Chemistry Module Perfect Score 2009 Scheme

BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER

MODUL PERFECT SCORE 2009

CHEMISTRY

ANSWERS

Answer Chemistry Perfect Score Module 2009

1

STRUCTURE PAPER 2 Answer 1 o (a) T C (b) t2 (c) Heat energy released to the surroundings is balanced by the heat release by the particles attract one another to form a solid (d)(i) molecules (ii) P: R

(e)(i) liquid and solid (ii)Solid (f) Sublimation

Mark 1 1 1 1

1+1

1 1 1 …….10

2. (a) bromine and phenol (b) Sodium chloride (c) liquid (d) Nickel (e)

1 1 1 1 1

(f) ion (g) (i) T1 (ii) Heat is absorbed by the particles/molecules is used to overcome the attraction forces between the particles/molecules in solid naphthalene. (ii) Become faster A representation of a chemical substance using letters for atoms and subscripts for each type of atoms present in the substance.

3 (a)

Flow hydrogen gas must through the combustion tube for a few minutes before heating/The flow of hydrogen gas must be continuous throughout the experiment

(b) (c )(i) (ii) (iii)

(d) (e)

Number of mole of copper = 1.62 64 = 0.025 mol Number of mole of oxygen = 0.40 16 = 0.025 mol Number of mole of copper : Number of mole of oxygen 0.025 : 0.025 The simplest ratio 1 : 1 The empirical formula of copper(II) oxide is CuO Iron oxide / Tin oxide / Lead(II) oxide / Silver oxide/ Aurum oxide / Burning of metal in excess oxygen

Answer Chemistry Perfect Score Module 2009

1 1 1 1 1 ……10 1

1

1 1 1 1 1 …..7

2

4

(a)

(i)

(b)

(ii) (i)

Zinc hydrochloric acid / suphuric acid Zn + 2HCl
ZnCl2 + H2

1 1 1 1 1 1

Empirical formula is MO2 ( c) (d)

(ii) MO2 + 2H2
M + 2H2O The air in the combustion tube must be displaced before lighting the hydrogen gas// The heating, cooling and weighing is repeated until a constant mass is obtained

1 1

No. Magnesium is more reactive than hydrogen.

1 1

5 (a) The chemical formula that shows the simplest ratio of the number of atoms of each type of elements in the compound 1 (b) (i) Mass of magnesium = (26.4-24.0)g =2.4 g Mass of oxygen = (28.0 – 26.4) g = 1.6 g 1 (ii) The number of moles magnesium atoms =2.4 = 0.1 24 The number of moles oxygen atoms =1.6 = 0.1 16 0.1 mole of magnesium combines with 0.1 mole oxygen. Therefore, 1 mole of magnesium atoms combines with 1 mole of oxygen atoms. 1 (iii) The empirical formula of magnesium oxide is MgO. 1 (iv) 2Mg + O2
2MgO 1 (c) To allow oxygen to enter the crucible for complete combustion to occur. 1 (d) (i) X oxide Dry hydrogen gas → Heat (ii) Collect the gas in a test tube Place a burning wooden splinter at the mouth of the test tube No pop sound. 6 (a)

(i) (ii)

(b)

(i) (ii) (iii)

(iv)

(c)

(i) (i)

1+1 1 1 1 …..11

2.8.4 Period : 3 Group : 16 17 18 35 X 17

1 1 1 1 1

-Element Y is more reactive than element Z. - The size of atom Y is smaller tha atom Z -The attraction forces between nucleus and valence electron of atom Y is stronger than atom Z -it is easier for atom Y to receive valence electron compare to atom X.

1 1 1

Ionic bond Has high melting and boiling point// Conduct electricity in aqueous solution and

1 1

Answer Chemistry Perfect Score Module 2009

1 1

1

3

molten state // Soluble in water // insoluble in organic solvent ….13

(d)

O

C

O

1 1 …………16

Correct number of atom C and O Correct number electrons and shells

7 (a) (b)

(i) (ii) (iii)

2.8.2 Ionic bond -Atom X releases 2 electrons to atom Y 2+ - to achieve octet electron arrangement / to form ion X . Each ion drawn correctly

1 1 1 1

2Y

X

(iv) (c)

(i) (ii) (iii)

Number of electrons for ion X and ion Y are correct Charge of ions and ratio of ion X to ion Y are correct Has high melting and boiling point// Conduct electricity in aqueous solution and molten state // Soluble in water // insoluble in organic solvent ZY2 12 + 2(16) / 44 Name any covalent compound

1 1 1 1 1 1

……10 8

(a)

1

(b)

Metal : P/Q/R/S/T Non-metal : V/U/W/X 2.8.5

(c)

R

1

(i)

Electronegativity increases from Q to U

1

(ii)

-

1

(d)

-

(g) (h)

1

Q+

1

(i)

X

1

(ii)

Atom X has achieved octet electron arrangement

1

2R

1

(e) (f)

All atoms have same number of shells/Number of proton in the nucleus increases from Q to U The nuclei attraction forces between electrons and nucleus increases from Q to U The atomic size decreases from Q to U The tendency to receive electrons increases from Q to U

1

+

U2

→ 2RU

1. form coloured ions 2. has more than one oxidation number 3. as catalyst 4. form complex ions [ any one]

Answer Chemistry Perfect Score Module 2009

1

…..11

4

9 (a)

(i) (ii)

(b)

(i) (ii)

2.6 Period : 2 Group : 16 Ionic

1 1 1….2 1 2+

+ P

Q

+ P

Number of electrons for ion X and ion Y are correct Charge of ions and ratio of ion X to ion Y are correct

1 1….2

(c)

Correct number of atom R and S Correct number electrons and shells

1 1….2

(d)

-

1

-

Compound in (b) cannot conduct electricity when in solid state but it can conduct electricity at molten or aqueous state . Compound in (c) cannot conduct electricity at any state Compound in (b) consist of ions while Compound in (c) consist of neutral molecules Ions in compound (b) are not free to move at solid state and the molecules in (c) Neutral molecules in compound (b) are not free to move at solid state and the molecules move freely at molten/aqueous state

@ - The melting point of compound (b) is higher than compound (c) - Compound in (b) consist of ions while Compound in (c) consist of neutral molecules - Ions in compound (b) are attracted by strong electrostatic forces while molecules in compound (c) are attracted by weak van der Waals forces - More heat energy is needed to overcome the strong electrostatic forces between ions in compound (b) compared to weak van der Waalls forces between molecules in compound(c)

1 1 1

1 1 1 1….4 ….12

10 (a) (b)(i) (b)(ii) (c)(i)

Q Ion solid state : Ions are not freely moving// ions are in a fixed position. molten state : Ion can move freely R : Gas T : Liquid

1 1 1 1…..2 1 1…..2

(c)(ii) 2

(c)(iii)

-

Van der waal/intermolecular forces between molecules are weak Small amount of heat is required to overcome it

1 1 .......10

Answer Chemistry Perfect Score Module 2009

5

ESSAY PAPER 2 QUESTION: 11 (a) (i) The number of neutrons : 18 and 20 respectively

2

(ii) Similarities

Differences

1. having the same proton number/number of electrons

1. different in the number of neutrons /different in the nucleon number

2. having the same valence electron/ having the same chemical properties

2. different in physical properties 4 1 1 1 1 …….4

(b)(i) 1. Nucleus contains 6 proton and 6 neutron 2. Electrons move around the nucleus 3. Two shells are filled with electrons 4. There are 6 valence electron// electron arrangement is 2.6 (ii) Comparison Proton number Number of valence electron Chemical properties Number of neutron//nucleon number Physical properties Standard representation of element

Diagram 1.2 6 4 similar 6//12 different different Any four

Another atom 6 4 similar 7//13 different different 4

(c) - Correct curve - Mark 71 on the y axes which is at the same level with the - X and Y axis with correct title and unit

1 1 1……3

Temperature/ ◦ C 100 71 60 Time/s (ii)

12

(a)

1 2. 3. 4.

Able to explain correctly 1. 2. 3. 4.

