Perfect Score Chemistry SBP 2012 - ANSWER

BAHAGIAN PENGURUSAN SEKOLAH BERASRAMA PENUH DAN SEKOLAH KLUSTER

JAWAPAN MODUL PERFECT SCORE 2012 CHEMISTRY [KIMIA]     

Set 1 Set 2 Set 3 Set 4 Set 5

1

JAWAPAN SET 1 PAPER 2 : STRUCTURED QUESTION SECTION A

No

Explanation

1(a)(i) The number of proton in the nuclues of an atom (ii) 18 (b) M and N, because both atoms have same proton number but different nucleon number/ the same the number of proton but different the number of neutron (c)(i) 2.8.6 (ii) Group 16 , Period 3 (iii) 6 valence electron , three shells fill / occupied with electron (d) M-

No

Explanation

2(a) 2.8.3 (b)(i) Z (ii) Z atom has achieve a stable octet electron arrangement. Atom does not donate or receive / share electron with other atom. (c)(i) VW (ii)

Total Mark

1 1 1 1

1 1 2

1 1 1 1 1

1 1 2

Total

9

Sub Mark

Total Mark

1 1 1 1

1 1 2

1

1

1 1 1 1 1

2

Total

10

1

Draw +

V

-

W

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

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

Sub Mark

W/Y U T

1 1 1

2

No

Explanation

Sub Mark

Total Mark

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

2.8.6 16 Atomic size decrease / become smaller The number of proton increase/ positive charge increase Force of attraction between neucleus and valence electron increase Has achieve a stable octet electron arrangement Atom does not donate, receive or share electron with other atom. Al / Aluminium 2Na + 2H2O 2NaOH + H2 Correct the formula reactant and product Balance equation

1 1 1 1 1

1 1 1 2

1 1

2

1

1 2

(c)

(d) (e)

No

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

(c)(i) (ii) (d) (e) (f)

Explanation

K J/ L / M J/L/M Atom of J/L/M is not stable// has 1/6/ 7 valence electron 2 atoms of J/L/M share a pair of electron To achieve duplet/octet electron arrangement 17 17, atom has 7 valence electron M,L, K K+ KM

No

5(a)(i) (ii) (iii) (iv) (v) (b)(i) (ii) (iii) (iv)

Explanation Magnesium / Mg Mg2+ Chlorine / Cl2 gas has free moving ions covalent compound T2 oC becomes slower

1 1 Total

10

Sub Mark

Total Mark

1 1 1 1

2 1 3

1 1 1 1 1 1 1

1 1 1 1 1

Total

11

Sub Mark

Total Mark

1 1 1 1 1 1 1 1 1

1 1 1 1 1 1 1 1 1

Total

9

3

No 6(a)(i) (ii) (b)(i) (ii) (iii) (iv) (v) (vi)

No 7(a) (b) (c)(i) (ii) (iii) (d)(i) (ii) (iii) (iv) (v)

Explanation 0.125 mol 0.125x 6.02 x 1023 // 1.7525 x 1023 molecules 0.125 x 44 = 5.5 g Chemical formula that shows simplest ratio of atoms of elements/each element in a compound Mg = 2.4g , O = 1.6g 1:1 MgO To allow oxygen enter the crucible Copper is less reactive towards oxygen than hydrogen

Explanation Chemical formula that shows actual number of atoms of elements/each element in a molecule Number of mole Method 1 Magnesium is reactive //Magnesium reacts easily with oxygen To allow oxygen/air to enter the crucible//To ensure magnesium reacts completely Mass of Pb = 49.68 g No.of mole of Pb = 0.24 mol Mass of oxygen = 3.84 g No. of mole of O = 0.24 mol Empirical formula = PbO

No

Explanation

8(a)(i)

Chemical formula that shows the simplest ratio of atoms of each element in the compound To dry hydrogen gas//To absorb water/moisture Copper = 8.00g Oxygen = 2.00g 8/64 : 2/16 = 1:1 CuO To avoid the oxidation of copper//To avoid the formation of copper oxide Repeat the process of heating, cooling and weighing until the mass of copper is constant/no change Magnesium is more reactive towards oxygen than hydrogen//Magnesium is placed above hydrogen in the reactivity series Iron oxide/Tin oxide/Lead oxide/Silver oxide

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

(ii)

Sub Mark 1 1 1 1

Total Mark 3

7

1+1 1 1 1 1 Total

10

Sub Mark 1

Total Mark 1

1 1 1

1 3

1 1 1 1 1 1 Total

10

Sub Mark 1

Total Mark 2

1 1 1 1 1 1

5

4

2

1 1

2

1 Total

10

4

SECTION B No 9(a)

(b) (i) (ii)

(iii)

Rubric 1.Number of proton for both isotopes is 6 2.Number of electron for both isotopes is 6 3.Nucleon number for first isotope is 12 4.Nucleon number for second isotope is 14 5.Both isotope has different physical properties 6.Both isotope has same chemical properties Naphthalene / acetamide 1. Water 2. Melting point of X below than boiling point of water / uniform heating From 0 to t1 minutes 1. In solid state 2. Particles are closely packed and in orderly arrangement 3. Kinetic energy of particles very low 4. Attraction force between particles very strong

Sub Mark 1 1 1 1 1 1

Marks

6 1

1 1

2

1 1 1 1

From t1 to t2 minutes 1. In solid and liquid state 2. Some particle are closely packed and in orderly manner and some particle are slightly lose packed and in orderly manner 3. Kinetic energy of particles constant

1

From t2 to t3 minutes 1. In liquid state 2. Particles are closely packed and in orderly manner 3. Attraction forces between particles become weaker

1 1 1

1. correct curve 2. Label freezing point ( 80 o C )

1 1

1 1 Max 9

(v)

Sample answer Temperature (oC)

80 oC

2 Time (min)

TOTAL

20

5

No

Explanation

Sub Mark

10 (a)

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

1 1 1

(b)(i)

Electron arrangement of chlorine atom is 2.8.7 // Chlorine atom has seven valence electrons. Chlorine atom needs one electron to achieve the octet electron arrangement. Thus, two chlorine atoms share one pair of electrons.

1 1

Covalent compound

1

Total Mark 6

1 3

Cl

Cl

C

Cl

1+1 Cl

(b)(ii)

(c)

Electron arrangement of magnesium atom is 2.8.2 Magnesium atom donate 2 electrons to achieve the stable octet electron arrangement A positive ion / magnesium ion / Mg2+ ion is formed // Mg → Mg2+ + 2e // [Diagram]

1

Electron arrangement of chlorine atom 2.8.7 chlorine atom accepts 1 electron to achieve the stable octet electron arrangement A negative ion / chloride ion / Cl ion is formed // Cl2 + 2e → 2Cl // [Diagram]

1

Mg2+ and Cl ions are attracted to each other by a strong electrostatic/ionic force. Naphthalene cannot conduct electricity/non-electrolyte. Naphthalene does not has ions // exist as molecule. Sodium chloride solution conduct electricity/ an electrolyte. In sodium chloride solution, sodium ions/Na+ and chloride ions/ Cl are free to move.

7

1 1

1 1

1 1 1 1 1 Total

4

20

6

No

Explanation

11(a)(i) Al3+ , Pb4+ (ii) Aluminium oxide Lead(IV) oxide (b)(i) (CH2O)n = 60 12n + 2n + 16n = 60 n= 2 Molecular formula = C2H4O2//CH3COOH (ii) CaCO3 + 2CH3COOH (CH3COO)2Ca + H2O + CO2 (c)(i) 1.Green solid turn Black 2. Lime water becomes cloudy (ii) CaCO3 CaO + CO2 (iii) 1. mol of calcium carbonate decomposed into 1 mol of calcium oxide and 1 mol of carbon dioxide 2. Calcium carbonate is in solid state, calcium oxide is in solid state and carbon dioxide is in gaseous state 1. No. of mole for CuCO3 = 12.4 / 124 = 0.1 mol (iv) 2. 1 mol of CuCO3 produces 1 mol of CuO Therefor No. of mole for CuO = 0.1 mol 3. Mass of CuO = 0.1 X 80 g = 8 g Mass of oxygen is 0.8g (v) Simplest mol ratio : Cu : O = 3.2/64 : 0.8/16 = 1 : 1

Sub Mark 1+ 1 1+1

Total Mark 4

1

3

1 1 2

2

1 1 1+1 1

2 2 2

1 1 1

3

1 1 1

2

Total

20

7

No

Explanation

12(a)(i) Cation Anion (ii) Iron(III) ion and chloride ion (iii) K2SO4 (b)(i) Lead(II) sulphate Pb(NO3)2 + K2SO4 PbSO4 + 2KNO3 (ii) 1 mol of lead(II) nitrate reacts with 1mol of potassium sulphate producing 1 mol of lead(II) sulphate and 2 mol of potassium nitrate. (iii) 1.No.of mol of K2SO4 = 10 x 0.5 = 0.005 mol 1000 2. 1 mol of K2SO4 producing 1 mol of PbSO4 No of mol PbSO4 = 0.005 mol 3. Mass of PbSO4 = 0.005 x 303= 1.52 g (c)(i) 1.No.of mole:C = 52.2/12 H = 13/1 O = 34.8/16 = 4.35 = 13 = 2.175 2. Mol Ratio:C : H : O 4.35 / 2.175 : 13.0 / 2.175 : 2.175 / 2.175 Simplest mol ratio:C : H : O 2 : 6 : 1 3. Empirical formula: C2H6O 4. (C2H6O)n = 46 (2 x12)n + (6 x1)n + (1x16)n = 46 5. n = 1 6. Molecular formula : C2H6O 7. Structural formula : H H H

C – C – OH H

Sub Mark 1 1 1+1 1 1 1+1 1

Total Mark 5

8

1

1 1+1 7 1

1 1 1 1 1

1

H Total

20

8

SECTION C No

13 (a)(i)

(ii)

(b) (c)

