Physical Sciences Grade 12 Paper 2 Gauteng Mock 2019 Memo

Downloaded from hlayiso.com PREPARATORY EXAMINATION 2019 MARKING GUIDELINES PHYSICAL SCIENCES: CHEMISTRY (PAPER 2) (10842) 13 pages 1

Downloaded from hlayiso.com 10842 / 19 QUESTION 1 1.1 A ✓✓ 1.2 D ✓✓ 1.3 D ✓✓ 1.4 B ✓✓ 1.5 D ✓✓ 1.6 C ✓✓ 1.7 D ✓✓ 1.8 A ✓✓ 1.9 C ✓✓ 1.10 C ✓✓ [20] QUESTION 2 2.1 2.1.1 A✓ (1) 2.1.2 A✓ (1) 2.1.3 C✓ (1) 2.1.4 B✓ (1) 2.1.5 D✓ (1) Marking guidelines: 2.2 H H HC H H C H • Correct ketone functional group ✓ H H H • Both side chains / branches correct on H C C C C C C H correct carbon atom ✓ H H O H H H • Whole structure correct ✓ 2,4-dimethylhexan-3-one (3) ✓ ✓ ✓ 2.3 2,3-dichloro-3-fluorobutanal (3) [11] 2

Downloaded from hlayiso.com 10842 / 19 QUESTION 3 3.1 3.1.1 Esters ✓ OR carboxylic acids (1) 3.1.2 Ketones ✓ (1) 3.2 3.2.1 Chain isomers: Same molecular formula, but different types of chains ✓✓ (2) 3.2.2 H H H ✓ H H H ✓ H C C C C H H C C C C H H H H H H H but-1-ene but-2-ene ✓ ✓ OR 1-butene OR 2-butene H H C H Marking guideline: ✓H One mark for whole structure H C C C H One mark for correct IUPAC name for each of the isomers H H 2-methylprop-1-ene ✓ OR methylpropene OR 2-methyl-1-propene (6) 3.3 3.3.1 Butan-1-ol ✓ and pentanoic acid ✓ (2) 3.3.2 Condensation ✓ OR esterification (1) 3.4 If answered as: Butyl pentanoate has a If answered as: Butyl butanoate has a higher boiling point than butyl butanoate. lower boiling point than butyl pentanoate. Both esters have the same type of Both esters have the same type of intermolecular forces (London forces). ✓ intermolecular forces (London forces). Butyl pentanoate has a longer chain Butyl butanoate has a shorter chain therefore stronger forces between the therefore weaker intermolecular forces molecules. ✓ between the molecules. More energy is required to overcome the Less energy is required to overcome the intermolecular forces between butyl intermolecular forces between butyl pentanoate ✓ therefore the boiling point butanoate. is higher. Marking criteria • same intermolecular forces must be mentioned. Their names need not be mentioned. ✓ • comparison chain length / molecular mass ✓ • energy required ✓ (3) 3

Downloaded from hlayiso.com 10842 / 19 3.5.1 Addition polymerisation: A reaction in which small molecules join to form very large molecules by adding on at double bonds. ✓✓ Condensation polymerisation: Molecules of two monomers with different functional groups undergo condensation reactions with the loss of small molecules, usually water. ✓✓ (4) 3.6.1 Polythene or Poliethene✓ (1) 3.6.2 Ethene ✓ (1) [22] QUESTION 4 4.1 Elimination / dehydrohalogenation / dehydrobromination ✓ (1) 4.2 • heat ✓ ✓ • Concentrated sodium hydroxide (NaOH) / Concentrated potassium hydroxide (KOH) / Concentrated strong base OR Hot ethanolic concentrated sodium hydroxide / potassium hydroxide / KOH / NaOH (2) 4.3 H Marking guidelines: H H H H C H • Correct hydroxyl functional group on C-2 ✓ H C C C C O H • Side chains / branch correct on correct carbon atom ✓ H H H H C H • Whole structure correct ✓ H • If condensed structural formula given -1 (3) 4.4 H2O / water ✓ (1) 4.5 Addition / Hydration ✓ (1) [8] 4

Downloaded from hlayiso.com 10842 / 19 QUESTION 5 5.1 To prevent loss of any solution / acid from the flask ✓ OR To allow gas to escape OR To prevent any solids / liquids getting in / out OR To prevent spurting NOTE: Answer must not be given in terms of preventing evaporation or condensation. (1) 5.2 change in mass 164,84 − 165,00 ✓ Reaction rate = = = –0,016 g·s-1 ✓ change in time 10,0 0 ✓ Accept positive sign or no sign in the answer. (Negative value indicates loss in mass during reaction.) (3) 5.3.1 MARKING CRITERIA: • Plotting of all points correctly ✓✓ (Allow 1 plotting error.) • Draw best fit line through the plotted points ✓ • Dotted line represents new line. (3) 5