(b)

Substance X in both solid and liquid state 1 heat energy is released 1 kinetic energy of particles decreases 1 They are closer to each other // Attraction force between the particles become stronger 1……4 20

Number of mole in 16 g of oxygen = 16/32 // 0.05 mole Volume occupied by 16 g of oxygen = 0.05 mole x 24 dm-3// 12 dm-3 Number of mole in 22 g of CO2 = 22/44 // 0.05 mole Volume occupied by 22g of CO2 = 0.05 moles x 24 dm-3 // 12 dm-3

1 1 1 1…..4

Able to determine the empirical formula and molecular formula of caffeine correctly Element C H N O Mass /g

0.48

0.05

Answer Chemistry Perfect Score Module 2009

0.28

0.16

1

6

Number of mole The simplest ratio

0.48/12 //0.04

0.05/1 //0.05

0.28/14 //0.02

0.16/16 //0.01

0.04/0.01 // 4

0.05/0.01 // 5

0.02/0.01 // 2

0.01/0.01 // 1

1

1 1 1 1 1…..8

Empirical formula = C4H5N2O [C4H5N2O ]n = 194 [ 97 ]n = 194 n = 194/97 // 2 Molecular formula = C8H10N4O2 (c)

1

Able to calculate the molar mass and the percentage of nitrogen by mass in each of the three fertilisers and choose the best fertiliser. 1 molar mass of ammonium sulphate = 132g/mol 2 percentage of nitrogen in ammonium sulphate = 28/132 x 100% // 21.2% 3 molar mass of urea = 60 g/mol 4 percentage of nitrogen in urea = 28/ 60 x 100% // 46.7% 5 molar mass of hydrazine = 32g/mol 6 percentage of nitrogen in hydrazine = 28/132 x 100% // 87.5% 7 Hydrazine has the richest source of nitrogen compares with other fertilizers. 8 The farmer should choose hydrazine

1 1 1 1 1 1 1 1…..8 20

13

(a)

(b)

1. The proton number is 11 // Number of proton is 11 2. Nucleon number is 23 // Atomic mass is 23 3. Number of neutron = 23-11 = 12 4. Nucleus contains 11p and 12n 5. Position of electron circulating the nucleus 6. Correct number shell consists of electron 7. Symbol of sodium is Na any 6

(i) (ii)

(c)

(i)

Formula that show simplest ratio number of atoms of each element in compound 1. Relative molecular mass for n(CH2O) = 180 // 12n + 2n + 16n = 180 2. n = 6 3. C6H12O6 1. 2. 3.

Element Mass/g No. of moles Ratio of moles/ Simplest ratio

Fe 2.80 2.80/56 = 0.05 0.05/0.05 = 1

4. Empirical formula = FeCl3 (ii)

1. 2. 3.

Formula of the reactants Formula of products Balance equation 2Fe

+

3Cl2

→

Cl 5.32 5.32/35.5 = 0.15 0.15/0.05 = 3

1 1 1 1 1 1.....6

1 1 1 1 ....4 1 1 1 1...4 1 1 1

2FeCl3

Answer Chemistry Perfect Score Module 2009

7

4. 5. 6.

No. of moles Fe = 2.80/56 = 0.05 mol No. of moles Cl2 = (0.05 x 3)/2 = 0.075 mol Volume of Cl2 = 0.075 x 24 = 1.8 dm3 / 1,800 cm3

1 1 1 ...6 20

14

(a)

Formula that shows the simplest ratio of the number of atoms for each element in the compound.

(b)

(c)

Element Mass (%)

C 92.3

H 7.7

Number of moles

92.3 = 7.7 12

7.7 = 7.7 1

Ratio of moles

1

1

1…1

1 1 1

Empirical formula : CH RMM of (CH)n = 78 [ 12 + 1]n = 78 13 n = 78 n = 6 Molecular formula : C6H6 Procedure: 1. Clean magnesium ribbon with sand paper 2. Weigh crucible and its lid 3. Put magnesium ribbon into the crucible and weigh the crucible with its lid 4. Heat strongly the crucible without its lid 5. Cover the crucible when the magnesium starts to burn and lift/raise the lid a little at intervals 6. Remove the lid when the magnesium burnt completely 7. Heat strongly the crucible for a few minutes 8. Cool and weigh the crucible with its lid and the content 9. Repeat the processes of heating, cooling and weighing until a constant mass is obtained Record all the mass 10. Results: Mass/g Crucible + lid x Crucible + lid + magnesium y Crucible + lid + magnesium oxide z Calculations:

1 1…5 1 1 1 1 1 1 1 1 1

1 1 1

Element Mass (g)

Mg y-x

O z-y

y−x 24

z−y 16

a

b

Number of moles Empirical formula: MgaOb / MgO Simplest ratio of moles

(d)

15

(a) (i)

1 1 14... Max 12

- Cannot separate copper from magnesium oxide - Cannot weigh copper Atom Y : 2.8.7

Answer Chemistry Perfect Score Module 2009

1 1 20 1

8

(ii)

15 (b)

Atom Z : 2.8.8.1

1

Group 17 Because atom Y has 7 valence electron Period 3 Because atom Y has three shells occupied with electrons

1 1 1 1….6

2Na + Y2 → 2NaY Correct formula of reactants and product correct Balance equation

1 1

→ NaY + NaOY + H2O Y2 + 2NaOH Correct formula of reactants and products correct Balance equation

1 1…4

Ionic compound/bond 1- The electron arrangement of atom P = 2, 8, 1 / 2. 8. 1 // The electron arrangement of atom Q = 2, 8, 7 / 2. 8. 7

1

2- to achieve the stable electron arrangement

1

3- atom of P donates / gives one electron to the atom of Q to form P+ // half equation

1

4 - atom of Q receives / accepts one electron from atom of P to form Q// half equation

1

+

-

5 -the P ion and Q ions are attracted by a strong electrostatic force to form ionic bond .

1

6- with formula PQ / correct diagram

1

Covalent compound / bond 7-The electron arrangement of atom R = 2, 6 / 2. 6

1

8-to achieve the stable electron arrangement

1

9-[atom R and atom Q share electrons]

1

10 - atom R contributes 4 electrons and atom Q contributes one electron

1

11- one atom R and 4 atom Q share 4 pairs of electrons

1

12- to form covalent compound with the formula RQ4 / diagram

1 Max 10

Answer Chemistry Perfect Score Module 2009

9

STRUCTURE PAPER 3 Explanation 16 (a)

(b)

Score

Suggested Answer (a) 0 s = 95.0 ° C, 30 s = 85.0 ° C, 60 s = 82.0 ° C, 90 s = 80.0 ° C, 120 s = 80.0 ° C,150 s = 80.0 ° C, 180 s = 78.0 ° C, 210 s = 70.0 ° C. Suggested Answer

3

3 Time (second) 0.0 30.0 60.0 90.0 120.0 150.0 180.0 210.0

Temperature (° C) 95.0 85.0 82.0 80.0 80.0 80.0 78.0 70.0

(c) (i) & (ii) [ Score 3 & 3 ]

Temperature / ° C

95 90

x x x

80.0

x

x

x

Freezing point

x x

70

60

30

60

90

Answer Chemistry Perfect Score Module 2009

120

150

180

210

Time / second

10

d(i) d(ii) e

f

Suggested Answer The constant temperature at which liquid becomes a solid Suggested Answer The heat released when the particles in the liquid arrange to form solid balanced by the heat loss to the surroundings. Suggested Answer The air trapped in the conical flask is a poor conductor of heat. The air helped to minimize heat loss to the surroundings. / to ensure uniform cooling. Suggested Answer Element Compound P, R Q, S

3 3

3 3

17 (a) Observation (i) White fume is released (ii) White solid is formed/The mass of crucible and its content increases.

Inference (i) Magnesium oxide is formed (ii) Magnesium reacts with oxygen

(b) The mass of crucible and lid = 25.35 g. The mass of crucible, lid and magnesium ribbon =27.75 g. The mass of crucible, lid and magnesium oxide when cooled = 29.35g (c) (i) (ii) (iii)

The mass of Mg= (27.75 -25.35)g =2.40g The mass of O2=(29.35-27.75)=1.60 g The number of moles Mg=0.1 mole The number of moles O = 0.1 mole The ratio of Mg : O = 1 : 1 The empirical formula is MgO.

(d) 0.1 mole of Mg reacts with 0.1 mole of O/1 mole of Mg reacts with 1 mole of O 18 (a)

Able to predict the manipulated variable, the responding variable and the constant variable completely. Manipulated variable : metals of Group 1 elements // sodium, lithium, potassium.

Able to state how to control the manipulated variables correctly

Responding variable: the reactivity of the reaction with water // the speed of movement on the water surface

Able to state correctly the way to control the manipulated variable To observe how fast the metals move on the surface of water.

Constant variable: size/mass of metals. Volume of water

Repeat the experiment by using the metals of sodium, lithium and potassium

Able to use the metal granules with the same size Use the metal granules with the same size.

18 (b) Able to state the relationship between the manipulated variable and the responding variable correctly..

18 (c) 18 (d)

Suggested answer: The reactivity of Group 1 elements increases going down the group. Able to arrange correctly the reactivity series of the metals according to descending order. Answer: potassium, sodium, lithium Able to classify the ions correctly. [to name or write all the formula of the ions correctly at the cations and anions group.] Answer: positive ion/ cation : sodium ion/ Na+, hydrogen ion/ H+ Negative ion/anion : hydroxide ion/ OH-

Answer Chemistry Perfect Score Module 2009

11

19(a)

19(b)

19(c)

[Able to state the aim of experiment accurately] To compare the reactivity of lithium, sodium and potassium based on the reaction with water and describing the effect of the solution towards the red litmus paper. Hypothesis Metals of lithium, sodium and potassium show different rate of reactivity with water and the solution formed turns red litmus paper to blue.