Explanation Y more reactive Atomic size of Y bigger than X // The number of shell occupied with electron atom Y more than X. The single valence electron becomes further away from the nucleus. the valence electron becomes weakly pulled by the nucleus. The valence electron can be released more easily. Name : Sodium 4Na + O2  2Na2O Chemical formulae Balance equation Put group1 metal into bottle that contain paraffin oil To avoid react with air Name : Sodium/any group 1 element Material : group 1 elements, water, Apparatus : forceps , knife, filter paper, basin, litmus paper. [procedure] 3. Pour some water into the basin 4. Group 1 elements is take out from paraffin oil using forceps 5. A small piece of group 1 element is cut using a small knife 6. Oil on group 1 element is dried using a filter paper 7. The group 1 element is placed in the basin contain water. 8. Put litmus paper into water [observation] 9. Color of red litmus paper turn to blue [chemical equation ] Sample answer 2 Na + 2 H2O  2NaOH + H2 Chemical formulae Balance equation

Sub Mark 1 1

Total Mark

5

1 1 1 1

1 1 1 1 1

3

2 max10

1

1 1 1 1 1 1

1

1 1 Total

20

9

No

14. (a)

(b)

(c)

Explanation

CO2 // any covalent compound covalent Intermolecular forces are weak Small amount of heat energy needed to overcomes the forces X = 2.1 X = 2.2 Y = 2.7 // Y = 2.6 // suitable electron aranggement Ionic bond to achieve octet electron arrangement One atom of X donates 1 electron to form ion X+ One atom of Y receives an electron to form ion YIon X+ and ion Y- are attracted together by the strong electrostatic forces material and apparatus; compound XY, Carbon electrode, cell, wire, crucible, bulb/ammeter/galvanometer Procedure A crucible is filled with solid XY Dipped two carbon electrode Connect two electrode with connecting wire with bulb Observed whether bulb glow Heated the solid XY in the crucible Observed whether bulb glow Observation Solid XY - bulb does not glow Molten XY - bulb glow

Sub Mark

Total Mark

1 1 1 1

4

1 1 1 1 1 1 1

7

1

1 1 1 1 1

1

Diagram

Functional diagram labeled 1 1 TOTAL

9 20 10

No 15 (a)

Explanation By burning metal P in the air/oxygen Because metal P is a reactive metal/ react actively with oxygen/place higher than hydrogen in the reactivity series

Sub Mark 1 1

(b)(i)

4 Element

Y

O

1. Mass of element (g)

2.07g

0.32g

2. Number of moles

2.07

0.32

207

16

0.01

0.02

1

2

Ratio of moles 3. Simplest ratio of moles

1

1 1

Empirical formula of copper oxide : YO2 (ii)

Total Mark 2

1. Collect the gas into a test tube 2. Put/place a lighted wooden splinter at the mouth of the test tube 3. no pop sound, 1. Weigh A crucible and its lid 2. Clean the magnesium ribbon with sandpaper. 3. Weigh the crucible with its lid and content 4. Heat the crucible without its lid with a strong flame 5. When the magnesium ribbon starts to burns cover the crucible with its lid. 6. Using a pair of tongs lift the lid a little at intervals (open once in a while) (to allow oxygen from the air to enter for the combustion of magnesium). 7. When the burning is complete, remove the lid and heat the crucible strongly. 8. Allow the crucible to cool to room temperature with its lid still on 9. Weigh the crucible with its lid and content again 10. Repeat the process of heating, cooling and weighing until a constant mass is obtained. Record the constant mass obtained. 11.Result: Description Mass of crucible + lid (a) Mass of crucible + lid + magnesium ribbon (b) Mass of crucible + lid + magnesium oxide(c)

1 1 1 1 1 1 1 1 1

3

1

1 1 1 1 1

Mass = ag =b g = cg

12.

.

Element Mass of element (g) Number of moles Ratio of moles Or: Simplest ratio of mole

Mg (b-a)

O (c-b)

(b-a) 24 x

(c-b) 16 y

1

1

13. Empirical formula of magnesium oxide: MgxOy / MgO

1

Max; 11 1 Total

20

11

No 16 (a)

(b)(i)

(ii)

(c)

Explanation 1. Name of the c ompound is sulphur trioxide 2. Sulphur trioxide is made up of two elements, which is sulphur and oxygen 3.One sulphur atom combine with three oxygen atoms Hydrochloric acid Calcium chloride n=2 1.No. of moles of HCl = 0.5 x 30 = 0.015 1000 2. 2 mol of HCl produces 1 mol of CO2 No. of moles of CO2 = 0.015/2 = 0.0075 mol 3. Volume of CO2 = 0.0075 x 24 = 0.18 dm3 1. Weigh a combustion tube with the porcelain dish in it and record the reading. 2. Add a spatulaful of metal oxide on to the porcelain dish and weigh the combustion tube again 3. Allow the dry hydrogen gas to flow .into the combustion tube for 10 minutes to expell ( remove) all the air in the tube. 4. Collect a sample of gas from the small hole of the tube. Introducing a burning splinter into the test tube. If there is no “pop” sound, then all the air has been removed. Burn the excess hydrogen gas. 5. Heat metal oxide strongly while the hydrogen gas is flowing. 6. Stop the heating when metal oxide turns completely from green to grey. 7. Continue the flow of hydrogen gas until the set of apparatus cools down to room temperature 8. Weigh the mass of the combustion tube with its content . 9. Repeat the heating, cooling and weighing process until a constant mass is obtained. 10 Result: Mass of combustion tube + porcelain dish = a g Mass of combustion tube + porcelain dish + Metal oxide (b) = bg Mass of combustion tube + porcelain dish + Metal M = cg 11. Element M O S1: Mass of M (c-a) (b-c) S3: No. of moles c-a b-c 56 16 S3: Ratio of moles x y S4: Simplest ratio of moles

1

Sub Mark 1 1

Total Mark 3

1 1 1 1 1

3

4

1 1 + 1` 1

Max:

1

11

1 1

1 1 1 1 1 1

1

1

12.Therefore: Empirical formula of copper oxide : M xOy/MO

1 Total

20

12

SET 2 SECTION A No

Explanation

Sub Mark

Total Mark

1(a) (b) (c) (i) (ii)

1 1

1 1

1 1 1 1

1 1 1

1 1 2

(f)(i) (ii)

Chemical to electrical energy Cu2+ , H+ , OH- and SO42zinc Zinc more electropositive than copper// zinc above copper in the Electrochemical series. Brown solid deposited // Copper becomes thicker. Cu2+ + 2e→ Cu Blue to colourless // Blue become fade /paler // The intensity of blue colour decreases The number of copper(II)ions decreases // The concentration of copper(II)ions decreases Cu , R , Q ,P 1.5V

No

Explanation

(d)(i) (ii) (e)

2(a)(i) Redox reaction // oxidation and reduction (ii) Zn + 2Fe3+  Zn2+ + 2Fe2+ (iii)

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

[Correct chemical formulae of reactants and products] [Balance the equation correctly.] Sodium hydroxide solution is added slowly into the solution until in excess. Green precipitate that is insoluble in excess sodium hydroxide solution is produced//White precipitate that is soluble in excess sodium hydroxide to produce colourless solution. Pink colouration / spot is observed Blue colouration / spot is observed When iron is in contact with zinc, iron does not rust. When iron is in contact with copper, iron rusts. Apply grease on the surface//apply paint on the surface // galvanising // tin plating

1 1

1

1 Total

1 10

Sub Mark

Total Mark

1

1 2

1 1 1

2

1

1 1 1 1 1

1 1 2

Total

10

1

13

SECTION B No

Explanation

3(a)(i) Cathode : hydrogen Anode : Chlorine Electrode Ions that are attracted Ions that are selectively discharged Reason

Half equation

(ii)

Cathode Mg2+ and H+

Anode Cl- and OH-

H+

Cl-

Sub Mark

Total Mark

1 1

10

1+1 1+1

H+ is lower in the electrochemical series. 2H+ + 2e→H2

Concentration of Clhigher than OH-

1+1

2Cl- → Cl2 + 2e

1+1

cathode : hydrogen anode : oxygen

1 1

2

(b) Aspect

Cell X

Cell Y

Types of cells

Electrolytic cell

Voltaic cell

1

Energy changes

Electrical to chemical

Chemical to electrical

1

Name of electrodes

Anode : Copper

Positive electrode : Copper Negative electrode: Magnesium

Cathode : Copper

Ions in the electrolyte Half equations

Observations

Cu2+ ,H+ ,SO2-4 , OH-

Cu2+ ,H+ ,SO2-4 , OH-

Anode : Cu→Cu2+ + 2e Cathode : Cu2+ + 2e→ Cu

Negative electrode : Mg→Mg2+ + 2e Positive electrode: Cu2+ + 2e→ Cu

Anode : Copper becomes thinner

Positive electrode: brown solid deposited//copper becomes thicker

8

1

1

Cathode :brown solid deposited//copper becomes thicker

1

1

1

Negative electrode : Magnesium become thinner.

1

Total

20

14

No

Explanation

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

Oxidation number of lead is + 4 Na2 O – Sodium oxide PbO2 – Lead(IV) oxide Metal P = Copper Metal Q = Zinc Experiment 1 Iron rust

Iron more electropositive than copper // Iron above copper in the electrochemical series Fe→ Fe2+ + 2e

Sub Mark

Total Mark

1 1 1 1 1 1

2 2 10

1 1 1+1

Experiment 2

(c)

Iron does not rust // The solution is alkaline Iron less electropositive than zinc // Iron below zinc in the electrochemical series. O2 + 2H2O + 4e→ 4OHSample answer Metal X : Magnesium Magnesium atom releases two electrons to form magnesium ion Magnesium is oxidized to magnesium ion/ undergoes oxidation Magnesium is a reducing agent Copper (II) ion accepts two electrons to form copper atom Copper (II) ion is reduced to copper atom/ undergoes reduction Copper (II) ion is an oxidizing agent

1 1 1+1 1

1

1 1 1 1 1 1

Max 5

Total

20

15

SECTION C No

5(a)

Explanation Electrode X : Hydrogen Electrode Y : Oxygen

(b)

Electrode

X

Y

Ions that are attracted to

Na+ , H+

SO42- , OH -

Name of ions that are selectively discharged

Hydrogen ion

hydroxide ion

Reason

-

Position of OH- is lower than SO42- in the electrochemical series

(c) [Sample answer : Sulphuric acid // any suitable solution ] 2H+ + 2e→ H2

(d)

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

Zinc plate

Sub Mark

Total Mark

1 1

2 5

1+1 1+1

1

1 1+1

3

1

10

Copper plate Porous pot

Zinc nitrate

1. 2. 3. 4.      

