1

1

1

1

1

6

Sample answer Fe

3+Fe2+ Magnesium as a reducing agent Add magnesium to iron(III) chloride solution Heat the mixture Filter the mixture Add sodium hydroxide solution Green precipitate is formed Fe

2+ Fe3+

Chlorine as an oxidizing agent Add chlorine water to solution containing Fe

2+

Stir the mixture Add sodium hydroxide solution Brown precipitate is formed

1

1

1

1

1

1

1

1

1

1

1

Max 10

Total 20

18

SET 3

SECTION A

No. Answer Mark 1 (a) (i) Solution in test tube C 1

(ii) 1. Solution in test tube A 2. Concentration of H+ ion in test tube A is the highest

1 1

(b) 1. Higher than pH value of 0.1 moldm-3

HCl // The pH is /5/6

2. Ethanoicacis is a weak acid// Etanoic acid ionizes partially in water to produce

low concentration of hydrogen ion

3. The lower the concentration, the higher the pH value

1 1

1

(c ) (i) Magnesium chloride and hydrogen 1

(ii) Mg + 2HCl MgCl2 + H2 1. Correct formula of reactant and product

2. Balanced equation

1 1

(iii) No of mole, HCl = 0.1 x 5 / 1000 = 0.0005 mol Based on balanced equation, 2 mol of HCl : 1 mol of H2 0.0005 mol of HCl : 0.00025 mol of H2 // mol of H2 = 0.005/2 = 0.0025 Volume of hydrogen gas = 0.00025 x 24 dm

3

= 0.006 dm3 // 6 cm

3

1

1 1

(d) White precipitate 1

TOTAL 13

No. Answer Mark 2 (a) (i) Solvent P: Water

Solvent Q: methyl benzene / propanone / suitable organic solvent

1 1

(ii) Effervescence / gas released // magnesium ribbon become thinner

1

(iii) 1. In the presence of solvent P/water , ethanoic acid ionize to form H+ ion.

2. H+ ion causes the ethanoic acid to show its acidic properties

3. In solvent Q, ethanoic acid exist as molecule// hydrogen ion does not present

1 1 1

(b) (i) 1. pH value increase / bigger 2. Concentration of acid is lower

1 1

(ii) (0.5)(V) = (0.04)(250) // V = 20 cm

3

1

1

V =0.04 250

0.5

19

3

(a) Alkali that ionize/dissociate completely in water to produce high concentration of

hydroxide ions. 1 1

(b) Alkaline / alkaline solution 1 1

(c) P: ion Q: molecule

1 1

2

(d) No Because there are no hydroxide ions in the solution// ammonia exist in the form of

molecule.

1 1

2

(e) (i) Colourless gas bubbles are released.// effervescence 1 1

(ii) Mg + 2HCl MgCl2 + H2 1. Correct formula 2. Balanced equation

1

1

2

(iii) Mol of Mg = 2.4/24 // 0.1 mol

Volume of H2 = 0.1 24 dm3 = 2.4 dm

3

1

1

2

Total 11

No. Answer Sub Mark Mark 4 (a)(i) Pink to colourless

1

1

(ii) 1. Correct formula 2. Balanced equation

H2SO4 + 2KOHK2SO4 + H2O

1 1

2

(iii)

ConcentrationKOH= 2 0.1 X 15

25

= 0.12 moldm

-3

1

1

2

(b)(i) Potassium sulphate 1 1 (ii) Mol of H2SO4=

0.1 X 15

1000 // 0.0015

1 mole H2SO4 produce 1 mole K2SO4 // 0.0015 mole H2SO4 produce 0.0015 mole K2SO4 Mass Salt X = 0.0015 X 174 = 0.261 g

1

1

1

3

I(i) 30 cm3 // double 1 1 (ii) 1. H2SO4 is diprotic acid while HNO3 is monoprotic acid

2. Number of mole H+ ion in HNO3 is half compare to H2SO4

1 1

2

Total 12

20

NO ANSWER MARK

5 (a) (i) Green

(ii) Double decomposition reaction/ precipitation reaction

1

1

(b) (i) carbon dioxide

(ii) CuCO3 CuO + CO2 1. Reactants and products are correct

2. Equation is balanced

(iii)

- Labelled diagram

- Functional

1

1

1

1

1

(c) (i) Sulphuric acid // H2SO4 1

(ii)

mol CuCO3 = 12.4/124 = 0.1 mol

ratio CuCO3 : CuSO4

1 : 1

Mol of CuSO4 = 0.1 x 135g

Mass = 13.5g

1

1

1

10

No. Answer Mark

6 (a) Mg + 2HCl MgCl2 + H2 1+1

(b) (i) 0.4/24 = 0.0167 mol 1

(ii) The number of mole of HCl = MV/1000 = 1x 50/1000 = 0.05 mol 1

(c) From the chemical equation 1 mol of magnesium produce 1 mol hydrogen

If 0.0167 mol produce 0.0167 mol hydrogen

Volume of hydrogen = 0.0167 x 24 dm3= 0.4 dm

3/ 400 cm

3

1

1

(d) I 400 /100 =4 cm3s

-1

II 400 /60 = 6.67 cm3s

-1

1

1

(e) As catalyst 1

(f) The temperatureof hydrochloric acid

The concentration of hydrochloric acid

1

1

TOTAL 11

Copper(II)

carbonate

Lime water

Heat

21

7 Marks

a i. Catalyst 11

ii. Change in total mass of conical flask and its content 1....1

b i. V cm3 // same volume

No. of moles of HCl used is the same // catalyst does not increase the

quantity of product

1

12

ii.

Same volume 1 Gradient of curve for Expt II steeper than Expt 1 -1

1

1.2

iii

.

1. Catalyst lowers the activation energy // provides an alternative reaction pathway that requires lower activation anergy

2. Frequency of effective between H+ and zinc atom increases 3. Rate of reaction increases

1

1

1..3

c.

1..1

10

Time / s

Volume of gas

released/ cm3 Key :

Experiment I :

Experiment II : ..