Downloaded from hlayiso.com 10842 / 19 5.3.3 The reaction will be faster. ✓ Increasing the temperature, increases the kinetic energy so that more particles have kinetic energy that is higher than the activation energy. ✓ More collisions with correct orientation ✓ More successful or effective collisions per unit time take place. ✓ (4) 5.4.1 2✓ (1) 5.4.2 State of division changed – Size of the zinc particles increased / surface area increased – originally lumps were used. ✓ Exp 1 powder and exp 2 lumps Accept: removed catalyst. (1) 5.4.3 0,5 g / half of the amount as in experiment 1. ✓ (1) [14] 6

Downloaded from hlayiso.com 10842 / 19 QUESTION 6 6.1 A catalyst was used in the experiment that produced the graph for experiment C. ✓ (1) 6.2 The concentration of the products in both graphs at equilibrium are the same and the concentration of the reactants at equilibrium are the same, ✓ but equilibrium was reached faster in Experiment C than in Experiment A. ✓ OR Equilibrium concentration is the same, equilibrium was not disturbed. OR Accept: A catalyst was added and the rate of both forward and reverse reactions were increased. (2) 6.3 EXOTHERMIC ✓ (1) 6.4 – negative marking from QUESTION 6.3 In experiment C the concentration of the reactants is higher than the concentration of the reactants in Experiment A when equilibrium is established. ✓  The reverse reaction is favoured, ✓ Endothermic reaction is favoured when the temperature is increased. ✓ Therefore reverse reaction is endothermic OR the forward reaction is exothermic. (3) 6.5 Use the data given to calculate the equilibrium constant at 500 o C. H2(g) I2(g) 2HI(g) Initial mol 0,5 0,5 0 Change in mol -0,2 -0,2 0,4 ✓ Mol at equilibrium 0,3 ✓ 0,3 0,4 ratio used ✓ Concentration at 0,3 0,3 0,4 equilibrium MARKING CRITERIA: Kc = HI2 ✓ • Change in mol of HI ✓ H 2 I 2  • Using ratio 1:1:2 ✓ (0,4)2 ✓ (positive marking) • Addition and subtraction to get the correct = (0,3)(0,3) mol at equilibrium / concentration at equilibrium ✓ = 1,78 ✓ • Correct Kc expression ✓ • Correct substitution of concentration at equilibrium values from this table✓ • Correct final answer ✓ No Kc expression, correct substitution Max ⅚ Wrong Kc expression Max 3/6 (6) 7

Downloaded from hlayiso.com 10842 / 19 OPTION 2 Concentration of I2 at equilibrium is 0,3 mol.dm-3. n n c (I2 ) initial = c (H2 ) initial = V V 0,5 0,5 = and = 1 1 = 0,5 mol  dm -3 = 0,5 mol  dm-3 H2(g) I2(g) 2HI(g) Initial 0,5 0,5 0 concentration Change in -0,2 -0,2 0,4 ✓ concentration concentration at 0,3 ✓ 0,3 0,4 ratio used ✓ equilibrium Kc = HI2 ✓ H 2 I 2  = (0,4)2 ✓ (positive marking) (0,3)(0,3) = 1,78 ✓ No Kc expression, correct substitution Max ⅚ Wrong Kc expression Max 3/6 (6) 6.6 Lower than ✓ positive marking from QUESTION 6.3. answer would then be: higher than (1) 6.7 + Kc = HI2 H 2 I 2  = (0,38) 2 (re-calculate equilibrium concentrations) Note: may use calculation (0,3)(0,3) or use explanation. = 1,6 The [product] is smaller and the [reactant] is higher at the higher temperature. ✓ This results in a lower Kc value at a higher temperature. ✓ Or Equilibrium is to the left and reverse reaction is favoured. (2) [16] 8

Downloaded from hlayiso.com 10842 / 19 QUESTION 7 7.1 Temperature = 25o C / 298 K ✓ Pressure = 101,3 kPa / 1,013 x 10 5 / 1 atm) ✓ [Cl-] = 1 moldm-3 ✓ (3) 7.2 7.2.1 Mn → Mn2+ + 2e- ✓✓ (2) Marking criteria: Mn  Mn2 + + 2e- (2 2) Mn  Mn2+ + 2e - ⇌ Mn (12) Mn2 + + 2e −  Mn (0 2) Mn ⇌ Mn2+ + 2e − (0 2) Ignore if charge is omitted on electron. If charge (+) is omitted on Mn2+ Max. ½ 7.2.2 Mn + Cℓ2 → 2Cℓ- + Mn2+ Marking criteria: ✓ Reactants ✓ Products ✓ Balancing (3) ✓ ✓ ✓ 7.3 Mn(s) / Mn 2+(aq)(1 moldm-3) // Cℓ2(g)(1 atm) / Cℓ-(aq)(1 moldm-3) / Pt NOTE: Do not penalise if phases and conditions are not included. (3) 7.4 MnCℓ2 / Mn(NO3)2 / MnSO4✓ (1) 7.5 Chlorine gas ✓ OR Cℓ2 (1) 7.6 E θcell = E reduction θ − E θoxidation ✓ Note: = +1,36 – (-1,18) ✓ Accept any other correct formula from the = +2,54 V ✓ data sheet. Any other formula using unconventional abbreviations, e.g. Ecell = EOA – ERA Followed by correct substitutions: Max 2 3 (3) [16] 9

