Chemistry Module Perfect Score 2009 Scheme

Answer Chemistry Perfect Score Module 2009 1

BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER

MODUL PERFECT SCORE 2009

CHEMISTRY

ANSWERS


STRUCTURE PAPER 2 Answer Mark 1

(a) ToC 1 (b) t2 1 (c) Heat energy released to the surroundings is 1 balanced by the heat release by the particles attract one another to form a solid 1 (d)(i) molecules

(ii) P: R 1+1

(e)(i) liquid and solid 1 (ii)Solid 1

(f) Sublimation 1 …….10

2. (a) bromine and phenol 1 (b) Sodium chloride 1 (c) liquid 1 (d) Nickel 1 (e)

1

(f) ion 1 (g) (i) T1 1 (ii) Heat is absorbed by the particles/molecules is used 1 to overcome the attraction forces between the particles/molecules in solid naphthalene. 1

(ii) Become faster 1 ……10

3 (a)

(b)

(c )(i)

(ii)

(iii)

(d) (e)

A representation of a chemical substance using letters for atoms and subscripts for each type of atoms present in the substance. 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 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

1 1 1 1 1 1 1 …..7


(a) (i) Zinc

hydrochloric acid / suphuric acid 1 1

(ii) Zn + 2HCl � ZnCl2 + H2 1 (b) (i)

Empirical formula is MO2

1

1

1

(ii) MO2 + 2H2 � M + 2H2O 1 ( c) 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

4

(d) 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)

1 + 1 (ii) Collect the gas in a test tube 1

Place a burning wooden splinter at the mouth of the test tube 1 No pop sound. 1

…..11

6 (a) (i) 2.8.4 1 (ii) Period : 3

Group : 16 1 1

(b) (i) 17 1 (ii) 18 1 (iii)

1 1

(iv) -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 1

(c) (i) Ionic bond 1 (i) Has high melting and boiling point// Conduct electricity in aqueous solution and 1

X 35

17

Heat

X oxide

Dry hydrogen gas

→


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

(d)

Correct number of atom C and O 1 Correct number electrons and shells 1 …………16

7 (a) 2.8.2 1 (b) (i) Ionic bond 1 (ii) -Atom X releases 2 electrons to atom Y

- to achieve octet electron arrangement / to form ion X2+. 1 1

(iii) Each ion drawn correctly 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

(iv) Has high melting and boiling point// Conduct electricity in aqueous solution and molten state // Soluble in water // insoluble in organic solvent

1

(c) (i) ZY2 1 (ii) 12 + 2(16) / 44 1 (iii) Name any covalent compound 1 ……10 8 (a) Metal : P/Q/R/S/T

Non-metal : V/U/W/X 1

(b) 2.8.5 1

(c) R 1

(d) (i) Electronegativity increases from Q to U 1

(ii) - 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

1

(e) Q+ 1

(f) (i) X 1

(ii) Atom X has achieved octet electron arrangement 1

(g) 2R + U2 → 2RU 1

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

1

…..11

O C O

X Y

2-


9 (a) (i) 2.6 1 (ii) Period : 2

Group : 16 1 1….2

(b) (i) Ionic 1 (ii)

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) - 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 1….4

….12

10 (a) Q 1 (b)(i) Ion 1 (b)(ii) solid state : Ions are not freely moving// ions are in a fixed position.

molten state : Ion can move freely 1 1…..2

(c)(i) R : Gas T : Liquid

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

+

P Q

2-+

+

P


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 (b)(i) 1. Nucleus contains 6 proton and 6 neutron 1 2. Electrons move around the nucleus 1 3. Two shells are filled with electrons 1 4. There are 6 valence electron// electron arrangement is 2.6 1 …….4 (ii) Comparison Diagram 1.2 Another atom

Proton number 6 6 Number of valence electron 4 4 Chemical properties similar similar Number of neutron//nucleon number 6//12 7//13 Physical properties different different Standard representation of element different different

Any four 4

(c) - Correct curve 1 - Mark 71 on the y axes which is at the same level with the 1 - X and Y axis with correct title and unit 1……3

(ii) 1 Substance X in both solid and liquid state 1 2. heat energy is released 1 3. kinetic energy of particles decreases 1 4. They are closer to each other // Attraction force between the particles become stronger 1……4 20 12 (a) Able to explain correctly

1. Number of mole in 16 g of oxygen = 16/32 // 0.05 mole 2. Volume occupied by 16 g of oxygen = 0.05 mole x 24 dm-3// 12 dm-3 3. Number of mole in 22 g of CO2 = 22/44 // 0.05 mole 4. Volume occupied by 22g of CO2 = 0.05 moles x 24 dm-3 // 12 dm-3

1 1 1 1…..4

(b)

Able to determine the empirical formula and molecular formula of caffeine correctly