Time/s

Volume of

carbon

dioxide /cm3

Experiment A

Experiment B

Experiment C

Graph 5

22

No. Answer Mark 8 (a) The amount heat change/released when 1 mol of copper is displaced by

magnesium from copper(II) sulphate solution 1

(b) Higher rate of reaction // Reaction is faster

1

(c) Correct formulae of reactants Correct formulae of products

Mg

2+ + Zn Mg + Zn2+

1 1

(d)(i) H = 50 x 4.2 x 5 J // 1050 J // 1.05 kJ (r: without unit)

1

(ii) n = // 0.025

1

(iii) H = //

0.025 mol CuSO4 produce 1050J 1 mol CuSO4 produce 1050 0.025 = - 42 000 J mol

-1 // - 42 kJ mol

-1 (r: without unit)

1

1

1

(e) Arrow upward with label energy and two levels Correct position of reactans and products Energy Cu

2+ + Mg

H = -42kJ mol-1 Cu + Mg

2+

1 1

(f) Reduce heat loss to surrounding. [r:prevent]

1

TOTAL 12

0.5 x 50

1000

1050

0.025

1.05

0.025

23

9(a) Heat released when 1 mol of metal/copper is displaced from its salt solution by a more electropositive metal/zinc

1

(b) To reduce heat lost to the surrounding 1

(c) Exothermic 1

(d)(i) Q= 50 X 4.2 X8 //1680 J //1.68 kJ 1

(ii) Mol CuSO4 = 50 x0.2 /1000 // 0.01 Heat of displacement= -1.68 0.01 // -168 kJ mol

-1 1 1

(e)

Arrow upword with label energy and two label - 1m Correct potion of reactant and product - 1m H = -168kJ mol-1 - 1m

1

1

1

(f)(i) Same//8 0 C 1

(ii) Heat produce/released is double

the heat released is distributed over a volume of solution which is doubled

1 1

12

10(a) - Heat released when one mol of propanol burnt in excess oxygen 1

(b) - The reaction is an exothermic reaction /heat released to the surroundings - 1 mole of propanol burnt in excess oxygen released 2015 kJ of heat energy

1

1

(c)(i) No. of mol of C3H7OH= 1.2/60 // 0.02 ..1 Heat released= 2015 X 0.02 =40.3 kJ//40300 J ..1

1

1

(ii) = 40.3 x 1000 // 40300 200 x 4.2 200 x 4.2 ..1 = 47.98 0C 1

1

1 (d ) Energy

C3H7OH + 9/2 O2 H = -2015 kJ mol -1

3CO2 + 4H2O

l. Arrow upwards with energy labeled and two energy level shown 2. Reactants and products are at the correct energy level 3. H = -2015 kJ mol-1 ( the value , unit and position)

1

1

1

(e) Use wind shield//use thin copper can//never use wire gauze//stir the water with

thermometer//weigh the spirit lamp immediately before and after burning 1

Energy

Zn 2+

+ Cu

//nSO4Z

Zn + Cu 2+

//Zn + CuSO4

H = -168 kJ mol-1

24

(f) 1.The heat of combustion of ethanol is higher 2.Number of carbon atoms per molecule in ethanol is higher 3. more carbon dioxide and water molecules are produced 4. more heat is released from the formation of bond in carbon dioxide and water

molecules max 3

1 1

1

1 max

3

TOTAL 14

SECTION B

No. Answer Sub Mark Mark 11 (a)(i) [Label of axes with units]

[All points are transferred correctly] [Correct shape of the graph and constant scale]

1 1 1

3

(ii) 2.5 cm3 (r: without unit) moles of Pb

2+ ions = // 0.0025

moles of I

- ions = // 0.005

0.0025 mol Pb

2+ : 0.005 mol I

- 1 mol Pb

2+ : 2 mol I

- Correct formulae of reactants and product Balanced equation Pb

2+ + 2I

- PbI2

1

1

1

1 1

1 1

7

(b)(i) Salt J : lead(II) nitrate // Pb(NO3)2 X oxide : lead(II) oxide // PbO Gas Y : nitrogen dioxide // NO2 Gas Z : oxygen // O2 Yellow precipitate : lead(II) iodide // PbI2

1 1 1 1 1

5

(ii) Nitrate ion Add sulphuric acid Add iron(II) sulphate solution Slowly and carefully add concentrated sulphuric acid Brown ring formed

1 1 1 1 1

5

TOTAL 20

1.0x 2.5 1000

1.0x 5 1000

25

No. Answer Sub

Mark Mark

12(a)(i) Carbonate ion // CO32-

1 1

(ii)

Salt S : Copper(II) carbonate // CuCO3

Compound T: Copper(II) oxide // CuO

Gas U: Carbon dioxide // CO2

Compound W: Copper(II) sulphate // CuSO4

1

1

1

1

4

(b)

1. Correct formulae of reactants and products

2. Balanced equation

CuO + H2SO4 CuSO4 + H2O

1

1

2

1. Add sodium hydroxide solution

2. Blue precipitate formed indicate the presence of Cu+ ion

3. Add hydrochloric acid

4. Add barium chloride solution

5. White precipitate formed indicate the presence of SO42-

ion

1

1

1

1

1

5

(d)(i) Salt X : Barium sulphate

Salt Y: Copper(II) nitrate

1

1

2

(ii) White precipitate

Double decomposition reaction

1

1

2

(iii)

Ba2+

+ SO42- BaSO4

Number of mol Ba2+

= 0.1 x 50

1000 // 0.005

1 mol Ba2+

produce 1 mol BaSO4 //

0.005 mol Ba2+

produce 0.005 mol BaSO4

Mass BaSO4 = 0.005 x 233 g // 1.165 g

1

1

1

1

4

TOTAL 20

26

No. Answer Sub Mark Mark 13 (a)(i) 1. copper(II) oxide and carbon dioxide

2. Correct formulae of reactants and products 3. Balanced equation

CuCO3CuO + CO2

1 1 1

3

(ii) 1. Yellow precipitate 2. Lead(II) iodide and potassium nitrate 3. Correct formulae of reactants and products 4. Balanced equation

Pb(NO3)2 + 2KI PbI2 + 2KNO3

1 1 1 1

4

(b) Test for cation

1. Pour aluminium nitrate solution and zinc nitrate solution into two different test tube

2. Add ammonia aqueous solution until excess 3. White precipitate insoluble indicate the presence of Al3+ 4. White precipitate dissolve in excess ammonia aqueous

indicate the presence of Zn2+

ion

Test for anion

5. Pour aluminium nitrate solution and zinc nitrate solution into two different test tube

6. Add sulphuric acid and iron(II) sulphate solution 7. Slowly and carefully add concentrated sulphuric acid 8. Brown ring formed indicate the presence of NO3

- ion

1

1 1

1

1 1 1 1

8

(c)(i) 1. Correct formulae of reactants and products 2. Balanced equation

CuO + H2SO4 CuSO4 + H2O

1 1

2

(ii)

Mol of H2SO4 = 1.0 x 25

1000 // 0.025

1 mole of H2SO4 produce 1 mole of CuSO4 // 0.025 mole of H2SO4 produce 0.025 mole of CuSO4 Mass of CuSO4 = 0.025 X [64 + 32 + 4(16)] = 4g

1

1

1

3

Total 20

27

No. Answer Mark

14 (a) (i) Size of the solid reactant

Concentration of the solution

Temperature of the reaction mixture

Catalyst

1

1

1

1 ..4

(ii) Temperature : 450-550oC

Catalyst : iron powder

Pressure : 200 atm

1

1

1...3

(b) (i) The axes are labeled together with its unit

The scale is correct

The points are transferred correctly

The curve is smooth

1

1

1

1..4

(ii) Average rate of reaction for experiment I = 26.0

210

= 0.12 cm3 s

-1

Average rate of reaction for experiment II = 26.0

150

= 0.17 cm3 s

-1

[correct unit]

