Physical Sciences Grade 12 Limpopo Prelim 2024 Memo

Downloaded from hlayiso.com NATIONAL SENIOR SERTIFICATE GRADE 12 PHYSICAL SCIENCES: CHEMISTRY (P2) MARKING GUIDELINES SEPTEMBER 2024 This marking guideline consists of a cover page and 10 pages. TOTAL: 150 TIME: 3 HOURS

Downloaded from hlayiso.com GRADE 12 TRIAL EXAMINATION – LIMPOPO 2024 SEPTEMBER PAPER 2: CHEMISTRY MARKS: 150 MARKING GUIDELINES QUESTION 1 1.1 A  (2) 1.2 C  (2) 1.3 D  (2) 1.4 B  (2) 1.5 B  (2) 1.6 C  (2) 1.7 A  (2) 1.8 A  (2) 1.9 D  (2) 1.10 A  (2) [20] QUESTION 2 2.1 Carbon has four valence electrons, so it can achieve a full outer (2) energy level by forming four covalent bonds (this property is known as tetravalency). Carbon can form single, double, or triple covalent bonds with other carbon atoms to produce long chain or ring structures. 2.2.1 CnH2n+2  (1) 2.2.2 (3) Marking criteria • Two methyl substituents • Chlorine substituent • Whole structure correct • • Chloor substituent  • Whole structure correct: 3⁄ 2 3

Downloaded from hlayiso.com 2.2.3 SATURATED  (1) 2.3.1 Compounds with the same molecular formula,  but different positions of the side chain/substituents/functional groups  on the parent chain. (2) 2.3.2 carbonyl  (1) 2.3.3 pentan-3-one / 3-pentanone (2) Marking criteria • Functional group and correct position i.e., 3  • Correct IUPAC name   Accept for ONE mark/Aanvaar vir EEN 2.3.4 Marking criteria • Correct functional group • Whole structure correct OR: Any correct structure of an aldehyde with five carbon atoms. OR OR 3

Downloaded from hlayiso.com (2) 2.4.1 TERTIARY  The C atom bonded to the functional group/hydroxyl (group)/-OH is (2) bonded to three other C atoms. /The C-atom bonded to the hydroxyl (group) has no hydrogen atoms.  2.4.2 2-methylpropan-2-ol / 2-methyl-2-propanol (2) Marking criteria • 2-methyl  • propan-2-ol / 2-propanol  • Any errors e.g., hyphens omitted and/or incorrect sequence: Max: 1⁄2 [18] QUESTION 3 3.1 Relative molecular mass/molar mass/Mr  (1) 3.2.1 Boiling point  (1) 3.2.2 Type of organic compound/homologous series/functional group (1) (alkane,alcohol,carboxylic acid) 3.3.1 LOWER THAN  (1) – 3.3.2 Negative marking from question 3.3.1 • 2-methylpropane is a spherical molecule, therefore smaller surface area  is presented to other molecules (sphere is 3D shape with lowest surface-to-volume ratio). • Less surface area at which van der Waals interactions with other molecules can occur  /fewer van der Waals interactions/fewer cohesive interactions. • Less energy required  to overcome cohesive forces as liquid hydrocarbon is vaporised. (3) 3.4 The pressure exerted by a vapour at equilibrium with its liquid in a closed system. (2 or 0) (2) 4

Downloaded from hlayiso.com 3.5 Ethanoic acid (C) – It is possible for 2 H-bonds (at C=O and O-H) to form between adjacent carboxylic acid molecules. Therefore H-bonds stronger  in carboxylic acid. Therefore, more energy required to overcome the stronger forces of attraction/to overcome the IMF BETWEEN molecules. OR propan-1-ol (B) – H-bonds weaker in alcohol  as it is only possible for 1 H bond at O-H to form between adjacent molecules of alcohol therefore less energy required  to overcome the strong forces of attraction/to overcome the IMF BETWEEN molecules. (3) [12] QUESTION 4 4.1.1 Substitution  / halogenation (1) 4.1.2 Substitution  (1) 4.1.3 Substitution  (1) 4.1.4 Substitution / dehydrohalogenation (1) 4.2 CH3CH2CH2Br  + H2O  → CH3CH2CH2OH  + HBr  (4) 4.3 (2) 3-C atoms in chain  Whole structure correct 4.4 H 2O  CO2  (2) 4.5 H H H H H (3) H H C C C Br + KOH(gek.) → H C C C + H2O + KBr H H H H H Marking criteria • Correct structural formula of 1-bromopropane  • Correct structural formula of prop-1-ene  • H2O + KBr  5