Element C H N O

Mass /g 0.48 0.05 0.28 0.16

1

Temperature/ ◦ C

100

71

60

Time/s


Number of mole

0.48/12 //0.04

0.05/1 //0.05

0.28/14 //0.02

0.16/16 //0.01

The simplest ratio

0.04/0.01 // 4

0.05/0.01 // 5

0.02/0.01 // 2

0.01/0.01 // 1

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

1 1 1 1 1 1 1…..8

(c) 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) 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

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

(b) (i) (ii)

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 1 1 1 ....4

(c) (i) (ii)

Element Fe Cl 1. Mass/g 2.80 5.32 2. No. of moles 2.80/56 = 0.05 5.32/35.5 = 0.15 3. Ratio of moles/

Simplest ratio 0.05/0.05 = 1 0.15/0.05 = 3

4. Empirical formula = FeCl3 1. Formula of the reactants 2. Formula of products 3. Balance equation 2Fe + 3Cl2 → 2FeCl3

1 1 1 1...4 1 1 1


4. No. of moles Fe = 2.80/56 = 0.05 mol 5. No. of moles Cl2 = (0.05 x 3)/2 = 0.075 mol 6. 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.

1…1

(b)

Empirical formula : CH

RMM of (CH)n = 78

[ 12 + 1]n = 78 13 n = 78 n = 6 Molecular formula : C6H6

Element C H

Mass (%) 92.3 7.7

Number of moles 12

3.92 = 7.7

1

7.7 = 7.7

Ratio of moles 1 1

1 1 1 1 1…5

(c) 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:

Element Mg O

Mass (g) y-x z-y

Number of moles 24

xy −

16

yz −

Simplest ratio of moles a b

1 1 1 1 1 1 1 1 1 1 1 1 1 1 14... Max 12

15 (a) (i) Atom Y : 2.8.7 1

(d) - Cannot separate copper from magnesium oxide - Cannot weigh copper

1 1

20

Empirical formula: MgaOb / MgO


Atom Z : 2.8.8.1 1 Group 17 1 Because atom Y has 7 valence electron 1 Period 3 1 Because atom Y has three shells occupied with electrons 1….6

(ii) 2Na + Y2 → 2NaY Correct formula of reactants and product correct 1

Balance equation 1 Y2 + 2NaOH → NaY + NaOY + H2O Correct formula of reactants and products correct 1

Balance equation 1…4

15 (b) 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 2- to achieve the stable electron arrangement 3- atom of P donates / gives one electron to the atom of Q to form P+ // half equation 4 - atom of Q receives / accepts one electron from atom of P to form Q- // half equation 5 -the P + ion and Q - ions are attracted by a strong electrostatic force to form ionic bond . 6- with formula PQ / correct diagram Covalent compound / bond 7-The electron arrangement of atom R = 2, 6 / 2. 6 8-to achieve the stable electron arrangement 9-[atom R and atom Q share electrons] 10 - atom R contributes 4 electrons and atom Q contributes one electron 11- one atom R and 4 atom Q share 4 pairs of electrons 12- to form covalent compound with the formula RQ4 / diagram

Max 10

1 1 1 1 1 1 1 1 1 1 1 1


STRUCTURE PAPER 3

Explanation Score

16 (a)

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.

3

(b) Suggested Answer

Time (second) Temperature (° C) 0.0 95.0

30.0 85.0 60.0 82.0 90.0 80.0

120.0 80.0 150.0 80.0 180.0 78.0 210.0 70.0

3

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

80.0

95

Temperature / ° C

Time / second

70

Freezing

point

30 60 90 210 180 150 120

60

90 x

x

x x

x

x

x

x


d(i) Suggested Answer The constant temperature at which liquid becomes a solid

3

d(ii) Suggested Answer The heat released when the particles in the liquid arrange to form solid balanced by the heat loss to the surroundings.

3

e 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.

3

f Suggested Answer Element Compound

P, R Q, S

3

17 (a)

Observation Inference

(i) White fume is released (ii) White solid is formed/The mass of crucible and

its content increases.

(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) The mass of Mg= (27.75 -25.35)g =2.40g (ii) The mass of O2=(29.35-27.75)=1.60 g (iii) 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. Responding variable: the reactivity of the reaction with water // the speed of movement on the water surface Constant variable: size/mass of metals. Volume of water

Able to state how to control the manipulated variables correctly Repeat the experiment by using the metals of sodium, lithium and potassium Able to state correctly the way to control the manipulated variable To observe how fast the metals move on the surface of water. 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..

Suggested answer: The reactivity of Group 1 elements increases going down the group.

18 (c) Able to arrange correctly the reactivity series of the metals according to descending order. Answer: potassium, sodium, lithium

18 (d) 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-


19(a) [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.

3

19(b) 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

19(c) 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

14(f) [Able to show the accurate tabulation of data with correct title.]