1

1

1

1..4

(iii) 1. The rate of reaction for Experiment II is higher than in Experiment I

2. The concentration of HCl in Experiment II is more/higher than in

Experiment I

3. The number of hydrogen ion/ H+ per unit volume of the solution in

Experiment II is more than in Experiment I

4. The frequency of collisions between hydrogen ion and calcium

carbonate in Experiment II is higher than in Experiment I

5. The frequency of effective collisions hydrogen ion and calcium

carbonate in Experiment II is higher than in Experiment I

1

1

1

1

1..5

TOTAL 20

28

No. Answer Mark

15 (a) Use magnesium powder

Increse the concentration sulphuric acid

Increase the temperature of the reaction mixture

Use copper(II) sulphate as catalyst [any 3 of the above]

1

1

1

1..3

(b) (i) Manganese(IV) oxide/ copper(II) oxide/ lead(IV) oxide 1. 1

(ii) 1. Changes / alters the rate of a reaction 2. Require in small amount 3. Does not increase the quantitiy of products formed 4. Remain chemically unchanged at the end of the reaction

[Any of the 2]

1

1

1

1..2

(iii) Conditions Justification

Vanadium(V) oxide As a catalyst // to increase the rate of reaction

450 500 oC Optimum temperature to maximise the amount of products at a resaonable rate of reaction

1 atm Not economical to use high pressure as the yield is

low

Any two of the conditions [2 + 2]

1+1

1+1

1+1

4

(c) 1. Rate of reaction decreases with time 2. Concentration of hydrochloric acid solution and 3. total surface area of calcium carbonate decreases with time

1

1

1..3

(d) Materials :20-volumes hydrogen peroxide, manganes(IV) oxide Apparatus : test-tubes, measuring cylinder, test-tube rack, glowing splinters

Procedure : 1. Measure 5 cm3 20-volumes of hydrogen peroxide solution and pour into

test-tube labeled A and B.

2. Add half spatula of manganese(IV) oxide into testt-tube B. 3. Insert a glowing splinter into test-tube A and B respectively 4. Record observations

Test-tube Observation

A (without catalyst) Glowing splinter does not relight

B (with catalyst) Glowing splinter relights

Conclusion :

The presence of catalyst manganese(IV) oxide increases the rate of

decompositionof hydrogen peroxide.

1..1

1

1

1

1..4

1..1

1..1

TOTAL 20

29

16 (a) (i) Characteristic

Figure 16.1(i)

Figure 16.2(ii)

Type of chemical

reaction Exothermic reaction

Endothermic reaction

Energy content of

reactants and

products

The energy content in

the reactants more

than the energy content

in the products

The energy content in the

reactants less than the

energy content in the

products

Ionic equation

Mg + Fe2+

Mg2+ + Fe

Ca2+

+ CO32-

CaCO3

1 + 1

1 + 1

1 + 1

(ii) Number of moles of FeSO4 = MV 1000 = (0.2)(50) = 0.01 mol 1000 Heat change = 0.01 x 200 kJ = 2 kJ // 2000 J Heat change = mc = 2000 (50)(4.2) = 9.5 oC

1

1 1

(b) 1. 1 mole of silver nitrate solution ionise to produce 1 mole of Ag+ ion and 1 mole of

sodium chloride solution ionise to produce 1 mole of Cl- ion

2. The heat of reaction of silver chloride is heat that released when 1 mole of Ag+ ion

react with 1 mole Cl- ion // Ag

+ + Cl

- AgCl

3. 0.5 mole of magnesium chloride ionise to produce 1 mole of Cl- ion

4. Number of mole of Cl- ion in 1 mole of sodium chloride same as number of mole

of Cl- ion in half mol of magnesium chloride

5. Half of the mole of magnesium chloride produces 1 mole of Cl- ion

6. Number of mole of Cl- ion in 1 mole of sodium chloride same as number of mole

of Cl- ion in half mol of magnesium chloride

1

1 1 1 1 1 Max 4

(c) (i) Heat change = mc = (100)(4.2)(42.2 30.2) = 5040 J / 5.04 kJ Number of moles of HCl / H

+ ion = (50)(2)

1000 = 0.1 mol Number of moles of NaOH / OH

- ion = (50)(2)

1000 = 0.1 mol The heat of neutralization = 5.04 0.1 H = - 50.4 kJ mol-1

1

1

1

(ii) 12.0 oC // same

Number of mole of hydrochloric acid and sodium hydroxide is double Number of mole of water produced is also doubled Heat released is also doubled The heat relased per unit volume of solution is the same // Heat released is

used to increase the volume of solution which is also double

1 1 1 1 1 max 4

30

SECTION C

No. Answer Sub Mark Mark 17 (a) 1. H2SO4

2. 1 mol acid ionises in water to produce 2 mol of H+ 3. HCl // HNO3 // H2SO4 4. Acid that ionises completely in water to produce high concentration of H+

1 1 1 1

4

(b) 1. Sodium hydroxide is strong alkali 2. Ammonia is weak alkali 3. Sodium hydroxide ionises completely in water to produce high

concentration of OH- ion

4. Ammonia ionises partially in water to produce low concentration of OH-

ion 5. Concentration of OH

- ion in sodium hydroxide is higher than in ammonia

6. The higher the concentration of OH- ion the higher the pH value

1 1 1

1

1 1

6

(c) 1. Volumetric flask used is 250 cm3 2. Mass of potassium hydroxide needed = 0.25 X 56 = 14 g 3. Weigh 14g of KOH in a beaker 4. Add water 5. Stir until all KOH dissolve 6. Pour the solution into volumetric flask 7. Rinse beaker, glass rod and filter funnel. 8. Add water 9. when near the graduation mark, add water drop by drop until the graduation

mark 10. Close the volumetric flask and shake the solution

1 1 1 1 1 1 1 1 1

1

10

Total 20

No. Answer Marks

18(a) (i) CuSO4 // FeSO4 // HgCl used as pesticides to destroy pests 1 + 1 2

(ii) Sodium chloride to add flavour to food // Sodium hydrogen carbonate used in baking cakes and breads

1 + 1

2

(iii) Calcium sulphates esquihydrate used to make cement casts to encase fractured bones.

Barium sulphate used on X-ray plates to identify tumours in the intestines.