3

3

Variables a) Manipulated variable :type of metals b) Responding variable : reactivity of reaction c) Constant variable : water and temperature

3 x1 2x1 1x1

19(d)

[Able to list the correct and complete substances and apparatus.] Substances and Apparatus Lithium, sodium and potassium metals with water, basin, knife, forceps, blue litmus paper and white tile.

3

19(e)

[Able to give all the procedures correctly, steps 1 - 7] 1. Lithium metal is cut into a small piece 2. The paraffin oil on the surface of the metal is wiped with the filter. 3. A basin is filled with water. 4. Lithium metal is put on the surface of the water with a pair of forceps. 5. Reactivity of the reaction is observed and recorded. 6. The experiment is repeated with sodium and potassium metals.

3

2

[Able to show the accurate tabulation of data with correct title.]

14(f)

Metals Lithium Sodium Potassium

Observations

Set€2€ 1

a) b) c) d)

e)

i) ii)

i) ii) iii) i) ii)

Negatively charged ion Sulphate ion, hydroxide ion

1 1

Electrical energy → chemical energy + + Na , H Hydrogen + 2H + 2e → H2 Put a burning wooden splinter at the mouth of the test tube ‘Pop’ sound Brown solid deposited Copper is below hydrogen in the electrochemical series. Copper ions are selectively discharged

1 1 1 1 1 1 1 1 1 11

Total 2

a) b)

i) ii)

c) d) e) f)

i) ii) i) ii) i) ii)

From magnesium to copper To allow movement of ions

1 1

Dilute nitric acid Chemical energy → Electrical energy Magnesium Magnesium is more electropositive than copper Copper electrode becomes bigger/thicker 2+ Cu + 2e → Cu Less than 2.7V The distance between zinc and copper is shorter than magnesium and copper in the electrochemical series. Total

1 1 1 1 1 1 1 1

Answer Chemistry Perfect Score Module 2009

10

12

3

a)

i) ii)

The pale green solution turn brown / yellow The colourless solution turn brown

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

Cl2 + 2Br → 2Cl + Br2 [Correct formula of reactants and products] [Balance] A substance that gain electrons from other substance 0 to -1 Iron (II) sulphate solution // Iron(II) ion sodium bromide solution // bromide ion 2Br → Br2 + 2e Oxidising agent [State the name of any oxidizing agent.Example: bromine]

-

b) c) d) e)

1 1

-

Total 4

a) b)

i) ii)

c) d) e)

1 1 1 1 1 1 1 1 1 11

Potassium chlorate (V) Oxidation

1 1

metals gain oxygen // metal atom release electron to form metal ion Zinc X, Z, Y [Functional diagram] [Labeled]

1 1 1 1 1

f) Y 4.8g 4.8 24

Mass Number of moles

O 8.0 – 4.2 g // 3.2 g 3.2 16

Empirical formula

1 1

YO Total

Essay (Section B) 5 (a) Oxygen

1 10

1

-

4OH → O2 + 2H2O + 4e Formula of reactant and products correct Balance (b)

(i) Substance X

1 1

…..3

Description

NaCl solid

No current Ions are not free to move

1 1

NaCl solution

Current produced There are free moving ions.

1 1

Naphthalene solid

No current There is no free moving ion// exist as molecule.

1 1 …..6

(ii) Electrolyte NaCl solution

Non-electrolyte NaCl solid Naphthalene solid

Answer Chemistry Perfect Score Module 2009

13

(c)

(d)

Electrolysis of dilute sodium chloride solution At anode, hydroxide ion is discharged to produce oxygen gas 2H2O + O2 + 4e 4OH Electrolysis of concentrated sodium chloride solution. At anode, chloride ion is discharged to produce chlorine gas 2Cl Cl2 + 2e

1 1 1 1 1 1

R, Q, P

1

More electropositive metal can displace less electropositive metal 1 ( from its salt solution) P is more electropositive because it can displace metal Q and R 1 Q is more electropositive than R because it can displace metal R.1

6

(a)

(b)

Sample answer. 2+ 2+ Mg + Cu → Mg + Cu Magnesium release electron to form magnesium ion Copper ion receive electron to form copper atom

1 1 1

Experiment I 2+ Iron nail is oxidized to form Fe ions Metal P speeds up the process of rusting Because iron is more electropositive than P 2+ Dark blue precipitate indicates the presence of Fe ions

1 1 1 1

Experiment II Metal Q is oxidized to form Q ions Because metal Q is more electropositive than iron Water and oxygen accept electron to become OH ions // 4OH → 2H2O + OH + 4e Pink colour of solution indicate the presence of OH ions Arrangement : metal Q , iron , metal P.

…..6

….4 20

…3

1 1

Any

1 1 1 8

……8

(c) [ Diagram : metal iron , water droplet ] [ Labeled : metal iron , water droplet , oxygen , 2+ electrons flow, Fe ions in water ] For iron to rust , oxygen and water must be present Iron atoms lose electrons/oxidized to form iron(II)ions // 2+ Fe→ Fe + 2e Electrons flow to the edge of the water Water and oxygen gains electrons and is reduced to hydroxide ion // O2 + 2H2O +4e→ 4OH 2+

The iron (II)hydroxide is then further oxidized to form hydrated iron(III)oxide / Fe2O3.xH2O / rust The presence of salts in the coastal breeze Increase the electrical conductivity // a better electrolyte

Answer Chemistry Perfect Score Module 2009

1 1 1 1 1

-

The Fe ions combine with the OH ions to form iron(II)hydroxide

(d)

1

1 1 max 7 ..…7 1 1

..…2 20

14

7

a)

In the molten state the potassium ions and iodide ions are free to move In the solid state the potassium ions and iodide ions are not free to move/ held together by strong electrostatic forces.

1 1….2

b) In terms of Type of cell Electrodes

Energy conversion Half equations at the anode Half equations at the cathode Observation at the anode Observation at the cathode c)

Cell A Electrolitic Cell Anode: Silver Cathode: Silver Electrical energy to Chemical energy Ag → Ag+ + e

Cell B Voltaic Cell Anode: Zinc Cathode: Silver Chemical energy to Electrical energy Zn → Zn2+ + 2e

Ag+ + e → Ag

Ag+ + e → Ag

Silver electrode become thinner Silver electrode become thicker

Zinc plate become thinner

1

2 1 1 1

1

Zinc plate become thicker

1….8

Procedure: 1. [name the electrolyte. Example, copper(II) sulphate solution] -3 2. 1.0 mol dm copper(II) sulphate solution is filled in a beaker until it is half full. 3. The impure copper plat is connected to the positive terminal and the pure copper is connected to the negative terminal of the batteries. 4. The switch is turned on for 30 minutes.

Impure copper plate

[Functional diagram] [Labeled]

1 1 1 1….4

Pure copper plate

Copper(II) sulphate solution 1 1….2

Anode: Half equation: Cu → Cu2+ + 2e Observation: The impure copper plate becomes thinner

1 1

Cathode: Half equation: Cu2+ + 2e → Cu Observation: The pure copper plate becomes thicker

1 1….4

Total

Answer Chemistry Perfect Score Module 2009

20

15

Essay (Section C) 8 (a)

(b)

(c)

[Able to name metal X and explain the redox reaction in terms of the transfer of electron.] 1. Magnesium/zinc/lead/iron/stanum r : aluminium 2. Magnesim atom releases two electrons to form magnesium ion. 3. Magnesium is oxidized to magnesium ion/ undergoes oxidation 4. Magnesium is a reducing agent 5. Copper (II) ion accepts two electrons to form copper atom 6. Copper (II) ion is reduced to copper atom/ undergoes reduction 7. Copper (II) ion is an oxidizing agent [Able to explain the neutralization reaction is not a redox reaction.] 1. Reaction between sodium hydroxide and hydrochloric acid 2. NaOH + HCl → NaCl + H2O 3 [Show the oxidation number of each elements of reactants and products] 4 No change in the oxidation number for each element [Able to describe an experimental to prove the transfer of electron from a reducing agent to an oxidizing agent]

1 1 1 1 1 1 1

..max 6

1 1 1 1 4

Sample answer Diagram: 1. Set up of apparatus must be functional 2. Label of electrolytes and electrodes

1 1

Procedure 3. Fill the U-tube half full with sulphuric acid and clamp it up vertically -3 4. Using a dropper, fill one arm with 1.0 mol dm acidified potassium manganate(VII) solution and the other arm with 1.0 mol dm-3 iron(II) sulphate solution 5. Dipped a carbon rod into each arm and (connect the carbon rods to the galvanometer using wires)/(complete the circuit) Observation 6. Galvanometer needle is deflected 7. The purple solution of acidified potassium manganate (VII) solution becomes colourless 8. The pale green of iron(II) sulphate solution turns brown/yellow Half equation 9. Half equation for oxidizing agent + MnO4 + 8H + 5e → 10. Half equation for reducing agent: Fe3+ + e Fe2+ →

Mn

2+

+

4H2O

1 1 1

1 1 1

1 1

10 20

9 (a)

(b)

3 Metal pairs Potential difference / V Negative terminal Copper/ Metal P 1.1 P Copper/ Metal Q 2.7 Q Copper/ Metal R 0.8 R The further apart the metal from copper in the electrochemical series, the greater the potential difference

Answer Chemistry Perfect Score Module 2009

3

16

9 (c) Manipulated variable: The metal as negative electrode.