Metals Observations Lithium Sodium

Potassium

2

Set�2�

1 a) i) Negatively charged ion 1 ii) Sulphate ion, hydroxide ion 1

b) Electrical energy → chemical energy 1 c) Na+ , H+ 1 d) i) Hydrogen 1 ii) 2H+ + 2e → H2 1 iii) Put a burning wooden splinter at the mouth of the test tube

‘Pop’ sound 1 1

e) i) Brown solid deposited 1 ii) Copper is below hydrogen in the electrochemical series.

Copper ions are selectively discharged 1 1

Total 11 2 a) From magnesium to copper 1 b) i) To allow movement of ions 1

ii) Dilute nitric acid 1 c) Chemical energy → Electrical energy 1 d) i) Magnesium 1 ii) Magnesium is more electropositive than copper 1 e) i) Copper electrode becomes bigger/thicker 1 ii) Cu2+ + 2e → Cu 1 f) i) Less than 2.7V 1 ii) The distance between zinc and copper is shorter than magnesium and copper

in the electrochemical series. 1

Total 10


3 a) i) The pale green solution turn brown / yellow 1 ii) The colourless solution turn brown 1

b) Cl2 + 2Br-→ 2Cl- + Br2 [Correct formula of reactants and products] [Balance]

1 1

c) A substance that gain electrons from other substance 1 d) i) 0 to -1 1 ii) Iron (II) sulphate solution // Iron(II) ion 1 e) i) sodium bromide solution // bromide ion 1 ii) 2Br- → Br2 + 2e 1 iii) Oxidising agent 1 iv) [State the name of any oxidizing agent.Example: bromine] 1 Total 11 4 a) Potassium chlorate (V) 1 b) i) Oxidation 1

ii) metals gain oxygen // metal atom release electron to form metal ion 1 c) Zinc 1 d) X, Z, Y 1 e) [Functional diagram]

[Labeled] 1 1

f) Y O Mass 4.8g 8.0 – 4.2 g // 3.2 g Number of moles 4.8

24

3.2 16

Empirical formula YO

1 1 1

Total 10 Essay (Section B) 5 (a) Oxygen 1 4OH-→ O2 + 2H2O + 4e Formula of reactant and products correct 1 Balance 1 …..3 (b) (i)

Substance X Description

NaCl solid

No current 1 Ions are not free to move 1

NaCl solution Current produced 1 There are free moving ions. 1

Naphthalene solid

No current 1 There is no free moving ion// exist as molecule. 1

…..6 (ii)

Electrolyte Non-electrolyte NaCl solution NaCl solid

Naphthalene solid


(c) Electrolysis of dilute sodium chloride solution 1

At anode, hydroxide ion is discharged to produce oxygen gas 1 4OH- 2H2O + O2 + 4e 1 Electrolysis of concentrated sodium chloride solution. 1 At anode, chloride ion is discharged to produce chlorine gas 1 2Cl- Cl2 + 2e 1 …..6

(d) 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 ….4

20 6 (a) Sample answer. Mg + Cu2+ → Mg2+ + Cu 1 Magnesium release electron to form magnesium ion 1 Copper ion receive electron to form copper atom 1 …3 (b) Experiment I Iron nail is oxidized to form Fe2+ ions 1 Metal P speeds up the process of rusting 1 Because iron is more electropositive than P 1 Dark blue precipitate indicates the presence of Fe2+ ions 1 Experiment II Metal Q is oxidized to form Q ions 1 Because metal Q is more electropositive than iron 1 Water and oxygen accept electron to become OH- ions // 4OH-→ 2H2O + OH- + 4e 1 Pink colour of solution indicate the presence of OH- ions 1 Arrangement : metal Q , iron , metal P. 1 Any 8 ……8 (c) [ Diagram : metal iron , water droplet ] 1 [ Labeled : metal iron , water droplet , oxygen , electrons flow, Fe2+ ions in water ] 1 For iron to rust , oxygen and water must be present 1 Iron atoms lose electrons/oxidized to form iron(II)ions // Fe→ Fe2+ + 2e 1 Electrons flow to the edge of the water 1 Water and oxygen gains electrons and is reduced to hydroxide ion // O2 + 2H2O +4e→ 4OH- 1 The Fe2+ ions combine with the OH- ions to form iron(II)hydroxide 1 The iron (II)hydroxide is then further oxidized to form hydrated iron(III)oxide / Fe2O3.xH2O / rust 1 max 7 ..…7 (d) The presence of salts in the coastal breeze 1 Increase the electrical conductivity // a better electrolyte 1 ..…2 20


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 Cell A Cell B

Type of cell Electrolitic Cell Voltaic Cell

Electrodes Anode: Silver Cathode: Silver

Anode: Zinc Cathode: Silver

Energy conversion

Electrical energy to Chemical energy

Chemical energy to Electrical energy

Half equations at the anode

Ag → Ag+ + e Zn → Zn2+ + 2e

Half equations at the cathode

Ag+ + e → Ag Ag+ + e → Ag

Observation at the anode

Silver electrode become thinner

Zinc plate become thinner

Observation at the cathode

Silver electrode become thicker

Zinc plate become thicker

1 2 1 1 1 1 1….8

Procedure: 1. [name the electrolyte. Example, copper(II) sulphate solution] 2. 1.0 mol dm-3 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.

