KZN-Physical-Sciences-Grade-11-November-2025-P2-and-Memo.pdf

) KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 11 MARKS: 150 TIME: 3 hours This question paper consists of 12 pages and 4 data sheets. Copyright reserved

Physiooouniamaded from Stanmorepgysics.com November 2025 2 INSTRUCTIONS AND INFORMATION A; 10. Copyright reserved This question paper consists of EIGHT questions. Answer ALL the questions in the ANSWER BOOK. Start EACH question on a NEW page in the ANSWER BOOK. Number the answers correctly according to the numbering system used in this question paper. Leave ONE line between two sub-questions, for example between QUESTION 2.1 and QUESTION 2.2. You may use a non-programmable calculator. You are advised to use the attached DATA SHEETS. Show ALL formulae and substitutions in ALL calculations. ne Round off your final ul t to a minimum of TWO decimal places. Give brief motivations, is ions et cetera where required. Write neatly and legi Please turn over

Physicalaoantalaaded from Stanmorapbysics.com November 2025 3 QUESTION 1: MULTIPLE-CHOICE QUESTIONS Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Choose the correct answer and write only the letter (A—D) next to the question number (1.1—1.10) in the ANSWER BOOK, for example 1.11 E. 1.1 The tendency of an atom to attract a bonded pair of electrons is known as ... A polarity. B electron affinity. C __ electronegativity. D__ ionisation energy. (2) 1.2 Which ONE of the following atomic combinations will have the shortest bond length? ‘ ‘jl A c-C SS B cC-O F | 1 Cc c=Cc oreph D CEC (2) 13 Which ONE of the following substances has a tetrahedral shape? A B c D COz SO2 HCN NHa* (2) 1.4 In which ONE of the following are |ON-DIPOLE forces present? A B c D NHs gas CO2gas LiC? aqueous solution MgCtesolid (2) Copyright reserved Please turn over

Prosiceiwmtetded from StanmoreWifysics.com November 2025 1.5 1.6 iG 1.8 One mole of a gas, SEALED in a container of volume V, exerts pressure p. The volume of the container is decreased to ‘V, while the temperature remains constant. What is the pressure now exerted by the gas? A ‘’p B 3p Cp D 15p (2) Consider the following statements about ideal gases: (i) The gas particles occupy no volume. (ii) | There are no intermolecular forces between the particles (iii) | The collisions between the particles are perfectly elastic Which of the above statements. are true? A (i)and (ii) only ws B (i) and (iii) only i > C (ji)and (ii) only D (i), (ii) and (iii) (2) Which ONE of the following solutions will have the greatest concentration of H* ions? A = 2mol-dm*NHs B 1,5 mol-dm* CHsCOOH C 1,5 mol-dm* H2SOs D = 1,5 mol-dm HCt (2) Consider the following reaction: HNOs (aq) + X — NO3 (aq) + HSO, (aq) In this reaction, X represents... A SOjacting as a base B s03° acting as an acid. C HeSOs acting as a base. D H2SOz4 acting as an acid. (2) Copyright reserved Please turn over

PhysicaDrowmdgended from Stanmorepseysics.com November 2025 5 1.9 The graph below shows the energy changes that take place during a chemical reaction. Course of reaction Which ONE of the following expressions corresponds to the activation energy of the reverse reaction? A Y-X BY C X-Z D Y-z (2) 1.10 Consider the following 10Cl + 2MnO« + 16H* = 2Mn?* + 8H20 + 5Cle A product of the reduction half reaction is ... A Mn* B Ck Cc H D Chk (2) [2x10 = 20] Copyright reserved Please turn over

