Download - Answer Chemistry Perfect Score & X A Plus Module 2013

@Hak cipta BPSBPSK/SBP/2013

1 Perfect Score & X A –Plus Module/mark scheme 2013

BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER

JAWAPAN MODUL PERFECT SCORE &

X A-PLUS 2013

CHEMISTRY

Set 1

Set 2 Set 3

Set 4 Set 5

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MODULE PERFECT SCORE & X A-PLUS 2013

SET 1 :THE STRUCTURE OF ATOM, PERIODIC TABLE OF ELEMENTS AND CHEMICAL BONDS

Question No Mark schemes Mark

1 (a) (i) Melting 1

(ii) Molecule 1

(b) The heat energy absorbed by the particles is used to overcome the forces of attraction

between the naphthalene molecules / particles.

1

(c) The particles move faster 1

(d) (i) X : electron Y : nucleus 1

(ii) Electron 1

(e) (i) W and X 1

(ii) W and X atom have different number of neutrons but same number of protons

Atom// Element W and X has different nucleon number but same proton number

1+1

Σ 10

Question No Mark schemes Mark

2 (a) No of electrons = 18, No of neutrons = 22 1+1

(b) (i) The total number of protons and neutrons in the nucleus of an atom 1

(ii) 40

(c) (i) 2.1 1

(ii)

X

(d) (i) W and Y 1

(ii) Atom W and Y have the same number of valence electrons 1

(iii) To estimate the age of fossils /artefacts. 1

Σ 10

X

X

e

3p

4n X e

e

X




Question No. Mark Scheme Marks

3 (a) (i) Total number of protons and neutrons in the nucleus of an atom 1

(ii) 35 – 18 = 17 1

(iii) shows nucleus and three shells occupied with electron

Label 12 proton, 12 neutron

1 +1

(iv) Number of electrons = 2 1 ...5

(b) (i) Liquid 1

(ii)

Q R

1+1

...3

(c)

1st mark - Label X and Y axis with correct unit

2 nd mark - Correct shape of curve

1+1

10

Temperature/oC

Time/s

67

90



4 a) (i) F 1

(ii) Atom F has achieve stable/octet electron arrangement // has 8 valence electron 1

b) (i) 2D + 2H2O 2DOH + H2

Correct reactant & correct product Balance equation

1

1

(ii) The nuclei attraction towards the valence electrons is weaker in atom G. More easier for atom G to lose / release an electron to form a positively charged ion.

1+1

c) (i) Covalent bond 1

(ii)

1 1

(iii) Cannot conduct electricity at any state/ low melting and boiling point/.... 1

(d) Show coloured ion//formed complex ion//has various oxidation number//act as

catalyst 1

11

5 (a) Increasing of proton number. 1

(b) (i) Na/sodium, Mg/magnesium .... 1

(ii) Atomic size decreases across the period // Period 3. 1

(iii) 1. Number of protons in atom increases when across the period.

2. Force of attraction between nucleus and electrons in the shell is stronger.

1+1

..4 (c) Chlorine more reactive than bromine

Size of chlorine atom is smaller than bromine atom

Chlorine atom is easier to receive one electron

1+1

(d) Al3+

1

(e) (i) Ionic compound 1

(ii)

1+1

11

Y Y x X x

x

x x x

E E Y



6 (a)

P : liquid Q : solid R : gas 1 +1+1

(b) (i) 1. P can be change to Q through freezing process. 2. When the liquid cooled, the particles in liquid lose energy and move slower.

3. As temperature drops, the liquid particles attract tone another and change

into solid

1 1

1

(ii) 1. P can change to R through boiling. 2. When liquid is heated, the particles of the liquid gain kinetic energy and

move faster as the temperature increase

3. The particles have enough energy to overcome the forces between them and gas is formed

1 1

1

(iii) 1. R can be change to P through condensation process. 2. When the gas cooled, the particles in gas lose energy and move slower.

3. Particles attract one another and change into liquid

1 1

1

(c) (i) 1. Uniform scale for X-axis and Y-axis and labelled/size of graph plotted ¾ of

graph paper. 2. Tranfer of point

3. Smooth curve

1 1 1

(ii) 1. Dotted line on the graph from the horizontal line to Y-axis at 80oC.

2. Arrow mark freezing point at 80oC

1 1

(iii) 1. Heat released to sorrounding

2. Is balanced when particles comes together to form a solid

1 1

(iv) Supercooling 1

20



Question No. Mark Scheme Mark 7 (a) (i) Atom R is located in Group 17, Period 3 1 + 1

(ii) Electron arrangement of atom R is 2.8.7.

Group 17 because it has seven valence electron. Period 3 because it has three shells filled with electron

1

1 1

(b) (i) Atoms P and R form covalent bond. To achieve the stable electron arrangement, atom P needs 4 electrons while atom R needs one electron. Thus, atom P shares 4 pairs of electrons with 4 atoms of R, forming a molecule with the formula PR4 // diagram

1 1 1

1

1

(ii) Atom Q and atom R form ionic bond. Electron arrangement for atom Q is 2.8.1 and electron arrangement for

atom R is 2.8.7// Atom Q has 1 valence electron while atow R has 7

valence electron To achieve a stable (octet ) electron arrangement, atom Q donates 1 electron to form a positive ion// equation Q Q

+ + e

Atom R receives an electron to form ion R

-//equation

and achieve a stable

octet electron arrangement. R + e R

-

Ion Q

+ and ion R

- are attracted together by the strong electrostatic forces

to form a compound with the formula QR// diagram

1

1

1

1 1

1

R

R

R

R

P

Q R

+ - -



Question

No Mark scheme Mark

8 (a) 12 represents the nucleon number. 6 represents the proton number.

1 1

(b) Able to draw the structure of an atom elements X. The diagram should be able to show the following informations: 1. correct number and position of proton in the nucleus/ at the centre of the atom. 2. correct number and position of neutron in the nucleus/ at the centre of the atom. 3. correct number and position of electron circulating the nucleus 4. correct number of valence electrons Sample answer:

or

1

1

1 1

e-

e- e

-

e-

e-

e-

e- e

-

e-

e-

e-

11p

12n √1 √2

√3

√4

e-

e- e

-

e-

e-

e-

e- e

-

e-

e-

e-

11p + 12n



(c) (i) Atoms W and Y form covalent bond. To achieve the stable electron arrangement, atom W contributes 4 electrons while atom Y contributes one electron for

sharing. Thus, atom W shares 4 pairs of electrons with 4 atoms of Y, forming a molecule with the formula WY4 // diagram

1 1 1 1 1

(ii) Atom X and atom Y form ionic bond. Electron arrangement for atom X is 2.8.1 and electron arrangement for atom Y is 2.8.7 To achieve a stable (octet )electron arrangement, atom X donates 1 electron to form a positive ion // equation X X

+ + e

Atom Y receives an electron to form ion Y-//equation

and achieve a stable octet

electron arrangement. Y + e Y

-

Ion X+ and ion Y

- are attracted together by the strong electrostatic forces to

form a compound with the formula XY// diagram

1

1

1

1 1

1

(d) The melting point of the ionic compound/ (b)(ii) is higher than that of the covalent

compound/ (b)(i) . This is because in ionic compounds oppositely ions are held by strong electrostatic forces. High energy is needed to overcome these forces. In covalent compounds, molecules are held by weak intermolecular forces. Only a little energy is required to overcome the attractive forces. OR The ionic compound/(b)(ii) conducts electricity in the molten or aqueous state whereas the covalent compound/(b)(i) does not conduct electricity. This is because in the molten or aqueous state, ionic compounds consist of freely

moving ions carry electrical charges. Covalent compounds are made up of molecules only

1

1

1 1 1

or 1 1 1 1 1

20

X Y

+ - -

Y

Y

Y

W Y



9 (a) (i)

1. Correct number of shells and valence electrons

2. Black dot or label Q at the center of the atom

1 1

(ii) 1. Group 14

2. There are 4 valence electrons 3. Period 2

4. Atom consists of 2 shells occupied with electrons

1 1 1 1

(b) (i) 1. Floats and moves fast on the water

2. ‘Hiss’ sound occurs 3. Gas liberates / bubble

[any two]

1 1

(ii) 2Q + 2H2O 2QOH + H2 1. Correct reactant and product

2. Balanced equation

1 1

(c) (i) Compound X Sharing electron between atom B and A

1 1

(ii) Choose any one ionic compound and any one covalent compound.

Melting/boiling point

Ionic compound Covalent compound

1. 2.

3.

High force of attraction between oppositely charged ions are

strong. more heat energy needs to overcome the forces.

low force of attraction between molecules are weak. less heat energy needs to

overcome the forces.

1 1 1 1

Electrical conductivity


4. 5.

Conduct in molten state

or aqueous solution. The free moving ions are

able to carry electrical

charges.

Not conduct electricity.

Neutral molecules are not

able to carry electrical

charges.

1 1 1

1

Solubility


6

7

Soluble in water.

Water molecule is

polar solvent.

soluble in benzene/ toluene /

any organic solvents. The attraction forces between

molecules in solute and

solvent are the same.

20

Q



10

(i)

Compound formed between X

and Y

Molecule formed between Z and

Y

Types of

chemical

bonds

Ionic bond is formed because X

atom donates electrons and Y

atom receives electrons to achieve stable octet electron

arrangement/involve transfer

electron

Covalent bond is formed because

Z and Y atoms share the

electrons to achieve stable electron arrangement //

Inovelve sharing of electron

Boiling

point and

melting point

High because a lot of heat

energy needed to overcome the

strong electrostatic forces between ions

Low because less heat energy is

needed to overcome the weak

forces of attraction between molecules

2

2

(b)

1.Correct electron arrangement of 2 ions 2.Correct charges and nuclei are shown

3. X atom with an electron arrangement of 2.8.2 donates 2 valence electrons to achieve the stable octet electron arrangement, 2.8. X

2+ ion is formed //

X X2+

+ 2e-

4. Y atom with an electron arrangement of 2.6 accept 2 electrons to achieve the

stable octet electron arrangement, 2.8. Y2-

ion is formed // Y + 2e

- Y

2-

5. The oppositely-charged ions, X2+

and Y2-

are attracted to each other by a strong electrostatic force.

6. An ionic compound XY is formed

1 1

1 1

1

1

1

X

X

X

X

X

X

X

X

X

X

X X

X

X

X

X

X

X

X

2+ 2-

X2+

Y2-

X

X

X

X

X

X

X

X

X

X

X X

X

X

X

X

X

X

X



(c)

1. A crucible is filled with solid P until it is half full. 2. Two carbon electrodes are dipped in the solid P and connected to the batteries

using connecting wire.

3. Switch is turned on and observation is recorded. 4. The solid P is then heated until it melts completely.

5. The switch is turned on again and observation is recorded. 6. Steps 1 to 5 are repeated using solid Q to replace solid P.

7. Observations: P does not light up the bulb in both solid and molten states.

Q lights up the bulb in molten state only.

P: naphthalene // any suitable answer Q: lead(II) bromide // any suitable answer

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

20

11 (a) (i) Z : 2.8.7 X : 2.4

1 1 ..2

(ii)

Z atom has 7 valence electrons needs one electron X atom has 4 valence electrons ,hence it needs 4 more electron each atom achieves stable octet electron arrangement share electrons between them four Z atoms , each contributes 1 electron // [ diagram one X atom contributes 4 electrons //[diagram] - four single covalent bonds are formed - the molecular formula is XZ4

- diagram [ no. of electrons in all the occupied shells in the X and Z atoms - correct] [ sharing of 4 pairs of single covalent bonds between 1 X atom and 4 Z atoms ]

1 1 1 1 1

1 1 1 1

1

..10

(iii) Colourless liquid 1

b) [Procedures of the experiment] eg. 1. Add a quarter of spatula of YZ solid and add into a test tube. 2. Pour 2-5 cm

3 of distilled water into the test tube containing theYZ2

3. Stopper the test tube and shake well. 4. Repeat Steps 1 to 3 using [ named organic solvent eg ether ] 5. Observe the changes and record them in a table

1 1 1 1 1

. [Results] Eg Solvent Observation Distilled water Colourless solution obtained [named organic solvent] e.g ether

Solid crystals insoluble in

liquid

[Conclusion] eg ZY is insoluble in organic solvent/[named organic solvent] but soluble in water.

1

1

..7



No Explanation Sub Total

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

electron atom Y more than X. 1

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

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

1 1 1

3

(b) Put group1 metal into bottle that contain paraffin oil Group 1 metal readily reacts with air/moisture in atmosphere/ water

1 1

2

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

1

1

[procedure] 3. Pour some water into the basin 4. Group 1 metal is take out from paraffin oil using forceps 5. A small piece of group 1 element is cut using a small knife 6. Oil on group 1 element is dried using a filter paper 7. The group 1 element is placed in the basin contain water. 8. Dip a red litmus paper into water

1 1 1 1 1 1

Max 5 [observation] 9. Color of red litmus paper turn to blue

1

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

Chemical formulae Balance equation

1 1

Total 20

No Explanation ` Total

13. (a) Glucose // naphthalene // any solid covalent compound covalent

Intermolecular forces are weak

Small amount of heat energy needed to overcomes the forces

1

1

1

1

4

(b) X = 2.1 X = 2.2

Y = 2.7 // Y = 2.6 //

1. Suitable electron aranggement

2. Ionic bond

3. to achieve octet electron arrangement

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

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

6. Ion X+ and ion Y

- are attracted together by the strong

electrostatic forces

1

1

1

1

1

1

1

7

(c) material and apparatus;

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

bulb/ammeter/galvanometer

1



Procedure

A crucible is half fill with solid XY powder

Dipped two carbon electrode

Connect the electrodes with connecting wire to the battery and

bulb

Observed whether bulb glow

Heated the solid XY in the crucible

Observed whether bulb glow

Observation

Solid XY - bulb does not glow

Molten XY - bulb glow

Diagram

Functional diagram

Labeled

1

1

1

1

1

1

1

1

9

TOTAL 20

SET 1:CHEMICAL FORMULAE AND EQUATIONS

Question No Mark scheme Mark

1

(a)

Molar mass is the mass of a substance that contains one mole of the substance. Example : Molar mass of one mole of magnesium is 24gmol

-1 .

1

(b)

Substance Molar mass / gmol

-1

N2 14x2 = 28

CO2 12+2(16) = 44

H2S 2(1)+ 32 = 34

H2O 2(1)+16 = 18

4

(c)

Mole of water = 0.9/ 18 = 0.05

Number of molecules = 0.05 x 6.02 x 1023

= 0.3 x 1023

// 3 x 1022

Mole of carbon dioxide = 2.2 / 44 = 0.05

1

1

1 1



Number of molecules = 0.05 x 6.02 x 1023

= 0.3 x 1023

// 3 x 1022

Number of molecule is simmilar

1

2 (a) (i) Volume CO2 = 0.1 mol x 24dm3mol

-1 = 2.4 dm

3 1

(ii) Mass of CO2 = 0.1 mol x 44 gmol-1

= 4.4 g 1

(iii) Number of molecules = 0.1 mol x 6.02 x 1023 1

(iv) Number of atoms = 6.02 x 1022

x 3 = 1.806 x 10

23 1+1

(b) (i)

Heating, cooling and weighing processes are repeated a few times until a

constant mass is obtained.

(ii)

Compound Anhydrous CoCl2 H2O

Mass/g (34.10-31.50)g = 2.60 g

(36.26-34.10)g = 2.16 g

Number of moles 2.60/130 = 0.02 2.16/18 = 0.12

Ratio of moles 0.02/0.02 = 1 0.12/0.02 = 6

Simplest ratio of moles 1 6

1 mole of CoCl2 combines with 6 moles of H2O Therefore, the molecular formula of hydrated cobalt(II) chloride crystal is CoCl2.6H2O. Hence, the value of x in CoCl2.xH2O is 6.

