Physical Sciences Grade 12 Paper 2 Gauteng Mock 2018 Memo

Downloaded from hlayiso.com PREPARATORY EXAMINATION 2018 MARKING GUIDELINES PHYSICAL SCIENCES: CHEMISTRY (PAPER 2) (10842) 10 pages

Downloaded from hlayiso.com 10842 / 18 GAUTENG DEPARTMENT OF EDUCATION PREPARATORY EXAMINATION PHYSICAL SCIENCES (Paper 2) MARKING GUIDELINES QUESTION 1 1 A  2 B  3 C  4 B  5 C  6 A  7 A  8 C  9 D  10 D  [20] QUESTION 2 2.1 2.1.1 Compounds which have the same molecular formula but different structural formula. (two  or zero) (2)   2.1.2 Methyl methanoate (2) 2.1.3 O H Marking guidelines: H O C C H  functional group correct  H  whole structure correct  (2) 2.1.4 Ethanoic acid (1) 2.1.5 Carboxylic acids (1) 2.2 2.2.1 CnH2n (1)  2.2.2 2,5-dimethylhept-3-ene (2) 2

Downloaded from hlayiso.com 10842 / 18 2.3 H Marking guidelines: H C H  OH on middle C  H H  complete structure correct  H C C C H H H O H (2) 2.4 Marking guidelines: H H H H  functional group H C O C C C H H H H O  H (1) [14] QUESTION 3 3.1 (Alcohols are flammable.) Do not bring it close to flames. (1) 3.2 3.2.1 H H H H OH H H C C C  OH H C C C H H H H H H H propan-1-ol  propan-2-ol  (4) 3.2.2 propan-1-ol (1)  of the –OH group in a chain will influence the boiling point, 3.2.3 The position because of the role of the intermolecular forces in branched structure. OR For a fair test, all alcohols need to be primary alcohols. (1) 3.3 3.3.1 The melting point of butan-1-ol will be higher.  Reasons:  The hydrogen bond between butan-1-ol molecules are stronger than between the butan-2-ol molecules, because the hydroxyl group is more exposed in the butan-1-ol than in the butan-2-ol, which causes the hydrogen bond to have a stronger influence.   More energy is required to weaken IM Forces between the butan-1-ol molecules, causing a higher melting point.   Straight chains have a higher melting point than branched chains. OR  Butan-1-ol has a larger surface area (chain length)  Therefore stronger IMF  More energy needed to overcome IMF Any TWO reasons (3) 3

Downloaded from hlayiso.com 10842 / 18 3.3.2 Butan-2-ol  Reasons:  If the hydroxyl group is on a terminal carbon atom, the intermolecular force (hydrogen bond) is stronger than when the –OH group is on the second carbon atom, where it is screened and has a smaller influence on the energy needed to cause a phase change (boil) in the second isomer, the butan-2-ol.   The boiling point of butan-2-ol will be lower and therefore the vapour pressure of butan-2-ol will be higher than the butan-1-ol. Less energy required to overcome the hydrogen bonds and to weaken the bonds.  Any TWO reasons (3) 3.3.3 Increase  (1) 3.3.4  Van der Waals forces (London forces) increase with an increase in molecular mass / chain length / size of molecule.   The longer the chain, the higher the boiling point, the stronger the London forces.  (2) 3.4 3.4.1 Polymerisation  (1) 3.4.2 H C CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH3 3 n Ethene.  (1) 3.4.3 Any ONE use: Packaging material  Squeeze bottles Electrical insulation Industrial protective clothing Toys, etc. (1) [19] QUESTION 4 4.1 4.1.1 Compound X = C 3H5Br  (1) 4.1.2 Addition / hydrohalogenation  (1) 4.1.3 Propan-2-ol  (1) 4.1.4 An alcohol; where the hydroxyl group is attached to a carbon atom which is bonded to two other carbon atoms.  (1) 4.1.5 Prop-1-ene OR propene  (1) 4.1.6 Dehydrohalogenation  (1) 4

Downloaded from hlayiso.com 10842 / 18 4.2 4.2.1 Elimination  (1) 4.2.2 Hydration  (1) 4.2.3 Hydrohalogenation  (1) 4.2.4 H2SO4  (1)  4.2.5 2-bromo-2-methylpentane  (2) [12] QUESTION 5 5.1 5.1.1 Use powdered CaCO3 instead of lumps  Use of a more concentrated HCℓ solution  Heat the reaction mixture  Add a suitable catalyst Any THREE ways (3) 5.1.2 Measure rate of CO2 formed by measuring the volume produced at certain time intervals.  Measure the rate at which the mass decreases, by placing the reaction container on a sensitive mass-meter and record the decrease in the mass per unit time.  (4) 5.1.3 It is not a closed system.  The CO2 gas escapes from the reaction.  (2) 5.2 5.2.1 5 cm3 / min OR 5 cm3 / min–1  (1) 5.2.2 Steeper gradient  (1) 5.2.3 The rate of production of hydrogen gas will be faster per unit time  because the reaction proceeds at a higher rate at a higher temperature.  (2) 5.2.4 The rate of production of hydrogen gas will increase as the length / surface area of the magnesium ribbon increases.  DO NOT ACCEPT: Graph is directly proportional; does not go through the origin. (2) [15] 5

