Physical Sciences Grade 12 P2 Limpopo Mock 2023 Paper & Memo

Downloaded from Stanmorephysics.com REPUBLIC OF SOUTH AFRICA DEPARTMENT OF EDUCATION NATIONAL SENIOR CERTIFICATE eS SSS SSSR EEE THEE Seeseeesesay errr rrrrt —_——— eSseeeeaey Seeeeeeseeeeees= seeeeeeest MARKS: 150 aonT TIME: 3 hours EPHSCP2 This question paper consists of 20 pages including DATA SHEETS 7 Copyright reserved Please turn over

Dowploadedefrom Stanmorephysics. com LimpopoDoE/September 2023 _ 10. iB INSTRUCTIONS AND INFORMATION | uestion paper consists of nine questions. Answer ALL the questions. EACH question on a NEW page. Number your answers correctly according to the numbering system used in this question paper. Leave ONE line between two sub-questions, e.g., between QUESTION 2.1 and QUESTION 2.2. A non-programmable calculator may be used. Appropriate mathematical instruments may be used. Show ALL formulae and substitutions in ALL calculations. Round off your FINAL numerical answers to a minimum of TWO decimal places. Give brief motivations, discussions, etc. where required. You are advised to use the attached data sheets. Write neatly and legibly. Copyright reserved Please turn over

DOMTEAGA from Stanmorephysics.com|mropedoe/September 2023 QUESTION 1: MULTIPLE-CHOICE QUESTIONS Four options are provided as possible answers to the following questions. Choose the answer and write only the letter (A-D) next to the question number (1.1-1.10), for example 1.10 E E of the following pairs represents compounds that are isomers? C. CH3—CH=CH2 and CH2=CH-CHs3 D. CH3sCOOCH3 and CH3CH2COOH (2) 1.2 Which ONE of the following organic reactions will take place only when exposed to light? A. CH2CH2 +H2> CH3CHs B. CH3CH3— CH2CH2 +H C. CH3CH3+Cla> CH3CH2Cl +HCl D. CH2CH2 + Cla> CH2CICH2Cl (2) 1.3 What is the name of the reaction between an unsaturated hydrocarbon and hydrogen gas? A. — halogenation B. hydrogenation C. dehydrohalogenation D. hydration (2) 1.4 When an alkane, containing n carbons, undergoes combustion in excess oxygen, the mole ratio of the combusted products, COz : H20, will be equal to: A. nin B. n:n+2 C. n:in+1 D. n:2n @) Please turn over Copyright reserved

Downtsdlder#ram Stanmorephygies.com 1.5 Ammonia gas NH:(g) and oxygen gas, O2, react in a closed container at a constant temperature according to the following equation below. 4NH3(g) + 502(g) = 4NO(g) + 6H20(g) (AH <0) ONE of the combinations of pressure and temperature will result in ease in the yield of nitrogen monoxide NO(g)? Pressure Temperature A. decrease decrease B. decrease increase C. increase increase D. increase decrease (2) 1.6 Dinitrogen tetra oxide,N2O4(g) decomposes to produce nitogen dioxide, NO2(g), in a closed container according to the following balanced equation: N2O4(g) 2 2NO2(g) AH = -xkJ Consider the following statements regarding this equation: (i): The forward reaction is endothermic. (ii): The temperature of the container decreases. (iii): For each mole of NO2(g) which is formed, xkJ of energy is released Which ONE of the following statements are/is true about this reaction? A. (i) only B. (i) and (ii) only C. (ii) and (iii) only D. (i), (ii) and (iii) (2) Copyright reserved

Rovere le from Stanmorephysics.com LimpopoDoE/September 2023 437 Four gases, P, Q, R, and S are in equilibrium in a closed container. The diagram below shows their relative concentrations over a period of time. The equilibrium position is altered at t1 by the addition of more of gas Q. P Q R eg 2 : 8 ' eee eee Ss : > ti t The equilibrium reaction is shown by: A. P+QSR+S B. Q+SSP+R Cc. P+SSQ+R D. QSP+R+S (2) 1.8 The ionisation constant,Kw, for pure water at 25°C is: A. 14 B. 7 Cc. 1x107 D. 1x10-4 (2) 1.9 The electrode potential of the Ag*/Ag-electrode is determined with the aid of a standard H*/He, Pt-electrode as reference electrode. The equation for the half-reaction which occurs at the ANODE of this cell, is: A. Ag—Ag*+e B. Ag*+e—Ag C. H2-— 2H* + 2e D. 2H*+2e— He (2) Copyright reserved