Downloaded from hlayiso.com 10842 / 19 QUESTION 8 8.1 8.1.1 P✓ (1) 8.1.2 P is connected to the positive terminal of the battery. ✓ (1) 8.2 8.2.1 Ni2+(aq) ✓ Note: (aq) may be omitted. (1) 8.2.2 Q✓ (1) 8.2.3 Q ✓ (1) 8.2.4 Ni2+ + 2e- → Ni ✓✓ Marking criteria: Ni  Ni2+ + 2e- (2 2) Ni2+ + 2e - ⇌ Ni (12) Ni2+ + 2e −  Ni (0 2) Ni ⇌ Ni2+ + 2e − (0 2) Ignore if charge is omitted on electron. If charge (+) is omitted on Ni2+ Max. ½ (2) 8.2.5 Remains the same ✓ (1) 8.2.6 For each mole or atom of nickel oxidised at the anode, a mole or atom of nickel is reduced at the cathode. ✓✓ Or Rate of oxidation equals the rate of reduction. (2) [10] 10

Downloaded from hlayiso.com 10842 / 19 QUESTION 9 9.1.1 A substance that can act as either an acid or a base ✓✓ (2) 9.1.2 H 2O ✓ (1) 9.1.3 n = cV ✓ = (0,05)(0,036) ✓ = 1,8 x 10-3 mol H2SO4 ✓ 1,8 x 10-3 mol H2SO4 neutralised 1,8 x 10-3 mol Na2CO3 in 25 cm3 + positive marking from QUESTION 9.1.3 (3) 9.1.4 m = nM ✓ Marking criteria: = 1,8 x 10-3 ✓ x 106 ✓ ✓ formula = 0,1908 g ✓ ✓ ratio of mol ✓ 106 ✓ correct answer Also accept: nb c V = b b na c a Va 1 c b 25 = 1 0,05 x 36 c b = 0,072 mol  dm-3 m ✓ c= MV 0,072 == mm ✓ ✓ 0,072 106x x0,025 0,25 106 m = 1,908 g m = 0,1908 g ✓ (4) 9.1.5 Positive marking from Question 9.1.2 Marking criteria: % Na2CO3 = Actual mass x 100 ✓ correct calculation of mass Original mass ✓ substitution of 5,13 = (10)(0,0018)(106) = 1,908 g Na2CO3 ✓ ✓ correct answer 1,908 = ✓ x 100 5,13 = 37,19 % ✓ Or 25 cm3 has 0,1908 g✓ 250 cm3 has 10✓ x 0,1908 g = 1,908 g ✓ 1,908 % Na2CO3 = x 100 ✓ = 37,19 % ✓ 5,13 (3) 11

Downloaded from hlayiso.com 10842 / 19 9.2 CO32- + H2O ⇌ HCO-3 + OH- ✓reactants; ✓products OR CO2- + 2H O ⇌ H CO + 2OH- 3 2 2 3 (2) [15] QUESTION 10 10.1 10.1.1 Total percentage mass fertiliser in the bag ✓ (1) 10.1.2 Nitrogen, ✓ phosphorous ✓ OR potassium (Any two) (2) 10.1.3 2✓ %N= x 40 ✓ 11 = 7,27 % ✓ (3) 10.2 10.2.1 Ammonium nitrate or NH4NO3 ✓ (1) 10.2.2 Haber process (1) 10.2.3 (NH4)2SO4 ✓ (1) 10.2.4 Ostwald process ✓ OR catalytic oxidation of ammonia (1) 10.2.5 So that plants can absorb them from the soil ✓ (1) 10.2.6 Eutrophication ✓ Any given examples that apply to rivers and dams. Note: do not accept red tide. (1) 10.2.7 Nitrogen ✓ and hydrogen ✓ (2) 10.2.8 Sulphur dioxide ✓ (1) 10.2.9 Enhance growth of crops / plants to produce more food for humans ✓ Production of fertiliser results in job creation. ✓ Selling of fertilisers stimulates the economy. ✓ (Any relevant positive impact) (3) [18] TOTAL: 150 12