Answer to Sec 3 Chemistry GE EOY 2007

1. Answer to Sec 3 Chemistry GE EOY 2007 Paper 1 1. C 2. A 3. C 4. B 5. C 6. B 7. C 8. C 9. B 10. D 11. D 12. C 13. B 14. C 15. C 16. C 17. A 18. C 19. B 20. A 21. B 22. B 23. C 24. B 25. C 26. D 27. C 28. C 29. B 30. B

2. Section B, B1 • Isotopes are atoms of the same element with the same number of protons but different number of neutrons/nucleon number/mass number. • The relative atomic mass of an element is the (weighted) average mass of one atom of the element when compared with 1/12 of the mass of a 12C atom. • % of 57Fe = 100-5.8-91.6 = 2.6% • Relative atomic mass of Fe =

3. Question B1 • The chemical properties are the same. • The chemical behaviour of atoms of an element is due to the number of valence electrons the atom possess. • Since atoms of all the iron isotopes have the same number of valence electrons, the chemical properties are expected to be the same.

4. 1s 2s 2p 3s x Y Z • Question B2 • i) Element G • ii) Element A/B/E • iii) Element F/J • D: 1s22s22p63s23p6 d) 3s 2pz 2py 2px

5. Read Q – “WF6 has low mp and bp, hence it must be simple molecular and bonds between atoms must be covalent Read Q – CaF2 has high mp and bp, hence it must be an ionic compound - 2+ X X Ca X X F 2 X X o x • Question B3

6. Question B3 (Cont’d) b) Structure: Tungsten (VI) fluoride has a simple molecular structure/exists as simple/discrete molecules. This can be deduced from the low melting and boiling point which is typical of compounds with simple molecular structure. Bonding: Atoms of tungsten and fluorine are held by strong covalent bonds formed by sharing of electrons. • Displacement • WF6 + 3Ca  3CaF2 + W

7. Question B4 • As we go across period 3, the no.of protons in each element increases, leading to an increase in nuclear charge. • However, the valence electrons are added to the same shell, hence shielding effect remains almost constant. • Hence the valence electron is more strongly held by the nucleus and more energy is needed to remove 1 valence electron from the atom and IE generally increases across the period.

8. Water out Liebig condenser Water in Q B5 a) • Miscible liquid mixture • The components of the mixtures are easily decomposed. OR • The quantity of the mixture is very small. • The boiling point of the liquids are too close/too high.

9. Question B5 • d) Distance moved by solvent = 5.3 cm • Distance moved by solute S = 4.2 cm • Rf = 4.2/5.3 = 0.79 • Substance T • Substances P, R and S

10. Question B6 (a) Empirical formula : MgCO3 Let the molecular formula be (MgCO3)n Mr of (MgCO3)n = 84 n x [24 + 12 + 3(16)] = 84 n = 1 Hence molecular formula is MgCO3

11. Question B6 (b) • To make magnesium carbonate from magnesium nitrate, precipitation method should be used. • The other reagent to be used is aqueous sodium carbonate (or any carbonate of Gp I) • Steps • Mix equal volumes of aqueous sodium carbonate and aqueous magnesium nitrate in a beaker and stir. • Filter the mixture and collect magnesium carbonate as residue. • Wash the residue with copious amounts of deionized water. • Dry the residue between pieces of filter paper.

12. Question B6 • Sodium nitrate or excess magnesium nitrate/sodium carbonate • Question B7 (not important) • Aluminium is a metal that is higher than metals like iron in the reactivity series. • While iron corrodes easily in the presence of moist air, Al does not. • A thin layer of aluminium oxide which forms on the Al protects it from corrosion as this layer is impervious (non-porous) to water and oxygen. • i) good conductor of heat or has low density • ii) Light weight/has low density/good weight to strength ratio • good conductor of electricity

13. Section C, Question C1 • Co-ordinate bond/dative bond Need to show the charge and arrow which represents dative bond Trigonal pyramidal tetrahedral Need to show lone pair of e- on P

14. Question C1 (c) Oxygen is more electronegative than hydrogen. Hence when a covalent bond is formed between an atom of oxygen and hydrogen, the electrons are attracted towards the oxygen atom, setting up a partial negative charge around the oxygen atom. A polar molecule is one with regions of partial positive charge and a partial negative charge set up OR Presence of dipole moments due to the electronegativity difference between the atoms in a molecule.

