Physical Sciences Grade 12 P2 2017 Memo

Education KwaZulu-Natal Department of Education REPUBLIC OF SOUTH AFRICA * PHYSICAL SCIENCES P2 (CHEMISTRY) * : PREPARATORY EXAMINATION ‘ : MEMORANDUM : " SEPTEMBER 2017 : rrr a NATIONAL SENIOR CERTIFICATE GRADE 12 — | | TIME: 3 hours 150 | MARKS: This memorandum consists of 9 pages.

enyscalscencee ~«=~d DOWNIoaded, from hlayiso.com P2 Preparatory Examination September 2017 NSC - Grade 12 QUESTION 1: MULTIPLE CHOICE 11 Dy (2) 12 Byy (2) 13. AvY OR C (2) 14. Dvv (2) 15 Cvv (2) 16 Byy (2) 17 AVY (2) 18 Dyvv (2) 19 Avv (2) 1.10 Cvv (2) [20] QUESTION 2 2.1 2-bromo-5-chloro-2,5-dimethyl” heptane “ (2) 2.2 C,Ho, ¥ (1) 23° oH HCIH HHH (3) Potopoto toa Y correct structural formula H C-C-C-C-C-C-C-H 4 od roto rod Y Clin correct position HoH H HHH Y ethyl group in correct position H-G—H H —¢ -H H 2.4 Triple bondv/ Accept Alkynes (1) 25 Oo H (2) UN 1 vv C-O-C-H i ! 4 4 Accept any correct formula except the molecular 2.6 Ketones ¥ (1) [10] Copyright reserved Please turn over

pnysical sciences DOWNloaded-fssom hlayiso.com P2 Preparatory Examination September 2017 NSC - Grade 12 QUESTION 3 3.1. CY OR Ethanoic acid (1) C 3.2 Ethanoic acid has the strongest intermolecular forces “ More energy is required to overcome the intermolecular forces “ least amount will evaporate “ (3) 33 BY (1) C@ Ethanal has the weakest intermolecular forces ¥ (3) Least energy is required to overcome the intermolecular forces v Ethanal has the lowest boiling point and therefore the highest vapour pressurev [8] QUESTION 4 4.1 H,SO,/ H;PO,/Heat” (1) 4.2 Pt/Pd/Niv (1) 43 CsH, + 50,———* 3CO, + 4H,0 V reactants/left hand side (3) If any structural formulae used 2. ” products/right hand side 3 balancing 44 HHH H HH (3) ' b= 4 Hc 4 ' é é 4 reactants/left hand side H= ¢ ae + c rot lv products/right hand side H H cl H structural formulae lf condensed formulae used 2 3 4.5 Sulphuric acid v (1) 46 HoH OH OW WOH ” correct functional group 2) H~G~G-G-O-C-C~C~G-H whole structure correct H HH H H H 4.7. Propyl butanoate “YY Any correct name given for 4.6 (2) [13] QUESTION 5 51 A molecule that consists of a large number of atoms. Vv” (2) 5.2 Polymerisation v (1) 53 HH H (2) H H [5] Copyright reserved Please turn over

physical Scenes 9 DOWNIoaded, from hlayiso.com P2 Preparatory Examination September 2017 NSC - Grade 12 QUESTION 6 6.1 How does the rate of a reaction change with time? vv (2) Any given investigative question 6.2. Time’ Mark based on given investigative question. (1) 6.3 (2) [Time (s) 0 1 2 3 4 5 6 7 8 9 Mass of flask and its contents (g) | 178 |176,2|174, 1|172,7|172,3|172,2/172,1|172,1|172,1|172,1 Mass of CO, (g) produced (g) 0 | 1,8 | 3,9 | 5,3 | 5,7 | 5,8 | 5,9 | 5,9 | 59 | 59] w 64 Graph of mass of CO, produced vs time ~ Correct shape im TPP Tae ~Y plotting of all points — Clee] [Appropriate scale and labels on both ete 1 laxes . ~ Correct heading 8 But plotting 5 or less points subtract one g ark. 2 6) oO 1 2 324 5 6 7 8 9 Time (s) 6.5 The reaction has reached completion. v (1) OR One of the reactants is finished. ~ 6.6 m y n= 58, 44 = 0,13 mol 1 mol occupies 24,46 dm?® 0,13 mol will occupy 3,18 dm? v Accept range 3.18 — 3.28 @) 6.7 Graph N has a steeper gradient vand finishes at the same point as the (2) original graphw 6.8 Higher temperature, molecules have greater kinetic energy “ (3) More effective collisions per unit time Reaction rate increases “ [19] Copyright reserved Please turn over

Physical Sciences Downloaded-:20 m h layiso . com P2 Preparatory Examination September 2017 NSC - Grade 12 QUESTION 7 7.1 CALCULATIONS USING NUMBER OF MOLES BEREKENINGE WAT GETAL MOL GEBRUIK Mark allocation/Puntefoekenning: e Calculating number of moles of H2, Cl, and HCI. ¥ Molar ratio. “ Number of moles at Equilibrium. “ Dividing by 0,5 to get the concentration. v K, expression. ¥ e Substitution in the K, expression “ e Calculating concentration of Cl, ¥ ¢ — Substitution in n= cV.v e Substitution in m=nM “ e _ Final answer” 7.1 H+ ~~) Ch = 2HCI (10) Initial mass 40g 355g Og Calc no. of Initial n (5 5 0) | “mol React/Prod Cx Xx 2x_) | ~ Ratio In at equilibrium] (5 - x 5-x ax Ji v ‘Conc. at 5-x 5-x 2x ae equilibrium 05 05. 05 v dividing by 0,5 x, = omy INo K, expressi t substituti 9 GJEb1 lo K, expression, correct substitution max 2 (> Wrong KC expression max 60 = 05 y 10 ven 5x 5-x CYC) 0,5 0,5 x = 3,97 Cone.Cl, at equilibrium = rad n=C-V _ 5-397 95 0,5 = 1,03 mol m=n-M = 1,03-71” = 73,139 % Copyright reserved Please turn over