1 1 1 1….4

[Functional diagram] [Labeled]

1 1….2

c)

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

1 1 1 1….4

Total 20

Impure copper

plate

Pure copper

plate

Copper(II)

sulphate solution


Essay (Section C)

9 (a)

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

3

(b) The further apart the metal from copper in the electrochemical series, the greater the potential difference

3

(a) [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

1 1 1 1 1 1 1

..max 6

(b) [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

1 1 1 1

4

8

(c) [Able to describe an experimental to prove the transfer of electron from a reducing agent to an oxidizing agent] Sample answer Diagram: 1. Set up of apparatus must be functional 2. Label of electrolytes and electrodes Procedure 3. Fill the U-tube half full with sulphuric acid and clamp it up vertically 4. Using a dropper, fill one arm with 1.0 mol dm-3 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 → Mn2+ + 4H2O 10. Half equation for reducing agent: Fe2+ → Fe3+ + e

1 1 1 1 1 1 1 1 1 1

10 20


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 3 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.

3

10 (b) Able to write all the inferences correctly. Anode : Copper electrode dissolves / ionizes to form copper(II) ions Cathode : Cu2+ ions are selectively discharged to form copper atom Copper(II)sulphate solution : The concentration of the blue Cu2+ ions remains unchanged.

3

10(c) Able to write the half-equation for oxidation and reduction correctly. Answer: Oxidation: Cu � Cu2+ + 2e Reduction : Cu2+ + 2e� Cu

3

10 (d) Able to give the operational definition accurately. Sample Answer: copper anode will dissolve in copper (II) sulphate solution when an electric current passes through it.

3

10 (e) Able to predict the product form at anode accurately. Sample answer: Oxygen gas

3

10 (f) Able to classify all the ions that are found in the copper(II) sulphate solution accurately. Sample answer: Positive ion: Copper ion / Cu2+ and hydrogen ion / H+ Negative ion: hydroxide ion / OH- and sulphate ion / SO4

2-

3


Question Number

Rubric Score

11(a)

[ Able to give the aim of the experiment correctly ] Example : To construct the electrochemical series based on the potential differences between metals

3

11(b)

[ 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

3

11(c)

[ Able to give the hypothesis correctly] Example : The distance between two metals increase/decrease, the potential difference will increase/ decrease

3

11(d)

[ 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

3

11(e)

[ 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

3

11(f)

[ 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) Negative terminal Mg and Cu Al and Cu Zn and Cu Pb and Cu Fe and Cu

2


Question number

Rubric Score

12(a)

Able to state the problem statement correctly Answer: Does the contact of iron with metal X inhibit rusting? / Does the contact of iron with metal Y increase rusting?

3

12(b)

Able to state three variables correctly Answer: Manipulated variable Types of metal Responding variable Rusting or iron Controlled variable Iron nails// jelly solution

3

12(c)

[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

3

12(d)

[Able to state the materials and apparatus correctly] Answer: Apparatus : Test tube, test tube rack Materials : Iron nails, metal X, metal Y, agar-agar, potassium hexacyanoferate (III) and fenolfthalein

3

12(e)

Able to state 6 steps: 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

3

12(f)

[Able to tabulate the data with the following aspects

• Correct titles

• Complete pairs of metals] Answer

Pairs of metal Intensity of blue colour

Intensity of pink colour

Nail

Nail + X

Nail + Y

2


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

b. Pink solution to colourless solution 1

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

d. H2SO4 + 2NaOH � Na2SO4 + 2H2O 1

e. (i) Average volume = (22.40 + 22.30 + 22.00)÷3 = 22.30 cm3 1

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

f. (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

2 a. (i) Yellow 1

(ii) Red 1

(iii) Orange 1

b. 15.00 cm3 1

c (i) H2SO4 + 2KOH � K2SO4 + 2H2O 1

(ii) H+ + OH- � H2O 1

d. 0.1 x 20 = 0.067 moldm-3 2 x 15.00 1 e. (i) Yellow 1

(ii) Red 1

f. 30 cm3 1 10 3 (a) Neutralisation 1

(b) To ensure all nitric acid is completely reacted 1 (c) ZnO + 2HNO3 � Zn(NO3)2 + H2O 1 (d)

[functional apparatus] 1 [label] 1

(e) Number mole of nitric acid = 2(50) = 0.1 mol 1 1000

Mass of salt = 0.1 x 189 = 9.45 g 1 1

(f) Zinc carbonate, 1 Zinc 1

10

zinc oxide


4 (a) Silver chloride 1

(b) White 1 (c) 3.0 cm3 1 (d) (i) 0.5(6.0) = 0.003 mol 1

1000 (ii) Number of mole of sodium chloride = 3.0(1) = 0.003 mol 1 1000

0.003 mol of sodium chloride reacts with 0.003 mol of silver nitrate 1 Therefore 1 mol of sodium chloride reacts with 1 mol of silver nitrate. 1