15. Question C1 d) The intermolecular forces of attraction between methanol molecules is hydrogen bonds while that between oxygen molecules is induced dipole-dipole forces. Hydrogen bonds are stronger than induced dipole-dipole forces, hence more energy is required to overcome the forces of attraction between methanol molecules. e)

16. Question C2 • Ammonia gas • NH4+ (aq) + OH- (aq)  NH3 (g) + H2O (l) • Aqueous NaOH test of A did not produce any precipitate, meaning that the cation should be either Na+/K+ (Gp I) or NH4+ • Upon warming, the mixture produced ammonia gas, showing that the cation must be NH4+ • Warming the solution with aq NaOH and Al foil produced ammonia gas implies that NO3- is present. • Aq NaOH test of B produced a green ppt, which on warming turned reddish-brown, hence the cation should be Fe2+ • Adding acidified barium nitrate to solution B produced a white ppt, showing that the anion in B is SO42-

17. Question C2 • c) cont’d • A is therefore ammonium nitrate or NH4NO3 • B is iron (II) sulphate or FeSO4 • Filtration • Add an excess of aqueous sodium or potassium carbonate/sodium or potassium hydroxide to precipitate out the iron (II) ions as insoluble iron (II) carbonate or iron (II) hydroxide. • Filter the mixture and add dilute nitric acid to the residue. • FeSO4 + 2NaOH  Fe(OH)2 + Na2SO4 • Fe(OH)2 + 2HNO3 Fe(NO3)2 + 2H2O

18. Question C3 Either • Sulphur dioxide dissolves in moisture in the air to form sulphurous acid which is further oxidised to form sulphuric acid. • SO2 + H2O  H2SO3 Or 2H2SO3 + O2 2H2SO4 • The moisture will condense into clouds and fall as acid rain which corrodes concrete/limestone/metal structures or damage vegetation. • Flue gas desulphurization • The acidified potassium dichromate (VI) turns from orange to green.

19. Question C3A EITHER d) No. of mol of K2Cr2O7 = 1 mol of Cr2O72- reacts with 3 mol of SO2 0.0005 mol of Cr2O72- react with 0.0005x3=0.0015 mol of SO2 Volume of SO2 = 0.0015 x 24 = 0.0360 dm3 or 36.0 cm3 e) % of SO2 in air =

20. Question C3A EITHER • Carbon dioxide is emitted largely from vehicles, power stations from the complete combustion of carbon-containing fuel. • It is a greenhouse gas which traps the infra-red radiation from the sun and thus reduces the amount of heat escaping into space. This causes average temperature on Earth to rise or global warming, which in turn leads to melting of polar ice caps/rising of sea levels/drastic change in climate.

21. Question 3B OR • Mg + 2HNO3→ Mg(NO3)2 + H2 • No. of mol of HNO3 = c) No. of mol of Mg = 1 mol of Mg reacts with 2 mol of HNO3 0.0250 mol of Mg react with 0.0500 mol of HNO3 But no. of mol of HNO3 added initially = 0.0600 Hence HNO3 is in excess and Mg is the limiting reactant. 1 mol of Mg produces 1 mol of H2 0.0250 mol of Mg produce 0.0250 mol of H2 Volume of H2 = 0.0250 x 24 = 0.600 dm3

22. Question 3B OR d) No. of mol of H2 actually produced = 1 mol of H2 is produced by 1 mol of Mg 0.0230 mol of H2 is produced by 0.0230 mol of Mg e) % purity of Mg = • MgO (s) + 2H+ (aq) → Mg2+ (aq) + H2O(l)

23. C3B OR • 1 mol of Mg reacts with 2 mol of HNO3 • 0.0230 mol of Mg react with 0.0230x2 = 0.0460 mol of HNO3 • Mass of MgO reacted = 0.600 - 0.552 = 0.0480 g • No. of mol of MgO = 1 mol of MgO reacts with 2 mol of HNO3 0.0012 mol of MgO reacted with 0.00120 x 2 = 0.00240 mol of HNO3. No. of mol of HNO3 left = 0.0600 – (0.0460 + 0.00240) = 0.0116 No. of mol = vol in dm3 x conc in mol/dm3 0.0116 = vol in dm3 x 1.00 Volume = 0.0116 dm3 or 11.6 cm3