(Accept other relevant answers)

1 + 1

2

(b) 1. Pour [20-100cm3] of zinc nitrate solution [0.1-1.0mol dm-3] into a

beaker

2. Add [20-100cm3] of sodium carbonate solution [0.1-1.0]mol dm

-3

3. Stir the mixture

4. Filter the mixture

5. wash the residue

6. Zn(NO3)2 + Na2CO3ZnCO3 + 2NaNO3 7. Pour [20-100cm

3][0.1-1.0mol dm

-3]sulphuric acid into a beaker

8. Add the residue/ zinc carbonate into the acid until in excess

9. Stir the mixtureand filter

10. Heat the filtrate until saturated

11. Cool the solution

12. Filter

13.dry the crystal by pressing between two filter paper

14. ZnCO3 + H2SO4 ZnSO4 + H2O + CO2

1

1

1

1

1

1

1

1

1

1

1

1

1

1 14

Total 20

31

Marks

19 a 2H+ + S2O3

2- S + SO2 + H2O All formulae correct - 1 Balanced equation - 1

2..2

b Rate of reaction = fixed mass of sulphur formed 1

per unit time 1

1

1..2

c

Factor Effect

Concentration of

hydrochloric acid /

sodium thiosulpate

solution

The higher the concentration of hydrochloric acid /

sodium thiosulpate solution, the higher the rate of

reaction.

Temperature of

hydrochloric acid /

sodium thiosulpate

solution

The higher the temperature of hydrochloric acid /

sodium thiosulpate solution, the higher the rate of

reaction

Any two of the factors and correct corresponding effect from above

1+1

1+1

4

d 1. Measure 50 cm3 of 1.0 mole dm-3 sodium thiosulphate solution using a 50 cm3 measuring cylinder and pour into a conical flask.

2. Place the conical flask on a piece of paper with the mark X. 3. Measure 5 cm3 of 1.0 mol dm-3 hydrochloric acid using another 10 cm3 measuring

cylinder.

4. Pour the acid into the conical flask quickly and carefully, at the same time start the stop-watch

5. Swirl the mixture in the conical flask quickly 6. Observe the X mark on the filter paper vertically above through the solution 7. Stop the stop watch and record the time taken when the mark X is not visible

through the mixture

8. Repeat step 1- 7 by using the volume of sodium thiosulphate solution, distilled water and acid as shown

Expt Volume of

Na2S2O3 solution

/ cm3

Volume of

distilled water/

cm3

Volume of

hydrochloric acid

/ cm3

Time taken

for mark X to disappear /

s

1 50 0 5

2 40 10 5

3 30 20 5

4 25 30 5

5 10 40 5

10. Plot a graph of concentration against time // 1/time

11. Graph

1

1

1

1

1

1

1

1

1

1

1

1/Time/s

Concentration

of

Na2S2O3,

mol dm-3-

32

12. Conclusion : As concentration of Na2S2O3 solution increases, the rate of reaction with

hydrochloric acid increases // The higher the concentration of Na2S2O3

solution, the higher the rate of reaction with hydrochloric acid

OR To investigate the effect of temperature on the rate of reaction

1. Measure 50 cm3 of 1.0 mole dm-3 sodium thiosulphate solution using a 50 cm

3 measuring cylinder and pour into a conical flask and record the

temperature of sodium thiosulphatesolution.using a thermometer.

2. Place the conical flask on a piece of paper with the mark X. 3. Measure 5 cm3 of 1.0 mol dm-3 hydrochloric acid using another 10 cm3

measuring cylinder.

4. Pour the acid into the conical flask quickly and carefully, at the same time start the stop-watch

5. Swirl the mixture in the conical flask quickly 6. Observe the X mark on the filter paper vertically above through the

solution

7. Stop the stop watch and record the time taken when the mark X is not visible through the mixture

8. Repeat step 1- 7 by using heating sodiumthiosulphate solution to 35oC, 40

oC, 45

oC, 50

oC respectively at step 2. All other condition remains

unchanged.

Expt Temperature of

Na2S2O3 solution

/ oC

Time taken for

mark X to disappear / s

Values of

1/time, s-1

1 25.0

2 35.0

3 40.0

4 45.0

5 50.0

9. Plot a graph of temperature of sodium thiosulphate against time // 1/time

10. Graph

11. Conclusion : As the temperature of Na2S2O3 solution increases, the rate of reaction with

hydrochloric acid increases // The higher the temperature of Na2S2O3

solution, the higher the rate of reaction with hydrochloric acid

1.12

1

1

1

1

1

1

1

1

1

1

1

1.12

Total 20

1/Time/s

Temperature

of

Na2S2O3, oC

33

No. Answer Marks

20 (a) (i) Experiment I hydrochloric acid Experiment II sulphuric acid

Mg + 2HCl MgCl2 + H2

1

2....3

(ii) The number of mole of HCl = MV/1000

= 1.0 x 50 = 0.05 mol

1000 or

The number of mole of H2SO4 = MV/1000

= 1.0 x 50 = 0.05 mol

1000

11

(iii) The rate of reaction is the change of volume of hydrogen gas per unit time 11

(b) (i)

1. Curve with label

2. Axis with title and correct unit

1

12

(ii) 1. Sulphuric acid in experiment II is diprotic acid, hydrochloric acid in experiment

I is monoprotic acid//Concentration of hydrogen ion, H+ in experiment II is

higher than experiment I

2. The number of hydrogen ion per unit volume in experiment II is higher than

experiment I

3. Frequency of collisions between hydrogen ions and magnesium atoms in

experiment II is higher than experiment I

4. Frequency of effective collisions between hydrogen ions and magnesium atoms

in experiment II is higher than experiment I

5. Rate of reaction in experiment II is higher than experiment I

1

1

1

1

15

(c) Diagram : Functional apparatus set-up

Label correctly

Procedure :

1. A burette is filled with water and inverted over a basin containing water. The burette is clamped vertically to the retort stand.

2. The water level in the burette is adjusted and the initial burette reading is recorded.

3. 50 cm3 of 0.2 moldm-3 hydrochloric acid / sulphuric acid is measured and poured into a conical flask

4. 4. 5 cm of magnesium ribbon are added into the conical flask 5. 5. Close conical flask immediately with the stopper fitted with

delivery tube.

6. At the same time the stopwatch is started shake the conical flask. 7. The burette readings are recorded at 30 second intervals for 5 minutes

Time/s 0 30 60 90 120 150 180

Volume of gas / cm3

1

1..2

1

1

1

1

1

1

1

Max.. 5

1

TOTAL 20

Experiment II

Time/s

Experiment I

Volume of hydrogen/ cm3

34

21 (a) (i)

(ii)

Heat change = mc = (25+25)(4.2)(33-29) = 445 J Heat of precipitation of AgCl = - 445 / 0.0125 = -35600 J mol

-1 // 35.6 kJ mol

-1

1. The position and name /formulae of reactants and products are correct.

2. Label for the energy axis and arrow for two levels are shown.

1

1

1

1 ... 4

(b) (i)

(ii)

1. Hydrochoric acid is a strong acid and ethanoic acid is a weak

acid.

