Physical Sciences Grade 12 Paper 2 Gauteng Mock 2015 Memo

Downloaded from hlayiso.com TENG PROVINCE GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATIONS 2015 MEMORANDUM SUBJECT: PHYSICAL SCIENCES P2 (10842)

Downloaded from hlayiso.com 10842/15 GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATIONS PHYSICAL SCIENCES (Second Paper) MEMORANDUM QUESTION 1 14 Cv (2) 1.2 BYY (2) 1.3 Dvv (2) 1.4 Cvv (2) 1.5 ByY (2) 1.6 AVY (2) 1.7 Avy (2) 1.8 AVY (2) 1.9 BYY (2) 1.10 Cry (2) [20]

1O842/15 QUESTION 2 2.7.4 Cr oD) 24.2 AY (1) 224 Bulanal (i) 2.2.2 2-methyl™ butan -2-o0l 2) OR 2-methyl-2-butanol 2.3.1 Waiter /H.0 “ qd) 2.3.2 | « Correct functional group “ e Methyl group on the second carbon e Whole siructure ” if HOG Notes: a Condensed or semi-structural formula: vA Aa) e Molecular formula: % (3)

2.3.3 2.4.1 2.4.2 Downloaded from hlayiso.com 10842/15 Addition / Hydration” (1) Process by which molecules of two monomers with different functional (2) groups~ react with the elimination/loss of water molecules. ¥ Fy (1) [13] QUESTION 3 3.1.1 Organic compounds having the same molecular formulaY with the same functional group situated at different positions. Yor same side chain at different positions OR Organic compounds with the same molecular formula, ¥ but different positions of the side chain / functional groups. “ an H—¢—C—C=C— H v H H (1) But -1-yne ¥ /1-butyne ¥ | H H (1) But-2-yne v /2-butyne

3.2.1 3.2.2 3.3.4 3.3.2 Downloaded from hlayiso.com 108425 ee ee HoH H HOH H H ia H-G—H | HoH H ReactantsY” Products (3) Condensed/semi-structural formulae: Max: ly Molecular formula; Of, Any additional reactant or products: Max: 24, Propene~ and methylpropane “ (2) The temperature Vat which the vapour pressure of a liquid is equal to the external atmospheric pressure “ OR The temperature Wat which a liquid changes phase into the gaseous phase. 1/2 (2) Propan -1-.ol has only one hydrogen bond “between its molecules. Thus, propan — 1 — ol has weaker intermolecular forces between its molecules. Vv Less energy is needed to overcome these intermolecular forces, “thus propan — 1 ~ ol has a lower boiling point. OR Propanoie acid has two hydrogen bonds between its molecules. The intermolecular forces between the propanoic acid molecules are thus stronger” ». more energy is needed to overcome these forces, “ thus propanoic acid has a higher boiling point. (3) [16]

Downloaded from hlayiso.com 10842/15 QUESTION 4 4.41 Q: Substitutionv /halogenation/chlorination (1) 4.1.2 R: Elimination/ dehydrohalogenation/ dehydrobrominationY (1) 4.2.1 e Functional group HH oH e Whole structure | | Vv correct H- r — H H H Accept -OH condensed in v structural formula (2) Condensed/semi — structural formulae Max: l/, Molecular formula of, 4.2.2 Propan — 2 — ol ¥ (1) 2-propanol 4.3.1 Hydrogen bromide v (1) v 4.3.2 H H H H \ lov | | a ep—e—H + H-Brv¥ —>y-Cc C—HY Hob W | | H H H e Functional group¥ e Whole structure correct” Accept HBr Ignore: = e Condensed/semi-structural formulae Max: 3/4 ¢ Molecular formula o%, Any additional reactant or products: | Everything correct, wrong balancing Max: 3/4 Max: 3/4 (4) [10]

Downloaded from hlayiso.com 108425 QUESTION 5 S11 Surace area/siate of division/reaction surface“ @ 5.1.2 There are more particles available for reaction.“ /more coll There are a greater number of elfective collisions” per un me “, reaction rate increases Q) isions, 5.1.3 Mass of 16% of original 2 g=0,3 gv Number of moles of CaCQs 0,3 g: A Po we M = 03 100” =0,003 mol. ¥ for both equations NicacOg: ANC? 1:2 0,003: 0,006“ ft Cret FT 0,2 = 9,006 Vv V =0,03 dm® ¥ (5) 5.2.1 Concentration” (1) 5.2.2 Equalio “ () 5.2.3 The same number of moles/mass/ number of particles of CaCO; ¥ (produce same number of moles of carbon dioxide). CaCOsis the limiting agent/Acid is in excess. (1)

Downloaded from hlayiso.com 10842/15 5.3.1 (Experiment) 6” (1) 5.3.2 3 Criteria for graph Mark 3 Correct shape for both graphs v 3 Graph of experiment 5 reaches 2 higher maximum/peak E value than experiment 6. v Ss Graph of experiment 6 shows more molecules of higher 5 kinetic energy v (3) g = xp 6 fy el € a3 2 kinetic energy [if no labels on axes max:2/3 | 5.4.1 exothermic’ (1) 5.4.2 The energy/potential energy of the products is lower than that of the reactants v OR (1) More energy released than absorbedY OR [18] Energy is released