Method to manipulate the variable: Replacing the negative electrode with different metals.

Responding variable: Potential difference/ Voltage

How the variable is responding: The voltmeter reading

Controlled variable: Concentration of copper(II) sulphate solution and the copper electrode

Method to maintain the controlled variable: - Use copper(II) sulphate solution of the same concentration - Use copper metal as positive electrode

3

(d)

Metal Q, P, R, copper

10 (a)

Able to write all observations at the anode , cathode and copper(II)sulphate solution correctly. Sample Answer: Anode: Size of anode decreased // anode become thinner. Cathode: Size of cathode increased/becomes bigger/thicker // cathode become thicker. Copper(II)sulphate solution: The intensity of the blue colour remain unchanged. Able to write all the inferences correctly. Anode : Copper electrode dissolves / ionizes to form copper(II) ions 2+ Cathode : Cu ions are selectively discharged to form copper atom Copper(II)sulphate solution : The concentration of the blue Cu2+ ions remains unchanged.

3

Able to write the half-equation for oxidation and reduction correctly. Answer: Oxidation: Cu
Cu2+ + 2e 2+ Reduction : Cu + 2e
Cu Able to give the operational definition accurately.

3

10 (b)

10(c)

10 (d)

3

3

3

Sample Answer: copper anode will dissolve in copper (II) sulphate solution when an electric current passes through it. 10 (e)

Able to predict the product form at anode accurately.

3

Sample answer: Oxygen gas 10 (f)

Able to classify all the ions that are found in the copper(II) sulphate solution accurately.

3

Sample answer: Positive ion: Copper ion / Cu2+ and hydrogen ion / H+ 2Negative ion: hydroxide ion / OH and sulphate ion / SO4

Answer Chemistry Perfect Score Module 2009

17

Question Number 11(a)

11(b)

11(c)

11(d)

11(e)

11(f)

Rubric

Score

[ Able to give the aim of the experiment correctly ] Example : To construct the electrochemical series based on the potential differences between metals [ Able to state All variables correctly ] Example : Manipulated variable : Pairs of different metals//Different types of metals Responding variable : Potential differences Constant variable : Concentration of Copper(II) sulphate / [ any suitable electrolyte]// positive terminal [ Able to give the hypothesis correctly] Example : The distance between two metals increase/decrease, the potential difference will increase/ decrease [ Able to give the list of the apparatus and materials correctly and completely] Example : List of apparatus and materials Copper strip, lead strip, iron strip, zinc strip, aluminium strip, magnesium strip, copper (II) sulphate solution//[ any suitable electrolyte] , sand paper, voltmeter, beaker, connecting wires with crocodile clip [ Able to state all procedures correctly ] Example : 1. Clean the metals with sand paper 2. Fill a beaker with copper(II)sulphate //[ any suitable electrolyte] solution 3. Dip the magnesium strip and copper strip into the copper(II)sulphate solution //[ any suitable electrolyte] 4. Complete the circuit//switch-on the circuit 5. Record the potential difference between the metals 6. Determine and record which metal strip is the negative terminal 7. Repeat steps 1 to 6 using other metals to replace magnesium strip [ Able to exhibit the tabulation of data correctly ] Tabulation of data has the following element : 1. 3 columns and 6 rows Example : Pair of metals Potential difference (V) Mg and Cu Al and Cu Zn and Cu Pb and Cu Fe and Cu

Answer Chemistry Perfect Score Module 2009

Negative terminal

3

3

3

3

3

2

18

Question number

Rubric

Score

Able to state the problem statement correctly 12(a) Answer: Does the contact of iron with metal X inhibit rusting? / Does the contact of iron with metal Y increase rusting? Able to state three variables correctly

3

12(b) Answer: Manipulated variable Types of metal Responding variable Rusting or iron

12(c)

3

Controlled variable Iron nails// jelly solution [Able to state the relationship between the manipulated variable and the responding variable correctly and with direction] Answer: When iron in contact with metal X, rusting of iron inhibit / When iron in contact with metal Y, rusting of iron increase [Able to state the materials and apparatus correctly]

3

12(d) Answer: Apparatus : Test tube, test tube rack Materials : Iron nails, metal X, metal Y, agar-agar, potassium hexacyanoferate (III) and fenolfthalein Able to state 6 steps:

3

12(e)

12(f)

Answer: 1, Iron nails, Metal X and Y are cleaned 2. Two iron nails are coiled with metal X and Y each 3. Three nails are put in to different test tube 4. Jelly solutions is poured into the test tube and covered the nail 5. The test tube left for a day 6. Any observation are recorded [Able to tabulate the data with the following aspects • Correct titles • Complete pairs of metals] Answer Intensity of blue Intensity of pink Pairs of metal colour colour Nail Nail + X

3

2

Nail + Y

Answer Chemistry Perfect Score Module 2009

19

Set€3€ 1

a. Certain volume of acid completely neutralises a given volume of alkali// Certain volume of acid at which change of colour of indicator when a drop of acid is added to alkali

1

Pink solution to colourless solution

1

c. Experiment 1 = 22.40 2=22.20, 3=22.00

1

b.

d. H2SO4 + 2NaOH
Na2SO4 + 2H2O e. (i) Average volume = (22.40 + 22.30 + 22.00)÷3 = 22.30 cm f.

2

1

(ii) Mol of sulphuric acid, H2SO4 = (22.30 X 1 )/1000 = 0.0223 mol

1

(i) Functional set-up: Conical flask, Burette

1

(ii) Label: (H2SO4, KOH and Phenolphthalein)

1

g. (i) Add drops of acid a little at a time - towards the end point

1

(ii) Conical flask with content - shaken during experiment

1 10

a. (i) Yellow

1

(ii) Red

1

(iii) Orange

1

3

b. 15.00 cm

1

c (i) H2SO4 + 2KOH
K2SO4 + 2H2O +

-

(ii) H + OH
H2O d. 0.1 x 20 = 0.067 moldm 2 x 15.00 e. (i) Yellow

1 1

-3

(ii) Red

1 1 1

3

3

1 3

f. 30 cm

1 10

(a) Neutralisation (b) To ensure all nitric acid is completely reacted (c) ZnO + 2HNO3
Zn(NO3)2 + H2O (d)

1 1 1

zinc oxide

[functional apparatus] [label] (e) Number mole of nitric acid = 2(50) = 0.1 mol 1000 Mass of salt = 0.1 x 189 = 9.45 g 1 (f) Zinc carbonate, Zinc

Answer Chemistry Perfect Score Module 2009

1 1 1 1 1 1 10

20

(a) (b) (c) (d)

Silver chloride White 3 3.0 cm (i) 0.5(6.0) = 0.003 mol 1000 (ii) Number of mole of sodium chloride = 3.0(1) = 0.003 mol 1000 0.003 mol of sodium chloride reacts with 0.003 mol of silver nitrate Therefore 1 mol of sodium chloride reacts with 1 mol of silver nitrate. + (e) Ag + Cl
AgCl (f) 1.5 cm (g)

4

1 1 1 1 1 1 1 1 1

Height of precipitate (cm)

1.5__________ Ι Ι Ι Ι 1.5

Volume of sodium chloride (cm3)

[label axis] [curve] (a) (i) Copper(II) sulphate (ii) Heat the solution until saturated then cool Filter and then rinse the salt with distilled water Dry the salt using filter paper (b) (i) Blue precipitate Cannot dissolve in excess (ii) Cu(OH)2 (c) Ammonia (d) (i) White 22+ (ii) Pb + SO4
PbSO4 (iii) Filter the mixture

5

No. 6(a) (b) (c)

(d) (e) (f)

1 1 10 1 1 1 1 1 1 1 1 1 1 1 10

Scheme 1. x-axis and y-axis labelled with units 2. plots transferred correctly 3. smooth curve (i) Draw tangent at the point 40.0oC and calculate the gradient o (ii) Draw tangent at the point 60.0 C and calculate the gradient

Marks 1 1 1……3 1 1……2

1. Rate of reaction of Experiment 4 is higher than Experiment 2 2. Temperature in Experiment 4 is higher. 3. The kinetic energy of the ions is higher 4. The frequency of collisions between the H+ ions and the S2O32- ions increases 5. The frequency of effective collisions increases.

