Physical Sciences Grade 12 Paper 2 Mpumalanga Mock 2018 Memo

Downloaded from hlayiso.com NATIONAL SENIOR CERTIFICATE GRADE 12 PHYSICAL SCIENCES: CHEMISTRY (P2) MARKING GUIDELINES SEPTEMBER 2018 MARKS: 150 TIME: 3 hours This marking guidelines consists of 9 pages

Downloaded from hlayiso.com 2 Physical Sciences P2 Memorandum NSC MDE/September 2018 QUESTION 1 1.1 B  1.2 D  1.3 C  1.4 D  1.5 D  1.6 C  1.7 C  1.8 C  1.9 A  1.10 B  [20] QUESTION 2 2.1 A series of organic compounds that can be described by the same general formula OR in which one member differs from the next with a CH2 group.  ‘n Reeks organiese verindings met dieselfde algmene formule OF waarvan die lede verskilmet ‘n CH2 groep. (2) 2.2.1 U (1) 2.2.2 S (1) 2.2.3 P (1) 2.2.4 Q and S (2)   2.3.1 3,4-dibromo-2,2-dimethylpentane / 3,4-dibromo-2,2-dimetielpentaan (3)  2.3.2 Heksan-3-one / Heksan-3-oon (2) 2.4.1 1 mark for functional group 1 mark for whole structure  (2)

Downloaded from hlayiso.com 3 Physical Sciences P2 Memorandum NSC MDE/September 2018 2.4.2 1 mark for functional group 1 mark for whole structure  (2)  [16] QUESTION 3 3.1 The temperature where the vapour pressure is equal to the atmospheric pressure. Die temperatuur waar die dampdruk gelyk is aan atmosferiese druk. (2) 3.2.1 Homologous series/ Functional group / Homoloë reeks/ Funksionele groep (1) 3.2.2 The number of C-atoms/ The length of the carbon chain  Die aantal C-atome / Die lengte van die C-ketting (1) 3.3 A (1) 3.4.1 London forces/ Londonkragte (1) 3.4.2 Dipole-dipole forces  / Dipool-dipoolkragte (1) 3.5 A Lower boiling point / Laer kookpunt (2) 3.6 Compound C: Alcohol-One site for hydrogen bonding  Compound D: Carboxylic acid- Two sites for hydrogen bonding  More energy needed to overcome the intermolecular forces of D. Verbinding C:Alkohol-Een plek vir waterstofbinding Verbinding D: Karboksielsuur – Twee plekke vir waterstofbinding Meer energie word benodig om die intermolekulêre kragte van D te oorkom. (3) 3.7 LOWER THAN. / LAER AS (1) [13] QUESTION 4 4.1.1 Substitution.  / Substitusie (1) 4.1.2 Elimination / Dehydrohalogenation /Eliminasie / Dehidrohalogenering (1) 4.1.3 Addition / Hydrogenation / Addisie / Hidrogenering (1) 4.2.1 2-chloropropane  / 2-chloropropaan (2) 4.2.2 + KOH → + KCℓ + H2O    (4)

Downloaded from hlayiso.com 4 Physical Sciences P2 Memorandum NSC MDE/September 2018 4.3.1 An alcohol where the C-atom bonded to the –OH group, is bonded to two other C-atoms.  ‘n Alkohol waar die C-atoom wat gebind is aan die –OH groep, aan twee ander C-atome gebind is. (2)  4.3.2 Propan-2-ol (2) [13] QUESTION 5 5.1 Rate of reaction / Time Reaksietempo / Tyd (1) 5.2 Temperature / Temperatuur (1) 5.3 ΔV Rate =  Δt 0,06−0 =  60−0 2 30−0 = = 2 cm3∙s-1 MAX: 30−0 3 = 2 X 10-3 dm3∙s-1  (3) 5.4 LOWER THAN / LAER AS (1) 5.5 The small pieces of magnesium has a greater surface area than the one piece.  More particles are exposed.  More effective collisions per second.  Higher reaction rate. Die klein stukkies magnesium het ‘n groterkontakoppervlakte. Meer deeltjies is blootgestel. Meer effektiewebotsings per sekonde. Hoër reaksietempo. (3)

Downloaded from hlayiso.com 5 Physical Sciences P2 Memorandum NSC MDE/September 2018 5.6 Mg + 2HCℓ → MgCℓ2 + H2 25 g n=cXV Marking Rule = 0,1 X 0,1  Mol HCℓ m  Use of ratio 2:1 n= = 0,01 mole   Substitute 24 g.mol-1 M  1,04 mol  Subtraction: 25 = initial mol - mol used 24  Final answer = 1,04 mole  Range (24,8 -25) 0,005 mole 0,01 mole (ratio) 1,04 – 0,005 = 1,035 mole  m = n X M = 1,035 X 24 = 24,84 g (6) [15] QUESTION 6 6.1 HOMOGENEOUS / HOMOGEEN (1) 6.2 The reactants and products are in the same phase.  (2) Die reaktante en produkte is in dieselfde fase. 6.3 m 138,6  n= = = 0,9  M 154  Ratio : 1:3:1:1   Calculation of mole at equilibrium   Divide by 2   Correct Kc equation  Correct substitution in Kc equation   Answer = 4,92 (8) CS2 3Cℓ2 S2Cℓ2 CCℓ4 Mole initial 1,2 4 0 0 Mole react/form -0,9 -2,7 +0,9 +0,9 Mole equillibrium 0,3 1,3 0,9 0,9 [] 0,15 0,65 0,45 0,45 [S2Cℓ2][CCℓ4] Kc = [CS2][Cℓ2]3 = (0,45)(0,45) (0,15)(0,65)3 = 4,92