Copper(II) nitrate

[Functional set up of apparatus] [Label : Zinc , copper , zinc nitrate solution , copper(II) nitrate solution and porous pot] [Mark of negative electrode – zinc plate Mark of positive electrode – copper plate] [Mark of the electron flow from zinc to copper]

1 1

Put/fill/pour zinc nitrate solution a beaker. Put copper(II) nitrate solution into a porous pot. Put the porous pot into the beaker. Dip/put zinc plate into zinc nitrate solution. Dip/put copper plate into copper(II) nitrate solution. Connect the wire // complete the circuit.

1 1 1 1 1 1

1 1

Total

20

16

No

Explanation

6(a)

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

(b)

Reactivity of metals in descending order is Q, carbon, R, P Experiment I Reaction between carbon and oxide of metal P occurs carbon is more reactive than metal P Experiment II Reaction between carbon and oxide of metal Q does not occur, metal Q is more reactive than carbon Experiment III Reaction between carbon and oxide of metal R occurs carbon is more reactive than metal R Reaction between carbon and oxide of metal P more vigorous than reaction between carbon and oxide of metal R Metal P is less reactive than metal R.

Sub Mark

Total Mark

1 1 1 1 1

4

6

1

1

1

1 1

Sample answer

Max 10

Fe3+→Fe2+ Magnesium as a reducing agent Add magnesium to iron(III) chloride solution Heat the mixture Filter the mixture Add sodium hydroxide solution Green precipitate is formed

1 1 1 1 1 1

Fe2+ →Fe3+ Chlorine as an oxidizing agent Add chlorine water to solution containing Fe2+ Stir the mixture Add sodium hydroxide solution Brown precipitate is formed

1 1 1 1 1 Total

20

17

SET 3 SECTION A

1

No. (a) (i) (ii)

Answer Solution in test tube C 1. Solution in test tube A 2. Concentration of H+ ion in test tube A is the highest

(b)

1. Higher than pH value of 0.1 moldm-3HCl // The pH is ¾/5/6 2. Ethanoicacis is a weak acid// Etanoic acid ionizes partially in water to produce low concentration of hydrogen ion 3. The lower the concentration, the higher the pH value

1 1

Magnesium chloride and hydrogen Mg + 2HCl → MgCl2 + H2 1. Correct formula of reactant and product 2. Balanced equation

1

(c )

(i) (ii)

(iii)

(d)

Mark 1 1 1

1

1 1

No of mole, HCl = 0.1 x 5 / 1000 = 0.0005 mol Based on balanced equation, 2 mol of HCl : 1 mol of H2 0.0005 mol of HCl : 0.00025 mol of H2 // mol of H2 = 0.005/2 = 0.0025 Volume of hydrogen gas = 0.00025 x 24 dm3 = 0.006 dm3 // 6 cm3 White precipitate

1

1 1 1 13

TOTAL

2

No. (a) (i)

Answer Solvent P: Water Solvent Q: methyl benzene / propanone / suitable organic solvent

Mark 1 1

(ii)

Effervescence / gas released // magnesium ribbon become thinner

1

(iii)

(b)

(i)

1. In the presence of solvent P/water , ethanoic acid ionize to form H + ion. 2. H+ ion causes the ethanoic acid to show its acidic properties 3. In solvent Q, ethanoic acid exist as molecule// hydrogen ion does not present

1 1 1

1. pH value increase / bigger 2. Concentration of acid is lower

1 1

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

V=

0.04 × 250 0.5

1

V = 20 cm3

1

18

3 (a) (b)

(c) (d)

(e)

Alkali that ionize/dissociate completely in water to produce high concentration of hydroxide ions. Alkaline / alkaline solution P: ion Q: molecule No Because there are no hydroxide ions in the solution// ammonia exist in the form of molecule.

1

1

1 1 1 1 1

1

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

1

1

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

1 1

2

(iii)

1

Mol of Mg = 2.4/24 // 0.1 mol

2

2

Volume of H2 = 0.1  24 dm = 2.4 dm 3

2

3

1 Total

No. 4 (a)(i) (ii)

Answer Pink to colourless 1. Correct formula 2. Balanced equation

11

Sub Mark 1

Mark 1

1 1

2

H2SO4 + 2KOHK2SO4 + H2O (iii) ConcentrationKOH=

2 ×0.1 X 15 25

1

= 0.12 moldm-3

(b)(i) (ii)

Potassium sulphate 0.1 X 15 Mol of H2SO4= // 0.0015 1000

2

1

1

1

1 mole H2SO4 produce 1 mole K2SO4 // 0.0015 mole H2SO4 produce 0.0015 mole K2SO4

1

Mass Salt X = 0.0015 X 174 = 0.261 g I(i) (ii)

1

30 cm3 // double 1. H2SO4 is diprotic acid while HNO3 is monoprotic acid 2. Number of mole H+ ion in HNO3 is half compare to H2SO4 Total

1

3

1 1 1

1 2 12

19

NO 5

(a) (i) (ii)

ANSWER Green Double decomposition reaction/ precipitation reaction

MARK 1 1

(b) (i) carbon dioxide (ii) CuCO3 → CuO + CO2 1. Reactants and products are correct 2. Equation is balanced

1 1 1

(iii)

Copper(II) carbonate

Heat Lime water

1 1 1

- Labelled diagram - Functional (c) (i) Sulphuric acid // H2SO4 (ii) mol CuCO3 = 12.4/124 = 0.1 mol ratio CuCO3 : CuSO4 1 : 1 Mol of CuSO4 = 0.1 x 135g Mass = 13.5g

1 1 1 10

6

No. (a) (b) (i) (ii) (c)

(d) (e) (f)

Answer Mg + 2HCl → MgCl2 + H2 0.4/24 = 0.0167 mol The number of mole of HCl = MV/1000 = 1x 50/1000 = 0.05 mol

Mark 1+1 1 1

From the chemical equation 1 mol of magnesium produce 1 mol hydrogen If 0.0167 mol produce 0.0167 mol hydrogen Volume of hydrogen = 0.0167 x 24 dm3= 0.4 dm3/ 400 cm3

I 400 /100 =4 cm3s-1 II 400 /60 = 6.67 cm3s-1 As catalyst The temperatureof hydrochloric acid The concentration of hydrochloric acid TOTAL

1 1 1 1 1 1 1 11

20

7 a b

i. ii. i.

Catalyst Change in total mass of conical flask and its content V cm3 // same volume No. of moles of HCl used is the same // catalyst does not increase the quantity of product

Marks 1…1 1....1 1 1…2

ii. Volume of gas released/ cm3

Key : Experiment I : Experiment II :

… …..

Time / s

iii .

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

1 1….2

1. Catalyst lowers the activation energy // provides an alternative reaction pathway that requires lower activation anergy 2. Frequency of effective between H+ and zinc atom increases 3. Rate of reaction increases

1 1 1…..3

c. Volume of carbon dioxide /cm3

Experiment C Experiment A Experiment B

Graph 5

1…..1

Time/s

10

21

No. 8 (a) (b) (c)

Answer The amount heat change/released when 1 mol of copper is displaced by magnesium from copper(II) sulphate solution Higher rate of reaction // Reaction is faster Correct formulae of reactants Correct formulae of products

Mark 1 1 1 1

Mg2+ + Zn → Mg + Zn2+ (d)(i)

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

(ii) n =

0.5 x 50 1000

// 0.025

∆H =

1050 0.025

//

(iii)

1

1

1.05 0.025

1

0.025 mol CuSO4 produce 1050J 1 mol CuSO4 produce 1050 0.025 = - 42 000 J mol-1 (e)

//

1

- 42 kJ mol-1

(r: without unit)

1

Arrow upward with label energy and two levels Correct position of reactans and products

1 1

Energy Cu2+ + Mg

ΔH = -42kJ mol-1 Cu + Mg2+ (f)

Reduce heat loss to surrounding. [r:prevent]

1 TOTAL

12

22

9(a)

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

1

(b)

To reduce heat lost to the surrounding

1

(c)

Exothermic

1

(d)(i)

Q= 50 X 4.2 X8 //1680 J //1.68 kJ

1

Mol CuSO4 = 50 x0.2 /1000 // 0.01 Heat of displacement= -1.68 ÷ 0.01 // -168 kJ mol-1

1 1

(ii) (e)

Energy Zn + Cu 2+ //Zn + CuSO4

∆H = -168 kJ mol-1

Zn 2+ + Cu //nSO4Z 1 1 1

Arrow upword with label energy and two label - 1m Correct potion of reactant and product - 1m ΔH = -168kJ mol-1 - 1m (f)(i) (ii)

Same//8 0 C

 

1

Heat produce/released is double the heat released is distributed over a volume of solution which is doubled

1 1 12

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

(d )

- Heat released when one mol of propanol burnt in excess oxygen - The reaction is an exothermic reaction /heat released to the surroundings - 1 mole of propanol burnt in excess oxygen released 2015 kJ of heat energy No. of mol of C3H7OH= 1.2/60 // 0.02 …..1 Heat released= 2015 X 0.02 =40.3 kJ//40300 J …..1 Ө = 40.3 x 1000 200 x 4.2 Ө = 47.98 0C