Downlegdemetepm Stanmorephysics.com LimpopoDot/September 2023 NSC 1.10 An electrochemical cell, consisting of two standard half cells, is allowed to deliver current for a period. Which one of the following combinations best represents the change in CONCENTRATION of the two electrolytes r the period. (2) [20] B. > \ 5 | wt ad r- 7 oe o >t t Cc. D. —~ nh ~~ ns 2 by 5 ie £ a —pesess > t ~ if Please turn over

Phowshendad from Stanmorephysics.com QUESTION 2 (Start on a new page.) The table below shows six organic molecules A to F B. —— butane CH3-CH2-OH c D. H | ial Hi H CH3COOH i ' H-C-C—C—C—H Ded H " = H HO oH . 1 oto b ‘i methanal e-F ¥ 1 H H H 21 Define the term homologous series. 2.2 Write down the: 2.2.1 Structural formula of compound C 2.2.2 IUPAC name of compound D 2.2.3 Homologous series to which F belongs 2.2.4 Structural formula of the functional group of E 2.3 For compound A, write down: 2.3.1 The general formula. 2.3.2 The balanced equation for the oxidation reaction. Copyright reserved Please turn over (2) (2) (3) (1) (1) (1) (3)

Dowgloadeddeesn Stanmorephysics.com LimpopoDoE/September 2023 NSC 2.4 Compound B reacts with another organic compound from the table to form an ester. 2.4.1 Write down the LETTER of the other organic compound that reacts with B. (1) 4.2 Write down the NAME of the ester that forms. ria} 16) Qu N 3 (Start on a new page.) The table below shows data collected for four organic compounds, represented by the letters A, B, C and D, during a practical investigation: Organic compound a ant A propane 42 B pentane 36 c 2-methylbutane 27,8 D pentan-1-ol 137 3.1 Define the term vapour pressure. (2) 3.2 Which ONE of the compounds A or B has the higher vapour pressure? (1) 3.3 Consider compound C. 3.3.1. Is compound C a SATURATED or UNSATURATED hydrocarbon? (1) 3.3.2 Give a reason for your answer in QUESTION 3.3.1. (1) 3.3.3 Which type of intermolecular forces are mainly present between these molecules? (1) 3.4 Compounds B and C are structural isomers of each other. 3.4.1 Define the term structural isomer. (2) 3.4.2 Refer to structure, intermolecular forces, and energy in your explanation to explain why B has a higher boiling point than C (3) 3.5 Explain the difference in the boiling points of B and ‘efer to intermolecular forces and energy in your explanati (4) [15] Copyright reserved Please turn over

rowisleaded from Stanmoraphysics . ComimpopcdotSeptember 2023 NSC QUESTION 4 (Start on a new page.) The flow chart below shows the preparation of various organic compounds using but-1-ene as starting material. lnYaYer fea NaOH Reaction 1 A CH3sCH2CHCH2 Heat x HBr ——————— but-1-ene Reaction 3 Reaction 2 Substitution Secondary alcohol 3 ¥ H20 41 Using STRUCTURAL formulas, write down a balanced chemical equation for Reaction 1 (4) Write down the: 4.2. Name of the type of the reaction that occurs when but-1-ene is converted to compound X (1) 4.3 Write the structural formula and IUPAC name of the secondary alcohol, Y, that forms. (3) 44 Name the type of substitution reaction that occurs when compound X is converted to the secondary alcohol Y (1) 4.5 But-1-ene can be converted directly to the secondary alcohol, without the formation of the intermediate product X, with the help of a catalyst. 4.5.1 Except but-1-ene, write the NAME of the reactant needed for this (1) reaction. 4.5.2 Write down the FORMULA of the catalyst that can (1) 4.5.3 Name the type of reaction that will occur during this direct Copyright j Please turn over

Downlgadednézom Stanmorephysics.com LimpopoDoE/September 2023 46 conversion. (1) Instead of using water as a reactant in reaction 2 , concentrated sodium hydroxide is used, and the mixture is heated. 6.1 Write down the IUPAC name of the compound W that forms. (1) .6.2_ Name the TYPE of reaction that occurs. (1) [14] QUESTION 5 (Start on a new page.) 5.1 5.2 The collision theory can be used to explain how different factors affect the rate of a chemical reaction. 5.1.1. Name TWO conditions that determine whether a collision between two molecules will lead to a chemical reaction. (2) 5.1.2 In terms of the collision theory, explain why the rate of a chemical reaction increases with increasing temperature. (2) Curve R represents the Maxwell-Boltzman distribution curve for gas in a closed container at a certain temperature. Curve Q and S represents the curves of the same molecules at different conditions. 3 2 Ss [3 e Tee. S FF @# — Sm_]**s? BFK gs F8R BS Fe RS STIS me | > P Kinetic energy 5.2.1 Write down the NAME of the energy represented by P. (1) 5.2.2 Write down the change in the condition(s) whi ults in: (a) Curve Q (1) (b) Curve S (1) Copyright reserved Please turn over