Downloaded from hlayiso.com 10842 / 18 QUESTION 6 6.1 If a stress is applied to a system in equilibrium, the system will respond in such a way as to relieve the stress and restore the equilibrium under a new set of conditions.  (2) 6.2 6.2.1 A (1)  6.2.2 The forward reaction is endothermic. Decreasing the temperature from the equilibrium system favours the reverse reaction.  The reverse reaction is an exothermic reaction. Therefore more iodine and hydrogen molecules are formed and less HI will be in the reaction mixture.  (3) 6.3 6.3.1 Kc = [H3O+][CH3COO-]  = (1,34 x 10-3)( 1,34 x 10-3)  = 1,80 x 10-6  (3) 6.3.2 INCREASES.  (1) 6.3.3 When sodium ethanoate, CH3COONa, dissolves in aqueous ethanoic (acetic) acid, CH3COOH, it dissociates into Na + ions and acetate ions, CH3COO-.  The acetate ion increases the total acetate ion concentration in the solution.  The equilibrium re-establishes to the left and more ethanoic (acetic) acid forms,  decreasing the [H3O+] concentration, therefore the pH increases. (3) 6.4 6.4.1 0,5 mol of gas X2Y3  (1) 6.4.2 2X(g) 3Y(g) X2Y3(g) Initial no. of moles 4 4 0 No of moles formed 0 0 0,5  ratio No of moles used 1 1,5 0 used No of moles at equilibrium 3 2.5 0,5  Equilibrium concentration 1,5 1,25 0,25  ÷ 2 0,25 X 2 Y3  X2 Y3  Kc = 0,25 =  1,52 1,253 = 0,057  (6) [20] 6

Downloaded from hlayiso.com 10842 / 18 QUESTION 7 7.1 7.1.1 HSO 4  DO NOT ACCEPT: Hydrogen sulphate ion (1) 7.1.2 CN-  (1)   7.2 7.2.1 Na2CO3.10H2O + 2HCℓ  2NaCℓ + 11H2O +CO2 (2) 7.2.2 Methyl orange  (1)  / The equivalence point is in the pH 7.2.3 Strong acid is titrated with a weak base range (3 – 4,4) / Low pH / Acidic solution after titration end point. Any ONE answer will be correct (2) 7.2.4 Red to yellow.  (1) 7.2.5 Option 1: n cHCℓ =  V  nHCℓ = CV 24,8 Marking guidelines: = 0,1 x 1000  Calculate n(HCℓ)  = 2,48 x 10-3 mol HCℓ   Use formula n C=  1 mol Na2CO3 reacts with 2 mol HCℓ  V n (Na2CO3) in 500 cm3:  Use ratio, 1 : 2  2,48 x 10  x 500 3  Calculate n(Na2CO3)  =  Calculate m of Na2CO3  2 25 -3 = 2,48 x 10 mol Na2CO3  m n= M m = nM = 2,48 x 10-3 x 286 gmol-1 = 7,092 g  7

Downloaded from hlayiso.com 10842 / 18 Option 2: na Ca Va   nb Cb Vb 0,1 24.5  2  1000   1 Cb  25   1000  Cb = 0,0496 mol·dm-3  n C= V n = (0,0496)(0,5) = 0,0248 mol Na2CO3  m n= M m = nM = (0,0248) (286) = 7,093 g Na2CO3  (5) 7.2.6 Positive marking from question 7.2.6 % Na2CO3 in commercial washing soda: actual mass = x 100  theoretical mass 7,092 =  x 100 7,6 = 93,32 %  If answer of 7,093 g is used then the answer is 93,33 % (3) [16] 8

Downloaded from hlayiso.com 10842 / 18 QUESTION 8    8.1 8.1.1 Mg|Mg2+||Ag+|Ag (3) 8.1.2 Silver  (1) agent 8.1.3 E  = E  oxidising agent – E  reducing cell = 0,8 V –(-2,36 V)  = 3,16 V  (3) 8.1.4 25°C or 298 K  (1) 8.1.5 1 moldm-3  (1) 8.1.6 Option 1   The reaction rate is too low to supply enough charges to maintain a large enough current.  Option 2 P 6W I= = =2A V 3V The light bulb is manufactured to operate effectively if it is connected to a source of 3 V which delivers a current of 2 A. Although the potential difference is sufficiently large, the current is probably too small (due to a too large internal cell resistance). (3) [12] QUESTION 9 9.1 9.1.1 To the cathode  (1) 9.1.2 2H3O+ + 2e- → H2 + 2H2O  (2) 9.1.3 Chlorine (C2) gas  (1) 9.2 9.2.1 The covering of the surface or an object with a thin layer of a metal by means of electrolysis.  (2) 9.2.2 Any ONE use of electroplating.  Protection of the surface of a base metal.   Decorative purposes. (1) 9.2.3 Negative electrode / cathode.  (1) [8] 9

Downloaded from hlayiso.com 10842 / 18 QUESTION 10 10.1 10.1.1 Wheat  (1) 10.1.2 Ammonium nitrate / NH4NO3  (1) 10.1.3 Wheat needs a soil low in potassium, medium in nitrogen and high in phosphorous.  Nitrogen needs to be increased in the soil.  More nitrogen needs to be added and ammonium nitrate has the most nitrogen.  (3) 10.1.4 %nitrogen in NH NO = mass nitrogen 4 3 x 100  molar mass ammonium nitrate 28 = x 100 = 35%  80 (4) 10.1.5 A Nitrogen / N2  B Hydrogen / H2  C Nitrogen dioxide / NO2  D Ammonium nitrate / NH4NO3  E Ammonium sulphate / (NH4)2SO4  (5) [14] TOTAL: 150 10