//

40300 200 x 4.2

…..1 ……1

1 1 1

1 1 1 1

Energy C3H7OH + 9/2 O2

∆H = -2015 kJ mol -1

3CO2 + 4H2O

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

Use wind shield//use thin copper can//never use wire gauze//stir the water with thermometer//weigh the spirit lamp immediately before and after burning

1 1 1 1 23

(f)

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

1 1 1 1 max 3 TOTAL 14

SECTION B No. 11 (a)(i)

(ii)

Answer [Label of axes with units] [All points are transferred correctly] [Correct shape of the graph and constant scale]

Sub Mark 1 1 1

2.5 cm3 (r: without unit) moles of Pb2+ ions = 1.0x 2.5

1000

moles of I- ions = 1.0x 5

1000

Mark

3

1 // 0.0025

1

// 0.005

1

0.0025 mol Pb2+ : 0.005 mol I1 mol Pb2+ : 2 mol I-

1 1

Correct formulae of reactants and product Balanced equation

1 1

Pb2+ + 2I- → PbI2 7 (b)(i)

(ii)

Salt J : lead(II) nitrate // Pb(NO 3)2 X oxide : lead(II) oxide // PbO Gas Y : nitrogen dioxide // NO 2 Gas Z : oxygen // O2 Yellow precipitate : lead(II) iodide // PbI2

1 1 1 1 1

5

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

1 1 1 1 1

5

TOTAL

20

24

No. 12(a)(i) (ii)

Sub Mark 1

Answer Carbonate ion // CO32-

Mark 1

Salt S : Copper(II) carbonate // CuCO3 Compound T: Copper(II) oxide // CuO Gas U: Carbon dioxide // CO2 Compound W: Copper(II) sulphate // CuSO4

1 1 1 1

4

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

1 1

2

1 1 1 1 1

5

Salt X : Barium sulphate Salt Y: Copper(II) nitrate

1 1

2

(ii)

White precipitate Double decomposition reaction

1 1

2

1 1

(iii)

Ba2+ + SO42-→ BaSO4 0.1 x 50 Number of mol Ba2+ = // 0.005 1000 2+ 1 mol Ba produce 1 mol BaSO4 // 0.005 mol Ba2+ produce 0.005 mol BaSO4 Mass BaSO4 = 0.005 x 233 g // 1.165 g

(b)

CuO + H2SO4→ CuSO4 1. 2. 3. 4. 5. (d)(i)

+ H2O

Add sodium hydroxide solution Blue precipitate formed indicate the presence of Cu+ ion Add hydrochloric acid Add barium chloride solution White precipitate formed indicate the presence of SO42- ion

1 1 TOTAL

4 20

25

No. 13 (a)(i)

(ii)

(b)

Answer 1. copper(II) oxide and carbon dioxide 2. Correct formulae of reactants and products 3. Balanced equation CuCO3→CuO + CO2

1. 2. 3. 4.

Sub Mark 1 1 1

Yellow precipitate Lead(II) iodide and potassium nitrate Correct formulae of reactants and products Balanced equation Pb(NO3)2 + 2KI → PbI2 + 2KNO3

1 1 1 1

Mark

3

4

Test for cation 1. Pour aluminium nitrate solution and zinc nitrate solution into two different test tube 2. Add ammonia aqueous solution until excess 3. White precipitate insoluble indicate the presence of Al3+ 4. White precipitate dissolve in excess ammonia aqueous indicate the presence of Zn2+ ion

1 1 1 1

Test for anion

(c)(i)

5. Pour aluminium nitrate solution and zinc nitrate solution into two different test tube 6. Add sulphuric acid and iron(II) sulphate solution 7. Slowly and carefully add concentrated sulphuric acid 8. Brown ring formed indicate the presence of NO 3- ion

1 1 1 1

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

1 1

CuO + H2SO4 → CuSO4

8 2

+ H2O

(ii)

1 Mol of H2SO4 =

1.0 x 25 1000

// 0.025

1

1 mole of H2SO4 produce 1 mole of CuSO4 // 0.025 mole of H2SO4 produce 0.025 mole of CuSO4

1

3

Mass of CuSO4 = 0.025 X [64 + 32 + 4(16)] = 4g Total

20

26

14

No. (a)

(i)

(ii)

(b)

(i)

(ii)

Answer Size of the solid reactant Concentration of the solution Temperature of the reaction mixture Catalyst Temperature : 450-550oC Catalyst : iron powder Pressure : 200 atm  The axes are labeled together with its unit  The scale is correct  The points are transferred correctly  The curve is smooth

Mark 1 1 1 1 …..4 1 1 1…...3 1 1 1 1…..4

Average rate of reaction for experiment I =

1 1

26.0 210 = 0.12 cm3 s-1 Average rate of reaction for experiment II = 26.0 150 = 0.17 cm3 s-1 [correct unit] (iii) 1. The rate of reaction for Experiment II is higher than in Experiment I 2. The concentration of HCl in Experiment II is more/higher than in Experiment I + 3. The number of hydrogen ion/ H per unit volume of the solution in Experiment II is more than in Experiment I 4. The frequency of collisions between hydrogen ion and calcium carbonate in Experiment II is higher than in Experiment I 5. The frequency of effective collisions hydrogen ion and calcium carbonate in Experiment II is higher than in Experiment I TOTAL

27

1 1…..4

1 1 1 1 1…..5

20

15

No. (a)

(b) (i) (ii)

Answer Use magnesium powder Increse the concentration sulphuric acid Increase the temperature of the reaction mixture Use copper(II) sulphate as catalyst [any 3 of the above] Manganese(IV) oxide/ copper(II) oxide/ lead(IV) oxide

Mark 1 1 1 1…..3 1. …1

1. 2. 3. 4.

1 1 1 1…..2

Changes / alters the rate of a reaction Require in small amount Does not increase the quantitiy of products formed Remain chemically unchanged at the end of the reaction [Any of the 2]

(iii) Conditions Vanadium(V) oxide 450 – 500 oC 1 atm

Justification As a catalyst // to increase the rate of reaction Optimum temperature to maximise the amount of products at a resaonable rate of reaction Not economical to use high pressure as the yield is low

Any two of the conditions [2 + 2]

1+1 1+1 1+1 ……4

(c)

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

(d)

Materials :”20-volumes” hydrogen peroxide, manganes(IV) oxide Apparatus : test-tubes, measuring cylinder, test-tube rack, glowing splinters 1…..1

1 1 1…..3

Procedure : 1. Measure 5 cm3 “20-volumes” of hydrogen peroxide solution and pour into test-tube labeled A and B. 2. Add half spatula of manganese(IV) oxide into testt-tube B. 3. Insert a glowing splinter into test-tube A and B respectively 4. Record observations Test-tube A (without catalyst) B (with catalyst)

Observation Glowing splinter does not relight Glowing splinter relights

1 1 1 1…..4

1…..1

Conclusion : The presence of catalyst manganese(IV) oxide increases the rate of decompositionof hydrogen peroxide.

1…..1

20

TOTAL

28

16

(a)

(i) Characteristic Type of chemical reaction Energy content of reactants and products

Ionic equation

Figure 16.1(i)

Figure 16.2(ii)

Exothermic reaction

Endothermic reaction

1+1

The energy content in the reactants more than the energy content in the products

The energy content in the reactants less than the energy content in the products

1+1

Mg + Fe2+  Mg2+ + Fe

Ca2+ + CO32-  CaCO3 1+1

(ii)

(b)

(c)

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 mole of silver nitrate solution ionise to produce 1 mole of Ag+ ion and 1 mole of sodium chloride solution ionise to produce 1 mole of Cl- ion 2. The heat of reaction of silver chloride is heat that released when 1 mole of Ag + ion react with 1 mole Cl- ion // Ag+ + Cl-  AgCl 3. 0.5 mole of magnesium chloride ionise to produce 1 mole of Cl- ion 4. Number of mole of Cl- ion in 1 mole of sodium chloride same as number of mole of Cl- ion in half mol of magnesium chloride 5. Half of the mole of magnesium chloride produces 1 mole of Cl - ion 6. Number of mole of Cl- ion in 1 mole of sodium chloride same as number of mole of Cl- ion in half mol of magnesium chloride (i)

Heat change = mcθ = (100)(4.2)(42.2 – 30.2) = 5040 J / 5.04 kJ Number of moles of HCl / H + ion

ΔH (ii)

o

1 1

1

1 1 1 1 1

Max 4

1 = (50)(2) 1000 = 0.1 mol

Number of moles of NaOH / OH - ion = (50)(2) 1000 = 0.1 mol The heat of neutralization

1

= 5.04 0.1 = - 50.4 kJ mol-1

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

1

1 1 1 1 1 1 max 4

29

SECTION C No. 17 (a) 1. 2. 3. 4. (b)

(c)

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

Sub Mark 1 1 1 1

1. Sodium hydroxide is strong alkali 2. Ammonia is weak alkali 3. Sodium hydroxide ionises completely in water to produce high concentration of OH- ion 4. Ammonia ionises partially in water to produce low concentration of OH ion 5. Concentration of OH- ion in sodium hydroxide is higher than in ammonia 6. The higher the concentration of OH- ion the higher the pH value

1 1 1

Volumetric flask used is 250 cm3 Mass of potassium hydroxide needed = 0.25 X 56 = 14 g Weigh 14g of KOH in a beaker Add water Stir until all KOH dissolve Pour the solution into volumetric flask Rinse beaker, glass rod and filter funnel. Add water when near the graduation mark, add water drop by drop until the graduation mark 10. Close the volumetric flask and shake the solution

1 1 1 1 1 1 1 1 1

1. 2. 3. 4. 5. 6. 7. 8. 9.