(e) Ag+ + Cl- � AgCl 1 (f) 1.5 cm 1 (g)

Height of precipitate (cm) 1.5__________ Ι Ι Ι Ι 1.5 Volume of sodium chloride (cm3) [label axis] 1 [curve] 1 10 5 (a) (i) Copper(II) sulphate 1 (ii) Heat the solution until saturated then cool 1

Filter and then rinse the salt with distilled water 1 Dry the salt using filter paper 1

(b) (i) Blue precipitate 1 Cannot dissolve in excess 1

(ii) Cu(OH)2 1 (c) Ammonia 1 (d) (i) White 1

(ii) Pb2+ + SO42- � PbSO4 1

(iii) Filter the mixture 1 10

No. Scheme Marks 6(a) 1. x-axis and y-axis labelled with units

2. plots transferred correctly 3. smooth curve

1 1 1……3

(b) (i) Draw tangent at the point 40.0oC and calculate the gradient (ii) Draw tangent at the point 60.0oC and calculate the gradient

1 1……2

(c) 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 S2O3

2- ions increases

5. The frequency of effective collisions increases.

1 1 1 1 1…..5

(d) Na2S2O3 + H2SO4 → Na2SO4 + S + H2O + SO2 1 (e) Sulphur 1

(f) Concentration 1

13


Energy

7(a) 1. Functional apparatus set-up

2. Complete labels 1 1…..2

(b) 25 32

1

(c) Zn + 2H+ → Zn2+ + H2 1 (d) 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 1 1……4

(e) 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 1 1 1…..4

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

1 1 1….3 13

8 (a) Heat released when 1 mole of a metal is displaced from its salt solution by a more electropositive metal. 1 (b) Silvery solid formed// Colourless solution of silver nitrate becomes blue// Amount of copper powder decreases 1 (c) (i) No. of moles of Ag+ reacted = No. of moles of AgNO3 used = mv/1000 = 0.5(50)/1000 = 0.025 mol 1

(ii) Heat released = No. of moles of Ag x ∆H = 0.025 x 105 kJ = 2.625 kJ = 2625 J 1 (iii) Heat change = mcθ � 2625 J = 50 (4.2) θ � θ = 12.5 oC 1

(d) Assumptions - no heat loss to surrounding 1

- specific heat capacity of solution = specific heat capacity of water 1 - density of solution = density of water ( any two) 1

(e)

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

2 Label for the energy axis and arrow for the two levels are shown. 1

(f) Lower/smaller 1 The total surface area exposed to the air is larger 1 Heat is lost to the environment 1

(g) To ensure all the silver nitrate solution reacted completely 1 (h) Bigger / Higher because magnesium is more electropositive than copper. 1

15

= 0.78 cm3s-1

2 Ag+ + Cu

∆H = -105kJ -1

2 Ag + Cu2+


9 (a) Strong acid – nitric acid/ hydrochloric acid / sulphuric acid 1

Strong alkali – sodium hydroxide / potassium hydroxide 1 (b) Heat released when 1 mole of water is formed from the reaction between an acid and an alkali 1 (c) - It is an exothermic reaction // heat energy is released to the surrounding 1

- The total energy of reactants is higher than the products 1 - 57 kJ of heat energy is released when 1 mole of water is formed ( any 2) 1

(d) (i) No. of moles alkali used = mv/100 = 1(50)/1000 = 0.05 mol 1 (ii) Heat change = mcθ = (50+50)x4.2x 6.5 = 2730 J 1

Heat of neutralization = - 2730/0.05 = - 54600 J/ mol = 54.6kJ / mol 1 (e) (i) The heat of neutralization for Experiment I is higher than Experiment II 1

(ii) Ethanoic acid – weak acid, dissociates partially in water 1 Part of heat released in Experiment t II during neutralization is absorbed to dissociate further the molecules of ethanoic acid 1

(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

13 ESSAY SECTION B Question Number

Explanation Marks

10(a) Ionises completely in water to produce hydrogen ions

1 1..2

(b) 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

(c)(i)

Test Aqueous HCl solution Solution of HCl in

methylbenzene Universal indicator

Green to red No changes

Add zinc powde

Gas bubbles formed No changes

Add copper(II)oxide powder

Blue solution formed No changes

1+1 1+1 1+1….6

(ii) Aqueous hydrogen chloride solution In presence of water hydrogen chloride ionizes to produce hydrogen ions/H+

1 1 1 1…..4

(iii) Aqueous HCl HCl in methylbenzene Observation Ammeter needle

deflected Ammeter needle does not deflect

Explanation Consists of free moving ions

Consists of molecules // no ions

1 1 1 1…..4 20

Question Number

Explanation Marks

11 (a) • 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.

