Physical Sciences Grade 12 IEB Paper 2 2019 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2019 PHYSICAL SCIENCES: PAPER II MARKING GUIDELINES Time: 3 hours 200 marks These marking guidelines are prepared for use by examiners and sub- examiners, all of whom are required to attend a standardisation meeting to ensure that the guidelines are consistently interpreted and applied in the marking of candidates' scripts. The IEB will not enter into any discussions or correspondence about any marking guidelines. It is acknowledged that there may be different views about some matters of emphasis or detail in the guidelines. It is also recognised that, without the benefit of attendance at a standardisation meeting, there may be different interpretations of the application of the marking guidelines. IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 2 of 10 QUESTION 1 MULTIPLE CHOICE 1.1 A 1.2 A 1.3 D 1.4 C 1.5 D 1.6 B 1.7 A 1.8 D 1.9 C 1.10 A QUESTION 2 m 2.1 n= M n= ( 6, 5 ) ( 27 ) n = 0,24 mol 2.2 For Aℓ to be the limiting reagent, the amount of NaOH needed is: 6 n (NaOH needed) = (0,24) × 2 = 0,72 mol Amount of NaOH actually present: n = cV = (2,6)(0,4) =1,04 mol There is (more than) enough NaOH to fully react with the Aℓ OR Amount of NaOH present: n = cV = (2,6)(0,4) =1,04 mol IF NaOH were the limiting reagent, then the amount of Aℓ needed would be: 2 n(Al needed) = (1,04) × 6 = 0,347 mol There is not enough Aℓ to fully react with the NaOH OR Amount of NaOH present: n = cV = (2,6)(0,4) =1,04 mol Mole ratio of (Al) : n(NaOH) = 0,24 : 1,04 = 1 : 4,33 This is greater than the actual mole ratio of 1 : 3 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 3 of 10 OR Amount of NaOH present: n = cV = (2,6)(0,4) =1,04 mol 3 Maximum amount of H2 produced by Aℓ = 0,24 × 2 = 0,36 mol 3 Maximum amount of H2 produced by NaOH = 1,04 × 6 = 0,52 mol OR 3 Maximum amount of Na3AℓO3 produced by Aℓ = 0,24 × 2 = 0,36 mol 3 Maximum amount of Na3AℓO3 produced by NaOH = 1,04 × 6 = 0,52 mol NaOH would produce more product (if completely consumed), so must be in excess. 2 2.3 Theoretical n(Na3AℓO3) = (0,24) × 2 = 0,24 mol actual n(Na3AℓO3) = theoretical n(Na3AℓO3) × 92% 92 actual n(Na3AℓO3) = (0,24) × 100 = 0,2208 mol n (0,2208 ) c =V= (0,4 ) = 0,55 mol⋅dm–3 2.4 2.4.1 A reaction that transforms chemical potential energy into thermal energy. 2.4.2 The energy released is greater than the energy absorbed. 2.5 2.5.1 The sum of the kinetic energy (of the reacting particles) is more than the activation energy. 2.5.2 • A decrease in concentration means that there are less (solute) particles per unit volume • This results in fewer collisions (between reacting particles) per unit time • Thus, there are fewer effective (or successful) collisions per unit time • This results in a lower reaction rate 2.6 Remain the same IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 4 of 10 QUESTION 3 3.1 • As the hydrogen molecules approach each other (get close) • there is a dispersion of the electrons in each hydrogen molecule • resulting in induced dipoles forming • which are then able to attract each other OR • As a result of the random movement of electrons, a temporary (instantaneous) dipole is set up in one hydrogen molecule • dispersing the electrons in the other (neighbouring) hydrogen molecule • resulting in an induced dipole forming • the two temporary dipoles are then able to attract each other 3.2 3.2.1 80 s ΔV (5,9) 3.2.2 Rate = Δt = (80) = 7,38 × 10–2 dm3 ⋅s–1 3.2.3 Graph showing relationship between volume (of hydrogen gas) and time 6 5 Volume of H2 (dm3) 4 3 2 1 0 0 20 40 60 80 100 120 140 Time (s) 3.3 3S(s) + 6NaOH(aq)  Na2SO3(aq) + 2Na2S(s) + 3H2O(ℓ) IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 5 of 10 QUESTION 4 4.1 NO2 was removed (concentration of NO2 decreased) 4.2 • Stress: decrease in concentration of NO2 • Le Châtelier's principle predicts the system will respond to counteract the stress and so increase the concentration of NO2 • Thus, the forward reaction is (initially) favoured as it produces NO2 • resulting in a decrease in the amount of reactants (NO and O2) and an increase in the amounts of products (NO2) as seen in the graph 4.3 The same [NO2 ] 2 4.4 Kc = [NO ] [O2 ] 2 [NO2 ] 2 4.5 Kc = [NO ] [O2 ] 2 [NO2 ] 2 (256) = 2 3 5     3 3 [NO2 ] = 20, 6559 n ( NO= 2) = cV ( 20, 6559 )( 3=) 61,97mol 4.6 • Stress: increase in temperature • Le Châtelier's principle predicts the system will respond to counteract the stress and so decrease the temperature • Thus, the endothermic reaction is favoured as it consumes heat • From the graph, we see that there is an increase in the amount of reactants (NO and O2) and a decrease in the amount of products (NO2) • Thus, the reverse reaction has been favoured • Therefore, the reverse reaction is endothermic IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 6 of 10 4.7 Forward Reaction Rate Reverse 0 20 40 60 80 100 Time (min) 4.8 A bond that occurs between atoms within molecules. 