Mark

4

1 1 1

6

1 10 20

Total No. 18(a)

(b)

Answer (i) CuSO4 // FeSO4 // HgCl – used as pesticides to destroy pests (ii) Sodium chloride – to add flavour to food // Sodium hydrogen carbonate – used in baking cakes and breads (iii) Calcium sulphates esquihydrate – used to make cement casts to encase fractured bones. Barium sulphate – used on X-ray plates to identify tumours in the intestines. (Accept other relevant answers) 1. Pour [20-100cm3] of zinc nitrate solution [0.1-1.0mol dm-3] into a beaker 2. Add [20-100cm3] of sodium carbonate solution [0.1-1.0]mol dm-3 3. Stir the mixture 4. Filter the mixture 5. wash the residue 6. Zn(NO3)2 + Na2CO3ZnCO3 + 2NaNO3 7. Pour [20-100cm3][0.1-1.0mol dm-3]sulphuric acid into a beaker 8. Add the residue/ zinc carbonate into the acid until in excess 9. Stir the mixtureand filter 10. Heat the filtrate until saturated 11. Cool the solution 12. Filter 13.dry the crystal by pressing between two filter paper 14. ZnCO3 + H2SO4 ZnSO4 + H2O + CO2 Total

Marks 1+1 1+1

2 2

1+1

2 1 1 1 1 1 1 1 1 1 1 1 1 1 1

14

20

30

Marks 19 a b

2H + S2O3  S + SO2 + H2O All formulae correct - 1 Balanced equation Rate of reaction = fixed mass of sulphur formed 1 per unit time 1 +

2-

2…..2 1 1…..2

-1

c Factor Concentration of hydrochloric acid / sodium thiosulpate solution Temperature of hydrochloric acid / sodium thiosulpate solution

Effect The higher the concentration of hydrochloric acid / sodium thiosulpate solution, the higher the rate of reaction. The higher the temperature of hydrochloric acid / sodium thiosulpate solution, the higher the rate of reaction

1+1

1+1 …4

Any two of the factors and correct corresponding effect from above d

1. Measure 50 cm3 of 1.0 mole dm-3 sodium thiosulphate solution using a 50 cm3 measuring cylinder and pour into a conical flask. 2. Place the conical flask on a piece of paper with the mark “X”. 3. Measure 5 cm3 of 1.0 mol dm-3 hydrochloric acid using another 10 cm3 measuring cylinder. 4. Pour the acid into the conical flask quickly and carefully, at the same time start the stop-watch 5. Swirl the mixture in the conical flask quickly 6. Observe the “X” mark on the filter paper vertically above through the solution 7. Stop the stop watch and record the time taken when the mark “X” is not visible through the mixture 8. Repeat step 1- 7 by using the volume of sodium thiosulphate solution, distilled water and acid as shown Expt

Volume of Na2S2O3 solution / cm3

Volume of distilled water/ cm3

Volume of hydrochloric acid / cm3

1 2 3 4 5

50 40 30 25 10

0 10 20 30 40

5 5 5 5 5

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

1 1 1 1 1 1 1 1

Time taken for mark “X” to disappear / s 1

1

11. Graph

Concentration of Na2S2O3, mol dm-3-

1

1/Time/s

31

12. Conclusion : As concentration of Na2S2O3 solution increases, the rate of reaction with hydrochloric acid increases // The higher the concentration of Na2S2O3 solution, the higher the rate of reaction with hydrochloric acid

1….12

OR To investigate the effect of temperature on the rate of reaction 1. Measure 50 cm3 of 1.0 mole dm-3 sodium thiosulphate solution using a 50 cm3 measuring cylinder and pour into a conical flask and record the temperature of sodium thiosulphatesolution.using a thermometer. 2. Place the conical flask on a piece of paper with the mark “X”. 3. Measure 5 cm3 of 1.0 mol dm-3 hydrochloric acid using another 10 cm3 measuring cylinder. 4. Pour the acid into the conical flask quickly and carefully, at the same time start the stop-watch 5. Swirl the mixture in the conical flask quickly 6. Observe the “X” mark on the filter paper vertically above through the solution 7. Stop the stop watch and record the time taken when the mark “X” is not visible through the mixture 8. Repeat step 1- 7 by using heating sodiumthiosulphate solution to 35 oC, 40oC, 45 oC, 50 oC respectively at step 2. All other condition remains unchanged. Expt

1 2 3 4 5

Temperature of Na2S2O3 solution / oC 25.0 35.0 40.0 45.0 50.0

Time taken for mark “X” to disappear / s

1

1 1 1 1 1 1 1

Values of 1/time, s-1

1

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

10. Graph

Temperature of Na2S2O3, oC 1

1/Time/s 11. Conclusion : As the temperature of Na2S2O3 solution increases, the rate of reaction with hydrochloric acid increases // The higher the temperature of Na2S2O3 solution, the higher the rate of reaction with hydrochloric acid

Total

1….12

20

32

20

No. (a) (i)

Answer Experiment I – hydrochloric acid Experiment II – sulphuric acid Mg + 2HCl → MgCl2 + H2

Marks 1 2....3

(ii)

The number of mole of HCl = MV/1000 = 1.0 x 50 = 0.05 mol 1000 or The number of mole of H2SO4 = MV/1000 = 1.0 x 50 = 0.05 mol 1000 (iii) The rate of reaction is the change of volume of hydrogen gas per unit time (b) (i)

1…1

1…1

Volume of hydrogen/ cm3

Experiment II

Experiment I

Time/s

Curve with label 2. Axis with title and correct unit Sulphuric acid in experiment II is diprotic acid, hydrochloric acid in experiment I is monoprotic acid//Concentration of hydrogen ion, H+ in experiment II is higher than experiment I The number of hydrogen ion per unit volume in experiment II is higher than experiment I Frequency of collisions between hydrogen ions and magnesium atoms in experiment II is higher than experiment I Frequency of effective collisions between hydrogen ions and magnesium atoms in experiment II is higher than experiment I Rate of reaction in experiment II is higher than experiment I 1.

(ii)

1.

2. 3. 4. 5. (c)

Diagram : Functional apparatus set-up Label correctly Procedure : 1. A burette is filled with water and inverted over a basin containing water. The burette is clamped vertically to the retort stand. 2. The water level in the burette is adjusted and the initial burette reading is recorded. 3. 50 cm3 of 0.2 moldm-3 hydrochloric acid / sulphuric acid is measured and poured into a conical flask 4. 4. 5 cm of magnesium ribbon are added into the conical flask 5. 5. Close conical flask immediately with the stopper fitted with delivery tube. 6. At the same time the stopwatch is started shake the conical flask. 7. The burette readings are recorded at 30 second intervals for 5 minutes Time/s 0 30 60 90 120 150 180 Volume of gas / cm3

1 1…2 1

1 1 1 1…5 1 1…..2

1 1 1 1 1 1 1 Max.. 5 ……1

TOTAL 20

33

21

(a) (i)

Heat change = mc = (25+25)(4.2)(33-29) = 445 J

1

Heat of precipitation of AgCl = - 445 / 0.0125 = -35600 J mol-1 // 35.6 kJ mol-1

1

Energy

(ii)

AgNO3 + NaCl H = -35.6 kJ mol-1 AgCl + NaNO3 1. The position and name /formulae of reactants and products are correct.

1

2. Label for the energy axis and arrow for two levels are shown. (b) (i)

(ii)

Hydrochoric acid is a strong acid and ethanoic acid is a weak acid. 2. Hydrochloric acid ionised completely in water to produce higher concentration of H+ ion 3. Ethanoic acid ionised partially in water to produce lower concentration of H+ ion. 4. During neutralisation reaction, some of the heat released are absorbed by CH3COOH to ionise to produce hydrogen ion 1.

-

(c)

Sulphuric acid is a diproctic acid and hydrochloric acid is a monoprotic acid. Concentration/ number of hydrogen in sulphuric acid is double compared to that in hydrochloric acid The number water molecules produced in experiment III is compared experiment I

Apparatus : Polystyrene cup, thermometer, measuring cylinder. Materials : Copper (II) sulphate, CuSO4 solution, zinc powder. Procedures : 1. Measure 25 cm3 of 0.2 mol dm-3 copper (II) sulphate, CuSO4 solution and pour it into a polystyrene cup. 2. Put the thermometer in the polystyrene cup and record the initial temperature of the solution. 3. Add half a spatula of zinc powder quickly and carefully into the polystyrene cup. 4. Stir the mixture with the thermometer. 5. Record the highest temperature .

1 ..…. 4 1 1

1 1…Max 3

1 1 1 …… 3

1 1….2

1 1 1 1 1………5

Tabulation of data:

34

Initial temperature of CuSO4 solution (oC) Highest temperature of the reaction mixture (oC) Temperature change (oC)

1 2 2 - 1

1

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

1 1

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

1……. 3 Max : 10

TOTAL

No. 23 (a)

1. 2. 3. 4.

(b)

1. 2. 3. 4. 5. 6.

(c) 1. 2. 3. 4. 5.

(d)

20

Answer Exothermic reaction is a reaction that releases heat to the surrounding The total energy content of the products is lower than the total energy content of the reactants Endothermic reaction is a reaction that absorbs heat from the surrounding The total energy content of the products is higher than the total energy content of the reactants Heat energy is absorbed from surrounding //It is an endothermic reaction Total energy content of C and D/ product is higher than total energy content of A and B/ reactants When reaction occurs, the temperature of the mixture of solutions increases / becomes hot X kJ heat is absorbed when one mol A reacts completely with one mol of B. A reacts with B to form C and D //A and B are the reactants while C and D are the products (any 4 of the above) . 1 mole of silver nitrate solution ionise to produce 1 mole of Ag+ ion 1 mole of sodium chloride solution ionise to produce 1 mole of Cl - ion One of mole of potassium chloride also ionise to produce 1 mole of Cl - ion The heat of precipitation of silver chloride is heat that released when 1 mole of AgCl is formed from the reaction between Ag+ ion and Cl- ion // Ag+ + Cl-  AgCl Number of mole of AgCl produced in both reactions are the same, heat released are the same.