1 1 1 1 1


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

1 1 1 1 1…..10

11(b) • 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.

1 1 1 1 1 1…..6

11(c)(i) Lead(II) oxide Brown when hot, yellow when cold Carbon dioxide Turn lime water chalky

1 1 1 1….4 20

ESSAY SECTION C

12 (a) (i) Acid that will produce two moles of hydrogen ion, H + from one mole of the acid in water. H2SO4

1 1....2

(a)(ii) Acid that dissociates completely into hydrogen ion, H + in water. HCl

1 1....2

(b) 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 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 [ preparation of 1.0 mol dm

-3 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

contents into a 250 cm3 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. [ preparation of 0.1 mol dm

-3 KOH ]

[calculation] Volume of KOH is added 9. M1 x V1 = M2 x V2 V1 = M2 x V2 / M1 10. = 0.1 x 250 / 1 = 25 cm3 11. 25.0 cm3 of 1.0 mol dm-3 KOH is transfer to 250 cm3 using 25.0 cm3 pipette. 12. Distilled water is added to the volumetric flask until the

1 1 1 1 1 1 1 1 1 1


13 (a) [Name any insoluble salt] [Name any two suitable solution] [Write correct ionic equation] Example: Lead(II) sulphate Lead(II) nitrate solution and sodium sulphate solution Pb2+ + SO4

2− � PbSO4

1 1+1 1…. 4

(b) Test for Fe

2+ 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.

Inference: Fe2+ ion is present Test for SO4

2+ ion

• 5 cm3 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: SO4

2+ion is present.

1 1 1 1 1 1….6

(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

1 1 1 1 1 1 1 1 1 1….10

calibration mark.

1 1....12 20


(a) (i)

• Y-axes : energy and two different level of energy • The position of reactants and products correct

1 1….2

(ii) - reactants have more energy // products have less energy - energy is released during the experiment // this is exothermic reaction

1 1.…2

(b) (i)

- 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.

1 1 1 1 Max 3

14

(ii) - 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

1 1 1 1 Max 3

(c) - apparatus & materials : 2 m - procedures : 5 m - table : 1 m - calculation : 2 m Sample answer : Apparatus : thermometer, measuring cylinder Materials : calcium nitrate soln, sodium carbonate soln, plastic/ polystyrene cup

1 1…2

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 Tabulation of data :

Initial temperature of Ca(NO3)2 (oC) Ө1

Initial temperature of Na2CO3 (oC) Ө2

Average initial temperature (oC) (Ө1 + Ө2)/2 = Ө3 Lowest temperature of the mixture (oC) Ө4

1 1 1 1 1…5

energy

Zn + CuSO4

ZnSO4 + Cu

∆H = -152 kJmol-1


Change in temperature (oC) Ө3- Ө4 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) = x J heat of combustion of ethanol = x kJmol-1 0.05 = y kJmol-1

1 1 1….3 20


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 (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 Calculation : RMM of propanol C3H7OH = 60

12. The no. of mol of propanol burnt =60ba− = y mol

13. The released heat = mcθ = 100 x 4.2 x t = x J

14. The heat of combustion of propanol = y

x J mol-1 or

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)

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

1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1….14 1 + 1 1 1

16(a) Burette reading

24.10, 18.00, 13.00 3

(b) 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

3

(c)

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

3

(d)

Experiment II Because the curve in Experiment II is steepest//the gradient is higher

3


(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

3

(e)(ii)

Hypothesis

When the total surface area of calcium carbonate increases, the rate of reaction increases

3

(f) 0.00 cm3 3

(g)

Fast reaction

Slow reaction

Combustion Photosynthesis Neutralization Rusting Precipitation Fermentation

3

17 a. - Brown solid formed

- Blue solution of coper(II) sulphate becomes colourless - Amount of zinc powder decreases

b. Initial temperature : 30.0o C Highest temperature : 38.0 oC Temperature change : 8.0 oC

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 + Cu2+ � Cu + Zn2+

18 (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


(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 4. Extinguish the lamp when the temperature increase reaches 30oC. 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 Heptane Octane

Initial temperature of water/ oC

Highest temperature of water / oC

Increase in temperature/ oC

Initial mass of spirit lamp and content / g

Final mass of spirit lamp and content/ g

Mass of fuel( /alkane) used/ g


Set�4� No Marking scheme Mark 1(a) CnH2n + 2

n = 1, 2, 3, ..... 1

(b)

| | S : carbon-carbon double bond // - C = C – V : carboxyl group // - COOH

1

1

(c)

1+1

1+1

(d) (i) (ii) (iii)

Propyl propanoate Catalyst Sweet/pleasent/fragrance/fruity smell

1

1

1

(e) (i) (ii)