rysical Scenes DOWNloaded, from hlayiso.com P2 Preparatory Examination September 2017 NSC - Grade 12 7.2 When the equilibrium in a closed system is disturbed, the system will re- instate a new equilibrium by favouring the reaction that will oppose the disturbance. vv (2) 7.3 7.3.1 Decrease “ (1) 7.3.2 Increase “ (1) 74 The temperature of the system was increased, according to Le Chatelier’s principle, the system reacts by favouring the endothermic reaction.“ Hence, the reverse reaction is favoured.” (2) [16] QUESTION 8 8.1 8.1.1. Strong acid “ (1) 8.1.2 7 (5) f rl ined v CV, = ©, V3 Vol. of conc.HCl require (10\(x) = (0,35)(0,9" n=CV v x = 0,0125dm? = 0,25 x 0,5” OR = 12,5em°¥ =0,0125 v n “V7 0,0125 V = 0,0125 “dm? = 12,50m° 1. H 8.1.3 HCI tho _, H,0° + Cr 1 c vo 0 Vv 0,25 mol.dm* 0,25 mol.dm®? pH = -log [H,07} v = -log (0,25) = 0,60 , Copyright reserved Please turn over

Physical Sciences Downloaded..f£rom hlayiso.com P2 Preparatory Examination September 2017 NSC - Grade 12 8.2 8.2.1 Option 1 n(HCl) = cW= (0,25) (0,5¥ = 0,125mol n(NaOH) = cV = (0,2)(0,14¥ = 0,028mol n(HCl) reacted with (NaOH) = 0,028mol” n(HCl) reactedwith (CaCO,) = 0,125-0,028% = 0,097 mol n(CaCO,) = >* 0,09™ = 0,0485mol mass of CaCO, = 0M , = 0,0485x100 = 4,85¢% Option 2 Volume of HCI reacted with NaOH om, n aaa CM, OM, (0,25)V 4 a: ce y (0,2)(140 ~ 14 V. = 12cm? a Volume of HCI reacted with CaCO, Vici = 500 - 112 v = 388 cm? = 0,388 dm? No. of mol of HCI reacted with CaCO 3 n Cy n 0,25 = 0388 n = 0,097 mol No. of mol of CaCO reacted with HCI Net * "eaco, = 2:4 0,097 "caco, a = 0,0485 mol Mass of mol of CaCO; reacted with HCI me=n-M = 0,0485 -100% = 4,859 v (8) Copyright reserved Please turn over

nyseat sciences DOWNloaded, from hlayiso.com P2 Preparatory Examination September 2017 NSC - Grade 12 8.2.2 (3) % purity = mass of CaCO, 100 mass of sample =42 100 % 5 =97% ¥ [20] QUESTION 9 9.1 9.1.1 cathode v (1) 9.1.2 P to silver medal “ (1) 9.1.3. Electrode P is made of silver. Rate of oxidation is equal to the rate of reduction.v” (2) 91.4 Agv (1) 9.2 9.2.1 Temp =25°CVv Concentration of the electrolyte solution = 1 mol.dm?v (2) 9.2.2 Chemical energy to electrical energy “ (1) 9.2.3 Maintain electrical neutrality between the two half cells “ (2) Completes the circuit ~ 9.2.4 6 _ 8 . £e Eco = cathode E node 2,70" = 0,34°- E° anode Eo = -2,36V% anode The anode must be magnesium v (5) + marking 9.2.5 Mg > Mg” +2evv OR (2) 9.2.6 (3) vov v Mg/Mg""//Cu**/Cu. OR YIY?*#1Cu2*/Cu 9.2.7 Remain the same” (1) 9.2.8 Ag’ will form a precipitate with the Cl’. This decreases the concentration of the Cl in solution, but not the concentration of Cu2*v “ The reading on the voltmeter can only be affected by a change of concentrations of either Cu2* or Mg2*/ Y2+.” (3) QUESTION 10 10.1 10.1.1 Haber process (1) 10.1.2 Fractional distillation of air V (1) 10.1.3 Ostwald process ¥ (1) 10.1.4 Platinum v (1) 10.1.5 HOY OR (H20 +02) (1) Copyright reserved Please turn over

Downloaded. from hlayiso.com Physical Sciences pP2 Preparatory Examination September 2017 NSC - Grade 12 10.1.6 Enables plants to grow better and faster “ (2) Increases food production “ 10.1.7 Ammonium nitrate ” (1) 10.2 10.2.1 The ratio of N:P:K or primary nutrients in the bag “ Amount of fertilizer in the bag V (2) 10.2.2 or — S x22 = 9,68% v wp = 19x22 = 4,40% Vv 50 YN = Lox 22 = 7.92% v (3) OR 22% VV 10.2.3 Potassium” Shortage of potassium causes poor quality flowers and fruit (colour and taste) and make leaves have brown or yellow edges. ~ (2) [15] TOTAL MARKS: [150] Copyright reserved Please turn over