1

1

1

(iii) Percentage of water

= 6(18)2(35.5)59

)18(6

x 100%

= 108 x 100% = 45.4% 238

1

1

Total 10

3 (a) (i) concentrated sulphuric acid 1

(ii) zink and hydrochloric acid[ any suitable metal and acid ] 1

(iii) Zn + 2HCl ZnCl2 + H2

(b)

(i) Mole of oxygen = 46.35 - 45.15 16

= 1.2 = 0.075 16

1



(ii) Mole of copper = 45.15 - 40.35 64 = 4.8 = 0.075 64

1

1

(iii) Empirical formula = CuO

1

(c) (i) Collect the hydrogen gas in a test tube Put a burning wooden splinter at the mouth of the test tube ‘No pop sound ‘ produced.

1 1 1

(ii) To avoid the hot copper react with oxygen/air 1

(iii) Repeat heating, cooling and weighing processes until a constant mass obtained. 1

Total 11

4 (a) (i) Pb(NO3)2 1

(ii) AgCl 1

(b) (i) Pb2+

+ 2 Cl- PbCl2

Correct formula for reactants and product Balance ionic equation

1+1

(ii) Qualitative aspect : Lead(II) nitrate and sodium chloride are the reactants and lead (II) chloride and sodium nitrate are the products // Lead(II) nitrate solution reacts with sodium chloride solution to form lead(II)

chloride precipitate and sodium nitrate solution.

Quantitative aspect : One mole of lead(II) nitrate reacts with 2 mole sodium chloride to produce 1 mole of lead(II) chloride and 2 mole of sodium nitrate.

1

1

(c) (i) 2 Pb(NO3)2 2 PbO + 4NO2 + O2 1

Compound Colour of the

residue when

hot

Colour of the

residue when

cold

PbO

Brown

Yellow

Gases

Colour of the gas released

NO2

Brown

O2

Colourless

1

1

1

Total 10



No Explanation Mark

5 (a) (i) Al3+

, Pb4+ 1+ 1

(ii) Aluminium oxide Lead(IV) oxide

1 + 1

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

1

1 1

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

2

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

1 1

(ii) CuCO3 CuO + CO2

1 + 1

(iii) 1. 1 mol of copper(II) carbonate decomposed into 1 mol of copper(II) oxide and

1 mol of carbon dioxide 2. copper(II) carbonate is in solid state, copper(II) oxide is in solid state and

carbon dioxide is in gaseous state

1

1

(iv) 1. No. of mole for CuCO3 = 12.4 / 124 = 0.1 mol 2. 1 mol of CuCO3 produces 1 mol of CuO Therefor No. of mole for CuO = 0.1 mol 3. Mass of CuO = 0.1 mol X 80 g mol

-1 = 8 g

1 1

1

(v) Mass of oxygen is 0.8g Simplest mol ratio : Cu : O = 3.2/64 : 0.8/16 = 1 : 1

1 1

20

Mark

6

(a) (i) Empirical formula of a compound is a formula that shows the simplest whole number ratio of each atoms of each element in a compound.

1

(ii) (ii) Substance Empirical formula

C10H8 C5H4

H2SO4

H2SO4

1

1



(b) Element Carbon Hydrogen Oxygen

Percentage (%) 62.07 10.34 27.59 Mass/ g 62.07 10.34 27.59 Mole 62.07/12

= 5.17 10.34/1 = 10.34

27.59/16 = 1.72

Simplest mole ratio

5.17/1.72 = 3

10.34/1.72 = 6

1.72/1.72 =1

Empirical formula = C3H6O

n [C3H6O ] = 116 [ 3(12) + 6(1) + 16 ] n = 116 58 n = 116 n = 2 Molecular formula = C6H12O2

1

1

1

1

1

(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. 10.Record all the mass.

Tabulation of result :

Description Mass/ g Crucible + lid a Crucible + lid + magnesium b Crucible + lid + magnesium oxide c

10

1

Element Magnesium Oxygen Mass / g b-a c-b Mole b-a/ 24 c-b / 16 Simplest ratio of

mole x y

Empirical formula = MgxOy

1 1

1

Max

11

Total 20



No Sub T

7. (a)

1. Empirical formula is the chemical formula that shows the simplest ratio of

atoms of each element in the compound.

2. Molecular formula is the formula that shows the actual number of atoms of each element in the compound.

3. Example : empirical formula of ethene is CH2 and the molecular formula is

C2H4

1 1 1

3

(b)(i)

(ii)

Element Carbon Hydrogen Oxygen

Percentage 40.00 6.66 53.33

Number of moles

40

123.33

6.661

6.66

53.33

163.33

Ratio of moles 1 2 1

Empirical formula is CH2O

n(CH2O) = 180 12n + 2n + 16n = 180 30n = 180 n=6 molecular formula = C6H12O6

1

1

1

1 1

5

(c)(i)

Magnesium is more reactive than hydrogen//Position of magnesium is above

hydrogen in the reactivity series 1

(ii)

Lead(II) oxide / Stanum oxide / iron oxide / copper(II) oxide

(iii)

1. Clean [5 – 15] cm magnesium ribbon with sandpaper and coil it.

2. Weigh an empty crucible with its lid.

3. Place the magnesium in the crucible and weigh again. 4. Record the reading.

5. Heat the crucible very strongly. 6. Open and close the lid very quickly.

7. When burning is complete stop the heating 8. Let the crucible cool and then weigh it again

9. The heating, cooling and weighing process is repeated until a constant mass is

recorded. 10.

Description Mass(g)

Crucible + lid

Crucible + lid + Mg / Zn / Al

Crucible + lid + MgO / ZnO / Al2O3

10

Total 20



SET 2 :ELECTROCHEMISTRY

Question

No Mark scheme Mark

1(a) Electrical to chemical energy / Tenaga elektrik kepada tenaga kimia 1 (b) Pure copper / Kuprum tulen 1 (c) Cu

2+ and H

+ 1

(d)(i) (ii)

Become thicker / brown solid formed Bertambah tebal / pepejal perang terbentuk Cu

2+ + 2e Cu

1 1

(e) Blue solution remain unchanged // the intensity of blue solution is the same. Larutan biru tidak berubah // keamatan warna biru larutan adalah sama.

(i) the concentration of Cu2+

ions remains the same. kepekatan ion kuprum(II) tidak berubah

(ii) the rate of ionized copper at the anode same as the rate of discharged copper(II)

ion at the cathode . kadar pengionan kuprum di anode sama dengan kadar ion kuprum(II) dinyahcaskan di katod

1

1

1

(f) Oxidation / pengoksidaan Copper atom released electron to form copper(II) ion. Atom kuprum menderMarkan / membebaskan elektron menghasilkan ion kuprum(II).

1 1

(g) Electroplating of metal // extraction of metal Penyaduran logam // pengekstrakan logam

1

Total 11

2(a)(i) Chloride ion / Cl-, hydroxide ion / OH

-, sodium ion / Na

+ and hydrogen ion / H

+

Ion klorida / Cl-, ion hidroksida /OH

-, ion natrium , Na

+ dan ion hidrogen / H

+ 1

(ii) Cl-. The concentration of chloride ion is higher than hydroxide ion.

Cl-. Kepekatan ion klorida lebih tinggi daripada ion hidroksida

1 + 1

(iii) 2Cl- Cl2 + 2e 1

(b)(i)

Functional – 1 Label - 1

1 1

(ii) - place lighted splinter at the mouth of the test tube containing hydrogen gas

- “pop” sound produced - Letakkan kayu uji menyala ke dalam tabung uji berisi gas hydrogen

- Bunyi “pop” terhasil

1

1

(iii) - Sodium ion and hydrogen ions move to the cathode, hydrogen ion is selectively

discharged

- hydrogen ion is lower than sodium ion in the Electrochemical Series. - Ion natrium dan ion hydrogen bergerak / tertarik ke katod, ion hidrogen terpilih

untuk nyahcas / discas

- Ion hidrogen terletak di bawah ion natrium dalam Siri Elektrokimia

1 1

Total 11

Carbon electrodes

Elektrod karbon

Sodium sulphate

solution

Larutan natrium

sulfat A

Oxygen gas

Gas oksigen

Hydrogen gas

Gas hidrogen




Question

No Mark scheme Mark

3(a) Cu2+

, H+

1 (b) Carbon electrode which connect to copper electrode in cell A.

Because oxidation takes place Elektrod karbon yang disambung kepada elektrod kuprum dalam sell A Kerana proses pengoksidaan berlaku

1 1

(c)(i) X – silver electrode / elektrod argentum Y – impure silver electrode / elektrod argentum tak tulen

1 1

(ii) Ag+

+ e Ag 1 (d)(i) - The electrode become thinner

- Silver atom ionized / silver atom oxidized to form silver ion - elektrod seMarkin nipis - atom argentum mengion / atom argentum dioksidakan membentuk argentum ion.

1 1

(ii) Y : Ag Ag+ + e

Z : Ag+

+ e Ag 1 1

(e) The waste chemicals emitted contain poisonous heavy metal ions and cyanide ions / alter

the pH of water. Bahan buangan kimia dibebaskan mengandungi logam berat yang beracun dan sianid / mengubah nilai pH air

1

11

Question

No Mark scheme Mark

4(a)(i) Lead(II) ion// Pb2+

, bromide ion// Br-

Ion plumbum(II)// Pb2+

, ion bromida// Br-

1

(ii) Sodium ion // Na+, hydrogen ion// H

+, sulphate ion// SO4

2-, hydroxide ion//OH

-

ion natrium // Na+, ion hidrogen// H

+, ion sulfat // SO4

2-, ion hidroksida //OH

- 1

(b)(i) Lead / Plumbum 1

(ii) Pb2+

+ 2e Pb 1

(iii) Brown gas / Gas berwarna perang 1

(c)(i) hydroxide ion / ion hidroksida 1

(ii) Anode : Oxygen gas anod : Gas oksigen Cathode : hydrogen gas Katod : gas hidrogen

1

1

(iii) Sodium nitrate solution // sulphuric acid Larutan natrium nitrat // asid sulfurik (Any suitable electrolyte)

1

9



Rubric Mark

5(a) (i) Q, R, S , Cu 1 …. 1

(ii) positive terminal : Cu Potential difference : 0.7 V S is higher than Cu in the Electrochemical Series

1 1 1 ..... 3

(b) (i) positive terminal : copper / Cu Negative terminal : Metal P

(ii) metal P : Zinc / Zn // Magnesium/Mg (any suitable metal) Solution Q : Zinc sulphate // magnesium sulphate (any suitable electrolyte)

1 1 1

1

..... 4

(c) (i) anode : greenish yellow gas cathode : colourless gas (bubbles)

1 1 ….. 2

(ii) gas X : hydrogen gas Y : chlorine

1 1 ….. 2

(iii)

Anode Cathode

Ions move to / ion

attracted to

Hydroxide ion/OH-

Chloride ion/Cl-

Hydrogen ion/H+ ,

Potassium ion/K+

Ions selectively

discharged Cl

- H

+

Reason

Concentration Cl- higher

than OH-

Position of hydrogen

ion/H+ is lower than

potassium ion/K+ in the

Electrochemical Series. Half equation

2Cl- Cl2 + 2e 2H

+ + 2e H2

1+1

1+1

1+1

1+1 …. 8

Total 20

Question

No Mark scheme Mark

6(a) (i) Substance R : Glucose / ethanol (any suitable covalent compound) Substance S : Sodium chloride solution ( any salt solution / acid / alkali)

1 1

….. 2

(ii) 1. S conducts electricity but R does not 2. S has free moving ions // ions free to move 3. R consists of molecules / no free moving ions

1 1 1

….. 3

(b) (i) negative terminal : zinc positive terminal : copper

1 1

….. 2

(ii) 1. zinc electrode become thinner 2. Zn Zn

2+ + 2e

1 1

….. 2

(iii) 1. the potential difference decreases 2. iron is lower than zinc in the Electrochemical Series // iron is less electropositive than zinc // distance between iron and

1 1



copper is shorter than distance between zinc and copper in the Electrochemical Series

….. 2

(c) (i) Sample answer Lead(II) bromide / lead(II) iodide /sodium chloride/sodium iodide (any suitable ionic compound) r : substance that decompose when heated. Example : lead(II) nitrate, lead(II) carbonate

1

(ii)

Diagram: Functional Label

Observation: Anode : brown gas Cathode: grey solid

Half equation: Anode : 2Br

- Br2 + 2e

Cathode : Pb2+

+ 2e Pb

Product: Anode : lead Cathode : bromine gas

1 1

1 1

1 1

1 1

….. 8

Total 20

Question

No Mark scheme Mark

7(a)

Sample answer Silver nitrate solution

Functional – 1 Label - 1

Anode : Ag Ag

+ + e

Cathode : Ag+

+ e Ag

1

1 1 1 1 ….. 5

(b) 1. metal X is more electropositive than copper // X is higher than copper in the

Electrochemical Series 1

Carbon electrodes

Elektrod karbon

PbI2 // PbBr2 //

NaCl

Heat

Panaskan

Note :

Observations and half-equations are

based on the substance suggested.

Silver nitrate solution Iron spoon

Silver



2. atom X oxidises to X ion // atom X releases electron 3. copper(II) ion accepts electron to form copper 4. the concentration of copper(II) ion decreases 5. metal Y is less electropositive than copper // Y is lower than copper in the

Electrochemical Series

1 1 1 1 ….. 5

(c) Material 0.5 mol dm

-3 of P nitrate, Q nitrate, R nitrate, S nitrate solutions, metal P, Q, R and S

Apparatus Test tube, test tube rack, sand paper

Procedure

1. Clean the metal strips with sand paper

2. Pour 5 cm3 of P nitrate solution , R nitrate solution , S nitrate solution into different

test tubes.

