1. (B)
2. (C)
3. Redox not covered in sec 3, anyway, answer is (A)
4. (A)
5. given no. of moles of Al2O3 = 1.0 mole
mole ratio of Al2O3:O2 is 2:3
no. of moles of oxygen is 3/2 x 1 = 1.5 moles
mass = mole x Mr of (O2)
= 1.5 x 2(16)
= 48 g (B)
6. chlorine = 2,8,7
when chlorine gains 1 electron to form chloride ion, it becomes 2,8,8 and it has 3 SHELLS
fluorine = 2,7
when fluorine gains 1 electron to form fluoride ion, it becomes 2,8 and it has 2 SHELLS
Magnesium = 2,8,2
when magnesium lose 2 electrons to form magnesium ion, it becomes 2,8 and it has 2 SHELLS
Oxygen = 2,6
when oxygen gains 2 electrons to form oxide ion, it becomes 2,8 and it has 2 SHELLS
so form the largest radius.... which means greater 'circumference' = chlorine
(A)
7. chemical analysis not tested in sec 3, anyway, answer is (B)
8. for graphite, each carbon atom is COVALENT bonded to each other. However, the carbon atoms form a HEXAGON RING. Each ring is connected to another ring by weak VAN DER WAALS ( weak VDW) forces of attraction. Graphite is used as lubricant because its layers can SLIDE over each other. (D)
9. (D)
All these elements are from group 2. All have 2 electrons in the outermost shell.
10. (B)
11. REDOX not teated in sec 3, anyway, answer is (D)
12. option (A) look like group 1 metals , option (B) look like macromolecules. copper is a metal, so it will have HIGH MP, HIGH DENSITY, GOOD ELECTRICAL CONDUCTIVITY (D)
13. (B)
14. empirical formula is the SIMPLEST ratio but MOLECULAR FORMULA is the formula multiply by a factor. In order to calculate the Molecular Formula, we need the relative molecular masses. Eg: the Mr of methane, CH4 is 16 while the Mr of (CH4 )2 is 32. (B)
15. given conc of acid = 1.0 M and mass of CaCO3 = 1.0 g.
no. of moles of CaCO3 = 1/100 = 0.01 mol
mole ratio of HCl: CaCO3 is 2:1
no. of moles of HCl = 0.01 x 2 = 0.02
vol = 0.02 / 1.0 = 0.02 dm3 = 20 cm3 (B)
16. H2SO4 + MgO ---> MgSO4 + H2O ;
H2SO4 + MgCO3 ---> MgSO4 + H2O + CO2
so, in both reactions, water is formed. (C)
17. CHEMICAL ANALYSIS not tested in sec 3, anyway, the answer is (A)
18. (A)
19. option B is very interesting. Covalent compounds are made up by atoms.... but the BETTER word for ' Nature of Particles' should be MOLECULES. Anyway, the formula for option B is wrong too. answer(C)
20. you need knowledge from Chemical Analysis, not tested in sec 3. Anyway, option (D)
21. (C)
22. if X is noble gas, then X,Y,Z is consecutive. So, Y should be in group 1, Z should be group 2.(D)
23. IMPT
Cl2 (g) + 2KBr ( aq) ---> 2KCl (aq) + Br2 (aq) ; colour of KBr and KCl = colourless; but aqueous Br2 = reddish brown
Cl2 (g) + 2KI (aq) ---> 2KCl (aq) + I2 (aq) ; colour of KI and KCl = colourless; but aqueous iodine is dark purple or blue (D)
24. although both potassium (2,8,1) and calcium (2,8, 2) have more electrons than aluminium ( 2,8,3 ). But aluminium donates MORE electrons into the sea : Al (s) ---> Al3+ (aq) + 3e- so answer (A)
25. total no. of electrons share = no. of BONDS in the molecule: 8 bonds x 2 e- = 16 e- answer (C)
26. (C) reason: the formula should be XO
27. when ammnium chloride is heated, it forms ammonia gas and hydrogen chloride gas: NH4Cl (s) ---> NH3 (g) + HCl (g) option (B)
28. notice the no. of protons and elecytrons for both X and Y is not equal! by right, in an neutral atom, X should have 4 electrons and 4 protons while Y should have 20 protons and 20 electrons. So this shows that both X and Y lose electrons. Only Metals lose electrons. Hence option (D)
29. R has 4 electrons involved in bonding. Hence R is in group 4. option (B)