1 1 1 1

Na2S2O3 + H2SO4 → Na2SO4 + S + H2O + SO2 Sulphur Concentration

1 1 1 13

Answer Chemistry Perfect Score Module 2009

1…..5

21

7(a) (b) (c) (d)

(e)

(f)

8

(a)

1. Functional apparatus set-up 2. Complete labels 25 = 0.78 cm3s-1 32 Zn + 2H+ → Zn2+ + H2 1. Rate of reaction in Experiment II is higher than in Experiment I 2. In Experiment II copper(II) sulphate is a catalyst. 3. Catalyst will lower the activation energy. + 4. The frequency of effective collisions between zinc atoms and the H ions increases.

1 1…..2 1

1. Rate of reaction of Experiment III is higher than in Experiment I 2. Sulphuric acid in Experiment III is a diprotic acid (hydrchloric acid is a monoprotic acid) + 3. Concentration of H ions per unit volume in Experiment III is 2 times higher than in Experiment I. + 4. The frequency of effective collisions between zinc atoms and the H ions increases

1

(i) Hydrogen (ii) Bring a lighted wooden splinter to the mouth of the test tube Pop sound

1 1 1….3 13

Heat released when 1 mole of a metal is displaced from its salt solution by a more electropositive metal. Silvery solid formed// Colourless solution of silver nitrate becomes blue// Amount of copper powder decreases (i) No. of moles of Ag+ reacted = No. of moles of AgNO3 used = mv/1000 = 0.5(50)/1000 = 0.025 mol (ii) Heat released = No. of moles of Ag x ∆H = 0.025 x 105 kJ = 2.625 kJ = 2625 J o (iii) Heat change = mc θ
2625 J = 50 (4.2) θ
θ = 12.5 C Assumptions - no heat loss to surrounding - specific heat capacity of solution = specific heat capacity of water - density of solution = density of water ( any two)

(b) (c)

(d)

(e)

1 1 1 1 1……4

1 1 1…..4

1 1 1 1 1 1 1 1

Energy 2 Ag+ + Cu ∆H = -105kJ -1

2 Ag + Cu2+ 1.

The position and name/formulae for the reactants and products are correct Label for the energy axis and arrow for the two levels are shown.

1 1

(f) Lower/smaller The total surface area exposed to the air is larger Heat is lost to the environment (g) To ensure all the silver nitrate solution reacted completely (h) Bigger / Higher because magnesium is more electropositive than copper.

1 1 1 1 1

2

15

Answer Chemistry Perfect Score Module 2009

22

9

(a) Strong acid – nitric acid/ hydrochloric acid / sulphuric acid

1

Strong alkali – sodium hydroxide / potassium hydroxide (b) Heat released when 1 mole of water is formed from the reaction between an acid and an alkali (c) - It is an exothermic reaction // heat energy is released to the surrounding - The total energy of reactants is higher than the products - 57 kJ of heat energy is released when 1 mole of water is formed ( any 2) (d) (i) No. of moles alkali used = mv/100 = 1(50)/1000 = 0.05 mol (ii) Heat change = mc θ = (50+50)x4.2x 6.5 = 2730 J Heat of neutralization = - 2730/0.05 = - 54600 J/ mol = 54.6kJ / mol (e) (i) The heat of neutralization for Experiment I is higher than Experiment II (ii) Ethanoic acid – weak acid, dissociates partially in water Part of heat released in Experiment t II during neutralization is absorbed to dissociate further the molecules of ethanoic acid (f) The number of moles of water produced doubled, hence amount of heat energy released is doubled but the total volume of solution used also doubled, therefore the temperature increase remain the same

1 1 1 1 1 1 1 1 1 1 1 1 13

ESSAY SECTION B Question Number 10(a) (b)

(c) (i)

(ii)

(iii)

Question Number 11 (a)

Explanation

Marks

Ionises completely in water to produce hydrogen ions

1 1..2

Number of sulphuric acid = 0.5(50)/1000 = 0.025 mol Number of hydrochloric acid = 1.0(50)/1000 = 0.05 mol Both are strong acids ionizes completely in water To produce the same concentration of H+

1 1 1 1…..4

Test

Aqueous HCl solution

Universal indicator Add zinc powde Add copper(II)oxi de powder

Green to red

Solution of HCl in methylbenzene No changes

Gas bubbles formed

No changes

Blue solution formed

No changes

Aqueous hydrogen chloride solution In presence of water hydrogen chloride ionizes to produce hydrogen ions/H+ Aqueous HCl Observation Ammeter needle deflected Explanation Consists of free moving ions

HCl in methylbenzene Ammeter needle does not deflect Consists of molecules // no ions

1+1 1+1 1+1….6

1 1 1 1…..4 1 1 1 1…..4 20

Explanation • • •

• •

25 cm3 of sodium hydroxide solution is measured and poured into a conical flask. Two drops of phenolphthalein are added to the solution. A bureete is filled with nitric acid and the acid nitric is added slowly into the potassium hydroxide solution until the pink solution turned coloruless. The volume of acid added is calculated. The experiment is repeatedby using the same volume of sodium hydroxide and nitric acid without the phenolphthalein.

Answer Chemistry Perfect Score Module 2009

Marks 1 1 1 1 1

23

Question Number

Explanation

Marks

The salt solution formed is heated until saturated. The saturated is allowed to cool. The salt crystals are filtered and rinsed with distilled water. The salt crystals are dried by pressing them between sheets of filter papers. • NaOH + HNO3
NaNO3 + H2O • The copper(II) sulphate solution is added into sodium carbonate solution. • The mixture is filtered to obtain green precipitate, copper(II) carbonate. • The copper(II) carbonate is rinsed with water. • The copper(II) carbonate is added a little at a time into the dilute hydrochloric acid until some copper(II) carbonate solid no longer dissolved anymore. • The mixture is filtered to remove excess copper(II) carbonate. • The filtrate is copper(II) chloride solution. Lead(II) oxide Brown when hot, yellow when cold • • • •

11(b)

11(c)(i)

Carbon dioxide Turn lime water chalky

1 1 1 1 1…..10 1 1 1 1 1 1…..6 1 1 1 1….4 20

ESSAY SECTION C 12

(a) (i) (a)(ii) (b)

+

Acid that will produce two moles of hydrogen ion, H from one mole of the acid in water. H2SO4 Acid that dissociates completely into hydrogen ion, H + in water. HCl sodium hydroxide is a strong akali that undergoes complete dissociation in aqueous solution Ammonia is weak alkali that undergoes partial dissociation only The concentration of hydroxide ion in sodium hydroxide is higher than in ammonia Hence, the pH of sodium hydroxide is higher than that of ammonia.

1 1....2 1 1....2 1 1 1 1....4

(c)

[calculation] 1. Molar mass of KOH = 39+16+1 = 56 2. Mol KOH = 250 x 1.0/1000 = 0.25 3. Mass = mol x molar mass = 0.25 x 56 = 14.0 gram

1 1

-3

[ preparation of 1.0 mol dm KOH ] 4. Weigh exactly 14.0 g of KOH accurately in a weighing bottle. 5. Dissolve 14.0 g of KOH in a little water in a beaker 6. transfer the 3 contents into a 250 cm volumetric flask 7. Rinse the beaker with distilled water and transfer all the contents into the volumetric flask 8. Distilled water is added to the volumetric flask until the calibration mark. -3

[ preparation of 0.1 mol dm KOH ] [calculation] Volume of KOH is added 9. M1 x V1 = M2 x V2 V1 = M2 x V2 / M1 3 10. = 0.1 x 250 / 1 = 25 cm 3

-3

1 1

1 1 3

11. 25.0 cm of 1.0 mol dm KOH is transfer to 250 cm using 25.0 cm3 pipette. 12. Distilled water is added to the volumetric flask until the

Answer Chemistry Perfect Score Module 2009

1 1 1

1

24

calibration mark.

1 1....12 20

13 (a)

(b)

[Name any insoluble salt] [Name any two suitable solution] [Write correct ionic equation]

1 1+1 1…. 4

Example: Lead(II) sulphate Lead(II) nitrate solution and sodium sulphate solution 2− 2+ Pb + SO4
PbSO4 2+

Test for Fe

ion

Procedure I: • A few drops of sodium hydroxide solution are added into the salt solution of X until in excess. Observation : • Green precipitate cannot dissolve in excess sodium hydroxide solution Procedure II: • A few drops of ammonia solution are added into the salt solution of X until in excess. Observation : • Green precipitate cannot dissolve in excess ammonia solution.