3. Place a strip of metal P into each test tube 4. Record the observation after 5 minutes

5. Repeat steps 2 to 4 using strip of metal Q, R and S to replace metal P.

Observation

Metal Metal ion P Metal ion Q Metal ion R Metal ion S P X X X

Q / X X

R / / X

S / / /

1

1

1 1 1 1 1

1 1

Conclusion The electropositivity of metals increases in the order of P,Q,R,S

1 …..10

TOTAL 20

SET 2 :OXIDATION AND REDUCTION

Question

No Mark scheme Mark

1 (a) To allow the flow / movement / transfer of ions through it 1

(b) chemical energy to electrical energy 1

( c) mark at electrodes 1

Cell 1 Cell 2 Positive electrode

Negative electrode

Positive electrode

Negative electrode

Q P R S

(d)(i) magnesium more electropositive than copper // above copper in the Electrochemical Series

(ii) blue becomes paler / colourless 1

Concentration / number of Cu2+

ion decreases 1

(iii) Mg→ Mg2+

+ 2e 1

(iv) Oxidation 1

(e)(i) copper become thicker // brown solid deposited 1

(ii) zinc 1

(iii) zinc undergoes oxidation // zinc atom release electron to form zinc ion 1

11



Question

No Mark scheme Mark

2(a) A reaction which involves oxidation and reduction occur at the same time 1 (b) (i) green to yellow/brown 1

(ii) oxidation 1

(iii) Fe2+

→ Fe3+

+ e 1

(iv) 0 1 (c) (i) magnesium 1

(ii) Mg +Fe2+

→ Mg2+

+ Fe 1

(iii) +2 to 0 1

(d) 1. label for iron, water and oxygen

2. ionization of iron in the water droplet (at anode) 3. flow of electron in the iron to the edge of water droplet

Water droplet O2

Iron

1

1

1

11

3 (a) Reaction A : not a redox reaction Reaction B : a redox reaction

1 1

Reaction A: No change in oxidation number

Reaction B: Oxidation number of magnesium changes/increases from 0 to +2 // Oxidation number of zinc changes/decreases from +2 to 0

1

1.....4

(b) (i) Oxidation number of copper in compound P is + 2 Oxidation number of copper in compound Q is + 1

1 1.....2

(ii) Compound P : Copper(II) oxide Compound Q : Copper(I) oxide Oxidation number of copper in compound P is +2 Oxidation number of copper in compound P is +1

1 1 1 1.....4

(iii) Substance that is oxidised : H2

Substance that is reduced : CuO

Oxidizing agent : CuO

Reducing agent : H2

1 1 1 1.....4

(c) (i) X, Z, Y 1

Y : Copper Z : Lead X : Magnesium

1 1 1.....3

Fe Fe

2+ +2e

e e



2Mg + O2 → 2MgO // 2X + O2 → 2XO [Correct formulae of reactants and product] [Balanced equation]

1 1.....2

TOTAL 20

4 (a) (i) Iron(II) ion releases / loses one electron and is oxidised to iron(III) ion// Oxidation number of iron in iron(II) ion increases from +2 to +3. Iron(II) ion undergoes oxidation, Iron(II) ion acts as a reducing agent

1

1

(ii) Iron(II) ion receives/ gain one electron and is reduced to iron.// Oxidization number of iron in iron(II) iron decreases from +2 to 0. iron(II) ion undergoes reduction, Iron(II) ion acts as an oxidising agent

1

1

(b) eMgMg 22

Oxidation number of magnesium increases from 0 to +2 magnesium undergoes oxidation

CueCu 22 oxidation number of copper in copper(II) ion decreases from +2 to 0 copper(II) ion undergoes reduction

1 1 1

1 1 1

(c) At the negative terminal: Iron(II) ion release / lose one electron and is oxidised to iron(III) ion. Fe

2+ Fe

3+ + e

The green coloured solution of iron(II) sulphate turns brown. Fe

2+ act as a reducing agent.

At the positive terminal: Bromine molecules accepts electrons and is reduced to bromide ions, Br

- Br2 + 2e 2Br

- The brown colour of bromine water turns colourless. Bromine acts as an oxidising agent

1 1 1 1 1

1 1 1 1 1

20


5 (a) 1. Mg/Al/Fe/Pb/Zn

2. Magnesium undergoes oxidation as oxidation number of magnesium

increases from 0 to +2 and 3. Copper (II) oxide undergoes reduction as oxidation number of copper in

copper(II) oxide decreases from +2 to 0 4. Oxidation and reduction occur at the same time.

1 1 1 1

(b) Experiment I

1. Fe

2+ ion present

2. Metal X lower than iron in the Electrochemical Series // Metal X is less electropositive than iron

3. Iron atoms releases electrons to form iron(II) ions

1 1 1



Experiment II 1. OH

- ion present

2. Metal Y higher than iron in the Electrochemical Series // Metal Y is more electropositive than iron

3. Atom Y releases electrons to form Yn+

ions 4. Water and oxygen gain electron to form OH

- ion //

2H2O + O2 + 4e → 4OH-

1 1 1 1

Max 3

(c) Procedure

1. One spatula of copper(II)oxide powder and one spatula of carbon powder is

placed into a crucible 2. The crucible and its content are heated strongly

3. The reaction and the changes that occur are observed

4. Steps 1 to 3 are repeated by replacing copper(II)oxide powder with zinc oxide powder and magnesium oxide powder.

Observation

Mixture Observation

Carbon and

copper(II)oxide The mixture burns brightly. The black powder turns brown

Carbon and zinc

oxide The mixture glows dimly. The white powder turns grey.

Carbon and

magnesium oxide No Changes

1

1

1

1

1+1

Explanation Carbon can react with copper(II)oxide and zinc oxide Carbon more reactive than copper and zinc / carbon is above copper and zinc in

the Reactivity Series Carbon cannot react with magnesium oxide Carbon less reactive than magnesium / carbon is below magnesium in the Reactivity Series

1

1

1

1

20

6 Sample answer

(a) Magnesium/Aluminium/zinc/iron/lead 1 Magnesium dissolve//The blue colour of copper(II)sulphate solution become paler // brown solid deposited 1 Mg→Mg

2+ + 2e 1

Cu2+

+ 2e→ Cu 1 Oxidising agent- Cu

2+ ion / copper(II) sulphate 1

Reducing agent- Mg 1..6 (b) sample answer

Pb(NO3)2 + 2KI Pbl2 + 2KNO3 1

Oxidation number: +2 +5 -2 +1 -1 +2 -1 +1 +5 -2 1



no changes of oxidation number of all elements in the compounds of reactants and products. 1

Neutralization 1...4 (c ) sample answer

[Material : Any suitable oxidizing agent (example : acidified potassium manganate(VII) solution, acidified potassium dichromate(VI) solution, chlorine water, bromine water), any suitable reducing agent (example : potassium iodide solution, iron(II) sulphate solution) and any suitable electrolyte] 1 [ Apparatus : U-tube , carbon electrodes , connecting wires and galvanometer] 1 Diagram Functional 1 Labelled 1

Procedure 1 Sulphuric acid is added into a U-tube until 1/3 full 1 2 Bromine water is added into one end of the U-tube while potassium iodide solution is added into the other end of the U-tube 1 3 carefully 1 4 Two carbon electrodes connected by connecting wires to a galvanometer are dipped into the two solution at the two ends of the U-tube. 1 Observation The colour of bromine water change from brown to colourless// The colour of potassium iodide solution change from colourless to yellow/brown// The needle of the galvanometer is deflected 1

Oxidation reaction : Br2 + 2e→ 2Br

- 1

Reduction reaction: 2I- → I2 + 2e 1

Max : 10 20



SET 3 :ACIDS, BASES AND SALTS

Question

No Mark scheme Mark

1 (a)(i) Propanone / Methylbenzene / [any suitable organic solvent] 1

(ii) Water 1

(b)(i) Molecule 1

(ii) Ion 1

(c) 1. Beaker A : No observable change Beaker B : Gas bubbles released 2. H

+ ion does not present in beaker A but H

+ ion present in beaker B //

Hydrogen chloride in beaker A does not show acidic properties but hydrogen chloride in beaker B shows acidic properties

1

1

(d)(i) 1. Correct formula of reactants and products 2. Balanced equation

Mg + 2HCl → MgCl2 + H2

1 1

(ii) 1. Mole of HCl 2. Mole ratio 3. Answer with correct unit

Mole HCl =

// 0.005

2 mol HCl reacts with 1 mol Mg 0.005 moles HCl reacts with 0.0025 moles Mg

Mass Mg = 0.0025 x 24 // 0.06 g

1 1 1

TOTAL 10

Question

No Mark scheme Mark

2 (a)(i) Substance that ionize / dissociate in water to produce H+ ion 1

(ii) 3 1

(iii) 1. Concentration of acid / H+ ion in Set II is lower than Set I

2. The lower the concentration of H+ ion the higher the pH value

1 1

(iv) 1. Ethanoic acid is weak acid while hydrochloric acid is strong acid

2. Ethanoic acid ionises partially in water to produce low concentration of H+ ion

while

3. hydrochloric acid ionises completely in water to produce high concentration of H+

ion

1 1

1

(b)(i) 1. The pH value of sodium hydroxide in volumetric flask B is lower than A 2. Concentration of sodium hydroxide / OH

- ion in volumetric flask B

is lower than A

1

1




(ii) 1. Mole of NaOH 2. Mass of NaOH with correct unit

Mole NaOH =

// 0.005

Mass NaOH = 0.005 x 40 g // 0.2 g

1 1

(iii) 0.01 x V = 0.002 x 100 // 20 cm3

TOTAL 10

Question

No Mark scheme Mark

3 (a) Pink to colourless

1

(b) Potassium nitrate

1

(c)(i) HNO3 + KOH → KNO3 + H2O

1

(ii) 1. Mole of HNO3 // Substitution 2. Mole ratio

3. Concentration of KOH with

Mole HNO3 =

// 0.01

0.01 mole HNO3 reacts with 0.01 mole KOH

Molarity KOH =

mol dm

-3 // 0.4 mol dm

-3

1 1 1

(d)(i) 10 cm3 1

(ii) 1. Sulphuric acid is diprotic acid but nitric acid is monoprotic acid // 1 mole of

sulphuric acid produce 2 moles of H+

ion but 1 mole of nitric acid produce 1 mole of H

+ ion

2. Concentration of H+ ion in sulphuric acid is double compare to nitric acid

3. Volume of sulphuric acid needed is half

1

1 1

TOTAL 10

Question

No Mark scheme Mark

4 (a) Ionic compound formed when H+

ion from an acid is replaced by a metal ion or

ammonium ion 1

(b) Pb(NO3)2 1

(c) To ensure all the nitric acid reacts completely 1

(d)(i) 1. Correct formula of reactants and products 2. Balanced equation 2H

+ + PbO → Pb

2+ + H2O

1 1



(ii) 1. Mole of acid 2. Mole ratio 3. Answer with correct unit

Mole HNO3 =

// 0.05

0.05 moles HNO3 produce 0.025 moles salt G Mass of salt G = 0.025 x 331 g // 8.275 g

1 1 1

(e) 1. Add 2 cm3 dilute sulphuric acid followed by 2 cm

3 of Iron(II) sulphate solution

Slowly add concentrated sulphuric acid by slanted the test tube. Then turn it upright. 2. Brown ring is formed.

1 1

TOTAL

Question

No Mark scheme Mark

5 (a)(i) Salt W : Copper(II) carbonate Solid X : Copper(II) oxide

1 1

(ii) 1. Flow gas into lime water 2. Lime water turns cloudy / chalky

3.

1 1

(iii) Neutralisation

(iv) 1. Correct formula of reactants and products 2. Balanced equation CuO + 2HCl → CuCl2 + H2O

1 1

(b) Cation : Cu2+

ion // copper(II) ion Anion : Cl

- ion // chloride ion

1 1

(c)(i) Ag+ + Cl

- → AgCl

1

(ii) Double decomposition reaction

1

TOTAL

Question

No Mark scheme Mark

6 (a)(i) Green 1

(ii) Double decomposition reaction 1

(b)(i) Carbon dioxide 1

(ii) CuCO3 → CuO + CO2 1

(iii) 1. Functional apparatus 2. Label

1 1

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

Heat

Copper(II) carbonate

Lime

water



(ii) 1. Mole of CuCO3 2. Mole ratio 3. Answer with correct unit

Mole CuCO3 =

// 0.1

0.1 moles CuCO3 produces 0.1 mole CuO Mass CuO = 0.1 x 80 g // 8 g

1 1 1

TOTAL

7 (a) 1. Vinegar 2. Wasp sting is alkali

3. Vinegar can neutralize wasp sting

1 1 1

(b) 1. Water is present in test tube X but in test tube Y there is no water. 2. Water helps ammonia to ionise // ammonia ionise in water

3. OH- ion present

4. OH- ion causes ammonia to show its alkaline properties

5. Without water ammonia exist as molecule // without water OH- ion does not

present

6. When OH- ion does not present, ammonia cannot show its alkaline properties

1 1 1 1 1 1

(c) 1. Sulphuric acid is a diprotic acid but nitric acid is a monoprotic acid

2. 1 mole of sulphuric acid ionize in water to produce two moles of H+ ion but 1 mole

of nitric acid ionize in water to produce one mole of H+ ion

3. The concentration of H+ ion in sulphuric acid is double / higher

4. The higher the concentration of H+ ion the lower the pH value

1 1

1 1

(d)(i) 1. Mole of KOH

2. Molarity of KOH and correct unit

Mole KOH =

// 0.25

Molarity =

mol dm

-3 // 1 mol dm

-3

1 1

(ii) 1. Correct formula of reactants 2. Correct formula of products

3. Mole of KOH // Substitution 4. Mole ratio

5. Answer with correct unit

HCl + KOH → KCl + H2O

Mole KOH =

// 0.025

0.025 mole KOH produce 0.025 mole KCl Mass KCl = 0.025 x 74.5 g // 1.86 g

1 1 1 1 1

TOTAL 20



Question

No Mark scheme Mark

8 (a)(i) 1. PbCl2

2. Double decomposition reaction

1 1

(ii) Copper (II) chloride : Copper(II) oxide / copper(II) carbonate , Hydrochloric acid Lead (II) chloride : Lead (II) nitrate solution , sodium chloride solution ( any solution that contains Cl

- ion)

1 + 1

1 + 1

(b)(i) 1. S = zinc nitrate 2. T = zinc oxide

3. U = nitrogen dioxide 4. W = oxygen

1 1 1 1

(ii) 2Zn(NO3)2 2ZnO + 4NO2 + O2 1+1

(c)(i) 1. Both axes are label and have correct unit

2. Scale and size of graph is more than half of graph paper 3. All points are transferred correctly

1 1 1

(ii)

1

(iii) Mole Ba2+

ion =

// 0.0025

Mole SO4 2-

ion =

// 0.0025

Ba

2+ ion : SO4

2- ion

0.0025 : 0.0025 // 1 : 1

1

1

1

(iv) Ba2+

+ SO42-

→ BaSO4 1

TOTAL 20

Question

No Mark scheme Mark

9 (a) 1. HCl // HNO3 2. 1 mole acid ionises in water to produce 1 mole of H

+ ion

3. H2SO4

4. 1 mole acid ionises in water to produce 2 moles of H+ ion

1 1 1 1

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

-

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

- ion

5. Concentration of OH- ion in sodium hydroxide is higher than in ammonia

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

1 1 1 1 1 1

5



(c) 1. Volumetric flask used is 250 cm3

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

graduation mark 10. stopper the volumetric flask and shake the solution

1 1 1 1 1 1 1 1 1

1

TOTAL 20

Question

No Mark scheme Mark

10 (a)(i) Substance C : Glacial ethanoic acid Solvent D : Propanone [ or any organic solvent]

1 1

(ii) Solution E

1. Ethanoic acid ionises in water 2. Can conduct electricity because presence of freely moving ions

3. blue litmus paper turns to red because of H+ ions is present

Solution F

4. Ethanoic acid exist as molecules

5. Cannot conduct electricity because no freely moving ion 6. Cannot change the colour of blue litmus paper because no H

+ ion

1 1 1 1 1 1

(b) 1. Measure and pour [20-100 cm3] of [0.1-2.0 mol dm

-3]zinc nitrate solution into a

beaker

2. Add [20-100 cm3] of [0.1-2.0 mol dm

-3]sodium carbonate solution

3. Stir the mixture and filter 4. Rinse the residue with distilled water

5. Zn(NO3)2 + Na2CO3ZnCO3 + 2NaNO3

6. Measure and pour [20-100cm3]of [0.1-1.0mol dm

-3]sulphuric acid into a beaker

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

8. Stir the mixture and filter 9. Heat the filtrate until saturated / 1/3 of original volume

10. Cool the solution and filter

11. Dry the crystal by pressing between two filter papers 12. ZnCO3 + H2SO4 ZnSO4 + H2O + CO2

1

1 1 1 1 1 1 1 1 1 1 1

TOTAL 20



SET 3 :RATE OF REACTION

Question

No Mark scheme Mark

1(a)(i) Set II 1

(ii) Able to draw the graph with these criterion:

1 Labelled axis with correct unit 2. Uniform scale for X and Y axis & size of the graph is at least half of the graph

paper 3. All points are transferred correctly 4. Curve is smooth.