PhysicalSaayaiaded from Stanmorephysics.com November 2025 QUESTION 2 (Start on a new page.) Consider the following reaction of nitric acid (HNOs) with lithium oxide (Li20). 2HNOs (aq) + Li2zO (s)— 2LiNOs (aq) + H20 (2) AH = — 139 kJ-mol 2.1 Define the term ionic bond. (2) 2.2 Show that the bond between Li and O is ionic. (2) 2.3 Write down the name of the salt formed in this reaction. (1) 2.4 Consider the H2O molecule. 2.4.1 Write down the valency of the O atom. (1) 2.4.2 Draw the Lewis structure for the H2O molecule. (3) 2.4.3 What is the shape of the H20 molecule? (1) 2.4.4 NAME the jon that is formed when H:0 forms a dative covalent bond. | i (1) 25 The above reaction takes plage in a beaker. How will the temperature of the contents of the beaker be afiected? Choose from INCREASES, DECREASES or REMAINS THE SAME. Give a reason for the answer. (2) 2.6 Define bond length. (2) 2.7 The diagram below shows the bonds that form between the atoms in the HNOs molecule. fo} ll H-O-N=0 Consider the bonds between the following atoms: (a) H-O (b)N-O (c)N=O 2.7.1 Is the bond length larger in (a) or in (b)? Explain the answer. (3) 2.7.2 Is the bond energy greater in (b) or in (c)? Give a reason for the answer. (2) [20] Copyright reserved Please turn over

7 QUESTION 3 (Start on a new page.) The following table provides the bond energies between atoms. 3.1 3.2 3.3 3.4 pry oreinlo ded from Stanmorephysics.com Bond Energy (kJ-mol"') C=C 839 c=0 804 O=O 498 Cc-C 348 0-0 145 H-O 463 Cc-O 358 H-C 413 Consider the following reaction: CHs (g) + 202 (9), — CO2(g) + 2H20 (g) Define the term a ” energy. Determine: vg P| 3.2.1 The activati rgy for the reaction 3.2.2 The heatof reaction in kJ-mol* November 2025 (2) (3) (4) How will the value calculated in QUESTION 3.2.2 be affected if a suitable catalyst is used? Choose from INCREASES, DECREASES or REMAINS THE SAME. Give a reason for the answer. (2) 24 g of CHa reacts completely with excess O2 at STP. 3.4.1 Is there a NET ABSORPTION or a NET RELEASE of energy? (1) 3.4.2 Calculate the energy referred to in QUESTION 3.4.1. (3) 3.4.3 If 5,5 moles of gas is present when the reaction is complete, calculate the initial number of moles of O2 (g) used in the reaction. a Please turn over Copyright reserved

Physicaleemelgaded from Stanmorephysics.com November 2025 8 QUESTION 4 (Start on a new page.) The graphs below show the relationship between vapour pressure and temperature for CHCts, CC& and H20 respectively. Atmospheric pressure is 101,3 kPa. ¢ "TI }_| 60 H} 40 20 = a a 0 20 40 60 80 86100 Temperature (°C) 4.1 Define the term vapour pressure. (2) 4.2 Write down the: 4.2.1 Vapour pressure of CC& at 70°C (1) 4.2.2 Temperature at which the vapour pressure of CHCfs is 100 kPa (1) 4.2.3 Phase of CHC: at 65°C (1) 4.2.4 Boiling point of CC& (2) 4.3 Fully explain the difference in the boiling points of CC&4 and H20. (4) 44 Is CC& soluble in HzO? Choose from YES or NO. Give a reason for the answer. (3) [14] Copyright reserved Please turn over

Physicdeowmneraied from Stanmorepttysics.com November 2025 QUESTION 5 (Start on a new page.) An experiment was conducted to investigate the relationship between pressure and volume of a fixed mass of gas at a temperature of 273 K. The results obtained are shown in the graph below. 4 5.1 (3) 5.2 (4) 5.3 (3) 5.4 Draw a sketch graph of pressure (y-axis) vs volume (x-axis) for this experiment. No values or labels are required on the axes. (2) The experiment is repeated using a greater mass of gas at 273 K. The results obtained are plotted on the same system of axes. 5.5 aph r the greater mass? Explain the answer. (3) [15] Copyright reserved Please turn over