Downloaded from hlayiso.com 6 Physical Sciences P2 Memorandum NSC MDE/September 2018 6.4.1 DECREASES / NEEM AF (1) 6.4.2 INCREASES / NEEM TOE (1) 6.4.3 REMAINS THE SAME / BLY DIESELFDE (1) 6.5 If the concentration of S2Cℓ2 is increased, the system will according to Le Chatelier act in such a way to decrease the concentration of S2Cℓ2.  The reverse reaction will be favoured.  The amount of reactants will increase.  As die konsentrasie an die S2Cℓ2 verhoog word, sal die sisteem volgens Le Chatelier, sodanig reageer dat die konsentrasie weer sal verlaag. Die terugwaartse reaksie word bevoordeel. Die aantal mol reaktante sal toeneem. (3) 6.6 The pressure is decreased.  / Die druk word verlaag. (1) 6.7 INCREASE  / TOENEEM (1) 6.8 The rate of both forward and reverse reactions has increased.  (1) Die tempo van beide voorwaartse en terugwaartse reaskies het verhoog. [20]

Downloaded from hlayiso.com 7 Physical Sciences P2 Memorandum NSC MDE/September 2018 QUESTION 7 7.1 It dissociates completely in water and produce a high concentration of OH- ions.  DIt dissosieer volledig in water en vorm ‘n hoë konsentrasie OH- ione. (2) 7.2 [H3O+] = 0,4 mol∙dm-3 pH = -log [H3O+]  = -log (0,4) = 0,4  (3) 7.3 pH = 13 [H3O+] = 10-13  [H3O+][OH-] = 1 x 10-14 [OH-] = 10-1 mol∙dm-3 [NaOH] = 0,1 mol∙dm-3  (3) 7.4.1 BASIC  (1)   7.4.2 CO3 + H2O ⇌ HCO3- + OH- 2- (2) 7.5 Na2CO3 → CO2 Marking Rule V  Formula 𝑛 = V m  Substitute 22,4 dm3.mol-1  Ratio: 1:1  Subtitute 106 g.mol-1  Final answer OPTION 1 4,48 dm3 m=nXM V 4,48 = 0,2 X 106  n =   Vm 22,4 = 21,2 g 0,2 mole = 0,2 mole (ratio) 21,2 %= X 100 = 84,8%  (5) 25 OPTION 2 25 0,2 n= = 0,236 mole 0,236 X 100 = 84,75% 106 [16]

Downloaded from hlayiso.com 8 Physical Sciences P2 Memorandum NSC MDE/September 2018 QUESTION 8 8.1 Oxidation is an increase in oxidation number.  Oksidasie is ‘n toename in oksidasiegetal. (2) 8.2 Complete the circuit. / Voltooi die stroombaan, Maintain electrical neutrality./ Handhaaf elektriese eutraliteit. (2) 8.3.1 Fe3+ + 3 e- → Fe  (2)  8.3.2 Ti(s)│Ti3+ (aq)║ Fe3+ (aq)│Fe(s) OR Ti │ Ti3+ ║ Fe3+ │ Fe (3) 8.4 E°cell = E°reduction - E°oxidation  1,57  = -0,06 - E°oxidation E°oxidation = -1,63 V  (4) 8.5 0 (V)  (1) [14] QUESTION 9 9.1 The process where electrical energy is converted to chemical energy. (2) Die proses waar elektriese energie omgeskakel word in chemiese energie. 9.2.1 Ag → Ag+ + e-  (2) 9.2.2 Ag + Ag+ → Ag + Ag+  (2) 9.3 Directly proportional / Direk eweredig (1) 9.4.1 Q (1) 9.4.2 Equal to 17 g  (1) 9.4.3 m 17 n= =  = 0,157 mole  (2) M 108 9.4.4 Number of e- = 0,157 X 6,02 X 1023  = 9,45 X 1022  (2) [13]

Downloaded from hlayiso.com 9 Physical Sciences P2 Memorandum NSC MDE/September 2018 QUESTION 10 10.1.1 Fe / FeO  (1) 10.1.2 NH4NO3  (1)  10.1.3 4NH3 + 5O2 → 4NO + 6H2O (bal) (3) 10.2.1 X  (1) 10.2.2 X has the highest percentage of potassium that is needed for the growth of fruit.  X het die hoogste persentasie kalium wat nodig is vir die groei van vrugte. (1) 10.2.3 OPTION1 15+3+3 = 21 3 X 51 = 7,29% K 21 7,29 X 20 = 1,45 kg  (3) 100 OPTION 2 𝟓𝟏 X 20 = 10,2 kg 𝟏𝟎𝟎 3 X 10,2 = 1,46 kg [10] 21 TOTAL 150