2. Hydrochloric acid ionised completely in water to produce

higher concentration

of H+ ion

3. Ethanoic acid ionised partially in water to produce lower

concentration of H+ ion.

4. During neutralisation reaction, some of the heat released are

absorbed by CH3COOH to ionise to produce hydrogen ion

- Sulphuric acid is a diproctic acid and hydrochloric acid is a monoprotic acid.

- Concentration/ number of hydrogen in sulphuric acid is double compared to that in hydrochloric acid

- The number water molecules produced in experiment III is compared experiment I

1 1

1 1Max 3

1 1 1 3

(c) Apparatus : Polystyrene cup, thermometer, measuring

cylinder.

Materials : Copper (II) sulphate, CuSO4 solution, zinc

powder.

Procedures :

1. Measure 25 cm3 of 0.2 mol dm-3 copper (II) sulphate, CuSO4 solution and pour it into a polystyrene cup.

2. Put the thermometer in the polystyrene cup and record the initial temperature of the solution.

3. Add half a spatula of zinc powder quickly and carefully into the polystyrene cup.

4. Stir the mixture with the thermometer. 5. Record the highest temperature . Tabulation of data:

1

1.2

1

1

1 1 15

Energy

AgNO3 + NaCl

AgCl + NaNO3

H = -35.6 kJ mol-1

35

Initial temperature of CuSO4 solution (oC) 1

Highest temperature of the reaction mixture (oC) 2

Temperature change (oC) 2 - 1

Calculation : Number of mole of CuSO4 = MV/1000 = (0.2)(25)/1000 = 0.005 mol

Heat change = mc(2 - 1) = x J Heat of displacement = x / 0.005 kJ mol-1 = y kJ mol-1

1

1 1

1. 3 Max : 10

TOTAL 20

No. Answer Mark

23 (a) 1. Exothermic reaction is a reaction that releases heat to the surrounding

2. The total energy content of the products is lower than the total energy content of the

reactants

3. Endothermic reaction is a reaction that absorbs heat from the surrounding

4. The total energy content of the products is higher than the total energy content of the

reactants

1

1

1

1

(b) 1. Heat energy is absorbed from surrounding //It is an endothermic reaction

2. Total energy content of C and D/ product is higher than total energy content of A and

B/ reactants

3. When reaction occurs, the temperature of the mixture of solutions increases /

becomes hot

4. X kJ heat is absorbed when one mol A reacts completely with one mol of B.

5. A reacts with B to form C and D //A and B are the reactants while C and D are the

products

6. (any 4 of the above)

1

1

1

1

(c) .

1. 1 mole of silver nitrate solution ionise to produce 1 mole of Ag+ ion

2. 1 mole of sodium chloride solution ionise to produce 1 mole of Cl- ion

3. One of mole of potassium chloride also ionise to produce 1 mole of Cl- ion

4. The heat of precipitation of silver chloride is heat that released when 1 mole of AgCl

is formed from the reaction between Ag+ ion and Cl

- ion // Ag

+ + Cl

- AgCl

5. Number of mole of AgCl produced in both reactions are the same, heat released are

the same.

1

1

1

1

1

Max 4

(d) Materials : calcium nitrate solution, sodium carbonate solution

Procedures :

1. Measure 50 cm3 of 1.0 mol dm

3 calcium nitrate and 50 cm

3 of 1.0 mol dm

-3

sodium carbonate solution separately and poured into two different plastic cups

2. Measure and record the initial temperature of both solutions after 5 minutes.

3. Pour quickly and carefully calcium nitrate solution into the plastic cup that

contains sodium carbonate solution and

4. Stir the mixture

5. Measure and record the lowest temperature reached

Tabulation of data :

1

1

1

1

1

1

5 max 4

36

1

Calculation :

No. of moles of CaCO3 = No. of moles of Ca(NO3)2

= mv/1000 = 1.0(50)/1000 = 0.05 ..1

heat change = mc(4 3) ..1 = x kJ

heat of reaction = + x kJmol-1

..1 0.05

= + y kJmol-1

Initial temperature of Ca(NO3)2 / oC 1

Initial temperature of Na2CO3 / oC 2

Average initial temperature / oC (1 + 2)/2 = 3

Lowest temperature of the mixture / oC 4

Change in temperature / oC 3- 4

1

1

1

TOTAL 20

SET 4

SECTION A

No Answer Mark 1 (a) Compound that contains only carbon and hydrogen

And has double bonds between carbon carbon atoms 1 1

(b) Alkene 1

(c) Propene 1

(d) (i) Hydrogenation / Addition reaction 1

(ii)

1

(e)

(i)

C3H6 + 9/2 O2 3CO2 + 3H2O or 2C3H6 + 9O2 6CO2 + 6H2O

2

(ii) No. of mole of C3H6 =

42

1.2

= 0.05 Volume of gas CO2 = 0.05 x 3 x 24 = 3.6 dm

3

1

1

TOTAL 10

No Answer Mark 2 (a) Ethanol 1

(b) Hydroxyl group //OH 1

37

(c)

(i) (ii) (iii)

Oxidation Orange colour of potassium dichromate (VI) solution turns to green

1 1 1

(d)

(i) (ii) (iii) (iv)

Esterification Ethyl ethanoate Pleasant smell CH3COOH + C2H5OH CH3COOC2H5 + H2O

1 1 1 2

TOTAL 10

3 (a) (i) Fermentation 1

(ii) Ethanol 1

(iii)

1

(b) C2H5OH + 3O2 2CO2 + 3H2O 1+1

(c) (i) Ethene 1

(ii)

1+1

(d) Purple to colourless 1

(e) (i) esterification 1

(ii) Ethylethanoate 1

No. Explanation Mark 4

(a) (i)

Sulphuric acid

1

(ii)

Contact process 1

(iii) Sulphur trioxide 1

(iv)

Vanadium(V) oxide, 450oC 1+1

(v) Ammonium sulphate 1

OH

C

H

H C H

H

H

Ethene

gas

ethanol

H O

H C C O H

H

38

(b) (i) Composite material 1

(ii) Correct arrangement Correct label

1 1

(iii)

nC2H3Cl --( C2H3Cl )-- 1

(iv)

It has low thermal ex ansion coefficient // resistant to thermal shock 1

TOTAL 11

No. Explanation Mark 5

(a) (i)

Sodium /potassium salts of fatty acid

1

(ii)

Hydrophobic hydrophilic

1+1

(iii)

Saponification // Hydrolysis under alkaline condition 1

(iv) To reduce the solubility of soap// to precipitate out the soap 1

(b) (i) To relief pain 1

(ii) It causes internal bleeding /ulceration 1

(iii) Antibiotic 1

(iv)