1 1

1 1

(b)(i) Set I : 1.Tangen shown in graph correctly 2.Rate of reaction = 0.19 cm

3s

-1 ( +- 0.05)

Set II : 1.Tangen shown in graph correctly 2.Rate of reaction = 0.23 cm

3s

-1 (+- 0.05)

1 1

1 1

(ii) Add catalyst Increase the temperature Use smaller size/ metal powder Increases the concentration of acid// Double the concentration of acid but half volume

[Any two]

1 1

Question

No Mark scheme Mark

2 (a) 1. Correct formulae of reactants and product 2. Balanced equation CaCO3+ 2HNO3 → Ca(NO3) 2+ CO2 + H2O

1 1

(b)

Functional diagram

Label

1 1

(c) 1. Mole of nitric acid 2. Mole ratio 3. Answer with correct unit Number of moles of HNO3 = 0.2 X 50 = 0.01 mol // 1000 2 mol of HNO3 produce 1 mol of CO2 0.01 mol of HNO3 produce 0.005 mol of CO2

1 1 1

Nitric acid

Calcium carbonate

Water




Maximum volume of CO2 = 0.005 x 24 = 0.12 dm

3 // 120 cm

3

(d) Experiment I = 0.12 X 1000 // 0.2 cm3 s

-1 //

10 X 60 //0.12 //0.012 dm

3 min

-1 10 Experiment II = 0.12 X 1000 // 0.4 cm

3 s

-1 //

5 X 60 // 0.12 // 0.024 dm

3 min

-1 5

1

1

(e)(i) Rate of reaction in Experiment II is higher than I 1

(ii) - The size of calcium carbonate in Experiment II is smaller than Experiment I // calcium carbonate powder in Experiment II has a larger total surface area exposed to

collision than Experiment I. - The frequency of collision between between calcium carbonate and hydrogen ion in Experiment II is higher than Experiment I. - The frequency of effective collision s in Experiment II is higher than Experiment I

1

1

1

Question

No Mark scheme Mark

3 (a)

-Total surface area of smaller pieces wood is larger/bigger/ greater than the bigger

pieces of wood - More surface area exposed to air for burning

1 1

(b)(i)

1. Experiment II 2. Present of catalyst /manganase(IV) oxide in Experiment I

1 1

(ii) 1.Correct formulae of reactants and product 2.Balanced equation

2H2O2 → 2H2O + O2

1 1

(iii)

1. Arrow upward with energy label ,two levels and position of reactant and products are correct

2. Curve of Experiment I and experiment II are correct and label

3. Activation energy of experiment I and experiment II are shown and labelled

1 1 1

(c)(i) 1.Correct formulae of reactants and product 2.Balanced equation Zn + 2HCl ZnCl2 + H2

1 1

(ii) No. of mol HCl = 50 X 0.5 // 0.025 1000

1

Energy

2H2O2

Ea

2 H2O +

O2

Ea’



2 mol HCl : 1 mol H2 0.025 mol HCl : 0.0125 mol H2 Volume of H2 = 0.0125 x 24 // 0.3dm

3 // 300 cm

3

1

1

(iii) 1. Add excess zinc powder with 12.5 cm3 of 1 mol dm

-3hydrochloric acid .

2. At the same temperature

OR

1. Add excess zinc powder with 25 cm3 of 0.5 mol dm

-3hydrochloric acid

2. At the higher temperature //present of catalyst

1

1

1

1

(iv) 1. Rate of reaction using sulphuric acid is higher 2. The concentration of H

+ ion in sulphuric acid is higher

3. Maximum volume of gas collected is double

4. The number of mole of H+ ion in sulphuric acid is double

1 1 1 1

20

Question

No Mark scheme Mark

4 (a) 1. Temperature in refrigerator is lower than in cabinet

2. The activity of microorganisme (bacteria) in refrigerator is lower than in

refrigerator 3. The amount of toxin produced in the refrigerator is less then in the kitchen

cabinet.

1 1

1

(b)(i) 1. Correct formula of reactants and products 2. Mol of sulphuric acid

3. Mole ratio 4. Volume and ratio

Zn + H2SO4 ------- ZnSO4 + H2 No. Of mol H2SO4 = 1 X 50/1000 // 0.05 1 mol of H2SO4 : 1 mol of H2 0.05 mol of H2SO4 : 0.05 mol of H2

Volume of H2 = 0.05 x 24 dm

3 //1.2 dm

3 //0.05 x 24000//1200 cm

3

1

1

1

1

(ii) Experiment I = 1200 // 15 cm3 s

-1 80 Experiment II = 1200 // 7.5 cm

3 s

-1 160 Experiment III = 600 // 2.5 cm

3 s

-1 240

1

1

1

(iii) Exp I and II 1.Rate of reaction of Expt I is higher 2.The size of zinc in Expt I is smaller 3.Total surface area of zinc in Expt I is bigger/larger 4.The frequency of collision between zinc atom and hydrogen ion/H

+ in Expt I is

higher 5. The frequency of effective collision in Exp I is higher

1

1 1 1 1



Exp II and III 1. Rate of reaction in Expt II is higher 2.The concentration of sulphuric acid/ H

+ ion in Exp II is higher

3. The no. of H+ per unit volume in Expt II is higher/greater in Expt II//

4. The frequency of collision between zinc atom and H+ in Expt II is higher

5. The frequency of effective collision in Expt II is higher

1 1 1 1 1

20

Question

No Mark scheme Mark

5.(a) (i)

N2 + 3H2 ------- 2NH3 1 + 1

(ii) Temperature : 450 – 550 ˚ C Pressure : 200 – 300 atm Catalyst : Powdered iron// Iron filling

[ Any two]

1

1

(b)(i) Example of acid Sample answer : Hydrochloric acid / HCl// Sulphuric acid // Nitric acid Correct formula of reactant and product Balance Sample answer 2HCl + Mg → MgCl2 + H2

1

1

1

(ii) 1. Experiment I : 20 cm3 / 60 s // 0.33 cm

3s

-1 2. Experiment II : 20 cm

3 / 50 s // 0.4 cm

3s

-1 1

1

(iii) (Catalyst) Experiment 1:

1.Pour /measure (50-100) cm3 of (0.1-2 mol dm

-3 ) hydrochloric acid .

2.Add excess zinc powder/granules 3.Add a (2-5 cm

3 ) of copper(II) sulphate solution

4.At the same temperature Experiment II :

1. Pour /measure (50-100) cm3 of (0.1-2 mol dm


2. Add excess zinc powder/granule 3. At the same temperature

OR (Temperature) Experiment 1:


-3 ) hydrochloric acid

2. Heat acid to (30-80OC)

3. Add excess zinc powder/granule

Experiment II :



2. Without heating

3. Add excess zinc powder/granules

OR (Concentration) Experiment 1:



2. Add excess zinc powder/granules 3.At the same temperature

1

1

1

1

1

1

1

1

1

1

1

1

1



Experiment II :



2. Add excess zinc powder/granules

3. At the same temperature

OR (Size) Experiment 1:



2. Add excess zinc powder 3.At the same temperature

Experiment II :



2. Add excess zinc granule 3. At the same temperature

1

1

1

1

1

1

1

1

(iv) (Catalyst) 1.Catalyst/copper(II) sulphate is used in Experiment I 2. Catalyst/(copper(II) sulphate) lower activation energy (and provide an alternative

path) 3. More colliding particles / ions are able to achieve that lower activation energy. 4.The frequency of effective collision between magnesium atoms and hydrogen ion increases. 5. The rate of reaction of Experiment I is higher.

(Any 4) (Temperature)

1. Rate of reaction in Experiment I is higher. 2. The temperature of reaction in Experiment I is higher

3. The kinetic energy of particles increases in Experiment I // The particles move faster

4. Frequency of collision between magnesium atom and H+ ion in Experiment I is

higher 5. Frequency of effective collision in Experiment I is higher

(Any 4)

(Concentration)

1. Rate of reaction in Experiment II is higher 2. The concentration of acid in Experiment I is higher

3. The number of hydrogen ion per unit volume in Experiment II is higher

4. Frequency of collision between magnesium atom and H+ ion in Experiment I is higher

5. Frequency of effective collision in Experiment II is higher (Any 4) (Size) 1.Rate of reaction in Experiment I is higher 2.The size of magnesium in Experiment I is smaller 3.Total surface area of magnesium in Experiment I is bigger/larger 4.The frequency of collision between magnesium atoms and hydrogen ions in Experiment I higher 5.The frequency of effective collision between in Experiment I is higher (Any 4)

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

(v) The number of mol are same // The concentration and volume of acid are same 1



Question

No Mark scheme Mark

6.(a) (i)

1. First minute = 24/60 =0.4 cm3 s

-1 // 24 cm

3 min

-1

2. 2 nd

minute = 34-24/60 =0.167 cm3 s

-1 // 10 cm

3 min

-1

1

1

(ii) 3. rate in 1 st minute higher than 2

nd minute (vice versa)

4. concentration of sulphuric acid / mass of zinc decreases 1

1

(iii) All hydrogen ion from acid was completely reacts 1

(iv) A catalyst lower activation energy provide an alternative path More colliding particles /zinc atoms and hydrogen ions are able to overcome the lower

activation energy. The frequency of effective collisions between zinc atom and hydrogen ion in is higher.

(any 2 )

1

1

(b) - hydrogen and oxygen molecules collide - with correct orientation -total energy of particles higher or equal to activation /minimum energy

1 1 1

(Temperature)

Materials: 0.2 mol dm

-3 sodium thiosulphate, 1.0 mol dm

-3 sulphuric acid, a piece of white paper

marked ‘X’ at the centre. Apparatus: 150 cm

3 conical flask, stopwatch, 50 cm

3 measuring cylinder, 10 cm

3 measuring

cylinder, thermometer, Bunsen burner, wire gauze.

Procedure:

1.Using a measuring cylinder, 50 cm

3 of 0.2 mol dm

-3 sodium thiosulphate solution is

measured and poured into a conical flask.

2.The conical flask is placed on top of a piece of white paper marked ‘X’ at the centre. 3.5 cm

3 of 1.0 mol dm

-3 sulphuric acid is measured using another measuring cylinder.

4.The sulphuric acid is poured immediately and carefully into the conical flask. At the

same time, the stop watch is started 5.The mixture in a conical flask is swirled.

6.The ‘X’ mark is observed vertically from the top of the conical flask through the

solution. 7.The stopwatch is stopped once the ‘X’ mark disappears from view.

8.Step 1 – 7 are repeated using 50 cm

3 of 0.2 mol dm

-3 sodium thiosulphate solution at

40oC, 50

oC, 60

oC by heating the solution before 5 cm

3 of sulphuric acid is added in.

(Max 7) Conclusion When the temperature of sodium thiosulphate solution is higher , the rate of reaction is

higher

1

1

1

1

1

1

1

1

1

1

1



(Temperature)

Materials: 0.2 mol dm

-3 sodium thiosulphate, 1.0 mol dm

-3 sulphuric acid, water, a piece of white

paper marked ‘X’ at the centre.

Apparatus: 150 cm

3 conical flask, stopwatch, 50 cm

3 measuring cylinder, 10 cm

3 measuring

cylinder, wire gauze.

Procedure:

1.Using a measuring cylinder, 50 cm

3 of 0.2 mol dm

-3 sodium thiosulphate solution is

measured and poured into a conical flask.

2.The conical flask is placed on top of a piece of white paper marked ‘X’ at the centre.

3.5 cm

3 of 1.0 mol dm

-3 sulphuric acid is measured using another measuring cylinder.

4.The sulphuric acid is poured immediately and carefully into the conical flask. At the same time, the stop watch is atarted

5.The mixture in a conical flask is swirled. 6.The ‘X’ mark is observed vertically from the top of the conical flask through the

solution. 7.The stopwatch is stopped once the ‘X’ mark disappears from view. 8.Step 1 – 7 are repeated by adding 5 cm

3, 10 cm

3, 15 cm

3, 20 cm

3 and 40 cm

3 of

distilled water .(at the same time) maintaining the total volume of solution at 50 cm3

after dilution//table of dilution (Max 7)

Conclusion When the temperature of sodium thiosulphate solution is higher , the rate of reaction is

higher

1

1

1

1

1

1

1

1

1

1

1

SET 3 :THERMOCHEMISTRY

Question No

Mark scheme Mark

1 (a) Heat change /released when 1 mol copper is displaced from copper (II)

sulphate solution by zinc 1

(b) Blue to colourless 1

(c) (i) 50 X 4.2 X 6 J // 1260 J 1

(ii) (1.0 )(50) 1000

1

(iii) 1260 0.05

1

// 0.05

J // 25200 J mol-1



= - 25.2 kJ mol-1 1

(d) 1. Correct reactant and product 2. Correct two energy level for exothermic reaction 3. Correct value heat of displacement and unit

Sample answer Energy

1 1 1

(e) (i) 3°C 1

(ii)

Number of mole copper displaced is half Heat released is half / 1260 2

1 1

TOTAL 12


2 (a)

Heat of precipitation is the heat change when one mole of a precipitate is

formed from its solution. 1

(b)

To reduce heat loss to the surrounding. Reject : prevent

1

(c) Ag+

+ Cl- → AgCl 1

(d) (i) The heat released =(50 + 50) x 4.2 x 3.5 =1470 J

1

(ii) Number of moles of Ag+

= (50 x 0.5) = 0.025 mol 1000 Number of moles of Cl

- = (50 x 0.5) = 0.025 mol 1000

1

1

(iii)

0.025 mole of Ag+ reacts with 0.025 mole of Cl

- to form 0.025 mole of

AgCl Number of moles of AgCl = 0.025 mol

1

(iv) =

x 1470 J

=58 800 J Heat of precipitation of AgCl = -58.8 kJ mol

-1

1

1

(e)

(i)

Ag+ + Cl

-→AgCl ∆H = -58.8kJmol

-1 // AgNO3 + NaCl →AgCl + NaNO3 ∆H = -58.8kJmol

-1 1

J // 630 J

Zn + CuSO4 //Zn + Cu2+

∆H = - 25.2 kJmol-1

ZnSO4 + Cu //Zn2+

+ Cu



(ii)

1. Label axes 2. Energy levels of reactants and products correct with formula of

reactants and products 3. Heat of precipitation written

1 1

1

Total


3. (a) (i) Ethanol

1

(ii) 1260 kJ of heat energy is released when one mole of ethanol is burnt

completely in excess oxygen

1

(b)

(i) No of moles of alcohol = 0.23 / 46 = 0.005 mol

1 mol of alcohol burnt released 1260 kJ

Thus, 0.005 mol of alcohol burnt released 6.3 kJ

1

1

(ii) mc = 6.3 kJ

mc = 6.3 x 1000

= 6300/ 200 x 4.2

= 7.5 0 C

1

1

( c) Heat is lost to the surrounding // Heat is absorbed by the apparatus or

containers // Incomplete combustion of alcohol

1

(d)

(i)

1. Label axes 2. Energy levels of reactants and products correct with formula of reactants

and products

3. Heat of combustion written

1

1

1

C2 H5 O H + 3 O2

2 CO2 + H2 O

∆ H = - 1260 kJmol-1

Ag+ + Cl

-

∆H = -58.8kJmol-1

AgCl

Energy

Energy



(ii)