C3H7OH + 9/2O2 → 3CO2 + 4H2O // 2 C3H7OH + 9O2 → 6CO2 + 8H2O mol U = 11.2 // 0.19 // 0.2 60 mol CO2 = 3 x 0.19 // 0.57 No of molecule CO2 = 0.57 x 6.2 x 1023 // 3.534 x 1023

1

1

1

1

TOTAL 12

2- methylpropane

n-butane


2(a) H H

H ―C ― C ― O―H

H H

H O

H ― C ― C ― O―H

H

1 + 1

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

(d) (i) C2H5OH + CH3 COOH � CH3 COOC2H5 + H2O (ii) ethyl ethanoate (e) (i) Oxidation (ii) C2H5OH + 2[O] � CH3 COOH (f ) (i) fermentation (ii) C6H12O6 � 2C2H5OH + 2CO2

1 1

1 1

1 1

1 1

3. (a) (i) ester (ii) –COO- // -OOC- H

H ― C ― OH

H ― C ― OH

H ― C ― OH

H

HOOC(CH2)7CHCH(CH2)7CH3 (b) unsaturated; the molecule of olive oil contains C=C double bonds. (c) (i) Hydrogen, nickel, 180oC (ii) hydrogenation (addition reaction)

1 1

1

1

1 1

1+1 + 1 1

4 a

b (i)

(ii)

(iii)

c (i)

(ii)

(iii)

A compound that contain element of carbon

C2H4

ethene

double bond between carbon atoms

(a: C=C )

Bromine water

Halogenation// Addition of bromine

Brown bromine water decolourised

1

1

1

1

1

1

1


d (i)

(ii)

(iii)

C2H4 + H2O � C2H5OH

Alcohol

Temperature 300 oC // Pressure 60 atmosphere // concentrated

phosphoric acid

Total

1

1

1

10

5 (a) Ceramic 1 (b) (i) Lead glass 1 (b) (ii) High density / refractive index 1 (b) (iii) Use as prism 1 (c) (i)

H H C = C H Cl

1

(c) (ii) PVC does not rust 1 (d) (i)

1+1

(d) (ii) Bronze 1 (d) (iii) To make statues or medals 1

6(a) (i) Saponification (ii) To precipitate the soap. (iii) To remove the glycerol and excess sodium hydroxide solution. (iv) Concentrated potassium hydroxide solution (v)

(vi) It helps to suspend the grease particles. (b) (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.


No Marking scheme Mark 7 (a) i) Stimulant

ii) Antidepressant iii) Steroid

1 1 1

(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 Marking scheme Mark 8 (a) 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.

1 1 1 1 1 1 Max: 5

(b) Percentage of carbon in pentane, C5H12 = 5(12) 5(12) + 12(1) x 100% = 83.33% Percentage of carbon in pentene, C5H10 = 5(12) 5(12) + 10(1) x 100% = 85.71% 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 1 1 1 1 Max 5

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

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

1+1 1 1 1 1 1

H O H

H ― C ― C ― O ― C ― H

H H


(iv) Used as food flavouring / perfume / fragrance CH3 COOH + CH3OH � CH3 COOCH3 + H2O Catalyst: Concentrated sulphuric acid

1 1 1

TOTAL 10 9 (a) . Element C H O

% 52.2 13.0 34.8 No. of moles 52.2/12 13/1 34.8/16 Ratio of moles 4.35 13 2.175 Simplest ratio 2 6 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.

1

1

1

1

1..5

(b) • 50 cm3 ethanol and 25 cm3 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 + C2H5OH � CH3 COOC2H5 + H2O

1

1 1 1 1

1..6

(c)

Dehydration Alumina / unglazed porcelain chips, heat

1

1+1 ..3

(d) • 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.

1

1 1 1

1

Max: 4 (e) A long chain molecule that is formed by the joining together of smaller

molecules called monomers.

H H H H

n C == C --- C ― C ----

H H H H n

1

1..2 20


Question Number

Explanation Mark Σ

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

(c)(ii) • 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.

Arrangement of atoms – 1

Label - 1

1 1 1 1 1 1 Max: 5

Total 20

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

Zinc atom

Copper atom


Oleum is diluted with distilled water to form concentrated sulphuric acid 1..10 (b) Nitrogen gas is obtained from the atmosphere 1 Hydrogen gas is obtained from natural gas 1 Nitrogen gas is combined with hydrogen gas in the ratio of 1 : 3 by volume 1 Temperature: 450 - 550 ˚C 1 Pressure: 200 atm 1 Catalyst used: Iron powder 1..6 (c) Can withstand high temperature 1 Can withstand high pressure 1 Insulators of heat and electricity 1 Resistant to chemical reactions 1…4 20 No. Marking Criteria Mark Total

12 (a) Part X – hydrophobic/hydrocarbon Part Y – hydrophilic/ionic Parx X – dissolves in grease Part Y – dissolves in water

1 1 1 1

4

(b) 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 1 1 1 1 1

6 (c ) Patient X : Analgeis/anpirin

Patient Y: Antibiotic/penicillin/streptomycin Patient Z ; Psychotherapeutic / antidepressant