Physical Sciences P2 LimpopoDoE/September 2023 Downloaded from Stanmort Physics. com 5.3 A teacher uses the reaction between magnesium ribbon, Mg(s) and nitric acid, HNOs (aq), to investigate ONE of the factors that influence reaction rate. The balanced equation of the reaction between Mg(s) and HNOs (aq) is given below. 5.3.1 Mg(s) + 2HNO3(aq) —» Mg(NOs)2(aq) + H2(g) Write down an investigative question for this investigation. (3) The results obtained when using DILUTED nitric acid are shown on the graph below. §.3.2 5.3.3 5.3.4 Amount of magnesium (gram) Time (s) Which compound, Mg or HNOs, is in excess, and give a reason for your answer using the information on the graph. (2) Calculate the average reaction rate (in grams per second) during the first 30 s. (5) Copy the above graph in your answer book. On the same set of axes, use a DOTTED LINE to show the curve that will be obtained when CONCENTRATED nitric acid was used. No numerical values are required. (2) [19] Please turn over Copyright reserved

Dowgleadedefrom Stanmorephysics.com timpopodoE/September 2023 NSC QUESTION 6 (Start on a new page.) 6.1 6.2 graph below shows the effect of the temperature on the value of Ke he following reaction taking place in a closed container. N2O4(g) = 2NO2(g) Equilibrium constant (Kc) Temperature (K) 6.1.1 Write down the effect of increasing temperature has on the amount of NO2(g) formed. (1) 6.1.2 Which reaction was favoured due to an increase in temperature? Write only FORWARD or REVERSE. (1) 6.1.3 State Le Chatelier's Principle. (2) 6.1.4 Is the forward reaction EXOTHERMIC or ENDOTHERMIC? Explain your answer. (4) 6.1.5 Write down TWO changes, that can be introduced to decrease the rate of the forward reaction (2) 14 mol SOz (g) and x mole O2(g) is initially placed in an & container and sealed at a specific temperature. At eq SOs (g) was present in the container. If the value for the 1 following equilibrium at this temperature is 9, calculate x, the i itial.q antity O2 (g) that was placed in the container. Copyright reserved Please turn over

Downloaded from Stanmoreghysics.COM impopopoe/September 2023 NSC 2S02(g) + O2(g) = 2S0a(g) [16] QUESTION art on a new page.) 5 A | Del in acid according to Bronsted and Lowry (2) 7.2 Explain the difference between a WEAK ACID and a DILUTE ACID. (2) 7.3 When 500 cm Sisodium hydroxide solution, NaOH(aq) is added to 500 cmé of 0,25 mol « on ydrochloric acid solution HCl(aq), the temperature d the pH of the solution changes to 2,3. The reaction betwe on NaOH aq) and HCI (aq) is as follows: NaOH(aq) + HCI (aq) > NaCi(aq) + H20 (I) 7.3.1 Is the reaction between NaOH (aq) and HCI (aq) EXOTHERMIC (1) or ENDOTHERMIC? Calculate the: 7.3.2 Final concentration of the hydronium ions, H30*(aq) ions. (3) 7.3.3 Initial concentration of the NaOH (aq) solution in 500cm? (8) [16] ht Please turn over

Dowmplansdedesz0m Stanmorephysics.com LimpopoDoE/September 2023 QUES 8.1 8.2 8 (Start on a new page.) investigation to compare the relative reducing ability of copper, Cu(s) and ic, Zn (s) is carried out. The two metals are reacted with dilute sulphuric acid and the experiment set up as shown below. eS °<4-H,S0,(aq) |_| <T~ 4:50.29) Oo oO Zn powder Cu powder In test tube A, bubbles of gas are observed whilst in test tube B no bubbles are observed. 8.1.1 Write down the name of the gas produced in test tube A (1) 8.1.2 Refer to the relative reducing abilities of Cu and Zn to explain why no bubbles of gas are produced in test tube B (3) 8.1.3 Use the STANDARD REDUCTION POTENTIAL TABLE to write down a balanced equation of the reaction which takes place in test tube A (3) An electrochemical cell is set up using a redox reaction_given below. Mg(s) + 2AgNOs(aq) — Mg(NOs)2(aq) (s) 8.2.1 Write the cell notation of this cell. (3) Copyright reserved Please turn over

