1

2

3

4

5

6

7

8

9

10

B

D

B

A

D

C

C

A

D

B

11

12

13

14

15

16

17

18

19

20

B

B

C

C

D

D

C

A

C

A

21

22

23

24

25

26

27

28

29

30

C

C

A

C

B

B

C

A

A

D

31

32

33

34

35

36

37

38

39

40

D

A

D

A

D

A

A

A

B

C

Qn

Answers

A1

(a)   air {common mistakes: glass and limestone}

(b)  glass {not in syllabus}

(c)   brass {common mistake: air}

(d)   water/ salt

(e)   sugar/ limestone

A2

(a)   krypton

(b)  10

(c)   Molar mass of Ar = 40g

       Molar volume of Ar at r.t.p. = 24 dm3

       Density of Ar = 40/24 = .1.67g/dm3

(d)  215K {if unit given is in ^oC – no mark}

(e)  fractional distillation of liquid air

(f)  Argon is present in a larger quantity than other gases.

A3

(ai)             H H H  H                          H H  H  H

                   |    |   |    |                            |    |   |    |

              H-C-C=C-C-H                 H-C-C-C=C-H

                   |            |                             |    |

                  H          H                           H  H

                                                   {common mistake: but-1-ene}

(aii)   C₄H₈ + Br₂ ® C₄H₈Br₂

(aiii)  alkene: colourless; bromine: reddish-brown; reaction product: colourless

(aiv)  Alkene is unsaturated and a single product is formed.

(bi)    C₉H₁₂

(bii)   boiling point/ density/ viscosity

          {note: an increment of CH₂ in the molecular formula is not allowed as a physical property}

A4

(a)     P: anhydrous copper (II) sulphate; Q: water

         {common mistake: never mention oxidation state for Cu}

(bi)    barium sulphate

(bii)   H⁺ and SO₄^2- {CuSO₄ ®  CuO + SO₃, SO₃ + H₂O ® H₂SO₄}

(biii)  copper (II) oxide

          {common mistake: never mention oxidation state for Cu}

A5

(ai)    hydrogen; because it is lighter

(aii)   molecules move faster

(bi)   2H₂ + O₂ ® 2 H₂O

(bii)  water and oxygen

        {note: oxygen is in excess because 2 vol. of H₂ react with 1 vol. of O₂}

(c)    formation: In the car engine, the temperature is very high. The oxygen and nitrogen in the surrounding air react to form oxides of nitrogen.

         limiting its formation: The emission can be limited by installing a catalytic converter so that the oxides of nitrogen can be converted to harmless gases.

(d)    sulphur dioxide

A6

(a)    dilute sulphuric acid

(b)    Overall reaction: Zn + 2H⁺ ® Zn²⁺ + H₂

                                    {Zn – 2e ® Zn²⁺, 2H⁺ + 2e ® H₂}

        {common mistake: misread the question and gave half equations for the reactions at the electrodes}

(c)    A more reactive metal gives a higher meter reading.

(d)    2.72 – Magnesium, 2.00 – Aluminium, 1.10 – Zinc, 0.78 – Iron,

         0.00 - Copper 

Section B

B7

(ai) At the anode, for every 1 mole of O₂ formed, 4 moles of electrons is given out by OH^-. At the cathode, for every 1 mole of H₂ formed, only 2 moles of electron is taken in by H⁺. Hence, volume of O₂ formed is only half that of H₂/ volume of H₂ twice that of O₂.

       {Common mistake: using the formula of water to answer the question}

(aii)  SO₄^2-

(b)  At the anode, the product formed is oxygen gas and product formed at the cathode is hydrogen gas.

(ci)  Aluminium is manufactured by the electrolysis of molten aluminium oxide, using carbon as the electrode. The molten aluminium oxide is dissolved in molten cryolite to lower its melting point. During electrolysis, the following reaction occurs.

       At the anode: 2O^2- - 4e ® O₂

       At the cathode: Al³⁺ + 3e ® Al

       Aluminium is formed at the cathode. During electrolysis, the carbon anode must be changed periodically as oxygen given out will react with the carbon anode to form carbon dioxide gas.

(cii) making drink cans/ aluminium foil for wrapping food

        {Note: vague answers such as using aluminium to make windows or kitchens are not allowed}

(d)  Amphoteric oxide is a metal oxide that reacts with both acid and alkali to form salt and water.

B8

(ai)  burette and pipette

(aii) methyl orange; from yellow to orange

(aiii) aq: aqueous, l: liquid

(bi)  no. of moles of NaOH/ no. of moles of HCl = 1/1

        no. of moles of NaOH = 0.025 x 1 = 0.025 moles

        no. of moles of HCl = 0.025 moles

        volume of HCl = no. of moles/concentration = 0.025/2 = 0.0125 dm³

(bii) no. of moles of NaCl/ no. of moles of NaOH = 1/1

        no. of moles of NaCl = 0.025 moles

        mass of NaCl = 0.025 x 58.5 = 1.46g

(ci)   E is a metal because it gives away 2 electrons to form E²⁺O^2-

(cii)  Group II because it gives away 2 electrons.

B9

(ai)   To prevent steel alloy from rusting

(aii)  SnCl₄

(aiii) Covalent because SnCl₄ has a low boiling point.

(aiv) SnCl₄ ® Sn + 2Cl₂

        The oxidation state of Sn decreases from +4 (in SnCl₄) to 0 (in Sn) showing reduction. The oxidation state of Cl increases from -1 (in SnCl₄) to 0 (in Cl₂) showing oxidation. Since oxidation and reduction occur at the same time, this is a redox reaction.

       {note: must mention both reduction of tin and oxidation of chlorine to earn the marks}

(bi)  {not in syllabus} Diamond and Graphite

(bii) {not in syllabus} Diamond is used in saws and drills to cut hard solids. This is because diamond is very hard. It has a macromolecular structure. Each carbon atom is bonded to four other atoms tetrahedrally by strong covalent bond. Graphite is used as a lubricant. The carbon atoms are arranged in layers. These layers of atoms can slide over each other easily because of weak intermolecular forces and thus, graphite is greasy and soft. Or graphite can be used as electrodes because it contains electrons which are free to move to carry current.

B10

(ai)  Fine powder has a larger surface area. Reaction will be faster.

(aii) by measuring the decrease in volume of hydrogen with time

       {common mistakes: (1) mentioning hydrogen being evolved instead of decreasing (2) ignoring the time factor}

(aiii) because it is a transition metal

(b)    since one molecule of oil has only 1 double bond, one mole of the oil will react with 1 mole of hydrogen

         no. of moles of oil/ no. of moles of H₂ = 1/1

         no. of moles of H₂ = (60/1000)/24 = 0.0025 moles

         no. of moles of oil = 0.0025 moles

         molar mass of oil = mass/ no. of moles = 0.5/0.0025 = 200

         therefore, relative atomic mass of oil = 200

         {common mistake: fail to convert 24 dm³ to 24000 cm³ in calculations}

(ci)   Glycerol and soap

(cii)  Terylene

         {common mistake: giving ‘polyester’ as the answer}