1 1 1

3

(d) (i)

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 1

2

(ii)

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

3

(iii) 1.Drowsiness – 1m 2. poor coordination/light-headedness – 1m

2

7

TOTAL 20

13 (a) Sources of sulphur dioxide: Volcanic eruptions 1 Burning of fossil fuels 1 From industries manufacturing sulphur based products 1 [any two] Health hazards: Irritates the nose and eyes 1 Causes bronchitis and asthma 1 Formation of acid rain: Sulphur dioxide reacts with oxygen to form sulphur trioxide 1 2SO2 + O2 → 2SO3 1 Both oxides of sulphur dissolve in rain water to form sulphurous and

sulphuric 1


acids respectively 1 SO2 + H2O → H2SO3

SO3 + H2O → H2SO4 1 1

Effects of acid rain: Corrodes buildings and bridges 1 Damages vegetation 1..max

10 (b)

Tape a steel ball bearing to the brass block 1 Hang a weight of 1 kg at a specified height of 50 cm above the ball bearing 1 Drop the weight and allow it to hit the steel ball bearing 1 Use a caliper or ruler to measure the diameter of the dent made on the

brass block 1

Repeat the experiment to obtain another two readings so that an average value can be calculated

1

The whole experiment is repeated using copper block to replace brass block

1

Observation:

Diameter of dents (cm) Type of block

1 2 Average Brass Copper

1

Results and explanation: Diameter made by copper is larger than brass 1 Brass is harder than copper 1 The foreign atoms (zinc atoms) in brass prevent the layers of copper

atoms from 1

sliding past each other . 1..10 20 14 (a) Sulphur is burnt in the air to form sulphur dioxide. 1 S + O2 → SO2 1 Sulphur dioxide formed is then burnt in the air to form sulphur trioxide 1 2SO2 + O2 → 2SO3 1 Temperature: 450 - 550 ˚C 1 Pressure: 1 atm 1 Catalyst used: Vanadium(V) oxide, V2O5 1 Sulphur trioxide formed is dissolved in concentrated sulphuric acid to form

oleum 1

SO3 + H2O → H2S2O7 1 Oleum is diluted with distilled water to form concentrated sulphuric acid 1..10 (b) Nitrogen gas is obtained from the atmosphere 1 Hydrogen gas is obtained from natural gas 1 Nitrogen gas is combined with hydrogen gas in the ratio of 1 : 3 by volume 1 Temperature: 450 - 550 ˚C 1 Pressure: 200 atm 1 Catalyst used: Iron powder 1..6 (c) Can withstand high temperature 1 Can withstand high pressure 1 Insulators of heat and electricity 1 Resistant to chemical reactions 1..4 20


Question Rubric 15 (a)

[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 Action to be taken (i) Manupilated variable Vulcanized and unvulcanized

rubber// Mass of weight

(i) The way to manupilate variable Repeat by replacing vulcanized rubber with

unvulcanized rubber//Use weights with different masses

(ii) Responding variable Increase in length of rubber

strip//elasticity

(ii) What to observe in the responding variable To measure length of rubber strip

(iii) Fixed variable Initial 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 Marking scheme Mark 16 (a) How does the elasticity of vulcanized rubber differ from that of

unvulcanized rubber?

3 (b) (a) Variable:

Manipulated Variable: Types of rubber Responding variable: Length of rubber strip

Fixed variable: Size of rubber strip, Mass of weight

1 1

1

(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

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

3

No Marking scheme Mark 17 (a) To compare the cleansing power of soap and detergent in hard water. 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


(d) Beaker A B

Observation Only some greasy spots are removed

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

3

(f) 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.

3

No Marking scheme Mark 18 (i)

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

Problem statement: Iron rusts more easily than steel. Hypothesis: Iron rusts faster than steel. Material: Iron nail, steel nail, agar-agar solution, potassium hexacyanoferrate(III) solution. Apparatus: Test tubes. 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 The intensity of blue spots

A B

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.

3

3

3

3

3


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:

Diameter of the dent (cm) Material

Reading 1 Reading 2 Average (cm)

Iron block Steel block

No Marking scheme Mark 19 (i) Aim: To differentiate and identify hexan-1-ol, hex-1-ene and hexane

through chemical tests.

3

(ii) Variables

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

3

(iii)

Apparatus: Test tubes, test tube holder, dropper Materials: bromine water, 0.5 mol dm-3 potassium dichromate(VI)

solution, 1 mol dm-3 sulphuric acid, Liquids X, Y and Z

3

(iv) Procedure: 1. About 2 cm3 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

test tube followed by 1 cm3 of 1 mol dm-3 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

(v) Tabulation of Data:

Reagent

Liqiud

Acidified potassium

dichromate(VI) solution.

Bromine water Inference

X

Y

Z

3

Documents

Chemistry Module Perfect Score 2009 Scheme