PhysEabipereathed from Stanmorepiegsics.com November 2025 QUESTION 6 (Start on a new page.) 6.1 A 1,50 g sample of a compound containing carbon, hydrogen and oxygen was completely burned in air. The products formed were 1,74 g of carbon dioxide and 0,70 g of water. 6.1.1 Define the term empirical formula. 6.1.2 Determine, by calculation, the empirical formula of the compound. 6.1.3 If the ratio of empirical formula to molecular formula is 1:3, determine the molecular formula of the compound 6.2 An iron ore (impure sample) contains 30% Fe203. 6.2.1 Write down the oxidation number of iron in Fe2Os 6.2.1 Calculate the mass of Fe that can be obtained from 150 g of this ore. 6.3 In an experiment, a learner reacts 1,51 dm? of sulphuric acid, H2SO4, of concentration 0,43 mol-dm=,with solid lithium. The balanced equation for the reaction is: =} “ Hes. + 2Li — Li2SOs + He ae The graph below shows how the mass of lithium changes with time during the reaction: 4 ¥ 3 ” o oO = 1,24 > 0 Time (s) 6.3.1 Define the term limiting reagent. 6.3.2 Identify the limiting reagent in this reaction. Give a reason for the answer. 6.3.3 Calculate the value of Y as shown in the graph. Copyright reserved Please turn over (2) (7) (2) (1) (4) (2) (2) (7) [27]

Physidn@uenlended from Stanmorepkysics.com November 2025 11 QUESTION 7 (Start on a new page.) 7.1 Acetic acid (CHsCOOH) ionises in water according to the following balanced equation: CHsCOOH(2) + H2O0(t) = H30*(aq) + CHsCOO™ (aq) 7.14 Define an acid according to the Arrhenius theory (2) TA Write down the FORMULAE of TWO acids in this reaction. (2) 7.1.2 Is acetic acid a STRONG acid or a WEAK acid? Give a reason for the answer. (2) 7.2 A nitric acid solution, HNOs(aq), has a concentration of 0,15 mol-dm*. In a reaction, 31,3 cm* of the HNOs solution neutralises 24,5 cm? of a calcium hydroxide solution, Ca(OH)2(aq), according to the following balanced equation: 2HNOs(aq) + Ca(OH)2(aq) — Ca(NOs)2(aq) + 2H20(t) C24 Calculate the concentration of the Ca(OH)z solution. (5) 250 om: of the Nie ai NO: solution, with the same concentration of 0,15 mol-dm* is now rea pure Zn. Zn(s) + 2HNOs (aq) —> Zn(NOs)e (aq) + He (9) The pH of the final solution is 2,174. Assume that the volume of the solution does not change. 7.2.2 Calculate the mass of Zn that reacted. (8) 7.2.3 Determine the volume that the hydrogen gas produced will occupy at 25°C. Take molar gas volume as 24000cm*-mol" at 25°C (3) [22] Copyright reserved Please turn over

Physica eMalgaded from Stanmorephysics.com 12 November 2025 QUESTION 8 (Start on anew Page.) 8.1 A solution of iron (II) nitrate reacts with Ag (S) according to the following unbalanced ionic equation: At (s) + Fe** (aq) = At** (aq) + Fe (s) 8.1.1 Define the term redox reaction. (2) 8.1.2 Identify the reducing agent in the above reaction. (1) 8.1.3 Give a reason why the nitrate ion, NOs-, is a spectator ion in the reaction. (1) 8.1.4 Give a reason why the reaction is unbalanced. (2) 8.2 The incomplete ionic reaction, between dichromate ions ,Cr2O7*-, and tin (II) ions, Sn?*, in an acid medium is given below. = XN ei ee i | ~ ©r20745 + Sn? + Ht > Cr°* + Sn** 8.2.1 Write down the oxidation number of Cr in the Cr2O7?- ion. (1) 8.2.2 Write down ‘the reduction half-reaction. (2) 8.2.3 Using the Table of Standard Reduction Potentials, write down the balanced net ionic equation. (3) [12] TOTAL:150 Copyright reserved Please turn over