Materials : calcium nitrate solution, sodium carbonate solution Procedures : 1. Measure 50 cm3 of 1.0 mol dm3 calcium nitrate and 50 cm3 of 1.0 mol dm-3 sodium carbonate solution separately and poured into two different plastic cups 2. Measure and record the initial temperature of both solutions after 5 minutes. 3. Pour quickly and carefully calcium nitrate solution into the plastic cup that contains sodium carbonate solution and 4. Stir the mixture 5. Measure and record the lowest temperature reached Tabulation of data :

Mark 1 1 1 1 1 1 1 1

1 1 1 1 1 Max 4

1

1 1 1 1 1 5 max 4

35

Ө1 Ө2 (Ө1 + Ө2)/2 = Ө3 Ө4 Ө3- Ө4

Initial temperature of Ca(NO3)2 / oC Initial temperature of Na2CO3 / oC Average initial temperature / oC Lowest temperature of the mixture / oC Change in temperature / oC ……1

1 Calculation : No. of moles of CaCO3 = No. of moles of Ca(NO3)2 = mv/1000 = 1.0(50)/1000 = 0.05

……..1 1

heat change = mc(Ө4 – Ө3) = x kJ heat of reaction = + x kJmol-1 0.05 = + y kJmol-1

……..1 ……..1

1

TOTAL

20

SET 4 SECTION A

1

No (a) (b) (c) (d)

(i)

Answer Compound that contains only carbon and hydrogen And has double bonds between carbon – carbon atoms Alkene Propene Hydrogenation / Addition reaction

Mark 1 1 1 1 1

(ii)

1 (e)

(i)

C3H6 + 9/2 O2 → 3CO2 + 3H2O or

2

2C3H6 + 9O2 → 6CO2 + 6H2O (ii)

1

2 .1 42 = 0.05

No. of mole of C3H6

=

Volume of gas CO2

= 0.05 x 3 x 24 = 3.6 dm3

1 TOTAL

2

No (a) (b)

Answer Ethanol Hydroxyl group //OH

10 Mark 1 1

36

(c)

(i) (ii) (iii)

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

H

1 1 1

O

H CC O H H (d)

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

Esterification Ethyl ethanoate Pleasant smell CH3COOH + C2H5OH → CH3COOC2H5 + H2O

1 1 1 2 TOTAL

3

(a)

(i) (ii)

10

Fermentation Ethanol

1 1

H H

(iii)

H

C C H

1

H OH (b) (c)

C2H5OH + 3O2 (i)

→ 2CO2 + 3H2O

1+1

Ethene

1 ethanol

(ii)

Ethene gas

1+1

(d) (e)

No. 4 (a)

(i) (ii)

Purple to colourless esterification Ethylethanoate

(i)

Sulphuric acid

Mark 1

(ii)

Contact process

1

(iii)

Sulphur trioxide

1

(iv)

Vanadium(V) oxide, 450oC

(v)

Ammonium sulphate

1 1 1 Explanation

1+1 1

37

(b)

(i)

Composite material 1

(ii)

Tin atom

Correct arrangement Correct label

1 1

Copper atom (iii)

nC2H3Cl  --( C2H3Cl )--

1

(iv)

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

1

TOTAL

5

No. (a) (i)

Explanation Sodium /potassium salts of fatty acid

Mark 1

(ii)

1+1

(iii)

Hydrophobic Saponification // Hydrolysis under alkaline condition

(iv)

To reduce the solubility of soap// to precipitate out the soap

(b) (i)

11

hydrophilic 1 1

To relief pain 1

(ii)

It causes internal bleeding /ulceration

1

(iii)

Antibiotic

1

(iv)

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

1

(v)

Barbiturate/tranquilliser/ amphetamine / haloperidol

1 TOTAL

10

38

SECTION B 6 (a) Characteristics Same general formula

Explanation CnH2n + 1OH

1+1

successive member is different from each other by – CH2

Relative atomic mass is different by 14

1+1

Gradual change in physical properties // Melting / boiling point increase

Number of carbon atom per molecules increase // size of molecule increase

1+1

Similar chemical properties // oxidation produce carboxylic acid

Have same chemical/similar functional group

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

Have same chemical properties // have same functional functional group

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

1+1

1

C2H4O2

1

(ii) Carboxylic acid

2

1

React with carbonate to produce carbon dioxide (iii) 2 CH3COOH + CaCO3

1+1

→ (CH3COO)2Ca + H2O

1

2

1 1

2

+ CO2

Correct formula of reactants and products Balanced equation (c) The number of carbon atom The number of hydrogen atom

Type of covalent bond between // carbon/ Type of hydrocarbon Type of homologous series // // Name of compound

P 2

Q 2

4 6 number of hydrogen atom Q is higher Double bond / / Single bond/ / Unsaturated Saturated Alkene// Ethene //

Alkane // Ethane

CnH2n // C2H4

CnH2n+2 // C2H6

1 1

1

1

1 General formula// Molecular formula of the compound

Max 4 20

39

7

No (a) (i)

Answer

Mark 1

14.3 %

(ii) 1

Element Mass/ % No. of moles

C 85.7 85.7 = 7.14 12

H 14.3 14.3 = 14.3 1

2

Ratio of moles/ 7.14 14.3 =1 =2 Simplest ratio 7.14 7.14 3 Empirical formula = CH2 RMM of (CH2)n = 56 .............1 [(12 + 1(2)]n = 56 14n = 56 n = 56 14 = 4 ………..1 Molecular formula : C4H8 ………………..1

6 max 5

(iii)

1+1

1+1 But-2-ene

2-methylpropene

But-1-ene Max 4

[any 2] (iv)

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

=

4(12) x 100% 4(12)  8

48 x 100% 56 = 85.7% …………1 4(12) % of C in C4H10 = x 100% 4(12)  10 48 = x 100% 58 = 82.7% ………..1

=

(b)

(i) (ii)

(c)

(i) (ii)

Starch Protein H H CH3 H I I I I C = C– C = C I I H H 2-methylbut-1,3-diene or isoprene Rubber that has been treated with sulphur In vulcanised rubber sulphur atoms form cross-links between the rubber molecules These prevent rubber molecules from sliding too much when stretched TOTAL

.....3 1 1

1 1..2

1 1 1 20

40

8

(a)

(i)

(ii)

(iii)

(iv)

Hydrocarbon

Type of bond

Homologous series

General formula

A

covalent

alkane

CnH2n+2

3

B

covalent

alkene

CnH2n

3

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

1

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

6

1 1

3

1 1 1 1

Hydrocarbon B more sootiness. B has higher percentage of carbon by mass.

3

% of carbon by mass ;

(b)

Hydrocarbon A :

4(12) 4(12) + 10(1)

Hydrocarbon B :

4(12) 4(12) + 8(1)

× 100

× 100

// 82.76 %

1

// 85.71 %

1

4

Carboxylic acid X : 1

Propanoic acid

1

Alcohol Y:

1 Ethanol

1

4

TOTAL

20

41

9

(a)

(i) (ii)

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

(c)

(i)

(ii)

SO2 + H2O  H2SO3

1

   

Corrodes buildings Corrodes metal structures pH of the soil decreases Lakes and rivers become acidic [Able to state any three items correctly]  Oleum  2SO2 + O2  2SO3  Moles of sulphur = 48 / 32 =1.5  Moles of SO2 = moles of sulphur = 1.5  Volume of SO2 = 1.5  24 dm3 = 36 dm3

3 1 1 1 1 1 1

   

Pure metal are made up of same type of atoms and are of the same size. The atoms are arranged in an orderly manner. The layer of atoms can slide over each other. Thus, pure copper are ductile.

  

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

  

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.



4

6

1 1 1

1 1 1 1 Max:5 1 1 1 1

Zinc atom

Copper atom

1 Arrangement of atoms – 1; Label - 1 1 Max: 5 Total 20

42

No. 10 (a)

(b)

Explanation 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 (i)

(ii)

(iii)

Mark 1+1 1+1

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 Paracetamol

1 1

Codeine

1

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

1

(c) Type of food additives Preservatives

Examples

Function

Sugar, salt

2

Flavourings

Monosodium glutamate, spice, garlic Ascorbic acid

To slow down the growth of microorganisms To improve and enhance the taste of food To prevent oxidation of food To add or restore the colour in food

2

Antioxidants Dyes/ Colourings

Tartrazine Turmeric

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

1 1

TOTAL

20

43

SECTION C 11

(a)

(b)

(i)

X - any acid – methanoic acid Y - any alkali – ammonia aqueous solution

(ii)

1. 2. 3. 4. 5.

(iii)

Ammonia aqueous solution contains hydroxide ions Hydroxide ions neutralise hydrogen ions (acid) produced by activities of bacteria Alcohol Burns in oxygen to form carbon dioxide and water Oxidised by oxidising agent (acidified potassium dichromate (VI) solution) to form carboxylic acid

(i) (ii)

(iii)

1 1

Methanoic acid contains hydrogen ions Hydrogen ions neutralise the negative charges of protein membrane Rubber particles collide, Protein membrane breaks Rubber polymers combine together

Procedure: 1. Place glass wool in a boiling tube 3 2. Pour 2 cm of ethanol into the boiling tube 3. Place pieces of porous pot chips in the boiling tube 4. Heat the porous pot chips strongly 5. Heat ethanol gently 6. Using test tube collect the gas given off

1 1 1 1 1 5 max 4 1 1 1 1 1

6 max 5

Diagram: Glass wool soaked with ethanol

Heat

Porous pot chips

Heat

Water

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

1 1 1 1 20

44

12

(a)

(b)

(c)

(d) (i)

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

1

10H2O 1 1

3

Compound B & Compound D Same molecular formula / C4H8 Different structural formula

1 1 1

3

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

1 1 1 1

4

Any members of carboxylic acid and correct ester Example [Methanoic acid] [Prophylmethanoate]

1 1 1

1 (d) (ii)

Pour 2 cm3 of [methanoic acid] into a boiling tube Add 2 cm3 of propanol/compound E into the boiling tube Slowly/carefully/drop 1 cm3 of concentrated sulphuric acid Heated (with a small flame) the mixture Pour the mixture in a beaker that contain water Observation : formed liquid that fruity smell /float on water surface TOTAL

4

1 1 1 1 1 1

6 20

45

No.