1 1

1 1

Inference: 2+ Fe ion is present Test for SO42+ ion 3

5 cm of hydrochloric acid is added into the salt solution of X follow by 2 cm3 of barium chloride solution. Observation: • White precipitate is formed. •

Inference: 2+ SO4 ion is present. (c)

•

• • • • • •

• •

•

Chemicals : sulphuric acid and magnesium oxide / magnesium carbonate / magnesium 50 cm3 of 1.0 moldm-3 sulphuric acid is pour into a beaker and warmed carefully Magnesium oxide powder is added a little at a time into the acid using spatula. The mixture is stir well with a glass rod. Magnesium oxide powder is added continuously until some of it no longer dissolves. The mixture is filtered to remove the excess magnesium oxide. The filtrate is pour into an evaporating dish and heated gently to produce a saturated solution / heated until the filtrate is evaporated to about 1/3 of its original volume. The saturated solution is then allowed to cool to room temperature for crystallisation to occur. The magnesium sulphate crystals are filtered and dry by pressing them between a few pieces of filter paper. H2SO4 + MgO
MgSO4 + H2O

Answer Chemistry Perfect Score Module 2009

1 1….6

1 1 1 1 1 1 1 1 1 1….10

25

14

(a) (i)

energy

Zn + CuSO4 ∆H = -152 kJmol-1 ZnSO4 + Cu

(ii) (b) (i)

(ii)

(c)

• Y-axes : energy and two different level of energy • The position of reactants and products correct - reactants have more energy // products have less energy - energy is released during the experiment // this is exothermic reaction - HCl is strong acid // CH3COOH is weak acid - strong acid / HCl ionized completely // the degree of ionization of HCl / strong acid is 100% in water to produce + higher concentration of H // - CH3COOH / weak acid ionized partially // the degree of ionization of CH3COOH / weak acid is less than 100% in + water to produce very few H - when neutralization occurs, some of the heat released are absorbed by ethanoic acid / CH3COOH to break the O-H bonds in the molecules. - H2SO4 is diprotic acid // HCl is monoprotic acid - H2SO4 / diprotic acid produced two hydrogen ion / H+ when one mole of the acid ionized in water // - HCl / monoprotic acid produced one hydrogen ion / H+ when one mole of the acid ionized in water - when one mole of OH- react with two moles of H+ produced two moles of water, the heat of neutralization still the same as in experiment I because the definition of heat of neutralization is based on the formation of one mole of water - apparatus & materials :2m - procedures :5m - table :1m - calculation :2m Sample answer : Apparatus : thermometer, measuring cylinder Materials : calcium nitrate soln, sodium carbonate soln, plastic/ polystyrene cup Procedures : - measure 50 cm3 of 1.0 mol/ dm3 Ca(NO3)2 solution and poured into a plastic cup - measure 50 cm3 of 1.0 mol/ dm3 Na2CO3 solution and poured into another plastic cup - measure and record the initial temperature of both solutions after 5 minutes - pour quickly and carefully Ca(NO3)2 solution into the plastic cup that contains Na2CO3 solution and stir continuously - measure and record the lowest temperature reached

1 1….2 1 1.…2 1 1 1 1 Max 3 1 1 1

1 Max 3

1 1…2

1 1 1 1 1…5

Tabulation of data : o

Initial temperature of Ca(NO3)2 ( C) o Initial temperature of Na2CO3 ( C) o Average initial temperature ( C) o Lowest temperature of the mixture ( C)

Answer Chemistry Perfect Score Module 2009

Ө1 Ө2 (Ө1 + Ө2)/2 = Ө3 Ө4

26

Change in temperature (oC)

Ө3- Ө4 1

Calculation : No. of moles of CaCO3 = No. of moles of Ca(NO3)2 = mv/1000 = 1.0(50)/1000 = 0.05 heat change = mc(Ө4 - Ө3) = xJ

1

heat of combustion of ethanol = x kJmol-1 0.05 = y kJmol-1

Answer Chemistry Perfect Score Module 2009

1….3 20

27

15 (a) The heat of combustion of propanol is the heat energy released when 1 mole of propanol is burnt completely in the excess oxygen

1

(b) (i) 1. Functional set-up of apparatus 2. Labelled propanol, copper can, water 3. Measure (100 – 250) cm3 of water and pour into the copper can and initial temperature is recorded 4. Weigh the spirit lamp and its content 5. Light the spirit lamp to heat the water in the can and stir 6. Extinguish the spirit lamp when the temperature increase reaches 30˚C, record the maximum temperature of water reached 7. Weigh the spirit lamp with its remain. Result : 8. The initial mass of the spirit lamp + propanol = a g The final mass of the spirit lamp + propanol = b g 9. The mass of propanol burnt = (a-b) g 10. The initial temperature of water = t1˚C The maximum temperature of water = t2˚C 11. Increase in temperature of the water = (t2 – t1) = t˚C

1 a− b 60

1

= y mol

1

x -1 J mol or 14. The heat of combustion of propanol = y

1….14

Z kJ mol-1 (b) (ii) - use wind shields - Spirit lamp with its content is weighed immediately after the lamp is extinguished - Ensure the flame touches the bottom of the copper can - Stir continuously (either two answers)

1+1

(c) The heat of combustion of propanol is higher than the methanol because propanol contain more no. of carbon and hidrogen atoms per molecule.

(b)

(c)

(d)

Burette reading 24.10, 18.00, 13.00 Experiment I 32.00, 37.00, 40.50, 42.00, 42.00 Experiment II 28.00, 36.50, 41.00, 42.00, 42.00, 42.00 The graph consist of: 1. Both axis are labeled and with unit - y axis, volume of gas / cm3 - x axis, time/ s 2. All points are transferred correctly 3. Uniform scale 4. Best fit curve Experiment II Because the curve in Experiment II is steepest//the gradient is higher

Answer Chemistry Perfect Score Module 2009

1 1

1 1

13. The released heat = mc θ = 100 x 4.2 x t = xJ

16(a)

1 1 1

1 1

Calculation : RMM of propanol C3H7OH = 60 12. The no. of mol of propanol burnt =

1 1

1 1

3 3

3

3

28

(e)(i)

Manipulated variable Size of calcium carbonate /Total surface area of calcium carbonate Responding variable Rate of reaction Controlled variable Concentration and volume of hydrochloric acid Hypothesis When the total surface area of calcium carbonate increases, the rate of reaction increases

(e)(ii)

0.00 cm3

(f) (g)

17

18

a. -

3

3

3

Fast reaction

Slow reaction

Combustion Neutralization Precipitation

Photosynthesis Rusting Fermentation

3

Brown solid formed Blue solution of coper(II) sulphate becomes colourless Amount of zinc powder decreases

b.

Initial temperature : 30.0o C o Highest temperature : 38.0 C o Temperature change : 8.0 C

c.

Heat change = mc Ө = 50 (4.2) 8.0 = 1680 J

d.

It is an exothermic reaction // heat energy is released to the surrounding Zinc is more electropositive than copper // zinc is higher than copper in Electrochemical Series The total energy of reactants is higher than that of product

e.

The temperature increase will be 16.0 oC

f.

Zn + Cu

2+

2+


Cu + Zn

(a) Aim : To compare the heat of combustion of octane and heptane (b) Hypothesis : Heat of combustion of octane is higher than heptane (c) All variables : Manipulated : Type of fuel / alkane Responding : Heat of combustion Controlled : volume of water, size of copper can (d) Material : Octane, heptane, water Apparatus : spirit lamp, thermometer, copper can (any suitable container), measuring cylinder, weighing balance, wind shield

Answer Chemistry Perfect Score Module 2009

29

(e) Procedure : 1. Measure [100-200 cm3] of water and pour into the copper can and record the temperature 2. A spirit lamp is filled with heptanes and its mass is recorded 3. Adjust the height of the lamp so that the flame touches the bottom of the can. Light up the lamp o 4. Extinguish the lamp when the temperature increase reaches 30 C. Record highest temperature reached. 5. Weigh the spirit lamp with heptanes and record its mass 6. Repeat the experiment by using octane to replace heptanes. (f) Tabulation of data Type of fuel / alkane Initial temperature of water/ oC o Highest temperature of water / C o Increase in temperature/ C Initial mass of spirit lamp and content / g Final mass of spirit lamp and content/ g Mass of fuel( /alkane) used/ g

Answer Chemistry Perfect Score Module 2009

Heptane

Octane

30

Set€4€ No 1(a)

Marking scheme CnH2n + 2 n = 1, 2, 3, .....