Downloaded from hlayiso.com 4.6 Propan-1-ol / 1-propanol Marking criteria • Functional group and correct position, ie.1.  • Propanol  (2) 4.7 Esters  (1) 4.8 propylethanoate (2) [20] 6

Downloaded from hlayiso.com QUESTION 5 5.1 Activation energy  (1) 5.2.1 C  (1) – 5.2.2 Negative marking (1) Total area under graphs A and C approximately the same, but more molecules with greater average Ek therefore area X of graph C greater – more molecules with greater average Ek. 5.2.3 • Increased temperature therefore increased average kinetic energy of molecules. • More of these molecules will have sufficient kinetic energy  to react./ Ek≥activation energy • More effective collisions per second /unit time. • Increased rate of reaction.  (4) 5.3.1 A catalyst will speed up a chemical reaction  (2) without itself undergoing permanent change.  5.3.2 INCREASES  (1) 5.3.3 A catalyst will create an alternative path of lower activation energy  “line Y” moves to the left  – therefore area X increases in size. (2) [12] QUESTION 6 6.1 The change in concentration of reactants or products  (2) per unit time.  or any other acceptable definition(see Guideline) 6.2 It starts at zero in t0  and increases to time t1 (2) 6.3 Equilibrium  is reached. (1) 6.4 m t Criteria for graph Axes labelled correctly.  Gradient of graph initially high and then decreases with time.  7

Downloaded from hlayiso.com Graph ends parallel to x-axis to represent equilibrium.  (3) 6.5.1 Decreases  (1) 6.5.2 Decreases  (1) 6.5.3 Decreases  (1) 6.6 At lower T, the exothermic (forward) reaction is favoured.  Therefore, the reverse reaction is favoured,  reducing the concentration of the gas (products). If concentration of products decrease then Kc will also decrease (4) 6.7 𝐾𝑐 = [𝐶𝑂2 ] = 1,4 × 10−2  (2) [17] QUESTION 7 7.1.1 An acid is a proton (H+-ion) -donor.  (2) 7.1.2 It ionises to form 2 protons/2 mol H+-ions.  (1) OR It donates 2H+-ions per H2SO4 -molecule. 7.2.1 Amphiprotic (substance)/Ampholyte  (1) 7.2.2 H2CO3 (aq)  (1) 7.3.1 𝑛(𝐻2 𝐶𝑂3 ) = 𝑚  𝑀 27 = 84  = 0,32 𝑚𝑜𝑙 (0,3214285714 𝑚𝑜𝑙) 1 𝑛(𝐻2 𝑆𝑂4 ) = 2 𝑛(𝑁𝑎𝐻𝐶𝑂3 ) 1 = 2 (0,32)  = 0,16 𝑚𝑜𝑙 (0,1607142857 𝑚𝑜𝑙) 𝑛 𝑐=𝑉 0,16 6=  ( 6 and 0,16) 𝑉 (6) ∴ 𝑉 = 0,03 𝑑𝑚3  (30 𝑐𝑚3 / 0,027 𝑑𝑚3 / 27 𝑐𝑚3 ) 7.3.2 𝑛𝑎 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙) = 𝑛𝑎 (𝑓𝑖𝑛𝑎𝑙) 𝑐𝑎 𝑣𝑎 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙) = 𝑐𝑎 𝑣𝑎 (𝑓𝑖𝑛𝑎𝑙) 8