30. given X contains acid ---> from universal indicator, X should have a pH less than 7. As for the reaction between KI and Pb(NO3)2, it will give a yellow ppt of PbI2 :
2 KI + Pb(NO3)2 ---> 2 KNO3 + PbI2 ( PbI2 is yellow ) option (A)
SECTION B (STRUCTURED )
1a) ammonia b) Chlorine
c) Lead (II) oxide
d) Magnesium chloride
e) Brass
2(a) chemical analysis Not tested in sec 3
a) zinc hydroxide / white / soluble in excess giving colourless solution. [1.5]
iron(II) hydroxide / dark green / insoluble in excess [1.5]
b) (i) Magnesium, sodium, caesium and beryllium [½]
(ii) Sodium, aluminium, magnesium and chlorine [½]
(iii) Hydrogen, sodium and caesium [½]
(iv) Carbon and tin [½]
3 a) ionic / ionic / covalent / covalent / covalent / covalent [2]
b) giant / giant / simple molecular [2]
c) MgO is ionic and giant structure is strong and stable, hence it has higher
melting point. whereas the SO3 is covalent bonding and a simple
molecular structure. [2]
d) Na2O(s) + H2O(l) ---> 2NaOH(aq) [1]
SO3(g) + H2O (l) ---> H2SO4(aq) [1]
4 ( chemical analysis not tested in sec 3 )
a) A Copper(II) carbonate [1]
B Carbon dioxide [1]
C Copper(II) chloride [1]
D Silver chloride [1]
E Copper(II) nitrate [1]
b) CuCO3 (s) + 2HCl(aq) ---> CuCl2(aq) +CO2(g)+ H2O(l)
CuCO3(s) + 2H+ ---> Cu2+(aq) +CO2(g)+ H2O(l) [2]
5 (oxidation and reduxtion not tested in sec 3)
a) i) +6 ii) +3 [2]
b) Aluminium [1]
c) gain of oxygen or increase in oxidation state [1]
d) Cr2O3 is oxidizing agent. [1]
6 a) i) yes, they all have similar formulae
ii) yes, peroxides and super oxides are likely formed at the
elements below the group.
b) O- , O2-
c) i) H+(aq) + OH-(aq) ---> H2O(l)
ii) 2I-(aq) + Cl2(g) ---> 2Cl-(aq) + I2(s)
iii) Zn(s) + 2H+(aq) ---> Zn2+(aq) + H2(g)
7 a) i. P6O4 ii. P3O2 [2]
b) It is an non-metal oxide
it is covalent bonding
it has no free electron and no mobile ions when molten. [3]
SECTION C (FREE RESPONSE)
1 a) i) Mg(s) + S(s) ---> MgS(s) [1]
iii) Mg :24 + S: 32 à MgS : 56
3 4
12g 16g
Actual: 16g of Mg is used, Sulphur will be used up first in the reaction. [1]
Hence sulphur is the limiting agent. [1]
iv) S MgS
32 56
16g 28g 28g of MgS formed [1]
b) CaCO3 (s) + H2SO4(aq) à CO2(g) + CaSO4(s) + H2O (l) [1]
100 24000
y 1200 [1]
y = 100 x 1200/ 24000
= 5 g (CaCO3 is reacted) [1]
Percentage purity = 5/6 x 100%
= 83.3 % [1]
2 a) i) % of Na = 12.5/45 x 100%
= 27.8%
ii) Na H C O
27.8 12. 14.3 56.7
27.8 1.2 14.3 56.7
23 1 12 16
1.2 1.2 1.19 3.54
1 1 1 3
Empirical formula = NaHCO3
iii) Mr of empirical formula (NaHCO3 )= 23+1+12+ 16x3 = 84
Mr of molecular of NaHCO3 =84
n = 84/84 = 1, Hence, molecular formula = NaHCO3 [1]
b) i) NaHCO3(aq) + HCl à NaCl (aq) + CO2(g) [1]
ii) 84 24
45 x
Conc. Of HCl = 91.3/36.5
= 2.5 mol/dm3
mol of HCl used = 280/1000 x 2.5
= 0.7 mol
hence, vol of CO2 produced = 0.7 x 24dm3
= 16.8 dm3
c) [ chemical analysis, not in syllabus ]
Add equal volume of sodium hydroxide and small spatula of
aluminium powder into a sample solution. [1]
Warm till the aluminium dissolves. [1]
Effervescence and pungent smell gas produced turns red litmus
paper blue. [1]
3 a) methyl orange [1]
b) 22.8 cm3 21.5 cm3 28.7 cm3 [1]
√ √ [1]
c) 25.0 cm3 [1]
d) H2SO4 + 2NaOH à Na2SO4 + 2H2O [1]
e) average of sulphuric acid used = (22.8 + 21.5) / 2 = 22.2 cm3 [1]
mol of H2SO4 / mol of NaOH = 1 / 2
mol of H2SO4 = ½ x mol of NaOH
0.05 x 22.2/1000 = ½ x 25/1000 x mol of NaOH
mol of NaOH = 0.0444 mol/dm3
Disclaimer: only the MCQ answers are from miss pang, answers from section B and C are done by another teacher. Miss Pang is not responsible for the accuracy of the answers!