(b)

Mark 1

S

:

| | carbon-carbon double bond // - C = C –

1

V

:

carboxyl group //

1

- COOH

(c)

2- methylpropane

n-butane

(d)

(e)

1+1

1+1

(i)

Propyl propanoate

1

(ii)

Catalyst

1

(iii)

Sweet/pleasent/fragrance/fruity smell

1

(i)

C3H7OH + 9/2O2 → 3CO2 + 4H2O // 2 C3H7OH + 9O2 → 6CO2 + 8H2O

1

(ii)

mol U = 11.2 // 0.19 // 0.2 60

1

mol CO 2 = 3 x 0.19 // 0.57

1

No of molecule CO2 = 0.57 x 6.2 x 1023 // 3.534 x 1023

1

TOTAL

Answer Chemistry Perfect Score Module 2009

12

31

2(a)

H

H

H

H ―C ― C ― O―H H (b) (c)

O 1+1

H ― C ― C ― O―H

H

H 1 1

Acidified potassium dichromate(VI) solution// Acidified potassium manganate(VII) Effervescence occurs// bubbles

(d)

(i) C2H5OH + CH3 COOH
CH3 COOC2H5 + H2O (ii) ethyl ethanoate

(e)

(i) Oxidation (ii) C2H5OH + 2[O]
CH3 COOH

1 1

(f )

(i) fermentation (ii) C6H12O6
2C2H5OH + 2CO2

1 1

(i) ester (ii) –COO- // -OOC-

1 1

3. (a)

1 1

H H ― C ― OH H ― C ― OH 1 H ― C ― OH H HOOC(CH2)7CHCH(CH2)7CH3

1

(b) unsaturated; the molecule of olive oil contains C=C double bonds.

1 1

(c) (i) Hydrogen, nickel, 180oC (ii) hydrogenation (addition reaction)

4

1+1 + 1 1

a

A compound that contain element of carbon

1

b (i)

C2H4

1

(ii)

ethene

1

(iii)

double bond between carbon atoms

1

(a: C=C ) c (i)

Bromine water

1

(ii)

Halogenation// Addition of bromine

1

(iii)

Brown bromine water decolourised

1

Answer Chemistry Perfect Score Module 2009

32

d (i)

1

C2H4 + H2O
C2H5OH

(ii)

Alcohol

(iii)

Temperature 300 C // Pressure 60 atmosphere // concentrated

o

1 1

phosphoric acid

5 (a) (b) (i) (b) (ii) (b) (iii) (c) (i)

Total

10

Ceramic Lead glass High density / refractive index Use as prism

1 1 1 1 1

H

H

C=C H

(c) (ii) (d) (i)

Cl

PVC does not rust

1

1+1 (d) (ii) (d) (iii) 6(a)

(b)

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

Bronze To make statues or medals

1 1

Saponification To precipitate the soap. To remove the glycerol and excess sodium hydroxide solution. Concentrated potassium hydroxide solution

(vi) It helps to suspend the grease particles. (i) Hydrophobic part: CH3(CH2)nCH2  Hydrophilic part:

(ii) The formula mass of CH3(CH2)nCH2OSO3Na = 330 12 + 3 + (12 + 2)n + 12 + 2 + 16 + 32 + 3(16) + 23 = 330 15 + 14n + 133 = 330 14n = 182 n = 13 (iii) The cleansing action of detergents is more effective than soaps in hard water.

Answer Chemistry Perfect Score Module 2009

33

No 7

Marking scheme

Mark 1 1 1

(a)

i) ii) iii)

Stimulant Antidepressant Steroid

(b)

To stimulate positive emotion from the patience like self-confidence, more active and energetic

1

(c)

Usage of psychotherapeutic drugs can cause many side effects like addiction, fear, aggressiveness or death in a person.

1

(d)

Arthritis Asthma

1 1

Essay Section B No 8

(a)

Marking scheme A group of organic compound that has certain characteristics: Members of a homologous series can be represented by the same general formula. Members of a homologous series can be prepared by the same method. Members of a homologous series have similar chemical properties Successive members differ from each other by –CH2 unit. Members show a gradual change in their physical properties.

Mark 1 1 1 1 1 1 Max: 5

(b)

Percentage of carbon in pentane, C5H12

=

5(12) 5(12) + 12(1)

x 100%

1 1

= 83.33% Percentage of carbon in pentene, C5H10 =

5(12) 5(12) + 10(1) x 100% = 85.71%

(c) (i)

(ii)

1 1

Percentage of carbon by mass in pentene is higher than that in pentane, hence pentene burns with a more sooty flame than pentane

1 1 Max 5

W: propanoic acid; Z: ethyl methanoate W and Z have the same molecular formulae but different structural formulae. W has the carboxyl group as the functional group while Z has the carboxylate group as the functional group.

1+1 1

W dissolves readily in water whereas Z does not. W has a sour smell. Z has a fragrant fruity smell.

1 1

1

(iii)

H

O

H

H ―C ― C ― O ― C ― H H

1

H

Answer Chemistry Perfect Score Module 2009

34

(iv)

Used as food flavouring / perfume / fragrance

1

CH3 COOH + CH3OH
CH3 COOCH3 + H2O Catalyst: Concentrated sulphuric acid TOTAL 9

(a)

.

Element % No. of moles Ratio of moles Simplest ratio

C 52.2 52.2/12 4.35 2

H 13.0 13/1 13 6

O 34.8 34.8/16 2.175 1

Assume that the molecular formula is C2H6O. Given that the relative molecular mass is 46, n[ C2H6O] = 46 46n = 46 n=1 Therefore the molecular formula of compound X is C2H5OH. (b)

• • • • •

3

3

50 cm ethanol and 25 cm of ethanoic acid are added into a round-bottomed flask. 5 cm3 of concentrated sulphuric acid is added. A Liebig condenser is fixed to the round-bottomed flask. The mixture is heated under reflux for 30 minutes. The ester, ethyl ethanoate is distilled out from the mixture at its boiling point. CH3 COOH

(c)

+

C2H5OH

CH3 COOC2H5 + H2O




Dehydration Alumina / unglazed porcelain chips, heat

1 1 10

1 1 1 1 1..5 1 1 1 1 1

1..6

1 1+1 ..3

(d)

• • • • •

(e)

When ethanoic acid is added to latex, the H+ ions in the acid will neutralize the negative charges on the protein membrane of the rubber particles. As a result the rubber particles will collide with each other and break the protein membrane setting free the rubber polymer molecules which then coagulate. Coagulation can be prevented by adding an alkali.

A long chain molecule that is formed by the joining together of smaller molecules called monomers. H n

H

C == C H

H

1 1 1 1 1 Max: 4 1

H

--- C ― C ----

H

Answer Chemistry Perfect Score Module 2009

H

H

n

1..2 20

35

Question Number 10 (a)(i) (a)(ii)

(b)(i) (b)(ii) (b)(iii)

(c)(i)

Explanation

Mark Σ

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] • 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

• • • • • • •

3 1 1 1 1 1 1

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.

4

6

1 1 1 1 1 1 1

There are empty spaces in between the atoms. When a pure copper is knocked, atoms slide. Thus, pure copper are malleable.

Max:5 (c)(ii)

• • • •

1 1 1

Zinc. Zinc atoms are of different size, The presence of zinc atoms disturbs the orderly arrangement of copper atoms. This reduce the layer of atoms from sliding.

1

Zinc atom

Copper atom

Arrangement of atoms – 1

1

Label - 1

1 Max: 5 Total

11 (a)

Sulphur is burnt in the air to form sulphur dioxide S + O2 → SO2 Sulphur dioxide formed is then burnt in the air to form sulphur trioxide 2SO2 + O2 → 2SO3 Temperature: 450 - 550 ˚C Pressure: 1 atm Catalyst used: Vanadium(V) oxide, V2O5 Sulphur trioxide formed is dissolved in concentrated sulphuric acid to form oleum SO3 + H2O → H2S2O7

Answer Chemistry Perfect Score Module 2009

20

1 1 1 1 1 1 1 1 1 1

36

Oleum is diluted with distilled water to form concentrated sulphuric acid Nitrogen gas is obtained from the atmosphere Hydrogen gas is obtained from natural gas Nitrogen gas is combined with hydrogen gas in the ratio of 1 : 3 by volume Temperature: 450 - 550 ˚C Pressure: 200 atm Catalyst used: Iron powder Can withstand high temperature Can withstand high pressure Insulators of heat and electricity Resistant to chemical reactions

(b)

(c)

No. 12

(a)

(b)

Marking Criteria Part X – hydrophobic/hydrocarbon Part Y – hydrophilic/ionic Parx X – dissolves in grease Part Y – dissolves in water

Mark 1 1 1 1

1.The cloth in experiment II is clean whereas the cloth in Experiment I is still dirty. 2.In hard water,soap react with magnesium ion 3.to form scum 4.Detergent are more effective in hard water 5.Detergent does not form scum 6.Detergent are better cleansing agen then soap to remove oily stain.