When bacteria still remain, they become immune/ resistant to the antibiotic 1

(v) Barbiturate/tranquilliser/ amphetamine / haloperidol 1

TOTAL 10

Tin atom

Copper atom

39

SECTION B

6 (a) Characteristics Explanation

Same general formula CnH2n + 1OH

successive member is different from

each other by CH2 Relative atomic mass is different

by 14

Gradual change in physical

properties // Melting / boiling point increase

Number of carbon atom per

molecules increase // size of molecule increase

Similar chemical properties // oxidation produce carboxylic acid

Have same chemical/similar

functional group

Can be prepared by similar method // can be prepared by hydration of

alkene

Have same chemical properties // have same functional functional

group

1+1

1+1

1+1

1+1

1+1

(b) (i) (CH2O)n = 60 12 + 2 + 16)n = 60 n = 2

1

C2H4O2 1 2

(ii) Carboxylic acid 1

React with carbonate to produce carbon dioxide 1 2

(iii) 2 CH3COOH + CaCO3 (CH3COO)2Ca + H2O + CO2 Correct formula of reactants and products Balanced equation

1 1

2

(c)

P Q

The number of carbon atom 2 2

The number of hydrogen atom 4 6 number of hydrogen atom Q is

higher Type of covalent bond

between // carbon/ Type of

hydrocarbon

Double bond / /

Unsaturated Single bond/ /

Saturated

Type of homologous series //

// Name of compound

Alkene// Ethene //

Alkane // Ethane

General formula// Molecular formula of the

compound

CnH2n // C2H4

CnH2n+2 // C2H6

1 1

1

1

1

Max 4

20

40

No Answer Mark 7 (a) (i) 14.3 % 1

(ii)

Element C H

Mass/ % 85.7 14.3

1 No. of moles

12

7.85 = 7.14

1

3.14 = 14.3

2 Ratio of moles/ Simplest ratio 14.7

14.7= 1

14.7

3.14= 2

3 Empirical formula = CH2

RMM of (CH2)n = 56 .............1

[(12 + 1(2)]n = 56

14n = 56

n = 14

56

= 4 ..1 Molecular formula : C4H8 ..1

6 max 5

(iii)

[any 2]

1+1

1+1

Max 4

(iv) Compound M (Butene, C4H8) has a higher percentage of carbon atom in their molecule than butane, C4H10 .1

% of C in C4H8 = 8)12(4

)12(4

x 100%

= 56

48 x 100%

= 85.7% 1

% of C in C4H10 = 10)12(4

)12(4

x 100%

= 58

48 x 100%

= 82.7% ..1

.....3

(b) (i) Starch Protein

1 1

(ii) H H CH3 H I I I I C = C C = C I I H H 2-methylbut-1,3-diene or isoprene

1 1..2

(c) (i) Rubber that has been treated with sulphur 1

(ii)

In vulcanised rubber sulphur atoms form cross-links between the rubber

molecules These prevent rubber molecules from sliding too much when stretched

1

1

TOTAL 20

But-1-ene

But-2-ene

2-methylpropene

41

8 (a) (i)

Hydrocarbon Type of

bond Homologous

series General

formula

A covalent alkane CnH2n+2

B covalent alkene CnH2n

3

3

6

(ii) Carbon dioxide 2C4H10 + 13O2 8CO2 + 10H2O [Chemical formulae of reactants and products] [Balanced]

1

1 1

3

(iii) Hydrocarbon B. Hydrocarbon B is an unsaturated hydrocarbon which react with bromine. Hydrocarbon A is a saturated hydrocarbon which do not react with

bromine.

1

1

1

3

(iv) Hydrocarbon B more sootiness. B has higher percentage of carbon by mass. % of carbon by mass ; Hydrocarbon A : 4(12) 100 // 82.76 % 4(12) + 10(1)

Hydrocarbon B : 4(12) 100 // 85.71 % 4(12) + 8(1)

1 1

1

1

4

(b) Carboxylic acid X :

Propanoic acid Alcohol Y:

Ethanol

1

1

1

1

4

TOTAL 20

42

9 (a) (i)

(ii)

SO2 + H2O H2SO3

Corrodes buildings

Corrodes metal structures

pH of the soil decreases

Lakes and rivers become acidic [Able to state any three items correctly]

1

3 4

(b) (i)

(ii) (iii)

Oleum

2SO2 + O2 2SO3

Moles of sulphur = 48 / 32 =1.5

Moles of SO2 = moles of sulphur = 1.5

Volume of SO2 = 1.5 24 dm3

= 36 dm3

1 1 1

1 1 1 6

(c) (i) Pure metal are made up of same type of atoms and are of the same size.

The atoms are arranged in an orderly manner.

The layer of atoms can slide over each other.

Thus, pure copper are ductile.

There are empty spaces in between the atoms.

When a pure copper is knocked, atoms slide.

Thus, pure copper are malleable.

1 1 1

1 1 1 1 Max:5

(ii) Zinc.

Zinc atoms are of different size,

The presence of zinc atoms distrupt the orderly arrangement of copper atoms.

This reduce the layer of atoms from sliding.

Arrangement of atoms 1; Label - 1

1 1 1 1

1 1 Max: 5

Total 20

Zinc atom

Copper atom

43

No. Explanation Mark 10 (a) Examples of food preservatives and their functions:

Sodium nitrite slow down the growth of microorganisms in meat

Vinegar provide an acidic condition that inhibits the growth of microorganisms in pickled foods

1+1

1+1

(b) (i) No // cannot Because aspirin can cause brain and liver damage if given to children with

flu or chicken pox. // It causes internal bleeding and ulceration

1 1

(ii) Paracetamol Codeine

1

1

(iii) 1. If the child is given a overdose of codeine, it may lead to addition.

2. If the child is given paracetamol on a regular basis for a long time, it

may cause skin rashes/ blood disorders /acute inflammation of the

pancreas.

1

1

(c) Type of food

additives Examples Function

Preservatives Sugar, salt To slow down the growth of microorganisms

Flavourings Monosodium glutamate, spice,

garlic

To improve and enhance

the taste of food

Antioxidants Ascorbic acid To prevent oxidation of food

Dyes/ Colourings Tartrazine Turmeric

To add or restore the

colour in food

Disadvantages of any two food additives: Sugar eating too much can cause obesity, tooth decay and diabetes Salt may cause high blood pressure, heart attack and stroke. Tartrazine can worsen the condition of asthma patients

- May cause children to be hyperactive MSG can cause difficult in breathing, headaches and vomiting.