Physkepaaaag_ed from Stanmorephygics.com 13 DATA FOR PHYSICAL SCIENCES GRADE 11 PAPER 2 (CHEMISTRY) GEGEWENS VIR FISIESE WETENSKAPPE GRAAD 11 VRAESTEL 2 (CHEMIE) TABLE 1: PHYSICAL CONSTANTS/TABEL 1: FISIESE KONSTANTES November 2025 NAME/NAAM SYMBOL/S/IMBOOL VALUE/WAARDE Standard pressure e 5 Standaarddruk P 1,013 x 105 Pa Molar gas volume at STP apahie Molére gasvolume by STD Vim 22,4 dm*-mol Standard temperature Standaardtemperatuur Tw 273K Charge on electron : Lading op elektron is 1,6 x 107°C Avogadro's constant : Avogadro-konstante Na 6,02 x 1075 molt TABLE 2: FORMULAE/TABEL a m - N n=— n=— M Ny w 7 c=_— i ae V or/of n vs Ca hex ns m Cv, Ny PH = -log[H30*] PiV1= P2V2 PV =nRT Kw = [H30*][OH’] = 1 x 10-4 at/by 298 K Copyright reserved Please turn over

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Prtsonineeded from Stanmorephifsics.com TABLE 4A: STANDARD REDUCTION POTENTIALS TABEL 4A: STANDAARD-REDUKSIEPOTENSIALE Increasing oxidising ability/Toenemende oksiderende vermoé Copyright reserved Half-reactions/Halfreaksies £9 (v) Fxg)+2e = 2F +287 Co*+e = Co* +1,81 H,02+2H*+2e- = 2H,0 +1,77 MnO3 +8H*+5e° = Mn +4H,0 +151 Cl(g)+2e = 2Ch + 1,36 Cr0?" +14H* +620 = 2Ci*+7H,0 +1,33 Og)+4H*+4e° = 2H,0 + 1,23 MnOz+ 4H*+2e- = Mn*+2H,O +1,23 Pi +2e = Pt +1,20 Br(t)+2e = 2Br +1,07 NO} +4H'+3e = NO(g)+2H,0 + 0,96 Hg +2e- = H(t) + 0,85 Ag’+e = Ag + 0,80 NO3 +2H'+e = NOxg)+H,0 + 0,80 Fe*+e = Fe* +0,77 Ox(g) + 2H" + 2e°7 =) H202 +0,68 h+2e-e 2 + 0,54 Cu + 0,52 = S+2H,0 + 0,45 = 40H +0,40 = Cu +034 4 = SOXAg)+2H,0 +0,17 Ci +e Cur +0,16 Snt* + 2e- = Sn” +015 S+2H*+2e = H,S(g) +0,14 2H*+2e = Hg) 0,00 Fe*+3e = Fe —0,06 Pb*+2e- = Pb -0,13 Sn*+2e = Sn -0,14 NP? +2e = Ni -0,27 Co*+2e = Co - 0,28 Cd*+2e = Cd -0,40 Crete = Cr -0,41 Fe*+2e = Fe -0,44 Crt+3e = Cr -0,74 Zn*+2e = Zn -0,76 2H,0+2e == 2g) + 20H” 0,83 Cr*+2e = Cr -0,91 Mn?*+2e- = Mn -1,18 AM +3e = At - 1,66 Mg*+2e = Mg 2,36 Na’+e = Na -2,71 Ca*+2e « Ca 2,87 Sr+2e = Sr 2,89 Ba*+2e = Ba - 2,90 Cs*+e = Cs ~ 2,92 KY+e = K — 2,93 Lite - LU —3,05 November 2025 Increasing reducing ability/Toenemende reduserende vermoé Please turn over

