- (a) (i) Alkali metals
- Energy required to remove an electron from an atom of an element in its gaseous state.
- “P” has the smallest ionic radius therefore, the outermost electrons are most strongly attracted to the nucleus, hence more energy is required to remove this electron.
- Melts because the reaction is exothermic. Hissing sound because of the production of hydrogen gas. Float because it is less dense than water. Moves about due to propelling effect of escaping hydrogen.
- A strong base ionizes completely in water producing more OH– ions e.g. KOH and NaOH. A weak base ionizes slightly producing few OH ions e.g. NH4OH, Ca(OH)2 and Mg(OH)2.
- (i) Reaction between H+ ions from the acid and OH ions from bases to form 1 mole of water.
H+(aq) + OH(aq) H2O(l)
- Add 200cm3 of 2M nitric acid to 200cm3 of 2M sodium hydroxide. Heat the mixture so as to make it saturated / concentrated. Allow the mixture to cool for crystals to appear. Filter / decant to obtain the crystals.
- 2NaNO3(s) 2NaNO2(aq) + O2(g)
- (a) (i) Anhydrous calcium chloride / silica gel.P1
- To remove carbon (IV) oxide from airP1 or free air from CO2.
- 2Cu(s) + O2(g) 2CuO(S) P1
- Unreacted oxygen, noble gases e.g. Argon. P1
- It has a very low boiling point of -1960C.
- (i) Concentrated sulphuric VI acid / H2SO4 reject concentrated sulphuric acid.
(ii) KNO3(s) + H2SO4(l) HNO3(g) + NaHSO4(aq)
(iii) It is decomposed by lightP1 to brown nitrogen (IV) oxide water and oxygen. P1
- The reaction between copper metal and 50% nitric V acid produces colourless nitrogen II oxide gas which is further oxidized by atomospheric oxygen to brown fumes of nitrogen IV oxide.
- (a) Halogens
(b) They attract electrons into the outer energy level by its nuclear charge during reaction. The atomic radii increase down the group due to the successive additional energy level. Therefore the outer most energy level of iodine increasingly get further away from the nucleus than for bromine hence nuclear attraction for electrons grows weak for iodine than bromine.
(c) (i) NaOH(aq) and H2(g)
(ii) 2Cl–(aq) Cl2(g) + 2e
- (i) Temperature of 600C // heat.
(ii) NaClO3 : Cl2 = 1:3
Moles of NaClO3 = 1/3 x moles of Cl2
= 1/3 x 206,000
NaClO3 = 23 + 35.5 + 48 = 106.5g
Mass of NaClO3 = 2861 x 106.5 = 304.7kg
(e) (i) Manufacture of HCl acid.
(ii) Manufacture of Bleachants // or Bleaching agents.
(iii) Manufacture of plastics PVC.
- (a) (i) GRAPH OF COUNTS PER MINUTE AGAINST TIME FOR IODINE – 128
(iii) 132 + 1
(iv) 14 minutes + 1
(b) (i) Change in mass of isotopes. (1mk)
232 – 216 = 16
Number of He Particles = 16 = 4 He
2 4 2
4 – particles
Change in atomic number.
90 – (4 x 2) = 90 – 8 = 82 – 83 = -1
1 e => an Beta particle.
(ii) 232 216 4 0
TH Bi + 4 He e
90 83 2 -1
(c) (i) A – Beta (½ mks)
B – Gamma rays (½ mk)
C – Alpha (½mk)
(ii) I – Gamma rays (B) (½mk)
II – Beta particle (A) (½mk)
III – Alpha particle (C) (½mk)
- (a) ElectrolyticP1/Electrolysis
(b) Al2O3. 2H2OP1
(c) (i) Iron (III) oxideP1
Silicon (IV) oxideP1
(ii) NaOH is added to SiO2
(d) To reduce M.P of Al2O3 and to reduce on the cost.
- Oxygen produced at anode react with carbon to form CO2.
- It forms a protective layer of Al2O3 on its surface.
- (a) (i) A A2+(aq) + 2eP
B2+(aq) + 2e B(s)
(ii) A(s) + B2+(aq) A2+(aq) + B(s)
(iii) E red – E oxid
-0.13 – (-0.76) = + 0.63
(iv) Introduces K+ and NO3– replace the ions consumed at electrodes.
(v) Insoluble PbCl2 is formed.
(b) (i) Concentrated sodium chloride / brine.
(iii) – Soap
– bleaching agents
(iv) – Expensive
(v) 2Na/Hg + 2H2O 2NaOH + H2 + Hg
(vi) 24 litres 2 x 40g of NaOH
100C 100 x 2 x 40P
- (a) Cotton / silk / cellulose / sisal (any one) – 1mk)
(b) – Last longer / durable / stronger.
– Less easily attacked by acids and alkalis.
(c) (i) Hydrogen gas (1mk)
(ii) (a) CH3COOCH2CH3 – Ethylethanoate. (1mk)
(b) CH4 – Methane
Cl – C – C
(v) Tetrachloromethane / Chloroform