Bownleadedzfrom Stanmorephysics. com | impopoooe/September 2023 NSC 8.2.2 Write down the conditions needed for the cell to operate under standard conditions. (2) 8. Calculate the initial emf of the cell under standard conditions (4 T ll is now connected to a light bulb which is marked 3 V; 6 W. | ry, the bulb should glow, but in practice it doesn't. 8.2.4 Without any calculation, give an explanation for this. (1) The volume of the electrolyte in each standard half-cell is 0,4 dm?. 8.2.5 Calculate the maximum loss of mass that can occur at the anode. (6) [23] Copyright reserved turn over t Please turn

Physical Sciences P2 16 LimpopoDoE/September 2023 Downloaded from Stanmore}iiysics.com QUESTION 9 (Start on a new page.) The electrolytic cell shown below is used to purify IMPURE copper. Copper sulphate solution, CuSO«(aq), is used as the electrolyte. Electrode B contains 95,7% Cu by mass. Power A. source B Cu(s) — I—. 95,7% Cu(s) CuSO,(aq) 9.1 Is electrode A, the POSITIVE or NEGATIVE electrode? 9.2 Pure copper is used to make electrical wires. Give one reason why the copper must be purified first before it is used in electrical wires. 9.3 Will the mass of electrode B INCREASE or DECREASE when the cell is operating? Write down the relevant half-reaction to support your answer. 9.4 If 2 moles of electrons are transferred from the anode to the cathode during electrolysis, calculate the mass of the IMPURE copper sample. TOTAL: 150 Copyright reserved Please turn over (1) (2) (3) (5) (11]

Aawneaded, from Stanmorephysics. COM | mpopodoe/September 2023 NSC | gee | zee 47 ON PW Wi SZ JD 48 WO Wy ng GN NM eg UL tol | 208 40s = 008 6 26 46 % S6 | me 66 wa 66 06 oe. 1 ae oe . ’ ° -_+— -- ’ - St C26 69S LOL SOL | Oh OSB at tsi | ost Ce m1 GA WL 43° OH Ag Gh PD NZ WS) Wd PN 4g 89 ‘ oe) ee) ee ee ee ; = Ww oe fait — = , 8 (8 | ee a ar a Re ee Le Uy WHE IDs deat 6H Ay Ww 4 $0 ou OM Ke jJHe eT 88 S805 © st mt ct ee lo le 1 Sie le i ig ie is ieee ee Ae ce en en ee 8X | ReLzias< use Ua PI= By Pat ua nuh 1s OW = IN ZEA B48 5 9u & ms (c |iS 0S GF (fo oy Sy om ite iw ow io 6 oe ot | “yo oe 6 (sc ce (0 (So ‘sto 6s (0S 95 (SS 7S tS ‘Or “Sy OF | Gt SY JE eS sySemreoOs uZSNDS INZE9DE PSUWT IDE AT WSWTPOS ME wu st ime cc oe |e oe el le ae ae oe OM RR le o jsst (ce tu ow | | Bssewwoole dmanejed eploprueg [we | & w weSstd Ris gave conus eqnege engages cqenenedty BW <.eN S rr Sh in |e iis tlt ee | ee cls [sre] ON JEORNE9 bas yee a 79 j . fonortse ny ens 1.06 16 |e if [se |s_ | ez » ic | ; —o : a oH EEE | poturey T31N37S/A3M H = c= S sequany ywory fi ae (ma) A WOW ts (a) Ww ce 2 a <r |) | t z ‘ FININI 1A NVA 12BVL INFUTONId FIO °C TIGYLISLNINITI 4O IIL WOOWFd BHI °C IIGVL Please turn over Copyright reserved