1

Patient X : Analgeis/anpirin Patient Y: Antibiotic/penicillin/streptomycin Patient Z ; Psychotherapeutic / antidepressant

1 1 1

1..10 1 1 1 1 1 1..6 1 1 1 1…4 20

Total

4

1 1 1 1 1 6

(c )

(d) (i)

(ii)

(iii)

13 (a)

Precaution: 1.Take after food. 2. Swallowed with plenty of water Explain: 1. Acidic and can cause irritation of the stomach. 2. To avoid internal bleeding/ulceratiion [precaution – 1m] [explain – 1m ] 1.To make sure all the bacteria are killed / becomes ill again – 1m 2. bacteria become more resistant. – 1m 3.Need stronger antibiotic to fight the same infection – 1m 1.Drowsiness – 1m 2. poor coordination/light-headedness – 1m TOTAL

Sources of sulphur dioxide: Volcanic eruptions Burning of fossil fuels From industries manufacturing sulphur based products [any two] Health hazards: Irritates the nose and eyes Causes bronchitis and asthma Formation of acid rain: Sulphur dioxide reacts with oxygen to form sulphur trioxide 2SO2 + O2 → 2SO3 Both oxides of sulphur dissolve in rain water to form sulphurous and sulphuric

Answer Chemistry Perfect Score Module 2009

3

1 1

2

3 2

7 20

1 1 1

1 1 1 1 1

37

acids respectively SO2 + H2O → H2SO3 SO3 + H2O → H2SO4 Effects of acid rain: Corrodes buildings and bridges Damages vegetation

1 1 1 1 1..max 10

(b)

Tape a steel ball bearing to the brass block Hang a weight of 1 kg at a specified height of 50 cm above the ball bearing Drop the weight and allow it to hit the steel ball bearing Use a caliper or ruler to measure the diameter of the dent made on the brass block Repeat the experiment to obtain another two readings so that an average value can be calculated The whole experiment is repeated using copper block to replace brass block Observation: Type of block

1

Diameter of dents (cm) 2 Average

Brass Copper Results and explanation: Diameter made by copper is larger than brass Brass is harder than copper The foreign atoms (zinc atoms) in brass prevent the layers of copper atoms from sliding past each other . 14 (a)

(b)

(c)

Sulphur is burnt in the air to form sulphur dioxide. S + O2 → SO2 Sulphur dioxide formed is then burnt in the air to form sulphur trioxide 2SO2 + O2 → 2SO3 Temperature: 450 - 550 ˚C Pressure: 1 atm Catalyst used: Vanadium(V) oxide, V2O5 Sulphur trioxide formed is dissolved in concentrated sulphuric acid to form oleum SO3 + H2O → H2S2O7 Oleum is diluted with distilled water to form concentrated sulphuric acid Nitrogen gas is obtained from the atmosphere Hydrogen gas is obtained from natural gas Nitrogen gas is combined with hydrogen gas in the ratio of 1 : 3 by volume Temperature: 450 - 550 ˚C Pressure: 200 atm Catalyst used: Iron powder Can withstand high temperature Can withstand high pressure Insulators of heat and electricity Resistant to chemical reactions

Answer Chemistry Perfect Score Module 2009

1 1 1 1 1 1 1

1 1 1 1..10 20 1 1 1 1 1 1 1 1 1 1..10 1 1 1 1 1 1..6 1 1 1 1..4 20

38

Question 15 (a)

Rubric [Able to record all the six readings correctly.] Vulcanised rubber: 2, 4, 6 Unvulcanised rubber: 4, 8, 12

15(b)

[Able to relate between the manipulated variable and the responding variable.] Vulcanised rubber is more elastic than the unvulcanised rubber// Unvulcanised rubber is less elastic than the vulcanised rubber

15(c) Variable (i) Manupilated variable Vulcanized and unvulcanized rubber// Mass of weight (ii) Responding variable Increase in length of rubber strip//elasticity (iii) Fixed variable Initial length of rubber strip

Action to be taken (i) The way to manupilate variable Repeat by replacing vulcanized rubber with unvulcanized rubber//Use weights with different masses (ii) What to observe in the responding variable To measure length of rubber strip (iii) The way to maintain the controlled variable Use the same length of vulcanized and unvulcanized rubber strips

15(d)

[Able to make the correct inference] (i) Vulcanized rubber (ii) Presence of the sulphur cross links between the chain of rubber polymers in vulcanized rubber makes the small increase in length and can return to its original length after stretching.

15(e)

[Able to make an operational definition correctly:] Rubber that can stretch a bit and returns to its original length when not stretched.

No 16

Marking scheme How does the elasticity of vulcanized rubber differ from that of unvulcanized rubber? (a) Variable: Manipulated Variable: Types of rubber Responding variable: Length of rubber strip Fixed variable: Size of rubber strip, Mass of weight

Mark

(c)

Hypothesis: Vulcanized rubber is more elastic than unvulcanized rubber.

3

(d)

The elasticity of the rubber strip is shown by its ability to return to its original length after it is stretched.

3

(a) (b)

(e)

No 17

(a)

(b) Unvulcanized rubber: the minimum weight is 40g Vulcanized rubber could return to its original length even after the 50g weight was removed

3 1 1 1

3

Marking scheme To compare the cleansing power of soap and detergent in hard water.

Mark 3

(b)

The cleansing power of soap is weaker in sea water than the cleansing power of detergent.

3

(c)

(i) Soap and detergent (ii) Cleanliness of the clothes or amount of the greasy spots removed. (iii)Mass of soap and detergent dissolved in sea water.

1 1 1

Answer Chemistry Perfect Score Module 2009

39

(d)

Beaker

A Only some greasy spots are removed

Observation

B Most of the greasy spots are removed.

3

(e) The cleansing power of soap is weaker in sea water compare to detergent The cleansing power of soap is weaker in sea water compare to detergent (f)

No 18

(i)

3

Sea water contains magnesium and calcium ions. Soap particles form insoluble calcium and magnesium salt (called scum) with these ions. Detergent particles are not precipitated out by Ca2+ and Mg2+ ions present in sea water and will remain in the solution to do the cleansing job.

Marking scheme Problem statement: Iron rusts more easily than steel.

(ii)

Hypothesis: Iron rusts faster than steel.

(iii)

Material: Iron nail, steel nail, agar-agar solution, potassium hexacyanoferrate(III) solution.

3

Mark 3 3

3

Apparatus: Test tubes. (iv)

(v)

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. Prepare 5 % of agar-agar solution and add several drops of potassium hexacyanoferrate(III) solution to the agar-agar solution. 4. Pour the agar-agar solution into test tubes A and B until it covers the nails. 5. Leave for 1 day. 6. Both test tubes are observed to determine whether there is any blue spots formed or if there are any changes on the nails. Tabulation of data: Test tube A B

3

The intensity of blue spots 3

OR Based on the hardness of iron and steel. Problem statement: Iron is softer than steel. Hypothesis: The diameter of the dent of the steel is smaller than the diameter of the dent of iron. Material: Steel block, iron block. Apparatus: Ruler, 1 kg weight, retort stand and clamp, thread, steel ball, cellophane tape.

Answer Chemistry Perfect Score Module 2009

40

Procedure: 1. A steel ball is attached on the surface of the iron block using a cellophane tape. 2. The 1 kg weight is held 1 metre from the surface of the iron block. 3. The weight is then released. 4. The diameter of dent formed on the iron block is measured using a ruler. 5. Steps 2 to 4 are repeated on different surfaces of the iron block and the average diameter of dents is obtained. 6. The experiment is repeated by replacing the iron block with steel block. Tabulation of data: Material

Diameter of the dent (cm) Reading 1 Reading 2

Average (cm)

Iron block Steel block

No 19

(i)

Marking scheme Aim: To differentiate and identify hexan-1-ol, hex-1-ene and hexane through chemical tests.

Mark 3

(ii)

(iii)

Variables Manipulated Variable: Types of reagents Responding variable: Change in colour Fixed variable: hexan-1-ol, hex-1-ene and hexane

3

Apparatus: Test tubes, test tube holder, dropper Materials: bromine water, 0.5 mol dm-3 potassium dichromate(VI) -3 solution, 1 mol dm sulphuric acid, Liquids X, Y and Z

(iv)

(v)

Procedure: 3 1. About 2 cm of each liquid X, Y and Z are poured into three separate test tubes. 2. 1 cm3 of potassium dichromate(VI) solution is added into each 3 -3 test tube followed by 1 cm of 1 mol dm sulphuric acid and heat. 3. The mixture in each test tube is then shaken well. 4. The changes in each test tube are observed and recorded. 5. Steps 1 to 4 are repeated using 2 cm3 of bromine water to replace the acidified potassium dichromate(VI) solution.

3

3

Tabulation of Data: Reagent Liqiud

Acidified potassium dichromate(VI) solution.

Bromine water

Inference

X Y

3

Z

Answer Chemistry Perfect Score Module 2009

41