1. Label

2. Functional

1

1

(e) (i) - 2656 kJmol

-1 // 2500-2700 kJmol

-1

1

(ii) 1. The molecular size/number of carbon atom per molecule propanol is

bigger/higher methanol

2. Combustion of propanol produce more carbon dioxide and water molecules

3. More heat is released during formation of carbon dioxide and water

molecules

1

1

1

Total marks


4 (a) (i) Characteristic Diagram 4.1 Diagram 4.2 Change in

temperature Increase Decrease

Type of chemical

reaction

Exothermic reaction

Endothermic reaction

Energy content of reactants

and products

The total energy content of the reactants more than

the energy content of the

products

The total energy content of the reactants less than the

energy content of the

products

Amount of

heat absorbed

/realeased during

breaking of

bonds

Amount of heat absorbed

for the breaking of bond in

the reactant is less than heat released during

formation of bond in the

products

Amount of heat absorbed for

the breaking of bond in the

reactant is more than heat released during formation of

bond in the products

1 1

1+1

1+1

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

oC

1

1

1



(b) 1. Number of mole of Ag+ ion in both experiment

= 25 x 0.5 // 0.0125 mol

1000

2. Number of mole of Cl- ion in both experiment

= 25 x 0.5 // 0.0125 mol

1000

3. Number of mole of silver chloride formed is the same

4. Na+

ion and K

+ ion not involved in the reaction // Ag

+ ion

and Cl

- involved in the

reaction

1

1

1 1

(c) (i) Heat change = mcθ = (100)(4.2)(42.2 – 30.2) = 5040 J / 5.04 kJ

Number of moles of HCl / H

+ ion = (50)(2 = 0.1 mol

1000 Number of moles of NaOH / OH

- ion = (50)(2) = 0.1 mol

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

-1

1

1

1 1

(ii) Temperature change is 12.0 oC // same

Number of moles of sodium hydroxide reacted when hydrochloric acid or

sulphuric acid is used is the same // 0.01 mol Number of mole of water formed when hydrochloric acid or sulphuric acid used

is the same // 0.01 mol H

+ ion in excess when sulphuric acid is used

1

1

1 1

Total marks 20


5 (a) (i) Neutralisation//Exothermic reaction 1

(ii) Total energy content of reactant is higher than total energy content in

product 1

(iii) 1. The heat of neutralization of Experiment 1 is higher than Experiment 2

2. HCl is strong acid while ethanoic acid is weak acid 3. HCl ionises completely in water to produce high concentration of H

+ ion

4. CH3COOH ionizes partially in water to produce low concentration of H+

ion and most of ethanoic acid exist as molecules

5. In Expt 2,Some of heat given out during neutralization reaction is used to

dissociate the ethanoic acid molecules completely in water//part of heat that is released is used to break the bonds in the molecules of ethanoic

acid that has not been ionised

1 1 1

1

1

(b)

(i) No of mol acid/alkali= 50 X 1 /1000= 0.05 Q = ∆ H X no of mol = 57.3 X 0.05 = 2.865 kJ // 2865 J

1

1 1

(ii) 2865 = 100 X 4.2 X 0

θ = 2865 ÷ 420 1 1



= 6.8 oC ( correct unit) 1

(iii) 1. Some of heat is lost to the sorrounding

2. Heat is absorbed by polystyrene cup

1 1

(c )

A B The reaction is exothermic// Heat is released to the surrounding during the

reaction

The reaction is endothermic// Heat is absorbed from the surrounding

during the reaction

Heat released is x kJ when 1 mol

product is formed Heat absorbed is y kJ when 1 mol

product is formed. The total energy content in reactant is

higher than total energy content in

product

The total energy content in

reactant is lower than total energy

content in product The temperature increases during the reaction

The temperature decreases during the reaaction

Heat released during the formation of bond in product is higher than heat

absorbed during the breaking of bond

in reactant

Heat absorbed during the breaking of bond in reactant is higher than

heat released during the formation

of bond in product

1

1

1

1

1

TOTAL 20

6

(a) (i)

1. Y-axes : energy

2. Two different level of energy

1 1

(ii) 1. reactants have more energy // products have less energy 2.energy is released during the experiment // this is exothermic reaction

1 1

(b) No. of mol of H+ ion/OH

- = 1x50/1000// 0.05

Heat change = 100x 4.2 x7//2940 Joule//2.94 kJ Heat of neutralization= -2940/0.05 = -58800 J mol

-1//-58.8 kJ mol

-1

1 1 1 1

(c) 1. Heat of combustion of propane is higher 2. The molecular size/number of carbon atom per molecule propane is

bigger/higher 3. Produce more carbon dioxide and water molecules//released more heat energy

1 1 1

1. Methanol/ethanol/ propanol, Diagram: 2. -labelled diagram 3. -arrangement of apparatus is functional

1

1

1..3

energy

Zn + CuSO4

ZnSO4 + Cu

∆H = -152 kJmol-1



1. (100-250 cm

3 )of water is measured and poured into a copper can and the

copper can is placed on a tripod stand 2. the initial temperature of the water is measured and recorded 3. a spirit lamp with ethanol is weighed and its mass is recorded 4. the lamp is then placed under the copper can and the wick of the lamp is lighted

up immediately 5. the water in the can is stirred continuously until the temperature of the water

increases by about 30oC.

6. the flame is put off and the highest temperature reached by the water is

recorded 7. The lamp and its content is weighed and the mass is recorded

…. 8 max 4 Data

The highest temperature of water = t2 The initial temperature of water = t1 Increase in temperature, = t2 - t1 =

Mass of lamp after burning = m2

Mass of lamp before burning = m1

Mass of lamp ethanol burnt, m = m1 - m2 = m …..1

Calculation : Number of mole of ethanol, C2H5OH, n = m

46 ……1 The heat energy given out during combustion by ethanol = the heat energy absorbed

by water= 100x x c x J Heat of combustion of ethanol = m c KJ mol

-1

n

= -p kJ/mol …1

..4

..3

Total marks 20



Question No Mark scheme Mark 7 (a) (i) Heat change = mc = (25+25)(4.2)(33-29) = 445 J

Heat of precipitation of AgCl = - 445 / 0.0125 = -35600 J mol

-1 // 35.6 kJ mol

-1

1. The position and name /formulae of reactants and products are correct. 2. Label for the energy axis and arrow for two levels are shown.

1

1

1 1

(b) (i)

(ii)

1. HCl is a strong acid // CH3COOH is a weak acid. 2. HCl ionised completely in water to produce higher concentration of H

+

ion. // 3. CH3COOH ionised partially in water to produce lower concentration of

H+ ion.

4. during neutralisation reaction, some of the heat released are absorbed by

CH3COOH molecules to dissociate further in the molecules.

1. H2SO4 is a diprotic acid// HCl is a monoprotic acid. 2. H2SO4 produced two moles of hydrogen ion/H

+ when one mole of the acid

ionised in water // 3. HCl produced one mole of hydrogen ion/ H

+ when one mole of the acid

ionised in water. 4. When one mole of OH

- reacts with two moles of H

+ will produce one

mole of water, the heat of neutralisation is still the same as Experiment I because the definition of heat of neutralisation is based on the formation

of one mole of water.

4Max

3

4Max

3

(c) - apparatus and material : 2 marks - procedures : 5 marks - Table : 1 mark - Calculation : 2 marks

Sample answer: Apparatus : Polystyrene cup, thermometer, measuring cylinder. Materials : Copper (II) sulphate, CuSO4 solution, zinc powder.

Procedures :

1. Measure 25 cm3 of 0.2 mol dm

-3 copper (II) sulphate, CuSO4 solution and pour it

into a polystyrene cup.

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

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

4. Stir the reaction mixture with the thermometer to mix the reactants.

5. Record the highest temperature reached.

1 1

1 1 1 1 1

Energy

AgNO3 + NaCl

AgCl + NaNO3*

H = -35.6 kJ mol-1

* Accept ionic equation



SET 4 :CARBON COMPOUNDS

Question No Mark scheme Mark 1 (a)

Or

1

(b) C3H7OH + 9/2O2 3CO2 + 4H2O 1

(c) (i) Sweet/ pleasant smell /// fruity smell 1

(ii) Methanoic acid 1

(iii)

1+1

(d) (i) Oxidation 1

(ii) Orange colour of acidified potassium dichromate (VI) solution turns green 1

(iii) C3H7OH + 2[O] C2H5COOH + H2O 1

(e) C3H7OH C3H6 + H2O

(ii)

1+1

Tabulation of data: Initial temperature of CuSO4 solution (

oC) 1

Highest temperature of the reaction mixture (oC) 2

Temperature change (oC) 2 - 1

....1 Calculation : Number of mole of CuSO 4 = MV/1000 = (0.2)(25)/1000 = 0.005 mol ……1

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

-1 = y kJ mol

-1 …….1

TOTAL 20

propene

propanol

O H H H

H C O C C C H

H H H




Question No Mark scheme Mark 2 (a) (i) Fermentation 1

(ii) Ethanol 1

(iii)

1

(b) C2H5OH + 3O2 → 2CO2 + 3H2O 1+1

(c) (i) Ethene 1

(ii)

1

(d) Purple to colourless 1

(e) (i) Ethyl ethanoate 1

(ii) CH3COOH + C2H5OH CH3COOC2H5 + H2O 1+1

Question

No Mark scheme Mark

3 (a) Characteristics Explanation

Same general formula CnH2n + 1OH

successive member is different from each other by – CH2

Relative atomic mass is different by 14

Gradual change in physical

properties // Melting / boiling point increase

Number of carbon atom per

molecules increase // size of molecule increase

Similar chemical properties // oxidation produce carboxylic acid

Have same chemical/similar functional group

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

alkene

Have same chemical properties // have same functional group

1+1

1+1

1+1

1+1

1+1

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

1

C2H4O2 1

(ii) Carboxylic acid 1

React with carbonate to produce carbon dioxide 1

OH

C

H

H C H

H

H

H H

৷ ৷

C - C

৷ ৷

H H n



(iii) 2 CH3COOH + CaCO3 → (CH3COO)2Ca + H2O + CO2

Correct formula of reactants and products Balanced equation

1 1

(c) Compound P Q The number of carbon atom 2 2

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

Type of covalent bond

between // carbon/ Type of

hydrocarbon

Double bond / /

Unsaturated Single bond/ /

Saturated

Type of homologous series // // Name of compound

Alkene// Ethene //

Alkane // Ethane

General formula// Molecular formula of the

compound

CnH2n // C2H4

CnH2n+2 // C2H6

1 1

1

1

1

Max 4

20

Question No Mark scheme Mark 4 (a) (i) 14.3 % 1

(ii)

Element C H

Mass/ % 85.7 14.3

No. of moles

12

7.85 = 7.14

1

3.14 = 14.3

Ratio of moles/ Simplest ratio 14.7

14.7= 1

14.7

3.14= 2

Empirical formula = CH2

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

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

14n = 56

n = 14

56

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

1

1

1

6 max

5

(iii)

[any 2]

1+1

1+1

Max 4

But-1-ene

But-2-ene

2-methylpropene



(iv) Compound M (Butene, C4H8) has a higher percentage of carbon atom in their

molecule than butane, C4H10 …………….1

% of C in C4H8 = 8)12(4

)12(4

x 100%

= 56

48 x 100%

= 85.7% …………1

% of C in C4H10 = 10)12(4

)12(4

x 100%

= 58

48 x 100%

= 82.7% ………..1

.....3

(b) (i) Starch Protein / natural silk

1 1

(ii) H H CH3 H I I I I C = C – C = C I I H H

2-methylbut-1,3-diene or isoprene

1 1..2

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

(ii)

In vulcanised rubber sulphur atoms form cross-links between the rubber molecules These prevent rubber molecules from sliding too much when stretched

1

1

TOTAL 20


5 (a) (i)

Hydrocarbon Type of

bond Homologous

series General

formula

A covalent alkane CnH2n+2

B covalent alkene CnH2n

3

3

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

1

1 1

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

1

1

1



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

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

1 1

1

1

(b) Carboxylic acid X :

Propanoic acid Alcohol Y:

Ethanol

1

1

1

1

TOTAL 20

Question No Mark scheme Mark 6 (a) (i) X - any acid – methanoic acid

Y - any alkali – ammonia aqueous solution

1 1

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

1 1 1 1 1

5 max 4

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

1

1

(b) (i) Alcohol 1

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

form carboxylic acid

1 1

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

2. Soak the glass wool with 2 cm3 of ethanol

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

4. Heat the porous pot chips strongly

5. Heat glass wool gently



6. Using test tube collect the gas given off

Diagram:

[Functional diagram] [Labeled – porcelain chips, water, named alcohol, heat] Test: Add a few drops of bromine water Brown colour of bromine water decolourised

6 max

5

1 1

1 1

Total 20

Question

No Mark scheme Mark

7

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

1

1 1

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

1 1 1

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

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

1 1 1 1

(d)

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

1 1

1

1

Heat Heat

Glass wool

soaked with ethanol

Porcelain chips

Water



(d)

(ii) Pour 2 cm

3 of [methanoic acid] into a boiling tube

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

Slowly/carefully/drop 1 cm3 of concentrated sulphuric acid

Heat the mixture gently Pour the mixture in a beaker that contain water Observation : Colorless liquid with fruity smell is formed / Colorless liquid float on

water surface

1 1 1 1 1 1

TOTAL 20

Question

No Mark scheme Mark

8(a)

But-2-ene

2-methylpropene

1+1

1+1

(b)

(i)

(ii)

Propanoic acid Ethanol Chemical properties for propanoic acid:

1. React with reactive metal to produce salt and hydrogen gas 2. React with bases/alkali to produce salt and water

3. React with carbonates metal to produce salt, carbon dioxide gas and water 4. React with alcohol to produce ester

[any three]

Chemical properties for ethanol: 1. Undergo combustion to produce carbon dioxide and water

2. Burnt in excess oxygen to produce CO2 and H2O 3. Undergo oxidation to produce carboxylic acid / ethanoic acid

4. React with acidified K2Cr2O7 /KMnO4 to produce carboxylic acid / ethanoic acid

5. Undergo dehydration to produce alkene / ethene. [Any three answers]

1 1

1 1 1 1 1

1 1 1 1

1

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

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

Procedure:

1 1

1

1

C C C C H

H H H H

H

H H

C

C C C

H

H

H

H

H

H

H

H



1. Pour about [2 -5 cm3] of P into a test tube.

2. Add 4-5 drops of bromine water / acidified potassium manganate(VII) solution

and shake.

3. Observe and record any changes. 4. Repeat steps 1 to 3 by replacing P with Q

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

1

1 1

1 1

Max 6

20

SET 4 :MANUFACTURED SUBSTANCE IN INDUSTRY

Question No Mark scheme Mark 1

(a) (i)

Contact process

1

(ii)

Ammonia 1

(iii) Vanadium(V) oxide, 450 oC - 500

oC 1

(iv)

Ammonium sulphate 1

(v) 2NH3 + H2SO4 (NH4)2SO4 1+1

(b) (i) Composite material 1

(ii) Correct arrangement Correct label

1 1

(iii)

nC2H3Cl --( C2H3Cl )n 1

(iv)

It has low thermal expansion coefficient // resistant to thermal shock 1

TOTAL 11


2 (a) (i)

(ii)

SO2 + H2O H2SO3

Corrodes buildings

Corrodes metal structures

pH of the soil decreases

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

1

3 4

Tin atom

Copper atom




(b) (i)

(ii) (iii)

Oleum

2SO2 + O2 2SO3

Moles of sulphur = 48 / 32 =1.5

Moles of SO2 = moles of sulphur

= 1.5

Volume of SO2 = 1.5 24 dm3

= 36 dm3

1 1 1

1 1 1 6

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

The atoms are arranged in an orderly manner.

The layer of atoms can slide over each other.

Thus, pure copper are ductile.

There are empty spaces in between the atoms.

When a pure copper is knocked, atoms slide.

Thus, pure copper are malleable.

1

1 1

1 1 1 1 Max:5

(ii) Zinc.