10842/15 QUESTION 6 6.1 CALCULATIONS USING NUMBER OF MOLES: Mark allocation: 4 23 USING formula rx 30110 6,02.x10" @ e Change in n(NO) = Initial mol — equilibrium mol” e Using ratio NO(g) : Oo(g) : NO2(g) = 2: 1: 2% to determine change in n(©2) and change in n(NQ2) e Equilibrium [NO] = equilibriurn n(NO) /0,05 Equilibrium [Q2| = equilibrium n(Q,) /0,05 v Equilibrium [NOz] = equilibrium n(NO») /0,05 e® Correct K, expression (formulae in [ })~ * Subsiliution of concentrations and K, value into K, expression” 2 [0] at equilibrium” @ n(Oz) at equilibrium (x 0,05) v ® (Oz) initial = n at equilibrium + n change” GPTION 1 e 23 n(NOQ) = Sotto" = 5x 10° mol 6,02 x 10 NO) O2(g) NO2(g) Initial moles 1 0,5006”” Qo ; Change in moles 0,995v 0,4975 0,995 ¥ mole ratio Equilibrium roles 0,005 0,0031% 0,995 ¥ divide by 0,05 Equilibrium concentration | 0,1 0,0615 19,9 [No,F * [NoF To, | 644x105 = (19.9) (0,"0,] 10, ] = 0,0615 mol: dn? [9]

Downloaded from hlayiso.com 10842/15 6.2 Large K. means a high yield of products compared to reactantsY (1) 6.3 Darker” (1) 6.4 Temperature is decreased and the reaction will adjust to resist the change. The exothermicY reaction is favoured thus the forward reaction’ is favoured and the contents become darker.” (8) 6.5 No effect v a) [15] QUESTION 7 7A Brensted-Lowry base accepts a protony ¥ (2) 7.2.1 pH =-log[H30*]~ = -log(0,12)” = 0,92” (3) 7.22 Cty (1) 7.3.1 it can act as an acid or a base/ it can accept or donate an H* ionY (1) 7.3.2 HSO¢ + HO = H30* + so, v¥ reactants ¥ products ¥ balancing ignore > (3) 7.4 Drops of water will change the concentration of the acid/ make it more dilutew (1) 7.5.1 Option 1: Option 2: m m n=— = M “uv = 8,0 8.0 56Y =O 56 x 0,25 =0,14 mol V y n =0,57 mol-dm?v ory = 0,14 0,25¥ = 0,57 mol-dm’s ¥ (3)

~ Aa ~ Downloaded from hlayiso.com Cava Me % CyoVb Ay (dilute acid) G40) VW = Av (0,57)(25) 2 OR Gam 1x0,57 x 25% Caz 1x0,57 x 0,025” 2x40 v OR 2x 0,04 ¥v =0,178 moldm® “ = 0,178 mol-drn® 753 Concentrated acid th (moles concentrated) = nz (moles diluie) v cx Vj = Ce X Vo c; = 0,178 x (490+10) ¥ OR c= 0,178 x0,5 10 v 0,01 v = 8,9 moldm =“ = 8,9 mol-dm?/ QUESTION 8 8.1 Electrical energy to chemical energy” 8.2 Cathode At cathode, Ag’ / silver ions gain electrons” and are, reduced to Ag/ silver ¥ metal 8.3.1 Ag/silver “ a Ag’te=Ag (34) 83.2 Ag'+e — Agy¥ tee (0/ Ag > Agi+e (%5) Ag—Ag've (2) Ag=Ag™+e (94) 8.4.1 Electrode X becomes eroded/ smaller/ thinner.” 8.4.2 A (silver) layer forms on the medal.” 8.5 The rate at which Ag*/ silver ions are reduced at the cathode is equal to rate at which Ag metal/ silver is oxidised at the anode.v “ 10 10842N5 (4) (4) [22] (1) (1) (2) [14]

Downloaded from hlayiso.com 10842/15 QUESTION 9 9.1 Galvanicw cell (1) 9.2 There will no reading/ The reading will be zero/ OVY (1) 9.3 Temperature “ and the initial concentrations of the electrolytes Vv 9.4 9.5.1 9.5.2 9.6.1 9.6.2 9.7 But cannot give 25 °C and 1 mol-dm® because the investigation is not carried out under standard conditions because the emf values are not for standard conditions. (2) The different type of metal/the different haif-cells“” (1) Voltmeter’s terminals” have been connected incorrectly” OR Incorrect connection “ (+ to anode and — to cathode) ~ (2) Aluminium is a stronger reducing agent than zinc” and zinc is a stronger reducing agent than copper.” (2) Aluminium/ Atv (4) Zine/ Zn ¥ (1) 2AL (s) +3 Zn*(aq)¥—> = 2Af**(aq) + 3Zn(s)_ balancing ¥ ReactantsY Products” BalancingY Ignore = Marking rule 3.9: If only a reactant(s) followed by an arrow, or only a product(s) preceded by an arrow, is/are written, marks may be awarded for the reactant(s) or product(s). If only a reactant(s) or only a product(s) are written, without an arrow, no marks are awarded for the reactant(s) or product(s). Examples: Na + 3H2 v-> 2NH3” bal, ” No + He >Y yA — NH3”% % No + Ha % NHs % 4) 11

Downloaded from hlayiso.com 1OB42N5 QUESTION 10 10.1.1 lror oxide/iron/aluminium oxide“ ) 10.1.2 more collisions per unit time greater surlace aree/greater number of exposed particles. “ (2) 10.1.3 Increase the number of moles of Ne and He (reactants) favours the forward reaction” OR Increase pressure ~ and thus favours the forward reaction“ (2) 10.2.1 Calcium hydroxide increases the pH/alkalinity of the soil/ reduce the acidity of the soil.” Ammonium sulphate is a fertiliser. “ (2) 10.2.2 Ammonia gas is formed which will be released and reduce the concentration/number of moles/ of fertiliser.“ (1) 10.3 NO) =e 2” 230 mov M 24 MHNOs) = n(NO) = 30 mol” (3) m( HNO) = nM = (30)(63) = 1890 g ¥ C1] TOTAL: 150

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