4.9 Nonpolar (pure) covalent QUESTION 5 5.1 It dissociates completely. 5.2 A solution of known concentration. 5.3 n = cV n = (0,12)(0,4) n = 0,048 mol 5.4 Endothermic as the value of Kw is greater than 10−14. 5.5 Kw = [H3O +][OH –] (1,44 × 10–14) = [H3O +](0,24) [H3O+] = 6 × 10–14 mol⋅dm–3 5.6 Weak 5.7 ZF2 5.8 The reaction of an ion with water. 5.9 F − + H2O ⇌ HF + OH− 5.10 Alizarin yellow IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 7 of 10 QUESTION 6 6.1 Salt bridge 6.2 Any TWO of the following: Temperature of 25 °C 1 mol·dm−3 concentration of Sn2+ solution 1 mol·dm−3 concentration of Sn4+ solution 6.3 Any ONE of the following: It is inert It is solid It is conductive 6.4 The electrode where reduction takes place. 6.5 Pt 6.6 Sn4+ + 2e− → Sn2+ 6.7 A substance that donates electrons. 6.8 Zn 0 0 0 6.9 Ecell = Ecathode –Eanode 0 Ecell = (+0,15)–�–0,76� 0 Ecell = 0,91 V 6.10 Decreases 6.11 6.11.1 Electroplating 6.11.2 Cathode 6.11.3 • Ag+ is a much stronger oxidising agent than water • meaning Ag+ will be reduced predominantly 6.11.4 Silver 6.11.5 • Electrode X is a metal with positive kernels and delocalised electrons • These delocalised electrons are free to move and thus conduct electricity 6.11.6 Q = It Q = (3,2)(6 × 3600) Q = 69 120 C IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 8 of 10 Q 6.11.7 ne– = F ne– = ( 69120 ) ( 96500 ) ne– = 0,72 mol 6.11.8 • nAg = ne– = 0,72 mol • mAg = nM = (0,72)(108) = 77,76 g QUESTION 7 7.1 A concentrated aqueous solution of sodium chloride (NaCℓ). 7.2 Q 7.3 7.3.1 2Cℓ− → Cℓ2 + 2e− 7.3.2 • H2O has a more negative electrode potential than Cℓ− (OR H2O is a stronger reducing agent than Cℓ−) • Therefore, H2O would be more likely to be oxidised under standard conditions • However, the high concentration of Cℓ− ions increases the rate of its oxidation (OR makes its electrode potential more negative), making the oxidation of Cℓ− predominant 7.3.3 Oxygen / O2 7.4 • A membrane does not allow Cℓ− ions to pass through and so the NaOH solution will NOT be contaminated with Cℓ−/NaCℓ. OR • A diaphragm does allow Cℓ− ions to pass through and so the NaOH solution will be contaminated with Cℓ−/NaCℓ. 7.5 7.5.1 H2 7.5.2 Increase the surface area 7.6 • Both chorine and hydrogen have London forces only • Chlorine has more electrons • producing larger temporary dipoles • Therefore, chlorine has stronger London forces • resulting in more energy required to overcome the intermolecular forces and separate the particles in chlorine • thus, chlorine has a higher boiling point IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 9 of 10 QUESTION 8 8.1 A series of similar compounds that have the same functional group and have the same general formula, in which each member differs from the previous one by a single CH2 unit. 8.2 CnH2n+2 8.3 Alkenes 8.4 Saturated 8.5 H H C H H H H H H C C C C C H H H H H H C H H C H H 8.6 Double carbon-carbon bond 8.7 3−methylhexa−1,3−diene 8.8 Alkanes QUESTION 9 9.1 9.1.1 Substitution 9.1.2 Addition 9.2 9.2.1 Dehydrohalogenation 9.2.2 Hydration 9.3 Heptan−2−ol 9.4 9.4.1 C7H16 9.4.2 C3H6 IEB Copyright © 2019 PLEASE TURN OVER

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 10 of 10 9.5 Any one of the following: H H H H H H H H H H H H H H H C C C C C C C H H C C C C C C C H Br H H H H H H H Br H H H H H H H H H H H H H C C C C C C C H H H Br H H H H 9.6 This question is marked in relation to Q9.5. H H H H H H H If Q9.5 is 1−bromoheptane, then: H C C C C C C C H H H H H H If Q9.5 is 2−bromoheptane, then: H H H H H H H H H H H H H H H C C C C C C C H OR H C C C C C C C H H H H H H H H H H H H H H H H H H If Q9.5 is 3−bromoheptane, then: H C C C C C C C H H H H H H 9.7 9.7.1 C2H6O + 3O2 → 2CO2 + 3H2O 9.7.2 • The alcohol/carboxylic acid/ester is flammable • The water bath allows for gentle heating 9.7.3 O O H3C CH2 OH + H3C CH2 CH2 C OH → H3C CH2 CH2 C O CH2 CH3 + H2O OR CH3CH2OH + CH3CH2CH2COOH → CH3CH2CH2COOCH2CH3 + H2O OR CH3CH2OH + CH3(CH2)2COOH → CH3(CH2)2COOCH2CH3 + H2O 9.7.4 Ethyl butanoate 9.7.5 • Dehydrating agent • (Acid) catalyst Total: 200 marks IEB Copyright © 2019