Zinc atoms are of different size,

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

This reduce the layer of atoms from sliding.

Arrangement of atoms – 1; Label - 1

1 1 1 1

1

1 Max: 5

Total 20

Question No Mark scheme Mark 3 (a) Haber process

Iron

N2 + 3H2 2NH3

1 1

1+1

(b) Pure copper Bronze

Bronze is harder than pure copper

Tin atoms are of different size

The presence of tin atoms distrupt the orderly arrangement of copper

1

1+1

1 1

Zinc atom

Copper atom

Tin atom

Copper atom



atoms.

This reduce the layer of atoms from sliding.

1

1

MAX 6

Procedure: 1. Iron nail and steel nail are cleaned using sandpaper.

2. Iron nail is placed into test tube A and steel nail is placed into test tube

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

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

the nails. 4. Leave for 1 day.

5. Both test tubes are observed to determine whether there is any blue spots formed or if there are any changes on the nails.

6. The observations are recorded

Results:

Test tube The intensity of blue spots A High B Low

Conclusion: Iron rust faster than steel.

1 1+ 1

1

1 1

1 1

1

TOTAL 20

SET 4 :CHEMICALS FOR CONSUMERS

Question No Mark scheme Mark 1 (a) (i) To improve the colour of food

1

(ii) Absorbs water /inhibits the growth of microorganisms 1

(iii)

1. Preservative

2. Flavouring 1 1

(b) (i) Analgesic 1

(ii) To relieve pain 1

(c) (i) Saponification // alkaline hydrolysis 1

(ii)

Hydrophobic hydrophilic

1+1

(iii) Soap form scum/insoluble salts in hard water. 1

TOTAL 10




2 (a) Examples of food preservatives and their functions:

Sodium nitrite – slow down the growth of microorganisms in meat

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

1+1

1+1

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

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

1 1

(ii) Paracetamol

Codeine 1 1

(iii) 1. If the child is given a overdose of codeine, it may lead to addition. 2. If the child is given paracetamol on a regular basis for a long time, it

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

pancreas.

1

1

(c)

Type of food

additives Examples Function

Preservatives Sugar, salt To slow down the growth

of microorganisms Flavourings Monosodium

glutamate, spice,

garlic

To improve and enhance the taste of food

Antioxidants Ascorbic acid To prevent oxidation of

food Dyes/ Colourings Tartrazine

Turmeric To add or restore the

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

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

2

2

2

2

1 1

TOTAL 20

Question No Mark scheme Mark 3 (a) (i) Traditional medicines are derived from plants or animals.

Modern medicines are made by scientists in laboratory and based on

substances found in nature.

1 1

(ii) Type Modern medicine

Analgesics Aspirin Paracetamol Codein

Antibiotics Penicillin Psychotherapeutic Chloropromazin

Caffeina

1 1 1 1

1 1

MAX

5

(iii) Penicillin Cause allergic reaction, diarrhoea, difficulty breathing and easily bruising

1



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

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

1

1

(b) Hard water contains calcium ions and magnesium ions. Example : sea water

Procedure 1. 20cm

3 of hard water (magnesium sulphate solution) is poured into two

separate beakers X and Y.

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

and beaker Y.

3. A small piece of cloth with oily stains is dipped into each beaker. 4. Each cloth is washed.

5. The cleansing action of the soap and detergent is observed.

Results

Beaker Observation X The cloth is still dirty.

Y The cloth becomes clean.

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

1 1

1

1 1 1 1

1 1

1



SET 5 :PAPER 3 SET 1

Rubric Score

1(a)(i) Able to give correct observation Sample answer: Colourless solution formed//Aluminium oxide powder dissolved in nitric acid/sodium hydroxide solution.

3

Rubric Score

1(a)(ii) Able to give the correct inference.

Sample answer Aluminium oxide is soluble in nitric acid/sodium hydroxide solution//Aluminium oxide shows basic/acidic properties

3

Rubric Score

1(a) (iii) Able to give the correct property of aluminium oxide.

Answer: amphoteric

3

Rubric Score

1(b) Able to state the hypothesis correctly. Sample answer: When aluminium oxide dissolves in nitric acid, it shows basic properties,

when aluminium oxide dissolves in sodium hydroxide solution, shows

acidic properties.

3

Rubric Score

1(c) Able to state all the variables correctly.

Answer: Manipulated variable: type of solutions // nitric acid and sodium hydroxide solution Responding variable: solubility of aluminium oxide in acid and alkali//property of aluminium oxide Fixed variable: aluminium oxide

3

Rubric Score

1(d) Able to state the operational definition correctly.

Sample answer. When aluminium oxide solid is added into sodium hydroxide solution, the

solid dissolved.

3




Rubric Score

1(e)(i) Able to give the correct observations for both experiments. Red litmus paper turns blue Blue litmus paper turns red

3

Rubric Score

1(e)(ii) Able to classify all the oxides correctly. Acidic oxide Basic axide Carbon dioxide Phosphorous pentoxide

Magnesium oxide Calcium oxide

3

Rubric Score

2(a) Able to state the observation Sample Answer: 1. Iron glowed brightly 2. Iron ignited rapidly with bright flame. 3. Iron glowed dimly

3

Rubric Score

2(b) Able to state the observation and the way on how to control variable

Sample Answer : 1. change bromine with chlorine and iodine 2. Ignition or glowing of halogen 3. Use the same quantity of iron wool in each experiment.

3

Rubric Score

2(c) Able to state the correct hypothesis by relating the manipulated variable

and responding variable Sample Answer : 1. The higher the position of halogen in group 17 the higher the reactivity towards iron. 2. The higher the position of halogen in group 17 the greater the ignition

or glowing reaction with iron.

Rubric Score

2(d) Able to state the inference correctly. Sample answer: The solid of Iron(lll) bromide formed//Bromine combined with iron //Iron is oxidized by bromine//Bromine is reduced by iron

3

Rubric Score

2(e) Able to arrange the three position of halogen based on the reactivity

toward iron in ascending order Answer : Iodine. Bromine, Chlorine,

3

Rubric Score

3(a) Able to give the correct arrangement of the metals Answer: Magnesium, Y, copper

3



Rubric Score

3(b) Able to give the name of metal Y correctly.

Answer: Zinc//Iron//Lead

3

Rubric Score

3 (c) Able to give the three observations correctly. Answer:

1. Brown solid deposited

2. Blue solution turns light blue

3. Zinc strip becomes pale blue.

3

Rubric Score

4(a)

Able to give the problem statement correctly.

Sample answer: How is the effect of other metals on the rusting of iron when the metals

are in contact with iron.

3

Rubric Score

4(b) Able to state the three variables correctly.

Answer: Manipulated variable: Type of metals//Zinc and copper Responding variable: Rusting of iron Fixed variable: iron nail

3

Rubric Score

4(c) Able to state the hypothesis correctly.

Sample answer: When iron is in contact with a more electropositive metal/zinc, rusting

will not occur, when iron is in contact with less electropositive

metal/copper, rusting will occur.

3

Rubric Score

4(d) Able to list the apparatus and materials needed for the experiment. Apparatus: two test tubes, test-tube rack, Materials: hot agar-agar solution added with phenolphthalein and

potassium hexacyanoferrate(III) solution, iron nails, zinc strip, copper strip, sand paper.

3

Rubric Score

4(e) Able to give the procedures correctly

Sample answer:

1. Clean 2 pieces of iron nails, zinc strip and copper strip with sand

paper. 2. Coil the iron nails with zinc strip and copper strip each.

3. Put the iron nails into two different test tubes

4. Pour hot agar into each test tube until the iron nail is immersed. 5. Leave the apparatus for about 1 day and record the observations.

3



Rubric Score

4(f) Able to tabulate the data correctly

Answer:

Experiment Observation Iron nail coiled with zinc Iron nail coiled with copper

2

PAPER 3 SET 2

Rubric Score

1(a) Able to construct the table correctly with the following aspects:

Experiment Ammeter reading/A I 0.0 II 0.5 III 0.0

3

Rubric Score

1(b) Able to state the inference correctly.

Sample answer: Lead(II) bromide can conduct electricity in molten state//Naphthalene/Glucose

cannot conduct electricity in molten state

3

Rubric Score

1(c) Able to state the type of compound correctly

Answer: ionic compound

3

Rubric Score

1(d) Able to state all the three variables correctly:

Answer: Manipulated variable: type of compound Responding variable: ammeter reading//conductivity of electricity Fixed variable: state of compound//ammeter

3

Rubric Score

1(e) Able to state the hypothesis correctly.

Sample answer: Molten ionic compound can conduct electricity but molten covalent compound

cannot conduct electricity.

3

Rubric Score

1(f) Able to state the operational definition correctly. Sample answer: When carbon electrodes are dipped into molten lead(II) bromide, ammeter

shows a reading/ammeter needle deflects

3




Rubric Score

1(g) Able to explain the difference in conductivity of electricity in Experiment I and

II. Sample answer: In Experiment II, molten lead(II) bromide consists of free moving ions that

carry the electrical current, In Experiment I molten naphthalene consists of neutral molecules.

3

Rubric Score

1(h) Able to classify the substances correctly. Answer:

Substance can conduct electricity Substance cannot conduct electricity Carbon rod Copper(II) sulphate solution

Glacial ethanoic acid Molten polyvinyl chloride

3

Rubric Score

2(a) Able to give the correct value of the reading.

Answer: Final burette reading = 40.20 cm

3 Initial burette reading = 47.20 cm

3 X = 5.0 cm

3

3

Rubric Score

2(b) Able to draw the correct graph with the following aspects.

1. X –axis and y-axis with label and unit 2. Correct scale 3. Correct shape of graph

3

Rubric Score

2(c) Able to determine the correct mole ratio.

Answer: Ag

+ : Cl

- 1.0 x 5 : 1.0 x 5 1000 1000 0.005 : 0.005 1 : 1

3

Rubric Score

2(d) Able to write the ionic equation correctly. Answer: Ag

+ + Cl

- → AgCl

3

Rubric Score

2(e) Able to sketch the correct curve: Graph constant at V = 10 cm

3 3

Rubric Score



2(f) Able to classify the salts correctly.

Soluble salt Insoluble salt Potassium chloride Nickel nitrate Ammonium carbonate

Barium sulphate

3

Rubric Score

3. (a) Able to state the problem statement correctly.

Sample answer: What is the effect of size of zinc on the rate of reaction with sulphuric acid?

3

Rubric Score

3(b) Able to state the hypothesis correctly

Sample answer: When size of zinc is smaller, the rate of reaction is higher.

3

Rubric Score

3(c) Able to state the all the variables correctly

Answer: Manipulated variable: big sized granulated zinc and small sized granulated zinc Responding variable: rate of reaction Fixed variable: volume and concentration of sulphuric acid

3

Rubric Score

3(d) Able to list the necessary materials and apparatus needed.

Sample answer: Materials: big sized granulated zinc, small sized granulated zinc, 0.1 mol dm

-3

sulphuric acid, water. Apparatus: burette, conical flask, delivery tube with stopper, basin, retort, basin, weighing balance, stop watch, measuring cylinder.

3

Rubric Score

3(e) Able to list procedures for the experiment Sample answer.

1. [5-10] g of big sized granulated zinc is weighed and put into the conical flask.

2. Half filled a basin with water.

3. Fill burette with water and invert into the basin and record the initial reading.

4. Measure 50 cm3 of sulphuric acid and pour into the conical flask.

5. Stopper the conical flask and immediately start the stop watch.

6. Record the burette reading every 30 s intervals for 5 minutes.

7. Repeat the experiment by replacing the big sized granulated zinc with small sized granulated zinc.

3



Rubric Score

3(f) Able to tabulate the data with the following aspects:

Time/s 0 30 60 90 120 150 180 210 Burette

reading/cm3

Volume of

gas/cm3

2

PAPER 3 SET 3

RUBRIC SCORE

1(a) Able to record all the temperature accurately Sample answer :

Experiment 1

Initial temperature = 28.0 Highest temperature = 40.0 Change of temperature = 12.0 Experiment II

Initial temperature = 28.0 Highest temperature = 38.0 Change of temperature = 10.0

3

RUBRIC SCORE

1(b) Able to construct table accurately with correct title and unit Sample answer :

Temperature Experiment I Experiment II Initial temperature of mixture,

oC 28.0 28.0

Highest temperature of mixture, oC 40.0 38.0

Change of temperature, oC 12.0 10.0

3

RUBRIC SCORE

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

with direction correctly

Sample answer :

Manipulated variable : type of acid Responding variable : heat of neutralisation Direction : ?

The reaction between a strong acid and strong alkali produce a greater heat of

3



neutralization than the reaction between a weak acid and strong alkali.// The reaction between hydrochloric acid and sodium hydroxide produce a greater heat of neutralization than the reaction between ethanoic acid and sodium hydroxide// The heat of neutralization between a strong acid and a strong alkali is greater than the

heat of neutralization between a weak acid and a strong alkali

RUBRIC SCORE

1(d) Able to explain with two correct reasons

Sample answer :

This is to enable the change in temperature to be measured.

The change of temperature is needed to calculate the heat of neutralization

3

RUBRIC SCORE

1(e) Able to state the formula accurately

Sample answer :

Change in temperature = Highest temperature of mixture - initial temperature of

mixture

3

RUBRIC SCORE

1(f) Able to state three observation correctly

Sample answer :

1. A colourless mixture of solution is obtained 2. The vinegar smell of ethanoic acid disappears

3. The polystyrene cup becomes warmer

3

RUBRIC SCORE

1(g) Able to state three constant variables correctly

Sample answer :

1. The volumes and concentration of the acid and the alkali 2. The type of cup used in the experiment

3. The type of alkali

3

RUBRIC SCORE

1(h) Able to calculate the heat of neutralisation for experiment I and II correctly

Sample answer :

Experiment I

Heat released = mcƟ = 50 x 4.2 x 12 = 2520 J

3



Number of mole of sodium hydroxide = MV = 2.0 x 25/1000 = 0.05 mol 0.05 mole of sodium hydroxide releases 2520 J heat energy

1.0 mole of sodium hydroxide releases = heat released / number of mole = 2520 / 0.05 = 50400 J Heat of neutralisation = - 50.40 kJ/mol Experiment II

Heat released = mcƟ = 50 x 4.2 x 10 = 2100 J

Number of mole of sodium hydroxide = MV = 2.0 x 25/1000 = 0.05 mol 0.05 mole of sodium hydroxide releases 2100 J heat energy

1.0 mole of sodium hydroxide releases = heat released / number of mole = 2100 / 0.05 = 42000 J

Heat of neutralisation = - 42.0 kJ/mol

RUBRIC SCORE

1(i) Able to write the operational definition for the heat of neutralisation correctly. Able to

describe the following criteria

(i) What should be done (ii) What should be observed

Sample answer :

The heat of neutralization is defined as the temperature rises when one mole of water is

produced from reaction between acid and alkali

3

RUBRIC SCORE

1(j) Able to state the relationship between type of acid and value of heat of neutralization and

explain the difference correctly.

Sample answer :

1. The heat of neutralization of a weak acid by a strong alkali is less than the heat of neutralization of a strong acid by a strong alkali.

Explanation :

3



2. Experiment I uses a strong acid whereas Experiment II uses a weak acid.

3. During neutralization of a weak acid such as ethanoic acid, small portion of the heat released in experiment II is absorbed to help the dissociation of the ethanoic acid

molecules

RUBRIC SCORE

1(k) Able to predict the temperature change accurately

Sample answer :

Lower than 10

oC

3

RUBRIC SCORE

1(l) Able to classify the acids as strong acid or weak acid.