LimpopoDoE/September 2023 Bownloaded from Stanmorephysics.com TABLE 4A: STANDARD REDUCTION POTENTIALS TABEL 4A: STANDAARD-REDUKSIEPOTENSIALE Halt-reactions/Halfroaksios E° Ww FAg)*+2e 2F + 2,87 Co" +o # Co” + 1,01 H,O,¢ 2H" +2000 et) 2H,0 o177 aay MnO, +6H' +60 0 = Mn’ + 41,0 + 1,51 Clig)+2e « 2ct + 1,96 CO} + 14H" +60 2Cr eH, | +193 Oxg)* 4H +4e et 2H, + 1,23 MnO,* 4H" +20 Mn?’ +2H,O | + 1,23 t PI +2e oo Pt + 1,20 BrA(Q+2e + 266 +107 NO, + 4H" +30 = NO(g)+2H,0 | + 0,96 Hg" +2e = H(t) + 0,85 Ag’te =# Ag + 0,60 NO, +2H'+e & NO,g)+H,0 | +0,80 Fe" +e = Fe™ +077 8 Oxg)+2H"'+2e # H,0, + 0,68 5 +2e 2 2i + 0,54 $ Cu’+e = Cu + 0,52 SO,+4H'+4e = S+2H,O0 +045 3 2H,0+0,+4e = 40H" +0,40 x Cu*+2e = Cu + 0,34 SO? +4H'+2e & SOAg)+2H,0| +0.17 $ Cu" +e = Cu’ +0,16 Sn“ +20° = Sn” + 0,15 H 5 S+2H'+2e = H,S(g) +0.14 $ 2H’+2e H,(g) 0,00 = 2 Fe" +3e = Fe ~ 0,06 £ = Po" +2e = Pb ~0,13 5 = Sn +2e = Sn ~0,14 a 2 Ni" +2e & Ni ~ 0,27 4 Co" +2e = Co ~0.28 2 Tad z Ca" +2e 2 Cd - 0,40 3 2 Cr'+e 2 Cr” -0,41 2 zg Fe" +20 = Fe -0,44 o 6 Cr +3e # Cr ~0,74 & 2 Zn” +2e 2 Zn ~ 0,76 | 2HO+2e = HA9)+20H | - 0.83 4 Cr +2e = Cr -0,91 2 Mn" +2e = Mn ~ 1,18 aad £ AU’ +30° = Al ~ 1,668 Mg’ +20 = Mg ~ 2,38 ’ Na'+e # Na 2.71 Ca" +2e # Ca ~ 2,87 S?°+2e # SF ~ 2,89 Ba" +2e # Ba - 2,90 Cs’+e # Cs + 2,92 K'+e #& K ~ 2,93 liiete 2 ~ 3,05 Copyright reserved Please turn over

Dawaleadedefrom Stanmorepysics.com impopodot/September 2023 NSC TABLE 4B: STANDARD REDUCTION POTENTIALS TABEL 4B: STANDAARD-REDUKSIEPOTENSIALE Increasing oxidising ability/ Toenemende oksiderende vermoé Copyright reserved Half-reactions/Halfreaksios E° (v) Ute ww ~ 3,05 K'+e ¢ K ~2,93 Cs*+e@ # Cs 2,92 Ba" +20 = Ba ~ 2,90 S?'+2e = Sr - 2,89 Ca" +20 # Ca - 2,87 Na"+e = Ne -2,71 Mg" +2e = Mg - 2,36 AC’ +30" = At — 1,66 Mn” +2e" = Mn -1,18 Cr +2e = Cr ~ 0,91 2H,O+2e = H2(g) + 20H" - 0,83 Zn*+2e = Zn - 0,76 Cr'+3e = Cr - 0,74 Fe" +2e « Fe - 0,44 Cr*+e = Cr” -0,41 Ca” +2e = Cd -0,40 Co” +2e # Co - 0,28 N?’+2e = Ni -0,27 Sn” +2e = Sn ~ 0,14 Pb**+2e = Pb -0,13 Fe’ +3e = Fe - 0,06 2H’ +2e" = Hg) 0,00 S+2H'+2e = H2S(9) + 0,14 Sn“ +2e = Sn” +015 Cu" +e = Cu’ +0,16 SO} +4H"+2e == SOXg)+2H,0| +0,17 Cu*+2e = Cu + 0,34 2H,0+0,+4e = 40H +0,40 $0,+4H'+4e° = S+2H,0 +0,45 Cu'+e = Cu +0,52 h+2e 2 2 + 0,54 Og) + 2H" +2e = HO, + 0,68 Fe"+e = Fe” +077 NO; +*2H’+e = NOdg)+H,0 | + 0,80 Ag’+e # Ag + 0,80 Hg’ +2e" = Holt) + 0,85 NO, +4H"+3e" NO(g)+2H,0 | +0,96 Br{O+2e = 28 + 1,07 “+2e@ Pt + 1,20 MnO, + 4H" +2e = Mn +2H,O | +1,23 Og) + 4H’ +4e = 2H,0 + 1,23 CO} + 14H’ +607 + 2CP"+7H,O | + 1,33 Ct,(g)+2e = ‘ MnO, +8H"+S5e = H,O,+2H'+2e # Co*+e = Fg) + 2e Increasing reducing ability/ Toenemende reduserende vermoé Please turn over