2

2

2

2

1 1

TOTAL 20

44

SECTION C

11 (a) (i) X - any acid methanoic acid Y - any alkali ammonia aqueous solution

1 1

(ii) 1. Methanoic acid contains hydrogen ions 2. Hydrogen ions neutralise the negative charges of protein membrane 3. Rubber particles collide, 4. Protein membrane breaks 5. Rubber polymers combine together

1 1 1 1 1

5 max 4

(iii) Ammonia aqueous solution contains hydroxide ions Hydroxide ions neutralise hydrogen ions (acid) produced by activities of

bacteria

1

1

(b) (i) Alcohol 1

(ii) Burns in oxygen to form carbon dioxide and water Oxidised by oxidising agent (acidified potassium dichromate (VI)

solution) to form carboxylic acid

1 1

(iii) Procedure: 1. Place glass wool in a boiling tube

2. Pour 2 cm3 of ethanol into the boiling tube

3. Place pieces of porous pot chips in the boiling tube

4. Heat the porous pot chips strongly

5. Heat ethanol gently

6. Using test tube collect the gas given off

Diagram:

[Functional diagram] [Labeled porous pot, water, named alcohol, heat] Test: Put a few drops of bromine water Brown colour of bromine water decolourised

6 max 5

1 1

1 1

Total 20

Heat Heat

Glass wool

soaked with

ethanol

Porous pot chips

Water

45

12

(a) Carbon dioxide/ CO2 and water/ H2O Any one correct chemical equation Example 2C4H10 + 13O2 8CO2 + 10H2O Chemical formula of reactants balanced

1

1 1

3

(b) Compound B & Compound D Same molecular formula / C4H8 Different structural formula

1 1 1

3

(c) Pour compound A/B into a test tube Add bromine water to the test tube Test tube contain compound A unchanged Test tube contain compound B brown colour turn colourless

or Pour compound A/B into a test tube Add acidified Potassium manganate(VII) solution to the test tube Test tube contain compound A unchanged Test tube contain compound B purple colour turn colourless

1 1 1 1

4

(d) (i) Any members of carboxylic acid and correct ester Example [Methanoic acid] [Prophylmethanoate]

1 1

1

1

4

(d) (ii) Pour 2 cm3 of [methanoic acid] into a boiling tube

Add 2 cm3 of propanol/compound E into the boiling tube

Slowly/carefully/drop 1 cm3 of concentrated sulphuric acid

Heated (with a small flame) the mixture Pour the mixture in a beaker that contain water Observation : formed liquid that fruity smell /float on water surface

1 1 1 1 1 1

6

TOTAL 20

46

No. Mark Scheme Sub Mark Total

Mark

13(a)

But-2-ene

2-methylpropene // 2-methylprop-1-ene

1+1

1+1

4

(b)

(i)

(ii)

Propanoic acid Ethanol Chemical properties for propanoic acid:

1. React with reactive metal to produce salt and hydrogen gas 2. React with bases/alkali to produce salt and water 3. React with carbonates metal to produce salt, carbon dioxide

gas and water

4. React with alcohol to produce ester

[any three] Chemical properties for ethanol:

1. Undergo combustion to produce carbon dioxide and water 2. Burnt in oxygen to produce CO2 and H2O 3. Undergo oxidation to produce carboxylic acid / ethanoic acid 4. React with acidified K2Cr2O7 /KMnO4 to produce carboxylic

acid

5. Undergo dehydration to produce alkene / ethene.

[Any three answers]

1 1

1 1 1 1 1

1 1 1 1

1

2

6

(c) (i) P : Hexane Q : Hexene // Hex-1-ene

(ii) Reaction with bromine // acidified potassium manganate(VII) solution

Procedure: 1. Pour about [2 -5 cm3] of P into a test tube. 2. Add 4-5 drops of bromine water / acidified potassium

1 1

1

1 1

2

C C C C H

H H H H

H

H H

C

C C C

H

H

H

H

H

H

H

H

47

manganate(VII) solution and shake. 3. Observe and record any changes. 4. Repeat steps 1 to 3 by replacing P with Q

Observation: P : Brown/ Purple colour remains unchanged. Q : Brown/ Purple colours decolourise / turn colourless.

1 1

1 1

Max 6

Total mark 20

No. Explanation Mark 14 (a) Haber process

Iron

N2 + 3H2 2NH3

1 1

1+1

(b) Pure copper Bronze

Bronze is harder than pure copper

Tin atoms are of different size

The presence of tin atoms distrupt the orderly arrangement of copper

atoms.

This reduce the layer of atoms from sliding.

1

1+1

1

1

1

1 MAX

6

Procedure: 1. Iron nail and steel nail are cleaned using sandpaper. 2. Iron nail is placed into test tube A and steel nail is placed into test tube

B. 3. Pour the agar-agar solution mixed with potassium

hexacyanoferrate(III) solution into test tubes A and B until it covers

the nails. 4. Leave for 1 day. 5. Both test tubes are observed to determine whether there is any blue

spots formed or if there are any changes on the nails. 6. The observations are recorded Results:

Test tube The intensity of blue spots A High B Low

Conclusion: Iron rust faster than steel.

1 1+ 1 1 1 1

1 1

1

TOTAL 20

Tin atom

Copper atom

48

No. Explanation Mark 15 (a) (i) Traditional medicines are derived from plans or animals.

Modern medicines are made by scientists in laboratory and based on substances found in nature.

1 1

(ii)

Analgesics Aspirin

Paracetamol

Codein

Antibiotics Penicillin

Psychotherapeutic Chloropromazin

Caffeina

1 1 1

1

1 1

MAX

5

(iii) Penicillin

Cause allergic reaction, diarrhoea, difficulty breathing and easily bruising

Codein Cause addiction, drowsiness, trouble sleeping, irregular heartbeat and

hallucinations.

Aspirin Cause brain and liver damage if given to children with flu or chicken pox.

Cause internal bleeding and ulceration

1

1

1

(b) Hard water contains high concentration of calcium ions or magnesium ions. Example : sea water

Procedure 1. 20cm3 of hard water (magnesium sulphate solution) is poured into two

separate beakers X and Y. 2. 50 cm3 of soap and detergent solutions are added separately in beaker X

and beaker Y. 3. A small piece of cloth with oily stains is dipped into each beaker. 4. Each cloth is washed. 5. The cleansing action of the soap and detergent is observed.

Results

Beaker Observation

X The cloth is still dirty.

Y The cloth becomes clean.

Conclusion The cleansing action of detergent is more effective than soap in hard water

1

1

1

1

1 1 1

1 1

1

49

JAWAPAN SET 5

PAPER 3 SET 1

EXPLANATION SCORE

1. (a) (i)

[Able to record all reading accurately with unit] Sample answer Experiment Copper Bronze I 1.4 cm 1.1 cm II 1.5 cm 1.1 cm III 1.6 cm 1.2 cm

3

1(a)

(ii) [Able to construct a table to record the diameter of the dents and average

diameters for copper and bronze that contain: 1. correct title 2. Reading and unit Sample answer:

Material Diameter of the dents(cm) Average diameter,(cm) 1 2 3

X 1.4 1.5 1.6 1.5 Y 1.1 1.1 1.2 1.1

3

1.(b) [Able to state correct observation] Sample answer: The diameter of dents made on material Y is smaller than material X// The

diameter of dents made on material X is bigger than material Y

3

1.(c) [Able to state the inference correctly] Sample answer: Bronze is harder than copper// Copper softer than bronze

3

1.(d) [Able to state the correct operational definition for alloy] 1. what should be done and 2. what should be observe correctly Sample answer: A smaller dent is formed when the weight of 1 kilogram is dropped at height of 50

cm to hit the ball bearing which is taped onto the alloy block using cellophane tape

3

1.(e) [able to give all three explanations correctly] Sample answer: 1. atoms in copper are in orderly arrangement 2. atoms in bronze are not in orderly arrangement 3. layer of atoms in bronze difficult to slide on each other when force is applied

3

1.(f) [Able to state the relationship correctly between the manipulated variable and responding variable with direction] Sample answer: Alloy is harder than its pure metal.