Mark Scheme

Sub Mark

Total Mark

13(a)

H

H

H

H

H

C

C

C

C

H

But-2-ene 1+1

H

H

H H

H

C

C

C

H

C

H H

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

1+1 4

H (b)

(i)

(ii)

Propanoic acid Ethanol

1 1

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

1 1 1 1 1

2

[any three] Chemical properties for ethanol: 1. Undergo combustion to produce carbon dioxide and water 2. Burnt in oxygen to produce CO2 and H2O 3. Undergo oxidation to produce carboxylic acid / ethanoic acid 4. React with acidified K2Cr2O7 /KMnO4 to produce carboxylic acid 5. Undergo dehydration to produce alkene / ethene.

1 1 1

1 1 6

[Any three answers] (c)

(i)

(ii)

P : Hexane Q : Hexene // Hex-1-ene

1 1

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

1

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

2

1 1

46

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

1 1

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

1 1 Max 6

Total mark

No. 14 (a)

20

Explanation   

(b)

Haber process Iron N2 + 3H2 Pure copper

Mark 1 1 1+1

2NH3 Bronze

1 Tin atom

1+1

Copper atom

Bronze is harder than pure copper   

1

Tin atoms are of different size The presence of tin atoms distrupt the orderly arrangement of copper atoms. This reduce the layer of atoms from sliding.

1 1 1 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 A B

1 1+ 1 1 1 1

1 1 1

The intensity of blue spots High Low

Conclusion: Iron rust faster than steel. TOTAL

20

47

No. 15 (a) (i)

Explanation  Traditional medicines are derived from plans or animals.  Modern medicines are made by scientists in laboratory and based on substances found in nature.

Mark 1 1

(ii) Analgesics

Antibiotics

Aspirin Paracetamol Codein Penicillin

1 1 1 1

Psychotherapeutic

Chloropromazin Caffeina

1 1 MAX 5

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

1

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

1

1 Aspirin Cause brain and liver damage if given to children with flu or chicken pox. Cause internal bleeding and ulceration (b)

Hard water contains high concentration of calcium ions or magnesium ions. Example : sea water Procedure 1. 20cm3 of hard water (magnesium sulphate solution) is poured into two separate beakers X and Y. 2. 50 cm3 of soap and detergent solutions are added separately in beaker X and beaker Y. 3. A small piece of cloth with oily stains is dipped into each beaker. 4. Each cloth is washed. 5. The cleansing action of the soap and detergent is observed. Results Beaker X Y

Observation The cloth is still dirty.

1 1

1 1 1 1 1

1 1

The cloth becomes clean.

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

1

48

JAWAPAN SET 5 PAPER 3 SET 1

1. (a) (i)

EXPLANATION [Able to record all reading accurately with unit] Sample answer Experiment I II III

1(a) (ii)

Copper

Bronze

1.4 cm 1.5 cm 1.6 cm

1.1 cm 1.1 cm 1.2 cm

[Able to construct a table to record the diameter of the dents and average diameters for copper and bronze that contain: 1. correct title 2. Reading and unit Sample answer: Material Diameter of the dents(cm) Average diameter,(cm) 1 2 3 X 1.4 1.5 1.6 1.5 Y 1.1 1.1 1.2 1.1

SCORE

3

3

1.(b)

[Able to state correct observation] Sample answer: The diameter of dents made on material Y is smaller than material X// The diameter of dents made on material X is bigger than material Y

3

1.(c)

[Able to state the inference correctly] Sample answer: Bronze is harder than copper// Copper softer than bronze [Able to state the correct operational definition for alloy] 1. what should be done and 2. what should be observe correctly Sample answer: A smaller dent is formed when the weight of 1 kilogram is dropped at height of 50 cm to hit the ball bearing which is taped onto the alloy block using cellophane tape

3

1.(d)

3

1.(e)

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

3

1.(f)

[Able to state the relationship correctly between the manipulated variable and responding variable with direction] Sample answer: Alloy is harder than its pure metal. [Able to state all the three variables and all the three actions correctly] Sample answer:

3

1(g)

Names of variables (i) manipulated :

Bronze and copper (ii) responding: Diameter of dent

Action to be taken (i) the way to manipulate variable: Repeat experiment by replacing copper with bronze (ii) what to observe in the responding variable:

49

(iii) controlled: Mass of the weight // height of the weight // size of steel ball bearing.

The diameter of the dent formed on copper and bronze. (iii) the way to maintain the controlled variable: Uses same mass of weight // same height of weight // same size of steel ball bearing

EXPLANATION 2(a) [Able to state 4 inferences correctly] Test tube A B C D

3

SCORE 3

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

2(b) [able to explain a difference in observation correctly between test tube 1 and 2] Sample answer: Iron / Fe in test tube A rust / oxidized because iron is in contact with less electropositive metal, but iron in test tube B does not rust / oxidized because iron is in contact with less electropositive metal. 2(c) [Able to state the hypothesis correctly] Sample answer: When a more/ less electropositive metal is in contact with iron / Fe, the metal inhibits/ speed up rusting of iron.// When a more / less electropositive metal is in contact with iron/ Fe, rusting of iron is faster / slower// The higher /lower the metal in contact with iron/ Fe in electrochemical series than iron /Fe ,the rusting of iron/ Fe is slower / faster

3

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

3

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

3

2(f)

3

[able to classify all the three metals correctly] Metal that can provide sacrificial protection to iron zinc

Metal that cannot provide sacrificial protection to iron

3

copper

magnesium 50

2(g) [Able to compare the intensity of blue colour and relate the intensity of blue colour with the concentration of Fe2+ accurately ] Sample answer: The concentration of iron (II) ion is higher .The intensity of blue colouration after two days is higher.

EXPLANATION 3 (a)

3(b)

3 (c)

3(d)

[Able to make a statement of the problem accurately and must be in question form] Suggested answer: Does different type of alcohols affect the heat of combustions? // How does the number of carbon atom per molecule of alcohol affect the heat of combustion ? [Able to state all the three variables correctly] Suggested answer: Manipulated variable: Different types of alcohols//Different alcohols such as ethanol, propanol and butanol. Responding variable: Heat of combustion//Increase in temperature Fixed variable: Volume of water // type of container/ size of container [Able to state the relationship between manipulated variable and responding variable correctly] Suggested answer: When the number of carbon atoms per molecule of alcohol increases, the heat of combustion increases. [Able to state the list of substances and apparatus correctly and completely] Suggested answer: Ethonol, propanol, butanol, water, copper can, spirit lamp, thermometer, weighing balance, wooden block, tripod stand, wind shield, measuring cylinder.

3(e)

[Able to state a complete experimental procedure] Suggested answer: 1. [200 cm3] of water is poured into a copper can. 2. Initial temperature of the water is recorded. 3. A spirit lamp is half filled with ethanol. 4. Weight the spirit lamp with ethanol and record the mass 5. The spirit lamp is put under the copper can and ignites the wick immediately. 6. The water is stirred and the flame is put off after the temperature has increased by 30oC. 7. The highest temperature of the water is recorded 8. Immediately weight the spirit lamp and record the mass. 9. The experiment is repeated t by replacing ethanol with propanol and butanol.

3(f)

[Able to exhibit the tabulation of data correctly with suitable headings and units ] Types of Initial Highest Initial mass of Final mass of alcohols temperature/oC temperature/oC spirit lamp/g spirit lamp/g Ethanol Propanol Butanol

3

SCOR E

3

3

3

3

3

3

51

PAPER 3 SET 2 Question

1(a)

1(b)

1(c)

1(d)

1(e)

1(f)

1(g)

Rubric Able to record the burette readings accurately with 2 decimal places. Experiment I II III Initial burette 1.00 13.50 26.00 reading Final burette 13.50 26.00 38.50 reading Able to construct a table with the following information: 1. Accurate titles and units: 2. Burette readings and volume of acid used/cm3 Sample answer: Experiment I II Initial burette 1.00 13.50 3 reading/cm Final burette 13.50 26.00 reading/cm3 Volume of acid 12.50 12.50 used/cm3

3

3

III 26.00 38.50 12.50

Able to calculate correctly the molarity of acid with the following steps: Step 1: MaVa = 1 MbVb 1 Step 2: Ma = 1.0 x 25 12.5 Step 3: 2.0 mol dm-3 Able to give the volume and explaination correctly with following aspects: 1. 6.25 cm3 2. Sulphuric acid is a diprotic acid 3. Concentration of H+ ions is double Able to state the three variables correctly. Sample answer Manipulated variable: Type of acids//Hydrochloric acid, ethanoic acid Responding variable: pH values Fixed variable: Concentration of acids Able to state the hypothesis accurately. Sample answer When the concentration of hydrogen ion in acid is higher, , the pH value is lower// The higher the concentration of hydrogen ion, the lower the pH value Able to classify all the substances correctly. Sample answer: Substances with pH less than 7 Ethanoic acid Nitric acid

Score

3

3

3

3

Substances with pH more than 7 Ammonia solution Barium hydroxide

52

Question Rubric 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 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 2(c)

2(d)(i)

2(d)(ii)

2(d)(iii)

Able to name the alcohol and carboxylic acid correctly. Alcohol: Propanol Carboxylic acid: Butanoic acid 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 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 Able to predict and make explanations accurately Answer 1. Hexene 2. Percentage of carbon atoms per molecule hexene is higher than hexane

Question Rubric 3(a) Able to state the aim accurately Sample answer To compare the effectiveness of soap and detergent on cleansing action in hard water . 3(b) Able to state the three variables accurately. Answer Manipulated variable: Soap and detergent Responding variable: Effectiveness of cleansing action Fixed variable: Type of water//hard water 3(c) Able to state the hypothesis accurately with direction Sample answer

Score 3

3

3

3

3

3

Score 2 3

3

Detergent is more effective than soap in hard water 3(d)

Able to state the complete list of apparatus and material as follows Hard water, soap and detergent,2 beakers, 2 pieces of cloths stained with oil, galss rod