Sample answer :

Name of acid Heat of neutralization /kJmol

-1 Type of acid

Ethanoic acid - 50.3 Weak acid

Hydrochloric acid - 57.2 Strong acid

Methanoic acid - 50.5 Weak acid

3

RUBRIC SCORE

2(a) Able to record all the temperature accurately one decimal places.

Time 55.0 s at 30

oC

Time 48.0 s at 35oC

Time 42.0 s at 40oC

Time 37.0 s at 45oC

Time 33.0 s at 50oC

3

RUBRIC SCORE

2(b) Able to construct table accurately with correct title and unit

Sample answer :

Temperature/

oC 30 35 40 45 50

Time/s 55.0 48.0 42.0 37.0 33.0 1/time / s

-1 0.018 0.021 0.024 0.027 0.030

3

RUBRIC SCORE

2(c)(i) Able to draw the graph of temperature against 1/time correctly

i) Axis x : temperature / 0C and axis y : 1/time /1/s

ii) Consistent scale and the graph half of graph paper

iii) All the points are transferred correctly iv) Correct curve

3



RUBRIC SCORE

2(c)(ii) state the relationship between the rate of reaction and temperature correctly

The rate of reaction increases with the increase in temperature

3

RUBRIC SCORE

2(d

) Able to predict the time taken

From the graph, when temperature = 55

oC,

1/time = 0.033 s-1

Time = 1/0.033 = 30.3 s

3

RUBRIC SCORE

2(e)(i) Able to state all variables correctly Manipulated variable : Temperature of sodium thiosulphate solution Responding variable : Rate of reaction between sodium thiosulphate and hydrochloric

acid//time taken for the sign X disappear Constant variable : Concentration and volume of sodium thiosulphate solution and

hydrochloric acid

3

RUBRIC SCORE

2(e)(ii) Able to state how to manipulate one variable while keeping the other variables

constant.

Temperature is the manipulated variable. Heating sodium thiosulphate with several different temperatures by remaining the

3



concentration and volume of sodium thiosulphate and hydrochloric acid constant helps

maintain the responding variable.

RUBRIC SCORE

2(f) Able to give the hypothesis accurately

Manipulated variable : Temperature of sodium thiosulphate solution Responding variable : Rate of reaction between sodium thiosulphate and hydrochloric

acid//time taken for the sign X disappear The higher the temperature, the higher the rate of reaction is

3

RUBRIC SCORE

2(g) Able to state the relationship between temperature and the rate reaction in our daily lives correctly

The lower the temperature, the lower the rate of food turns bad

3

RUBRIC SCORE

3(a) Able to Marke a statement of the problem accurately and must be in question form

Does concentration of ions affect the product of electrolysis process at the anode?

3

RUBRIC SCORE

3(b) Able to state the relationship between manipulated variable and responding variable

correctly

The higher the concentration of ions at the anode, the higher its tendency to be

discharge.

3

RUBRIC SCORE

3(c) Able to state all the three variables correctly

Manipulated variables : concentration of sodium chloride solution Responding variables : product formed at anode Controlled variables : quantity of current, carbon electrodes

3

RUBRIC SCORE

3(d) Able to state the list of substances and apparatus correctly and completely Materials : 0.0001 mol dm

-3 sodium chloride solution, 2.0 mol dm-3 sodium chloride

solution. Apparatus : carbon electrode, electrolytic cell, test tubes, dry cell, blue litmus paper,

wooden splinter, Bunsen burner.

3



RUBRIC SCORE

3(e) Able to state a complete experimental procedure

1. Fill electrolytic cell with 0.0001 mol dm-3

sodium chloride solution. 2. Connect carbon electrodes to the power supply and ammeter. 3. Switch on the circuit for half hour. 4. Collect the gas at the anode and test with a glowing wooden splinter and a damp

blue litmus paper.

5. Repeat the step 1 to 4 by replacing 0.0001 mol dm-3 sodium chloride solution with

2.0 mol dm-3 sodium chloride solution.

3

RUBRIC SCORE

3(f) Able to draw a suitable table with title correctly

Solution Observation Product formed at

anode 0.0001 mol dm-3sodium chloride solution

2.0 mol dm-3sodium chloride solution

3

RUBRIC SCORE

4 (a)

Able to give the statement of problem correctly.

Sample answer: Does the type of electrode/anode affect the choice of ions to be discharged?

3

RUBRIC SCORE

4 (b) Able to state all variables correctly.

Sample answer:

Manipulated variable : Type of electrode/ anode Responding variable : Product formed at anode Controlled variable : Electrolyte

3

RUBRIC SCORE

4(c) Able to give the hypothesis accurately Sample answer:

Type of electrode/anode will influence the choice of ion to be discharged// type of

electrode/anode will produce different product.

3

RUBRIC SCORE

4(d) Able to list completely the materials and apparatus.

Sample answer: Materials:

3



1. copper(II) sulphate solution, (0.5 – 2.0) mol dm-3

//any suitable solution that match with metal plate used.

2. carbon rod

3. copper plate// any metal plate that match with a solution used. 4. wooden splinter// any suitable material used for testing a gas or any product at anode.

Apparatus: 1. electrolytic cell

2. battery 3. connecting wire

4. test tube

RUBRIC SCORE

4(e) Able to state all procedures completely and correctly.

Sample answer:

1. Fill the electrolytic cell (beaker) with half full of copper (II) sulpahate solution

(any suitable electrolyte that match with metal plate used).

2. A test tube filled with copper (II) solution is inverted on the anode carbon electrode.

3. Complete the circuit. 4. Electricity is flowed.

5. Record observation at anode..

6. Step 1-5 is repeated using copper plate

3

RUBRIC SCORE

4(f) Able to exhibit the tabulation of data correctly.

Sample answer: Type of electrode Observation Carbon Copper/any metal

2

PAPER 3 SET 4

Rubric Score

1(a) Able to state all the observations and inferences correctly

Sample answers:

Observations Inferences

1. Zinc electrode become thinner Zinc atom ionised to zinc ions//zinc atom

ionises 2. Brown deposite is formed at copper electrode//thicker

Copper atom is formed

3. Blue solution turn to colourless/ become paler // The intensity of blue solution decrease

Copper(II) ions is discharged to copper atom//concentration of copper(II) ion

decreases

6

Rubric Score



1(b) Able to state all the voltmeter readings accurately with unit

Sample answer:

Zinc and copper : 1.4 V P and copper : 0.8 V Q and copper : 2.8 V R and copper : 0.4 V

3

Rubric Score

1(c) Able to construct a table to record the voltmeter reading for each pair of metals

accurately Sample answer:

Pairs of metals Voltage / V

Zinc and copper 1.4 P and copper 0.8 Q and copper 2.8 R and copper 0.4

3

Rubric Score

1(d) Able to arrange all the metals in ascending order in electrochemical series

Sample answer: Copper, R, P, Zinc, Q

3

Rubric Score

1(e) Able to state the relationship between the manipulated variable and the responding variable with direction.

Sample answer: The further the distance between two/pair of metals in the electrochemical series the

higher/larger/bigger the voltage value.

3

Rubric Score

1(f) Able to state all the three variables correctly Sample answer: Manipulated variable : Pairs of metals Responding variable :Voltmeterreading/voltage/potential difference Constant variable : copper electrode, copper(II) sulphate solution

3

Rubric Score

1(g) Able to state the operational definition for the potential difference accurately Sample answer: The potential difference is the voltmeter reading when two different metals are dipped

in an electrolyte.

3



Rubric Score

1(h) Able to classify the cations and anions in copper(II)sulphate solution correctly

Sample answer:

Cations Cu

2+, H

+ anions SO4

2-, OH

-

3

Rubric Score

1(i) Able to predict the positive terminal and the voltage value correctly Sample answer:

Positive terminal Voltage /V

P 2.0

3

Rubric Score

1(i) Able to explain the relationship between the time for negative terminal to corrode and

the position in electrochemical series accurately

Sample answer: The distance between magnesium and copper in electrochemical series further//the

distance between zinc and copper in electrochemical series is closer

3

Rubric Score

2 (a) Able to state the inference correctly.

Sample answer: The reactivity (of alkali metals with oxygen) increase from lithium to potassium. //

Lithium, sodium and potassium / alkali metals show similar chemical in their reactions with oxygen.

3

Rubric Score

2 (b) Able to state the three variables correctly:

1. Method to manipulate variable. 2. The responding variable.

3. The controlled variable. Sample answer: (i) Use different types of (alkali) / (group 1) metals (ii) Reactivity of metals with oxygen // Vigorousness of the reaction between

metals and oxygen. (iii) Oxygen gas // size / mass of metal

3

Rubric Score

1 (c) Able to state the relationship correctly between the manipulated variable and the

responding variable. Sample answer: (The lower/higher the position of metal in)/(Going down/up) Group 1, the more/less

3



reactive is the metal in reaction with oxygen. // The lower/higher the metal in Group 1 the more/less reactive the reaction with oxygen.

Rubric Score

2 (d) Able to give the operational definition accurately by stating the following three

information.

- alkali metals - vigorously / more vigorous / reactive with oxygen

- more / highly reactive

Sample answer: An alkali metal that reacts more vigorously with oxygen is a more reactive metal.

3

Rubric Score

2 (e)(i) Able to state the position of metal X in Group 1 accurately.

Sample answer: Period 5/6/7

3

Rubric Score

2 (e)(ii) Able to arrange the metals in ascending order based on their reactivity.

Sample answer: Lithium, Sodium, Potassium, X // Li, Na, K, X

3

Rubric Score

2 (f) Able to state the relationship between the mass of sodium and the time taken for the metal to burn completely in oxygen gas.

- the higher the mass / the bigger the size - the longer the time taken

- burn completely Sample answer: The higher the mass of metals, the longer the time taken to burn completely. // The bigger the size of metals, the longer the time taken to burn completely.

3

Rubric Score

2 (g) Able to record all the readings with one decimal place accurately.

Sample answer:

10.1 , 10.6, 10.9

3



Rubric Score

2 (h) Able to state observations for blue and red litmus paper correctly. Sample answer: Solutions Red litmus paper Blue litmus paper Gas Jar I Turns blue No change Gas Jar II Turns blue No change Gas Jar III Turns blue No change

3

Rubric Score

2 (i) Able to write the two balanced chemical equations for the reaction accurately.

Sample answer : i. 4Na + O2 2Na2O and ii. Na2O + H2O 2NaOH Notes: Sodium can be replaced with any alkali metals from Table 1.

3

Rubric Score

2 (j) Able to classify all alkaline solutions into strong alkali and one weak alkali

correctly. Sample answer: Strong alkali : Sodium hydroxide / NaOH, Potassium hydroxide / KOH Calcium hydroxide / Ca(OH)2 Weak alkali : Ammonia solution/ NH3

3

Rubric Score

3(a) Able to give the statement of the problem accurately. Response is in question form.

Sample answer: Does the temperature of sodium thiosulphate solution affect the rate of reaction

between sodium thiosulphate solution and sulphuric acid? // How does the temperature of sodium thiosulphate solution affect the rate of reaction? between sodium thiosulphate solution and sulphuric acid?

3

Rubric Score

3 (b) Able to state the three variables correctly Sample answer: Manipulated variable: Temperature of sodium thiosulphate solution Responding variable : Rate of reaction // Time taken for mark ‘X’ to become invisible /disappear Constant variable: Volume and concentration of sodium thiosulphate/ sulphuric acid

/ size of conical flask

3

Rubric Score

3 (c) Able to state the relationship correctly between the manipulated variable and the

responding variable with direction. Sample answer:

3



The higher/lower the temperature of sodium thiosulphate solution, the higher/lower

the rate of reaction. // The higher/lower the temperature of sodium thiosulphate solution, the

shorter/longer the time taken for mark ‘X’ to disappear from sight/view //

Rubric Score

3(d) Able to give complete list of materials and apparatus Sample answer: Materials : Sodium thiosulphate solution, sulphuric acid. Apparatus : Conical flask, ,bunsen burner, measuring cylinder, stop-watch,

filter paper.

3

Rubric Score

3(e) Able to list all the steps correctly Sample Answer:

1. ‘X ‘mark is drawn on a piece of white/filter/ cardboard paper.

2. 50 cm3 of sodium thiosuphate solution [(0.01-1.0) mol dm

-3] is

measured with a measuring cylinder and is poured into a conical flask.

3. The solution is slowly heated until 30 oC.

4. 5 cm3 of hydrochloric acid [(0.1- 2.0) mol dm

-3] is measured with a

measuring cylinder and is added to the conical flask. A stop-watch is

started immediately.

5. The conical flask is swirled and is placed on a filter paper with a mark ‘X’.

6. The ‘X’ mark is observed vertically from the top through the solution.

7. The stop-watch is stopped immediately when the ‘X’ mark cannot be

seen. Time is recorded. 8. The experiment is repeated by using the sodium thiosuphate solution at

40 oC, 50

oC, 60

oC and 70

oC respectively.

3



Rubric Score

3 (f) Able to tabulate the data with following aspects

1. Correct titles with units 2. Complete list of temperatures

Sample answer:

Temperature (oC) Time (s)

30 40 50 60 70

2

PAPER 3 SET 5

Rubric Score

1(a) Able to state four observations correctly Sample answers:

Observations at anode Blue litmus paper : turn red then bleached / decolourise Glowing splinter : no change Blue litmus paper : no change Glowing splinter : is rekindled / relighted

3

Rubric Score

1(b) Able to state the colour change in the copper (II) chloride solution correctly Sample answer:

The intensity of the blue solution decreases / reduced // Blue colour of solution fades gradually // Blue solution becomes light blue

3

Rubric Score

1(c) Able to state all the variable and the action to be taken correctly

Sample answer:

Name of variables Action to be taken Concentration of copper (II) chloride

solution Change the concentration from 1.0 mol

dm-3

to 0.001 mol dm-3

Gas collected at anode

The change of damp blue litmus paper

and glowing splinter

Type of solution

Use the same copper (II) chloride

solution

6




Rubric Score

1(d) Able to state the relationship between the manipulated variable and the responding

variable with direction. Sample answer: The higher the concentration of ion in the solution in the electrolyte, the higher the chance the ion discharged at anode

3

Rubric Score

1(e) Able to classify the ions correctly -write the name or symbols of the ions. Sample answer:

Cations Anions Copper (II) ions, Cu

2+ Hydrogen ions, H

+ Hydroxide ions, OH

- Chloride ions, Cl

-

3

Rubric Score

2(a)

Able to state all the observation and inference correctly.

Sample answer:

Observation inference White fume is released White solid is formed The mass of crucible and its content

increases

Magnesium oxide is formed Magnesium reacts wth oxygen

6

Rubric Score

2 (b) Able to state all the masses accurately

Sample answer: The crucible and lid = 25.35 g The crucible, lid and magnesium ribbon = 27.75 g The crucible, lid and magnesium oxide when cooled = 29.35 g

3

Rubric Score

2 (c) (i) The mass of Mg = 27.75 – 25.35 = 2.4 g

(ii) The mass of Oxygen = 29.35 – 27.75 = 1.6 g

(iii) The number of mole of Mg = 2.4/24 = 0.1 mole The number of mole of O = 1.6/16 = 0.1 mole The ratio of Mg : O = 1 : 1 The empirical formula is MgO

3



Rubric Score

2 (d) 0.1 mole of Mg reacts with 0.1 mole of oxygen// 1 mole of Mg reacts with 1 mole of oxygen

3

Rubric Score

2 (e) Able to predict and give a reason for the prediction Sample answer:

Cannot because copper is a less electropositive metal. Copper cannot reacts with

oxygen gas to produce copper (II) oxide.