Physiaiasenesitzed from Stanmorephyssics.com TABLE 4B: STANDARD REDUCTION POTENTIALS TABEL 4B: STANDAARD-REDUKSIEPOTENSIALE Increasing oxidising ability/ Toenemende oksiderende vermoé + Copyright reserved Half-reactions/Halfreaksies E (V) litte = Li - 3,05 Kt+e = K - 2,93 Cst+e = Cs — 2,92 Ba*+2e = Ba - 2,90 Srt+2e- = Sr —2,89 Ca%+2e = Ca — 2,87 Nat+e- = Na -2,71 Mg**+2e- = Mg — 2,36 AG@+3e = At — 1,66 Mn?++2e- = Mn -1,18 Cr++2e- = Cr -0,91 2H20 + 2e- = He(g) + 20H- - 0,83 Zn*+2e- = Zn - 0,76 Cr*+3e- = Cr -0,74 Fe*+2e- = Fe - 0,44 Cr+e- = Cre -0,41 Cd?*+2e- = cd -0,40 a - 0,28 ‘ -0,27 = Sn - 0,14 = Pb -0,13 = Fe — 0,06 = HA9) 0,00 = HeS(g) +0,14 = Sn + 0,15 Cu+e- = Cur +0,16 so oa +4Ht+2e- = SO2(g)+2H20 +0,17 Cu*+2e- = Cu + 0,34 2H20 +O2+4e- = 40OH- + 0,40 SO2+4H*+4e- = S+2H20 +0,45 Cu*+e- = Cu +0,52 lo+2e = 2b +0,54 O2(g) + 2H* + 2e- = H202 + 0,68 Fe*+e- = Fe% +0,77 NO; +2H*+e- = NO2(g)+H20 +0,80 Agtt+e = Ag +0,80 Hg?*+2e- = Hag(t) + 0,85 NO; +4H*+3e- = NO(g)+2H20 + 0,96 Bro(t)+2e- = 2Br + 1,07 Ptt+2e = Pt + 1,20 MnOQ2+4H* + 2e- = Mn**+2H20 + 1,23 O2(g) + 4H*+4e- = 2H20 +1,23 C0? +14H*+6e- = 2Cr°*+7H20 + 1,33 Ch(g)+2e- = 2Ct + 1,36 MnO, +8H*+5e- = Mn’ +4H20 +151 H202+2H*+2e- = 2H20 41,77 Co%*+e- = Co* + 1,81 F.(g)+2e° = 2F- + 2,87 Increasing reducing ability/ Toenemende reduserende vermoé November 2025

| KWAZULU-NATAL PROVINCE EDUCATION REPUBLIC OF SOUTH AFRICA NATIONAL SENIOR CERTIFICATE GRADE 11 These marking guidelines consist of 10 pages.

PryPoantoaded from Stanmorephys QUESTION 1 1.1 CY¥y 1.2 : Dvv 1.3 Dyv 1.4 Cyvv 1.5 Byv 1.6 Dyvv 1.7 Cvv 1.8 Avy 1.9 Dyvv 110 Avv Copyright reserved Marking Guidelines ICS . CORMVAZULU-NATAL/NOVEMBER 2025 (2) (2) (2) (2) (2) (2) (2) (2) (2) (2) [2 x 10 = 20]

Phyponteoaded from Stanmorephysics. CORWAZULU-NATALINOVEMBER 2025 Marking Guidelines QUESTION 2 2.1. The force of attraction between oppositely charged ions,” formed as a result of the transfer of electrons.v (2) 22 ee = } Any one v AEN > 2.1 ¥ (2) 23 lithium nitrate v (1) 2.4.1 2 v (Accept: -2) (1) Boe Ox Marking criteria/Nasienriglyne H ¥ Correct number of electrons surrounding each atom. Y Single bond between O and each of the two H atoms. vy Whole structure correct (3) 2.4.3 Bent / Angular |v (1) 2.4.4 hydroniumv (1) 2.5 Increases. ¥ There is t) release of energy / the reaction is exothermic. ” (2) 2.6 The average distance between the nuclei of two bonded atoms. vv (2 or 0) (2) 2.74 (b) ¥ Nitrogen atoms are larger than hydrogen atoms. v The nuclei will be further apart. v (3) 272 (c) ¥ The bond length in (c) is shorter OR there is a stronger force of attraction wih the double bond v (2) [20] Copyright reserved

