100 x 100 = 55.6g ½ (1mk) 180

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1 TRANSNZOIA COUNTY JOINT EVALUATION TEST 2012 CHEMISTRY PAPER 2 233/2 MARKING SCHEME 1. a) A saturated solution is in that cannot dissolve anymore of solute at b) i) Scale 1mk Plotting 1mk Curve 1mk b) At 10ºC solubility = ( 80g /100g H 2 O) 80ºC solubility 136g / 100g H 2 O III i) If 180g solution contain 80g NaNO 3 and 100g of water Then 100g of H 2 O) = 100 x ½ = 44.4g NaNO 3 ½ ( ) = 55.6g H 2 O ½ OR 100 x 100 = 55.6g ½ (1mk) 180 i At 80ºC 148g NaNO 3 100g H 2 O? 55.6g 148 x 55.6 = g ½ (1mk) 100 iv) ( ) = of NaNO 3 v) Manufacture of NaCl and NaHCO from trona (1mk) Extraction of sugar 1

2 b) 2. a) b) i) i iv) v) vi) v i) At 10ºC solubility = ( ½ 80g / 100g of H2O) ( ½ ) 2mks) ZnCO3(s) + 2HCl(aq) ZnCl2(aq) + H2O(aq) + H2O(l) + CO2(g) To remove HCl fumes from the gas CO2. Some of the CO2(g) dissolves in water since is slightly soluble in water. NaOH(aq) + CO2(g) 2NaHCO3(aq) Heat NaHCO3 2NaHCO3(s) heat Na2CO3(s) + CO2(g) + H2O(g) NaHCO3 is less soluble at low temperature 1 mole NaOH (aq) 1000cm3 3? 30cm 30 cm3 x 1 mole = 0.03 moles 1000 cm3 1000cm3 1000g in 1000cm3-1mole ( 100g) HCl 50g x 1 mole = 0.05 moles 1000g 2

3 Mole ratio HCl NaOH I I Number of moles of HCl that reacted 0.05 moles moles = 0.02 moles i Base Acid X CO 3 HCl 1 2 ½ x moles 0.01 moles Moles = mass F.M 0.01 = 1g F.M = 1g F.m 0.01m = 100g iv) XCO 3 100g x = 40g 3. a) i) C 3 H 6 O Isomers i Q Esters R alkanoic acid / carboxylic acids iv) Sweet fruity smell b) i) Polymer i Making nylon ropes, nylon threads or cloths etc (any one for 1mk) c) A Alkynes B Alkanes C Alkenes Carbon dating in archeology - Detection of engine wear in an engine - Controlling thickness of paper and plastic 4. a) i) Q Bauxite (Al 2 O 3 ) R Aluminium hydroxide (AL(OH) 3 To dissolve the Al 2 O 3 ½ and precipitate the Fe2+ ions which are filtered out. i Heating to decompose the aluminium hydroxide to aluminium oxide. 2Al(OH) 3(s) heat Al 2 O 3(s) + 3H 2 O (g) iv) Cryolite lower the m.p of Al 2 O 3 (from 2015º down to 800ºC) b) i) Because oxygen produced at the anode oxidizes the carbon at the high temperature to form CO 2 / CO which are consumed. Na+ ions an more electropositive F- are more electronegative require more energy for reduction and oxidation respectively than Al 3+ and O 2- ions. i At anode 2O 2- (l) O 2(g) + 4e - At cathode Al 3+( aq) + 3e- Al (aq) ( penalize ½ for missing or wrong ss) 3

4 c) Aluminium has lower density hence light Aluminium is a better conductor of electricity d) Aluminium form an impervious layer less reactive oxide layer // Al 2 O 3 that coats the metal. 5. i) A (s) - 2e - Q 2- (aq) B 2+ (aq) + 2e- B (s) A (s) + B 2+ (aq) A 2- (aq) + B (s) i iv) E.m.f E R - E o Emf = ½ Per every mole of A 2+ (aq) going not solution, ½ moles of NO 3 leaves the salt bridge to balance the changes in cell B. ½ Where B 2+ is removed / discharged 2 moles ½ of K + enters A to balance the charges ½. v) Insoluble lead (II) chloride salt will be formed, hence no ions will be mobile hence no electrical conducting. b) i) Concentrated sodium chloride (Brine) i Titanium Manufacture of soaps / detergents - In beer industry - In paper industry (Any correct 2 uses) iv) - Mercury is expensive - Mercury is poisonous hence should not be left to get into the water system v) 2Na (s) + Hg (l) + 2H 2 O (l) 2NaOH (aq) + 2Hg (l) + H 2(g) b) NaOH : H 2 see equation in (v) 2moles 1 mole 100 litres 4 x 1667 x moles H2(g) 24 litres 1 moles H2(g) 100 Litres = moles 100 litres x 1 mole = moles 24 litres x 40g = g ( NaOH) 6. a) Alkali metal / alkali elements b) I G is more reactive than J Nuclear attraction in G weaker than in J electrons attracted more in J hence not easily lost II P is more reactive than N N is a smaller atom hence attracts incoming electrons more than N c) Atomic radius of K is larger than L. L has higher number of protons hence stronger attraction for the electrons in the energy levels reducing the size of atom. d) See grid e) Melting point in group 1 and 2 decreases down the group due to decrease in the strength of metallic bond with increase in atomic radius f0 KP 3 4

5 g) Ionic bond F is a metal and O is a halogen ( a non metal) F loses a valence electron, O gains the lost electron to form ionic bond. h) - In light bulb ( to create inert atmosphere - Advertisement light / street lights I H M i) To reduce /minimize / limit heat loss to the surrounding A marks the start of the experiment B Marks the end of the reaction // point of complete neutration i H 2 SO 4 : NaOH 1 mole 2 mole ½ x 1mole = 0.05 moles 1000 = mole In 1000cm 3 H 2 SO 4-1mole? mole moles x 1000 = 25cm 3 1 b) Catalyst has no effect on the position of the equilibrium 5