3

Rubric Score

2 (f) Able to classify the oxides into two groups, those which are basic oxides and those

which are acidic oxides correctly

Sample answer:

Basic oxides Acidic oxides

Magnesium oxide

Copper (II) oxide

Sulphur oxide

Carbon dioxide

3

Rubric Score

3(a)

Able to give the statement of the problem accurately. Response is in question form.

Sample answer How does ethanoic acid and ammonia solution affects the coagulation of latex?

3

Rubric Score

3(b) Able to state the three variables correctly Sample answer: Manipulated : ethanoic acid and ammonia solution Responding : coagulate / coagulation of latex Fixed : latex

3

Rubric Score

3(c) Able to state the relationship correctly

Ethanoic acid coagulates the latex while ammonia solution does not coagulate the

latex.

3

Rubric Score

3(d) Able to state the complete list of apparatus and material as follows.

Materials: ethanoic acid 0.5 mol dm

-3 and ammonia solution

Apparatus: Beaker, measuring cylinder, glass rod, dropper

3



Rubric Score

3(e) Able to list all the steps correctly

1. 10 cm3 of latex is poured into a beaker.

2. Ethanoic acid is added into the beaker using a dropper. 3. The mixture is stirred using glass rod.

4. The beaker is left aside.

5. The observation is recorded 6. Experiment is repeated using ammonia solution to replace ethanoic

acid.

3

Rubric Score

3(f) Able to tabulate the data correctly

Mixture Observation Latex + ethanoic acid Latex + ammonia solution

2

PAPER 3 SET 6

RUBRIC SCORE

1(a)(i) Able to record all reading accurately with units Sample answer :

Experiment Copper Bronze I 1.35 cm 1.20 cm II 1.60 cm 1.00 cm III 1.50 cm 1.20 cm

3

RUBRIC SCORE

1(a)(ii) Able to construct the table with correct label and unit

Sample answer :

Type of

blocks Diameter of dents (cm) Average diameter

of dents (cm) I II III Copper 1.35 1.60 1.50 1.48 Bronze 1.20 1.00 1.20 1.13

3

RUBRIC SCORE

1(b) Able to state the observation correctly and accurately

Sample answer : The average diameter of dents on bronze block is 1.13 cm and the average

diameter of dents on copper block is 1.48 cm// The size / diameter of dents on bronze block is smaller than size / diameter of

dents on copper block//

3




1(c) RUBRIC SCORE

Able to state the inference correctly and accurately Sample answer : Bronze is harder than copper // Copper is less harder than bronze

3

1(d) RUBRIC SCORE

Able to state operational definition correctly Sample answer : The smaller dent produced when 1 kg weight is dropped on the block.

3

1(e) RUBRIC SCORE

Able to explain the arrangement of particles in the materials correctly

Sample answer :

1. The atomic size of tin is bigger than copper // the atomic size of tin and copper are different.

2. The presence of tin atoms in bronze disrupts the orderly arrangement

of copper atoms. 3. Reduces / prevent the layers of atoms from sliding over each other

easily

3

1 (f) RUBRIC SCORE

Able to state the hypothesis correctly

Sample answer : Bronze is harder than copper // Copper is less harder than bronze

3

1 (g) RUBRIC SCORE

Able to state all three variables and all three action correctly

Sample answer :

Name of variables Action to be taken

(i) Manipulated variable:

Type of materials //

copper and bronze

(i) The way to manipulate variable:

Replace copper with bronze

(ii) Responding variable:

Diameter of dent

(ii) What to observe in the responding

variable:

The diameter of the dent formed on copper block and bronze block

(iii) Controlled variable:

Mass of weight // height

of the weight // size of steel ball bearing

(iii) The way to maintain the control variable:

Uses same mass of weight // same height of the weight // same size of ball

bearing

3



RUBRIC SCORE

2(a) Able to state 5 correct observations.

Sample answer

Test tube Observation 1 blue colour /solutions 2 High intensity of pink colour/ solutions 3 High intensity of blue colour /solutions 4 Low intensity of pink colour/ solutions 5 Low intensity of blue colour /solutions

3

RUBRIC SCORE

2(a) Able to state 5 correct inferences.

Sample answer

Test tube Inference 1

Iron(II) / Fe2+

ions formed / produced in the solutions // Iron / Fe rusted/corroded/oxidised

2 Iron(II) / Fe

2+ ions are not formed /produced in the solutions //

Iron / Fe does not rust/ corrode/oxidised Magnesium/Mg rusted/corroded /oxidised

3 Iron(II) / Fe

2+ ions formed / produced in the solutions //

Iron / Fe rusted/ corroded/ oxidised

4 Iron(II) / Fe

2+ ions are not formed /produced in the solutions //

Iron / Fe does not rust/ corrode/oxidised // Zinc/Zn rusted/ corroded / oxidised

5 Iron(II) / Fe

2+ ions formed / produced in the solutions //

Iron / Fe is rusted / corroded/ oxidised

3

RUBRIC SCORE

2(b) Able to explain a difference in observation correctly between test tube 2 and 3 Sample answer Iron/Ferum/Fe in test tube 2 does not rust/ corrode/ oxidised because ferum is

in contact with a more electropositive metal, but iron/Ferum/Fe in test tube 3 rusts/ corrodes/ is oxidised because ferum is in contact with a less

electropositive metal. // In test tube 2, magnesium is more electropositive than ferum/iron and and in test tube 3, copper is less electropositive the ferum/iron.

3

RUBRIC SCORE

2(c) Able to state the hypothesis correctly.

Sample answer When a more/less electropositive metal is in contact with iron/ferum/Fe, the

metal inhibits/(speeds up) rusting/corrosion of iron // If the metal in contact with iron is higher/lower than iron/ferum/Fe in

electrochemical series, the rusting/corrosion of iron is slower/faster //

3



RUBRIC SCORE

2(d) Able to state all the variables in this experiment correctly.

Sample answer (i) Manipulated variables : Type/different metal // position of metal in

electrochemical series (ii) Responding variable : Rusting / corrosion // presence of blue/pink colour (iii) Constant variable : Size/mass of iron nail // type of nail // clean iron nails //

temperature // medium in which the iron nail are kept

3

RUBRIC SCORE

2(e) Able to state the operational definition for the rusting of iron nail correctly. Sample answer Rusting occurs when iron nail is in contact with copper/tin /less electropositive metal and form blue colouration in potassium hexacyanoferrate(III) solution

3

RUBRIC SCORE

2(f) Able to classify all the metals correctly.

Sample answer

Metals that inhibit rusting Metals that speed up rusting Magnesium/Mg

Zinc/Zn Tin/Sn

Copper/Cu

3

RUBRIC SCORE

2(g)(i) Able to state the relationship between the time taken and the amount of rust

formed correctly.

Sample answer The longer the time taken, the greater/bigger/larger the rust formed // The

longer the time taken, more rust is formed // The rust formed is greater/bigger/larger, when the time taken is longer.

3

RUBRIC SCORE

2(g)(ii) Able to predict the time taken for the iron nail to completely rust correctly.

Answer Less than 5 days

3

RUBRIC SCORE

2(h)(i) Able to record the voltmeter readings correctly in one decimal place.

Answer

Pairs of metal

Positive terminal

Voltmeter reading

(V) Magnesium and iron Iron 2.0 Iron and copper Copper 0.8 Iron and zinc Iron 0.4 Iron and tin Tin 0.2

3



RUBRIC SCORE

2(h)(ii) Able to draw a labelled diagram accurately.

Sample Answer

3

RUBRIC SCORE

3 (a) Able to Marke a statement of the problem accurately and must be in

question form

Sample Answer Able to How do the heat of neutralisation for reactions between acids and

alkalis of different strengths differ?

3

RUBRIC SCORE

3 (b) Able to state all the three variables correctly

Manipulated Variable : different strength of acid // hydrochloric acid and

ethanoic acid Responding variable : the value of heat of neutralisation Fixed variable : volume and concentration of acid // volume and

concentration of alkali // polystyrene cup

3

RUBRIC SCORE

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

variable correctly

The value of heat of neutralisation for reaction between strong acid and

strong alkali is higher than of reaction between weak acid and strong alkali// The value of heat of neutralisation for reaction between hydrochloric acid

and sodium hydroxide is higher than of reaction between ethanoic acid and

sodium hydroxide

3

RUBRIC SCORE

3 (d) Able to state the list of substances and apparatus correctly and completely

Apparatus : Measuring cylinders, polystyrene cup with covers, thermometer Material : 2.0 mol dm

3 sodium hydroxide, 2.0 mol dm

3 ethanoic acid, 2.0

mol dm3 hydrochloric acid

3

V

Magnesium/Mg Iron/Fe

Dilute sulphuric

acid /H2SO4

Voltmeter



RUBRIC SCORE

3 (e) Able to state a complete experimental procedure


-3 sodium hydroxide solution, NaOH

solution using a measuring cylinder. Pour it into a polystyrene cup

with a cover.


-3 hydrochloric solution, HCl solution

using another measuring cylinder. Pour it into another polystyrene

cup with a cover.

3. Leave both the polystyrene cups on the table for 5 minutes. After 5 minutes, measure and record the initial temperatures of both the

solution.

4. Pour the hydrochloric acid, HCl quickly and carefully into the polystyrene cup containing sodium hydroxide solution.

5. Stir the mixture using the thermometer. 6. Record the highest temperature of the reaction mixture.

7. Repeat steps 1 to 5 using ethanoic acid to replace the hydrochloric

acid.

3

RUBRIC SCORE

3 (f) Able to tabulate the data correctly

Hydrochloric acid Ethanoic acid

Initial temperature of alkali, oC

Initial temperature of acid, oC

Highest temperature of the

reaction mixture, oC

3

PAPER 3 SET 7

RUBRIC SCORE

1 (a) Able to record all the readings accurately to two decimal points with units.

Sample answer: Activity I : 26.05 cm

3, 26.90 cm

3, 30.05 cm

3 Activity II : 13.30 cm

3, 25.85 cm

3, 38.45 cm

3

3

RUBRIC SCORE

1(b)

Able to construct a table containing the following information:

1. Headings in the table 2. Transfer all data from 1(a) correctly

3. With units Sample answer:

Titration

number Initial burette reading /

cm3

Final burette reading /

cm3

Volume of acid /

cm3

3



RUBRIC SCORE

1 0.80 13.30 12.50 2 13.40 25.85 12.45 3 25.90 38.45 12.55

RUBRIC SCORE

1(c) Able to show all the steps to calculate the concentration of sulphuric acid correctly. Sample answer: Step 1: Write the chemical equation: 2NaOH + H2SO4 Na2SO4 + 2H2O Step 2: Calculating the number of moles of sodium hydroxide Number of mol of NaOH : 0.1 x 25 // 0.0025 1000 Step 3: Calculating the concentration of sulphuric acid Concentration of H2SO4 : ( 0.0025 x 1000 ) // 0.1 mol/dm

3 12.50 x 2

3

RUBRIC SCORE

1(d) Able to state the colour change correctly Sample answer: Activity I : Pink change to colourless Activity II : Yellow change to orange

3

1(e) Able to state the correct type of acid in activity I and II and give the correct reason.

Sample answer: Type of acid : Activity I use monoprotic acid and Activity II use diprotic acid. Reason : The volume of acid used in activity I is twice with the volume of acid used in activity II.

3

RUBRIC SCORE

1(f) Able to state the colour change correctly Sample answer: Yellow change to orange and finally change to red

3

RUBRIC SCORE

1(g)

Able to predict the volume with the unit Sample answer: More than 25.00 cm

3 // 25.05 – 50.00 cm

3

3

1(h)

Able to state all the variable correctly Manipulated Variable : Type of acid uses // type of indicator Responding Vvariable : Volume of acid to neutralize 25.0 cm

3 of mol dm

-3 sodium

hydroxide solution // Change in the colour of the indicator. Fixed Variable : Concentration and volume of sodium hydroxide solution.

3

RUBRIC SCORE

1(i) Able to state the hypothesis (relate the manipulated variable with the responding

variable) correctly. Sample answer: If use different type of acid to neutralize 25.0 cm

3 of 1.0 mol dm

-3 sodium hydroxide

solution, the volume of acid use also different// Different indicator used in the titration create different colour.

3



RUBRIC SCORE

RUBRIC SCORE

1(j) Able to give the operational definition for the end-point of titration in activity I correctly. Able to describe the following criteria

(i) What should be done

(ii) What should be observed Sample answer: When hydrochloric acid is added to sodium hydroxide solution with phenolphthalein,

pink turns to colourless.

3

RUBRIC SCORE

1(k)

Able to classify all the acids into strong acid and weak acid correctly.

Sample answer:

Strong acid Weak acid Nitric acid Ethanoic acid

Phosphoric acid Ascorbic acid

3

RUBRIC SCORE

2(a) Able to state the inference accurately Sample answer When alcohol react with carboxylic acid, ester is formed//Esters have sweet pleasant smell property

3

RUBRIC SCORE

2(b) Able to construct a table correctly with the following information:

1. Columns with titles for alcohol, carboxylic acid, Ester

2. Name of all alcohols, carboxylic acid and ester

Alcohol Carboxylic acid Ester Methanol Ethanoic acid Methyl ethanoate Ethanol Propanoic acid Ethyl propanoate Propanol Methanoic acid Propyl methanoate

3

RUBRIC SCORE

2(c) Able to name the alcohol and carboxylic acid correctly. Alcohol: Propanol Carboxylic acid: Butanoic acid

3

RUBRIC SCORE

2(d)(i) Able to state the three variables correctly.

Sample answer Manipulated variable : Hexane and hexene Responding variable : Colour change of bromine water // colour change of

potassium manganate (VII) solution Fixed variable : Bromine water//acidified potassium manganate (VII) solution

3



RUBRIC SCORE

2(d) (ii)

Able to state the hypothesis accurately

Sample answer: Hexene declourised the brown colour of bromine water, hexane does not// Hexene

declourised the purple colour of acidified potassium manganate(VII) solution, hexane does not

3

RUBRIC SCORE

2(d)(iii) Able to predict and Marke explanations accurately

Answer

1. Hexene 2. Percentage of carbon atoms per molecule hexene is higher than hexane

3. Percentage of carbon in hexane = 72 x 100 84 = 85.71 %

4. Percentage of carbon in hexane = 72 x 100 86 = 83.72 %

3

RUBRIC SCORE

3(a) Able to state the problem statement accurately

Sample answer Are the effectiveness of the cleansing action of soap and detergent in hard water

different?

3

RUBRIC SCORE

3(b) Able to state the three variables accurately.

Answer Manipulated variable: Soap and detergent Responding variable: Effectiveness of cleansing action // the ability to remove

the oily stains on cloth Fixed variable: cloth with oily stains, hard water

3

RUBRIC SCORE

3(c) Able to state the hypothesis accurately with direction

Sample answer The cleansing action of a detergent is more effective in hard water than a soap

3

RUBRIC SCORE

3(d) Able to state the complete list of apparatus and material as follows

List of apparatus : 2 beakers, , glass rod List of material : Hard water, soft water, soap and detergent solution, 2 pieces of

cloths stained with oil

3

RUBRIC SCORE

3(e) Able to state procedures correctly as follows

1. [50 - 200] cm3 of hard water is poured into a beaker

2. Soap is added into the beakers

3



3. A piece of cloth stained with oil is immersed in the solution 4. The cloth is shaken/rubbed/stirred

5. Observation is recorded

6. Repeat steps 1 – 6 by using detergent .

RUBRIC SCORE

3(f) Able to tabulate the data correctly Sample answer

Type of cleaning agent Observation

Soap Detergent

3


Download - Answer Chemistry Perfect Score & X A Plus Module 2013

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