Pryponteoaded from Stanmorephysics. CORWAZULU-NATALINOVEMBER 2025 Marking Guidelines QUESTION 3 3a The minimum energy needed for a reaction to take place. v Vv (2) “Eegpsorves = (4 x C-H) + (2 x O = 0) = (4x 413) v+ (2 x 498) V = 2648 kJ-mol' (3) 3.2.2 POSITIVE MARKING FROM Q3.2.1: Ereieased = (2x C=O) + (4x O—H) = (2 x 804) V+ (4 x 463) v = 3460 kJ-mol" AH = DY Eabsorbed = DEreleased (4) = 2648 — 3460 v = —812 kJ-mol! v 3.3 REMAINS THE SAME ok A catalyst has no effect on / of reactants or productsv OR A catalyst only affects activation energy. d (2) 3.4.1 Net release. v (1) 3.4.2 POSITIVE MARKING FROM Q3.2.2 je =13 v 1 = 1,5 mol Energy released = 1,5x 812 kJ Vv = 1218 kJ v (3) roy 3.4.3. POSITIVE MARKING FROM Q3.4.2 No2z that reacted = 2 xncua that reacted = 2x1,5 = 3molv Ncoz that formed = 1 xncw4 that reacted = 1x1,5 = 1,5 molv nu20 that formed = 2 xncua that reacted = 2x 1,5 = 3 molv Nozin excess = 5,5-—(1,5+3) = 1molv Initial noz=3 +1 = 4molv (5) [20] Copyright reserved

Phyponteoaded from Stanmorepshysics. CORWAZULU-NATALINOVEMBER 2025 Marking Guidelines QUESTION 4 41 The temperature at which the vapour pressure of a liquid equals atmospheric pressure. Vv 4.2.1. 80 kPa ¥ 4.2.2 60°C v [Accept answers between 60°C and 62°C] 4.2.3 Gasv 4.24 75°Cvv 4.3 H20 has a higher boiling point than CClav Intermolecular forces between H2O molecules is hydrogen bonding, while intermolecular forces between CCl4 molecules is London/Dispersion forces.v Intermolecular forces in H2O are stronger / Intermolecular forces in CCl4 are weaker. Vv More energy is required to overcome the intermolecular forces in H20. ¥ wh 44 No. ¥ i ry OPTION 1: cclemo s are polar V while H2O molecules are non-polar” a OPTION 2: the imte r forces in H2O are stronger than in CCl4 Vv so they are not of com strength. v QUESTION 5 5; Marking quidelines If any of the underlined key words/phrases are omitted: minus 1 mark. Boyle’s Law v Pressure of an enclosed gas is inversely proportional to the volume it occupies at constant temperature. vv 5.2 OPTION 1: OPTION 2: 1 115-101,3v 115 =——G = 2 5,5 x 10 55- dv 5,5 v piVi = pxVo V V = 207,45 dm’v (115,5)(181,82) = (101,3)V2 ” V2 = 207,31 dm? ¥ Copyright reserved (2) (1) (1) (1) (2) (4) (3) [14] (3) (4)

Prywponieoaded from Stanmorepshysics. CORWAZULU-NATALINOVEMBER 2025 5.3 5.4 5.5 QUESTION 6 6.1.1 6.1.2 Marking Guidelines HighY The volume of the gas particles is no longer negligible Y OR Intermolecular forces become significant. Increasing the pressure does not decrease the volume as predicted by Boyle’s LawY (3) Marking criteria Correct shape (hyperbola) vv Vv (2) Av The gradient is proportional to‘the number of moles. A has a larger gradient, yY and tf e a larger number of moles of gas’. (3) = 15) The smallest whole n ratio of elements in a compound. v Y (2 OR 0) (2) 12 Mass of C = 77 x 1,74 v = 0,475g 2 Mass of H = 18 x 0,70 v = 0,078g Mass of O = 1,5 — (0.475 + 0,078) v = 0,947 g 12 5 ~ 16 = 0,04 = 0,078 = 0,059 C: H: © 0,04 : 0,078 : 0.059 2:4:3 Molecular formula: C2H4O3 ¥ (7) (C2H4Os)3” Molecular formula is CsH1200 W (2) Copyright reserved