Dowrdoadeet from Stanmorephysics. COM impopodoe/September 2023 NSC GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 12 VRAESTEL 2 (CHEMIE) DATA FOR PHYSICAL SCIENCES GRADE 12 PAPER 2 (CHEMISTRY) TABLE 1: rrsca CONSTANTSITABEL 1: FISIESE KONSTANTES NAME/NAAM SYMBOL/SIMBOOL | VALUE/WAARDE Gockeieion Na 6,02 x 10 mot" enue eleten e 16x10 paciiesteiead me 9,11 x 10" kg pear ay pal Vn 22,4 dm*-mol! Seventna pe 1,013 x 10° Pa prosreiless arbor gel Te 273K TABLE 2: FORMULAE/TABEL 2: FORMULES pL N ae n=— Ny n m Vv c=— orlof — _ Vv S MV L basa /. Ky =[H,O° [OH -]=1x10™ at/by 298K pH= -log[H30*] 0 _;70 _ 70 0 _ 70 _ 70 Feu E cathode Aa / Eva ~ Enatode E sit or/of 0 0 0 0 _7p0 _ pe Even = Ekibi ~ Eo sidation | Eva ia E reubsie E usidasie or/of 0 7) 0 0 _ 70 _ pe Een = EE xidising agent Leite agent | yer = E ctsideermiddel |i nickdel Fay Please turn over Copyright reserved

Downloaded from Stanmorephysics.com LIMPOPO PROVINCIAL GOVERNMENT REPUBLIC OF SOUTH AFRICA DEPARTMENT OF EDUCATION NATIONAL SENIOR CERTIFICATE This marking guideline consists of 9 pages including THE COVER PAGE Copyright reserved Please turn over

Downloaded from Stanmorephysics. com Physical Sciences P2 LimpopoDoE/September 2023 Noenrencnsnaay QUESTION 1 11 Dvv (2) 1.2 Cvv (2) 1.3 Byv (2) 14 Cvv (2) 1.5 Avy (2) 1.6 Cvv (2) 17 Byv (2) 1.8 Dvv (2) 1.9 Cvv (2) 1.10 Avy (2) [20] QUESTION 2 2 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. vv (2) 2:2 H | I Functional group v H-E-G Whole structure v H O-H correct. (2) 2:22 1,1-dimethylv propanv -1-olv / 1,1-dimethyl-1-propanol (3) 2.2.3 ketone Vv (1) 2.2.4 , —C—Hv (1) 2.3.1 CnHan+2 ¥ me (1) — 2.3.2 2C4H10 + 1302 V—> 8CO2 + 10H20 VY hi bal. v (3) d—L 2.4.1 Cv (1) 2.4.2 ethylv ethanoate v (2) Copyright reserved Please turn over

Downloaded from Stanmorephysics. com Physical Sciences P2 QUESTION 3 3.1 3.2 3.3.1 3.3.2 3.3.3 3.4.1 3.4.2 3.5 Copyright reserved ee The pressure exerted by a vapor at equilibrium with its liquid in a closed system. Vv ¥ AY SATURATED ¥ Only single bonds between C- atoms.” London forces Vv / dispersion forces / induced dipole forces Compounds with the same molecular formula, “ but different structural formulae. ~ e 2-methylbutane is a spherical molecule that offers a smaller surface area to other molecules.v// Pentane is a linear molecule which offers a larger surface area to other molecules. e Less/smaller surface area where intermolecular forces (London forces) can interact with other molecules.~// Greater/larger surface area where intermolecular forces (London forces) can interact with other molecules. e Less energy required to overcome the intermolecular forces. v// More energy required to overcome the intermolecular forces. e pentan-1-ol (D): H-bonds are stronger” than the weaker London forces vin (B), alkanes. e Therefore, more energy is required to Yovercome the stronger intermolecular forces in (D). e Consequently (D) has a higher boiling point. v OR e Pentane (B): London forces are weaker in alkanes than the stronger H-bonds in alcohols in (D). e Less energy is required to overcome the forces of attraction in (B). Consequently (B) has a lower boiling point. Please turn over LimpopoDoE/September 2023 [16] (2) (1) (1) (1) (1) (2) (3) (4)

Downloaded from Stanmorephysics. com Physical Sciences P2 LimpopoDoE/September 2023 ae [15] QUESTIO 4.1 vv vv H H H a H H H H oro G- eH + H-Br > H-C-6-C-C-n HH H Br HH (4) 4.2 addition / hydrohalogenation ¥ (1) 4.3 butan-2-ol v/ 2-butanol Hydroxyl group on the 2" v carbon. Whole structure correct. v (3) 4.4 hydrolysis ¥ (1) 4.5.1 Water v (1) 4.5.2 H2SO4 ¥/ HCl / HsPO4 (1) 4.5.3 hydration Y (accept addition) (1) 4.6.1 but-2-een / 2-buteen v (1) 4.6.2 dehydrohalogenation v / elimination (1) faa? [14] QUESTION 5 - 5.1.1 Sufficient kinetic energy (molecules move test Toh) during the collisions. ¥ Molecules must be correctly orientated. v (2) 5.1.2 Increased temperature: Copyright reserved Please turn over