3

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 beaker 3. A piece of cloth stained with oil is immersed in the solution 4. The cloth is shaken/rubbed/stirred

3

53

3(f)

5. Observation is recorded 6. Repeat steps 1 – 5 by using detergent . Able to tabulate the data correctly Sample answer Type of cleaning agent Observation

3

Soap Detergent

PAPER 3 SET 3 Question Rubric 1(a)(i) Able to record the thermometer reading accurately to 1 decimal place with unit. Answer Initial temperature = 30.0 oC Highest temperature = 60.0 oC (a)(ii) Able to state one observation accurately Sample answer Thermometer reading increases//Temperature increases (a)(iii) Able to state the inference accurately. Sample answer Heat energy is released //The reaction is exothermic reaction (b)(i) Able to state all the mass of alcohols accurately to 2 decimal places with unit Answers 1.55, 2.23, 3.56, 4.01 (b)(ii) Able to tabulate the initial mass, final mass and mass of alcohols accurately with units Sample answer: Alcohol Initial mass/g Final mass/g Mass of alcohol/g Methanol 354.9548 353.4012 1.55 Ethanol 342.0201 339.7892 2.23 Propanol 364.4303 360.8702 3.56 Butanol 332.9891 328.9790 4.01 (c)

(d)(i)

(d)(ii)

Able to calculate the heat of combustion of methanol correctly with the following steps: 1. Heat change = 200 x 4.2 x 30 J = 25200 J 2. No of mole of methanol = 1.55 ÷ 32 = 0.048 mol 3. Heat of combustion = - 525kJ mol-1 Able to state the variables correctly Sample answer Manipulated variable Type of alcohols// Methanol, Ethanol, Propanol, Butanol Responding variable Heat of combustion Fixed variable Volume of water Able to state the hypothesis accurately with the manipulated variable related to responding variable Sample answer

Score 3

3

3

3

3

3

3

3

54

(e)

(f)

(g)

When the number of carbon atoms per molecule alcohol increases, the heat of combustion increases Able to predict the heat of combustion for pentanol correctly Sample answer 2350 kJ mol-1// (2300 – 2400)kJ mol-1 Able to state the three reasons correctly Sample answer 1. Some of the heat energy is released to the surrounding 2. Some of the heat energy is absorbed by the copper can 3. Incomplete combustion of ethanol Able to classify all the compounds correctly Sample answer Hydrocarbon Non-hydrocarbon Hexene Propanoic acid Methane Ethanol

Question 2 Question Rubric 2(a) Able to state the aim of experiment accurately Sample answer To investigate the effect of type of electrode on the selection of ions to be discharged at the anode/ on the product formed at the anode. 2(b) Able to state all the variables Sample answer Manipulated variable Type of electrodes//Carbon electrodes and copper electrodes Responding variable Product formed at anode Fixed variable Electrolyte 2(c) Able to state the hypothesis accurately Sample answer When carbon electrodes are used, bubbles/oxygen gas released at anode, when copper electrodes are used, anode becomes thinner/ionised 2(d) Able to list completely the materials and apparatus as the following

3

3

3

Score 2

3

3

3

Copper(II) sulphate solution (0.5 – 2.0) mol dm-3, copper rods, carbon rods, electrolytic cell, battery, connecting wires, test tube 2(e)

2(f)

Able to state the all the following procedures 1. Half filled the electrolytic cell/beaker with copper(II) sulphate solution 2. A test tube filled with copper(II) sulphate solution is inverted over the anode carbon electrode 3. Both electrodes are connected to the batteries using connecting wires//Complete the circuit 4. Record the observations at the anode 5. Repeat steps 1-4 by using copper electrodes Able to draw a table for tabulation of data correctly with the following Type of electrodes Carbon Copper

3

3

Observation at anode

PAPER 3 SET 4 55

Q1 (a)

Rubric Able to state the two observation correctly

Score 3

Sample answer: 1. Burns very rapidly// burns vigorously 2. Produces white fumes (b)

(c)

Able to state the relationship correctly Sample answer: Going down the Group 1 element, the reactivity towards oxygen increases.

3

Able to give the inference correctly

3

Sample answer: Alkaline solution produced. (d)

Able to predict the position of X metal in the periodic table correctly

3

Sample answer: Between potassium and lithium// below lithium and above potassium Able to classify all the ions correctly (e)

Sample answer: Cation Lithium ion // Li+ Hydrogen ion // H+

Question Number

Anion Hydroxide ion // OH-

Rubric

3

Score

[Able to write potential difference with one decimal correctly]

Example

3

Pair of metals 2(a)

2(b)

2(c) 2(d)

Cu and M Cu and J Cu and Q

Potential difference(V) 1.10 1.85 0.40

Positive terminal L L Q

[Able to give the hypothesis accurately] Example: The further the distance between two metals in the electrochemical series, the higher the potential difference / voltage [Able to arrange the metals in the electrochemical series in descending order correctly] Answer: : J, M, Cu, Q [Able to classify the metals correctly]

3

3 3

56

Answer: : More electropositive : J, M Less electropositive : Cu, Q

[Able to state three observations correctly]

2(e)

Suggested answer: 1 Metal J dissolved // becomes thinner//smaller 2 Copper becomes thicker//bigger//brown solid deposited 3 Blue solution becomes light blue /fading in colour//intensity of blue colour of copper(II) sulphate decreases

Question Rubric 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? //

3

Score

3

How does the temperature of sodium thiosulphate solution affect the rate of reaction? between sodium thiosulphate solution and sulphuric acid? 3 (b)

Able to state the three variables correctly Sample answer: Manipulated variable: Temperature of sodium thiosulphate solution Rate of reaction // Time taken for mark „X‟ to become invisible /disappear

3 (c)

3

Constant variable: Volume and concentration of sodium thiosulphate/ sulphuric acid / size of conical flask Able to state the relationship correctly between the manipulated variable and the responding variable with direction. Sample answer: 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 //

3

The increase/decrease in temperature of sodium thiosulphate solution will increase/decrease the rate of reaction. // When the temperature of sodium thiosulphate solution increase /decrease, the rate of reaction will increase/decrease. 57

3(d)

Able to give complete list of materials and apparatus Sample answer: Materials: Sodium thiosulphate solution, sulphuric acid.

3

Apparatus: Conical flask, ,bunsen burner, measuring cylinder, stop-watch, filter paper. 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 (f)

3

Able to tabulate the data with following aspects 1. Correct titles with units 2. Complete list of temperatures Sample answer: Temperature (oC)

Time (s)

2

30 40 50 60 70

58

SET 5 PAPER 3 Question No. 1

.(a)

(b)

(c)

(d)

(e)(i)

(e)(ii)

Rubric

[Able to state the inference correctly] Sample answer: Ethanoic acid can show acidic properties in water // Ethanoic acid cannot show acidic properties without water [Able to give the explaination correctly with the following 3 points] 1.With water ethanoic acid can ionised 2. To produce H+ ion 3. H+ ion is responsible for the acidic property [Able to give the three variables correctly] Manipulated variable: type of solvent// water and propanone Responding variable: colour change to blue litmus paper Fixed variable: blue litmus paper [Able to state the hypothesis correctly that relates the manipulated variable to responding variable] Sample answer: When ethanoic acid dissolved in water, blue litmus paper turns red// When ethanoic acid dissolved in propanone, no change to blue litmus. [Able to give the three observations correctly] Test tube I: No change Test tube II: Red litmus paper change to blue Test tube III: No change [Able to construct the table correctly with the following information: 1. Column / rows with title 2. With correct observations I II III Experiment Dry Ammonia Aammonia ammonia dissolved in dissolved in water dry propanone Observation

(f)

Red litmus No change paper turns blue [Able to classify the substances correctly with the correct titles] Substance that change blue litmus paper to red: Phosphoric acid, oxalic acid, sulphurous acid Substance that change red litmus paper to blue Calcium hudroxide, barium hydroxide, ammonia aqueous, barium hydroxide

Score

3

3

3

3

3

3

No change

3

59

Question No. 2 (a)

Rubric [Able to measure all the height of precipitate accurately with one decimal place.] Answer: Test tube Height of precipitate

(b)

(c)

(d)

(e)

(f)

Score

1 2 3 4 5 6 7 0.5 0.9 1.3 1.6 2.0 2.0 2.0 [0.4[0.8[1.2[1.5[1.9[1.9[1.90.6] 1.0] 1.4] 1.7] 2.1] 2.1] 2.1] Able to draw the graph correctly with the following information: i. Axis- x : volume of barium chloride / cm3 and axis -y : height of precipitate/ cm ii. Consistent scale and the graph half of graph paper iii. All the points are transferred correctly iv. Correct curve

Able to state the volume and calculate the number of mol correctly Answer: 1. 5 cm3 2. No. of mole = 0.5 x 5 // 0.0025 mol 1000 Able to write the ionic equation correctly. Sample answer: Ba2+ + CrO42-  BaCrO4

Able to give the meaning of the precipitation reaction correctly. Sample answer: Yellow precipitate is formed when barium chloride solution reacts with potassium chromate (VI) solution. Able to classify all the salts correctly Sample answer: Soluble salts

Sodium carbonate, Na2CO3 Magnesium nitrate,Mg(NO3)2

3

3

3

3

3

Insoluble salts

Lead(II)sulphate, PbSO4 Silver chloride, AgCl

3

60

Question No.3 (a)

(b)

(c)

(d)

(e)

Rubric

Score

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

Able to state the three variables correctly Sample answer: Manipulated : ethanoic acid and ammonia solution Responding : coagulate / coagulation of latex Fixed : latex Able to state the relationship correctly Ethanoic acid coagulates the latex while ammonia solution does not coagulate the latex. 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 Able to list all the steps correctly

3

3

3

3

1. 10 cm3 of latex is poured into a beaker. 2. ethanoic acid is added into the beaker using adropper. 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. (f)

Able to tabulate the data correctly Mixture Latex + ethanoic acid Latex + ammonia solution

Observation

2

END OF QUESTION MARKING SCHEME

61