Downloaded from hlayiso.com ∴ (6)𝑣𝑎 = (0,1)(1) ∴ 𝑣𝑎 = 0,02 𝑑𝑚3 (20 𝑐𝑚3 / 0,0167 𝑑𝑚3 / 16,7 𝑐𝑚3 ) (2) 7.3.3 Shows end point (of titration).  /Shows when neutralisation occurs. (1) 7.3.4 Marking criteria: • Substitute initial [acid] and volume • Substitute initial [base] and volume • Use ratio 1 : 2 • Initial mole acid – mole acid reacted • Substitute volume acid + volume base • pH formula • Substitute 2 x ca in pH formula • Final answer: 1,44 𝑛𝑎 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙) = 𝑐𝑎 𝑣𝑎 = (0,1)(25 × 10−3)  = 2,5 × 10−3 𝑚𝑜𝑙 𝑛𝑏 (𝑟𝑒𝑎𝑐𝑡𝑒𝑑) = 𝑐𝑏 𝑣𝑏 = (0,1)(30 × 10−3 )  = 3 × 10−3 𝑚𝑜𝑙 𝑛𝑎 1 = 𝑛𝑏 2 1 1 ∴ 𝑛𝑎 (𝑛𝑒𝑢𝑡𝑟𝑎𝑙𝑖𝑠𝑒𝑑) = 𝑛𝑏 = (3 × 10−3 ) = 1,5 × 10−3 𝑚𝑜𝑙 2 2 𝑛𝑎 (𝑙𝑒𝑓𝑡) = 𝑛𝑎 (𝑖𝑛𝑖𝑡𝑖𝑎𝑙) − 𝑛𝑎 (𝑛𝑒𝑢𝑡𝑟𝑎𝑙𝑖𝑠𝑒𝑑) = 2,5 × 10−3 − 1,5 × 10−3  = 1 × 10−3 𝑚𝑜𝑙 𝑛 𝑐𝑎 = 𝑉 1 × 10−3 = (25 × 10−3 + 30 × 10−3 ) = 0,018 𝑚𝑜𝑙. 𝑑𝑚−3 𝑝𝐻 = −𝑙𝑜𝑔[𝐻3 𝑂 + ]  = −log [2(0,018)] = 1,44  (8) [22] 9

Downloaded from hlayiso.com QUESTION 8 8.1 Pb  (lead) (1) 8.2 Zn → Zn2+ + 2e-  (2 or 0) (2) 8.3 𝜃 𝜃 𝜃 𝐸𝑐𝑒𝑙𝑙 = 𝐸𝑜𝑥𝑖𝑑𝑖𝑠𝑖𝑛𝑔 𝑎𝑔𝑒𝑛𝑡 − 𝐸𝑟𝑒𝑑𝑢𝑐𝑖𝑛𝑔 𝑎𝑔𝑒𝑛𝑡  = −0,13  − (−0,76)  = 0,63 (𝑉)  (4) 8.4 • Temperature  (2) • (initial) concentration of the electrolytes.  8.5.1 EQUAL TO  (1) 8.5.2 Area/size of electrodes has no effect on the emf of a cell.  / It is (2) still a standard cell. 8.6.1 The cell has internal resistance.  (1) 8.6.2 The emf decreases as the concentration of Pb2+(aq) decreases.  / The cell is running flat as the electrolyte concentration in the Pb cell decreases. (1) [14] QUESTION 9 9.1 A substance of which the aqueous solution contains ions.  / (2) A substance that dissolves in water to give a solution that conducts electricity. 9.2 Endothermic  (1) 9.3 Bracelet  (1) 9.4.1 REMAINS THE SAME  – (1) 9.4.2 The rate of oxidation of copper at the anode is equal  (2) to the rate of reduction of copper (II) ions at the cathode.  9.4.3 Cu → Cu2+ + 2e-  (2) Notes: 𝐶𝑢 ⇌ 𝐶𝑢 2+ + 2𝑒 − (1⁄2) 𝐶𝑢 2+ + 2𝑒 − → 𝐶𝑢 (0⁄2) 𝐶𝑢 2+ + 2𝑒 − ← 𝐶𝑢 (2⁄2) 𝐶𝑢 2+ + 2𝑒 − ⇋ 𝐶𝑢 (0⁄2) 10

Downloaded from hlayiso.com 9.5 𝑚 𝑛(𝐶𝑢) = 𝑀 0,86 =  63,5 = 0,0135 𝑚𝑜𝑙  (0,01354330709 𝑚𝑜𝑙) 𝑛(𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠) = 2𝑛(𝐶𝑢) = 2(0,0135) = 0,027 𝑚𝑜𝑙 𝑁 𝑛=  𝑁𝐴 𝑁 (6) 0,027 = 6,02 × 1023  ∴ 𝑁 = 1,63 × 1022  𝑒𝑙𝑒𝑐𝑡𝑟𝑜𝑛𝑠 [15] TOTAL: 150 11