3

1(g) [Able to state all the three variables and all the three actions correctly] Sample answer:

Names of variables Action to be taken (i) manipulated : Bronze and copper

(i) the way to manipulate variable: Repeat experiment by replacing

copper with bronze (ii) responding: Diameter of dent

(ii) what to observe in the responding

variable:

50

The diameter of the dent formed on

copper and bronze. (iii) controlled: Mass of the weight // height of the

weight // size of steel ball bearing.

(iii) the way to maintain the controlled

variable: Uses same mass of weight // same

height of weight // same size of steel

ball bearing

3

EXPLANATION SCORE

2(a) [Able to state 4 inferences correctly]

Test tube Inference A Iron (II) /Fe2+ ions formed / produced // iron / Fe rusted / oxidized B Iron (II) /Fe2+ ions are not formed / produced // iron / Fe does not

rusted / oxidized C Iron (II) /Fe2+ ions are not formed / produced // iron / Fe does not

rusted / oxidized D Iron (II) /Fe2+ ions formed / produced // iron / Fe rusted / oxidized

3

2(b) [able to explain a difference in observation correctly between test tube 1 and 2]

Sample answer: Iron / Fe in test tube A rust / oxidized because iron is in contact with less electropositive

metal, but iron in test tube B does not rust / oxidized because iron is in contact with less

electropositive metal.

3

2(c) [Able to state the hypothesis correctly] Sample answer: When a more/ less electropositive metal is in contact with iron / Fe, the metal inhibits/

speed up rusting of iron.// When a more / less electropositive metal is in contact with iron/ Fe, rusting of iron is

faster / slower// The higher /lower the metal in contact with iron/ Fe in electrochemical series than iron

/Fe ,the rusting of iron/ Fe is slower / faster

3

2(d) [able to state all the variable in this experiment correctly] Sample answer: (i) manipulated variable: Type of metals/copper, magnesium and zinc (ii) responding variable: Rusting // presence of blue colour (iii) constant variable: Size/mass of iron nail // type of nail // medium in which iron nail are kept// temperature

3

2(e) [able to state the operational definition for the rusting of iron nail correctly ] 1. What should be done and 2. what should be observe correctly Sample answer: Blue colouration is formed when iron nail is in contact with copper/tin/less

electropositive metal and immersed in potassium hexacyanoferrate (III) solution.

3

2(f) [able to classify all the three metals correctly]

Metal that can provide sacrificial

protection to iron Metal that cannot provide sacrificial

protection to iron zinc

magnesium copper

3

51

2(g) [Able to compare the intensity of blue colour and relate the intensity of blue colour with

the concentration of Fe2+ accurately ] Sample answer: The concentration of iron (II) ion is higher .The intensity of blue colouration after two

days is higher.

3

EXPLANATION SCORE

3

(a) [Able to make a statement of the problem accurately and must be in question form] Suggested answer: Does different type of alcohols affect the heat of combustions? // How does the number of carbon atom per molecule of alcohol affect the heat of

combustion ?

3

3(b) [Able to state all the three variables correctly] Suggested answer: Manipulated variable: Different types of alcohols//Different alcohols such as ethanol, propanol and butanol. Responding variable: Heat of combustion//Increase in temperature Fixed variable: Volume of water // type of container/ size of container

3

3

(c) [Able to state the relationship between manipulated variable and responding variable

correctly] Suggested answer: When the number of carbon atoms per molecule of alcohol increases, the heat of

combustion increases.

3

3(d) [Able to state the list of substances and apparatus correctly and completely] Suggested answer: Ethonol, propanol, butanol, water, copper can, spirit lamp, thermometer, weighing

balance, wooden block, tripod stand, wind shield, measuring cylinder.

3

3(e) [Able to state a complete experimental procedure] Suggested answer:

1. [200 cm3] of water is poured into a copper can. 2. Initial temperature of the water is recorded. 3. A spirit lamp is half filled with ethanol. 4. Weight the spirit lamp with ethanol and record the mass 5. The spirit lamp is put under the copper can and ignites the wick immediately. 6. The water is stirred and the flame is put off after the temperature has increased

by 30oC.

7. The highest temperature of the water is recorded 8. Immediately weight the spirit lamp and record the mass. 9. The experiment is repeated t by replacing ethanol with propanol and butanol.

3

3(f) [Able to exhibit the tabulation of data correctly with suitable headings and units ] Types of

alcohols Initial

temperature/oC Highest

temperature/oC Initial mass of

spirit lamp/g Final mass of

spirit lamp/g Ethanol Propanol Butanol

3

52

PAPER 3 SET 2

Question Rubric Score

1(a)

Able to record the burette readings accurately with 2 decimal places. Experiment I II III Initial burette

reading 1.00

13.50

26.00

Final burette

reading

13.50

26.00

38.50

3

1(b)

Able to construct a table with the following information: 1. Accurate titles and units: 2. Burette readings and volume of acid used/cm3

Sample answer: Experiment I II III Initial burette

reading/cm3

1.00

13.50

26.00

Final burette

reading/cm3

13.50

26.00

38.50

Volume of acid

used/cm3

12.50 12.50 12.50

3

1(c) Able to calculate correctly the molarity of acid with the following

steps:

Step 1: MaVa = 1

MbVb 1

Step 2: Ma = 1.0 x 25

12.5

Step 3: 2.0 mol dm-3

1(d) Able to give the volume and explaination correctly with following

aspects: 1. 6.25 cm3 2. Sulphuric acid is a diprotic acid 3. Concentration of H+ ions is double

3

1(e) Able to state the three variables correctly. Sample answer Manipulated variable: Type of acids//Hydrochloric acid, ethanoic acid Responding variable: pH values Fixed variable: Concentration of acids

3

1(f) Able to state the hypothesis accurately. Sample answer When the concentration of hydrogen ion in acid is higher, , the pH value

is lower// The higher the concentration of hydrogen ion, the lower the pH

value

3

1(g) Able to classify all the substances correctly. Sample answer:

Substances with pH less than 7 Substances with pH more than 7 Ethanoic acid Nitric acid

Ammonia solution Barium hydroxide

3