Downloaded from Stanmorephysics. com Physical Sciences P2 5.2.1 5.2.2 5.3.1 5.3.2 5.3.3 5.3.4 Copyright reserved Keomemerannan LimpopoDoE/September 2023 More molecules move fast enough or have sufficient Ex. ~ There are more effective collisions per unit time v / Ek = activation energy. a (2) —_ Activation energy ¥ (1) (a) Increase in the concentration of one or both reactants. W (1) ! (b) Increase in temperature. v (1) How will a change in e Identify dependent and concentration — independent variable.v v affect the reaction rate? ¢ Aska question (?) about . OF , | the relationship between What is the relationship between dependent and the concentration and reaction independent variable. ¥ rate? (3) HNOs V/ Nitric acid The magnesium is used up. / Magnesium is the limited reagent. Vv (2) Opsie 1 Opsie 2 An = 1,0 — 0,8 v= 0,2 mol n= n= M M at _m =m n= M t= 0,8 =7¥ 0,2=— neh 24 m=24g “m=19,2g ~m=4 8g ‘ Am . Sin Gem. reaksietempo = =~ Gem. reaksietempo = ae _ 19,2 - 24v =) ~ 30-07 = 0,16 g.s71¥ = 0,16 g.st¥ (5) e Steeper slope below original graph.~ e Intercept x-axis earlier. v Please turn over

Downloaded from Stanmorephysics.com Physical Sciences P2 6 QUESTION 6 6.1.4 6.2 Copyright reserved LimpopoDoE/September 2023 NSC/Memorandum £ = ry oe c iT ES SE 2 Ss ° E ¢ 30 Time (s) (2) [19] Increases Vv (1) Forward v (1) 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) Increase in temperature increases Kc. V Increase in Kc indicates that the forward reaction has been favoured. ¥ Increase in temperature favours the endothermic reaction. ¥ Therefore, the forward reaction is endothermic. (4) Add a catalyst. Vv Decrease pressure OR Increase the volume of the container. ¥ (2) 2SO2 Q2 2SO3 Initial mol 8 y 0 Mol reacted -2x -x +2x Vv Mol at eq 2 y-3 vv 6 [ ]ateq 1 ple} 3y¥' 2) 2 _ [so3]2 Ke “Ronee correct Ke expression (3)2 a =~.” correct substitution (1)2(y-3) 2 Y =5mol v (6) Please turn over

Downloaded from Stanmorephysics.com Physical Sciences P2 7 QUESTION 7 7.1 7.2 —— 7.3.1 Cie 7.3.3 QUESTION 8 8.1.1 Copyright reserved LimpopoDoE/September 2023 NSC/Memorandum [16] or A proton donor (H* ion donor).v (2) WEAK ACIDS: ionizes incompletely in water to form only a few H,O ions.” DILUTED ACIDS: contains a large amount of water added to it.” (2) EXOTHERMIC v (1) PH = -log[H30*] ¥ 2,3 V= -log[H30*] «[HsO*] = 10-25 = 0,005 mol.dm? v (3) n(HCl) initial = cVv = 0,25(0,5) v = 0,125 molv n(HCl) after addition = cV = 0,005(1) v = 0,005 molv Reaction: HCl + NaOH — NaCl + H20 Number of moles of HCI reacted with NaOH: = 0,125 — 0,005 = 0,12 mol ¥ So 0.12 mol of NaOH reacted with 0.12 mol of HCI: n [NaOH initial = V = Oy 0,5 = 0,24 mol.dm-3¥ (8) [16] HydrogenY (1) Please turn over

Downloaded from Stanmorephysics. com Physical Sciences P2 LimpopoDoE/September 2023 mrmuanageeanea 8.1.2 In terms of the reducing agent: e Cuis a weaker reducing agentY than H2Y and will not reduce H* (to Hz). v (3) —, In terms of the oxidizing agent: Le H* is a weaker oxidising agent than Cu2* and will not oxidise Cu (to Cu?*). (NOTE: Compare the two reducing agents in the two half reactions OR the two oxidizing agents in the two half reactions.) OR H* (H2SOx) is a weaker oxidizing agent than Cu (to Cu*). (NOTE: No marks if referring to the relative positions on the table.) v vv 8.1.3 Zn + 2H* Zn? + He or Zn + H2SO4 > ZnSO4 + H2 (3) 8.2.1 Mg | Mg2* v ||v Ag* |Ag v (3) 8.2.2 25°C v/298K and1moldm? Vv (2) 8.2.3 E®cell = E®cathode — E®°anode Vv v v = 0,8 - (-2.36) = 3,16VY (4) 8.24 my The fi manufactured to work effectively when connected toa source that can deliver a current of 2 A. This cell produces a large enough potential difference, but the current is (1) probably too small v due to a very large internal resistance. 8.2.5 Anode: Mg is oxidized and therefore forms the The amount of AgNOs3 available determif 18S the anode (Mg) will go into solution. 1 mol Mg reacts with 2 mol Ag* n(AgNOs) = cV¥ = 1(0,4) v = 0,4 mol Copyright reserved Please turn over