Prydpoasetoaced from Stanmorephysics. CORWAZULU-NATALINOVEMBER 2025 6.2.1 6.2.2 6.3.1 6.3.2 6.3.3 Marking Guidelines +30 M if Fe20: ail, 150 v lass of Fea! 3 = 700 * = 45g 112 =— Vv: v Mass of Fe 160 x45 =31,5gv Limiting reagent is the reactant that is used up completely in a reaction. vv OR Limiting reagent determines the maximum amount of product formed in a reaction. H2S04 v Lithium was left over after the reaction OR All the H2SO4 was used up during the reaction. v 7 calculate no fo mol of HxSO.” of mol of Li used. ~ calculate mass of Li used / initial mass of Li. ss of Li used to fi al mass of Li OR Addition of moles of Li used (f) Final answer. Initial mass of Li = 9,1 + 1,24 vv) =" Ya) czy ’ 0,43 = al y(b) Nu2so4 = 0,65 mol H2SO4 1,3=5 yo m= 9,1 g Liused Y=10,34g (0 Copyright reserved (1) (4) (2) (2) (7) [27]

Prydpoasetoaced from Stanmorephysics. CORWAZULU-NATALINOVEMBER 2025 Marking Guidelines QUESTION 7 7.1.1 Anacid is a substance that produces hydrogen ions (H+)/hydronium (2) ions (H3O*) when it dissolves in water. vv 7.1.2. CHsCOOH V and H30* v (2) 7.1.2 Weak acid. v It ionises incompletely/partially in water to form a low (2) concentration of H3O* ions. v 24 OPTION 1 CeVe _ MB Cava A Ca(24.5)v _ 1) (0,15)(31,3)v 2 OPTION 2 n =v = Vv “0313 n=4,7x 10° tary jorephysies.com HNOs : Ca(O 2:1v 0:3 mol Ca(OH)2 (5) Copyright reserved

PryDownloaded from Stanmoreph ysics. COWKWAZULU-NATAL/NOVEMBER 2025 Marking Guidelines 7.2.2 Marking criteria: (a) Substitution into formula n = cV to calculate initial concentration of HNOs. (b) Substitution of pH value (2,174) in pH formula. (c) Substitution into formula n = cV to calculate final concentration of H3O*. (d) Apply ratio nina(H3O*): nrina(HNOs) =1:1 to calculate final moles of HNOs. (e) Subtraction: nreactes (HNO3) = Mfinai (HNO3) - Mmitias (HNOs). (f) Apply ratio Nreacted (ZN) = % Nreacted (HNOs) to calculate Nreacted (ZN). (g) Substitute in formula m = nM to calculate mass of Zn that reactd. (h) Final answer. Ninitia((HNOs) = cV = 0,15 x 0,25 V = 0,0375 mol pH = —log[H30*] 2,174 = —log[H30*] v) [HsO*] = 6,7 x 10° mol-dm*? Nfina(H3O*) =cV = 6,.7.x103 x 0,25.v= 1,675 x 10° mol Nfina(HNOs) = 1, 103 mol v7“) Nreacted(HNOs) = 0,0: x 10° ve) Nreacted (Zn) m = 0,018 x65 v =1,16g v") [ 7.2.3 Nproduced (H2) = % Nreacted (HNOs) OR A x Mreddes (Zn) = % (0,0358) v OR 1(0,018) = 0,018 mol V = nVm = 0,018 x 24 000 v = 429,6 cm v (0,4296 dm?) Copyright reserved (8) (3) [22]

Prypomaveaded from Stanmorephysics. CoRWAZULU-NATALINOVEMBER 2025 Marking Guidelines QUESTION 8 8.1.1 Arreaction in which electrons are transferred. Vv (2) 8.1.2 AlY (1) 8.1.3. NOs ions do not undergo oxidation nor reduction Vv OR The oxidation numbers stay the same. (1) 8.1.4 The net charge on the left-hand side / for the reactants is not equal to the net charge on the right-hand side / for the products. vv (2) 8.2.1 +6Vv (1) 8.2.2 Cr207*- + 14H*+ 6e Y— 2Cr3* +7H20 Y (2) 8.2.3. 3Sn?* — 3Sn** Cr2O7? + 14H+ Cr2O072-+14H*+3Sr 3Sn**+7H20 VY —_ Balancingv (3) [12] TOTAL:150 Copyright reserved