Downloaded from Stanmorephysics. com Physical Sciences P2 LimpopoDoE/September 2023 a «.n(Mg) = % n(Ag*) = %(0,4) v = 0,2 mol “. maxmimum loss in mass = nMv = 0,2(24) v = 4,8 g Mgv QUESTION 9 9.1 NEGATIVE ¥ 9.2 To improve electrical conductivity.v ~ 9.3 Decrease Vv Cu — Cu* +2e Vv 9.4 n(Cu) = (2) v Cu?* + 2e — Cu =1 mol m(Cu) = nM = 1(63,5) v =63,5g 9 i _™Mpure_ % purity = Ree x 100 63,5” = Bere x 100 = 66,35 g¥ TOTAL: 150 Copyright reserved (6) [23] (1) (2) (3) (5) [11] Please turn over

Downloaded from Stanmorephysics.com pea LIMPOPO . PROVINCIAL GOVERNMENT REPUBLIC OF SOUTH AFRICA DEPARTMENT OF EDUCATION PAPER 2: THE PAPER SHOULD BE MARKED OUT OF 140. QUESTION 1 1.6 Removed. - 2 marks QUESTION 2 2.2.2 2-methylbutan-2-ol Marking criteria: Correct stem : butan-2-ol ¥ Correct substituent : 2-methylY IUPAC name correct including hyphensY (3) QUESTION 3 3.3.2 No multiple bonds between C- atoms. ¥ (1) QUESTION 4 4.6.1 Correct spelling is but-2-ene QUESTION 5 5.3.1 Accept any of the following factors for the independent variable: e Change the surface area of Mg. = — Qa! e Add a catalyst. e Change in the temperature for the reactants. E.g How will a Change the surface area of Mg affect the reaction (2) rate? Vv v

Downloaded from Stanmorephysics.com 5.3.2 Removed. - 2 marks 5.3.3 Average rate = — = 08 ¥—1,0 ¥ 30-0 v = 6,67 x103g-st v (5) QUESTION 6 6.1.4 Endothermic. ¥ e Increase in temperature increases Kc ¥ e Increase in Kc indicates that the forward reaction is favoured ¥ e Increase in temperature favours the endothermic reaction’ (4) 6.1.5 e Decrease pressure ¥ * Decrease temperatureY (2) 6.2 Removed - 6 marks QUESTION 7 7.2. Weak acids ionize incompletely in water to form low concentration of H3s0* ¥ Dilute acids contain large amount of water. (2) 7.3.3. POSITIVE MARKING FROM 7.3.2 Marking criteria: e Formula n(HCl)initial = CV YW e Substitution of c (HCI) initial = 0,25 and V=0,5 v e Substitutionofc (HCl) after addition = 0,005 and V=1 ¥ e Calculation of n(HCl)reacted ¥ e Ratio NaOH: HCl ¥ e Substitution in the formula for [NaOH] ~ ean e Final answer ¥

Downloaded from Stanmorephysics.com n( HCl) initial =c Vv = 0,25 (0,5) ¥ = 0,125 mol n( H@fyafter addition =cV = 0,005 (1) v = 0,005 mol ¥ Nu of moles of HCl reacted with NaOH : = 0,125 - 0,005 = 0,12 mol ¥ Ratio: 0,12 mol HCI : 0,12 mol NaOH ¥ [NaOH] initial = c 0,12 =—Vv = (8) = 0,24 mol.dm?v QUESTION 8 8.1.2 Ho is a stronger reducing agent Vv than Cu Y and Cu will not reduce H* (to Hz) ACCEPT Zn is a stronger reducing agent than Cu, ~ therefore Zn will be oxidized. Cu is a weak reducing agent, V than Zn _ therefore will not undergo oxidation and will not produce gas. “ (3) 8.2.2 Temperature of 25°C / 298K Vv Concentration of 1mol: dm? “ (2) 8.2.4 OR Voltage drop due to the internal resistance. ¥ (1) 8.2.5 MARK ALLOCATION Remove 1 mark from n(AgNOs) = cV allocate it in the ratio

Downloaded from Stanmorephysics.com QUESTION 9 94 n(Cu) =1/,(2)¥ =1 mol m ( =nM = 1( 63,5) ¥ =63,5 g POSITIVE MARKING % purity 3—2"*- x 100 impure 63,5 ¥ = x 100 95,7 ¥ =66 35g Vv TOTAL MARKS : 140 (5)