FS-Physical-Sciences-Grade-12-March-2023-QP-and-Memo.pdf

education Department of Education FREE STATE PROVINCE TIME: 2 HOURS This paper consists of 12 pages and three information sheets. Copyright reserved Please turn over

Downloaded>fsom ‘Stanmorephysics. com FS/March 2023 Grade 12 INSTRUCTIONS AND INFORMATION 1. 10. 11. 12. Write your name and other information in the appropriate spaces on the ANSWER BOOK. This question paper consists of EIGHT questions. Answer ALL 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 pocket calculator. You may use appropriate mathematical instruments. You are advised to use the attached DATA SHEETS. Show ALL formulae and substitutions in ALL calculations. Round off your FINAL numerical answers to a minimum of TWO decimal places where applicable. Give brief motivations, discussions, et cetera where required. Write neatly and legibly. Copyright reserved Please turn over

Downloaded>fcom ‘Stanmorephysics. com FS/March 2023 Grade 12 QUESTION 1 Four options are provided as possible answers to the following questions. Each question has only ONE correct answer. Choose the answer and write down only the letter A, B, C or D next to the question number (1.1-1.10) in your ANSWER BOOK. 1.1 1.2 1.3 A suitcase is at rest on a table. Which ONE of the following is the reaction force to the weight of the suitcase, as described by Newton’s Third Law? A Force of the table on Earth B Force of suitcase on Earth Cc Force of the suitcase on the table D Force of the table on the suitcase (2) A horizontal force F is applied to a crate causing it to move over a rough, horizontal surface as shown below. F crate The kinetic frictional force between the crate and the surface on which it is moving depends on ... A the applied force F. B how fast the crate is moving on the surface. Cc the upward force exerted by the surface on the crate. D the surface area of the crate in contact with the floor. (2) Object P exerts a gravitational force F on object Q when the distance between their centres is r. The distance ris now DOUBLED. Which ONE of the following represents the gravitational force that P now exerts on Q? A “AF BF Cc 2F D 4F (2) Copyright reserved Please turn over

Downloaded>fcom ‘Stanmorephysics. com FS/March 2023 Grade 12 1.4 An object of mass m moving at velocity v collides head-on with an object of mass 2m moving in the opposite direction at velocity v. Immediately after the collision the smaller mass moves at velocity v in the opposite direction and the larger mass is brought to rest. Refer to the diagram below. Ignore the effects of friction. BEFORE COLLISION AFTER COLLISION Vv vy + « v7=0 m 2m m 2m OO OO OO LO Which ONE of the following is CORRECT? TOTAL MOMENTUM TOTAL KINETIC ENERGY A Conserved Conserved B Not conserved Conserved Cc Conserved Not conserved D Not conserved Not conserved (2) 1.5. A ball is thrown vertically upwards. Which ONE of the following physical quantities has a non-zero value at the instant the ball changes direction? A Velocity B Momentum Cc Acceleration D Kinetic energy (2) 1.6 Aballis released from rest from a certain height above the floor and bounces off the floor a number of times. The position-time graph represents the motion of the bouncing ball from the instant it is released from rest. D > time (s) When ignoring air resistance, which point (A, B, C or D) on the graph represents the position time coordinates of the maximum height reached by the ball after the SECOND bounce? (2) Copyright reserved Please turn over

Downloaded>fsom ‘Stanmorephysics. com Grade 12 1.7 1.8 1.9 Which ONE of the following momentum versus time graphs represents the FS/March 2023 motion of an object that starts from rest and moves in a straight line under the influence of a constant net force? A Impulse is equal to the ... A B Cc D p t B 4 p L____, t D pt —___» final momentum of a body. initial momentum of a body. Change in momentum of a body. rate of change in momentum of a body. Which ONE of the following combinations correctly indicates the STRONGEST intermolecular forces found in ethanol, ethanoic acid and ethyl ethanoate respectively? (2) (2) ETHANOL ETHANOIC ACID ETHYL ETHANOATE Hydrogen bonds Dipole-dipole forces Hydrogen bonds Hydrogen bonds Hydrogen bonds Dipole-dipole forces Hydrogen bonds Hydrogen bonds Hydrogen bonds oO} a] wD > Dipole-dipole forces Copyright reserved Hydrogen bonds Dipole-dipole forces Please turn over

Downloaded>fsom ‘Stanmorephysics. com FS/March 2023 Grade 12 1.10 The following is the structural formula for an organic molecule. if C—C—H x= xI—o—z | Cc Jef ce H H—-C—H H Which one of the following is the correct IUPAC name of this organic molecule? A 1-bromo-2-chloro-3-methylbutane B 4-bromo-3-chloro-2-methylbutane Cc 2-methyl-3-chloro-4-bromobutane D 2-methyl-4-bromo-3-chlorobutane (2) [20] QUESTION 2 Two blocks of masses 4 kg and 8 kg respectively are connected by light, inextensible string. A second light, inextensible string attached to block 4 kg block, runs over a frictionless pulley. A constant horizontal force, F, pulls the second string as shown in the diagram. The magnitude of the tension between the two blocks is 120 N. Ignore the effects of air resistance. F 4kg 120N 8kg 2.1. State Newton’s second law of motion in words. (2) 2.2. Drawa labelled free body diagram showing all the forces acting on the 4 kg block. (3) 2.3 Calculate the magnitude of force F applied on the system when it is accelerating. (5) Copyright reserved Please turn over

Downloaded: fcom ‘Stanmorephysics. com FS/March 2023 Grade 12 [10] QUESTION 3 Two metal balls A and B are rolling along in a horizontal straight line towards each other ina closed system. Ball A with a mass of 0,75 kg is rolling ata speed of 4 m-s™. Ball B with a mass of 1,25 kg collides head on with ball A at a speed of 3.m:s‘'. After collision ball A rolls in the direction opposite to its initial direction at a speed of 2,5 m-s". 4ms' 3m-s1 a, _ 3.1. Calculate the change in momentum experienced by ball A due to the collision. (4) 3.2 Use the change in momentum of ball B to calculate the velocity of ball B after collision. (4) 3.3. What is the net change in momentum for the whole system (ball A and ball B)? (1) 3.4 Calculate the magnitude of the average force that ball A and ball B exert on each other during collision if the two balls are in contact for 0,2 s. (3) 3.5 Is the collision ELASIC or INELASTIC? Explain the answer by means of calculations. (5) [17] Copyright reserved Please turn over

Downloaded>fsom ‘Stanmorephysics. com Grade 12 FS/March 2023 QUESTION 4 4.1. The graph below shows the velocity-time graph for the ball that is dropped and bounces. Ignore air resistance. Velocity-time graph I 8} ---------------- B 6 4 2 ® ' € 0 J 1 n { } Dp = ‘lA! T T i T > 0,2 0,4 0,6 018 1,071,2 1,4 t(s) -2 ! Aon nomen enn ra -6 -8 Two learners argue about the ball in the above scenario. One learner says the ball is a projectile, while the other says it is not. 4.1.1 Define the term projectile. (2) 4.1.2 Describe the motion of the ball between points A and B on the graph above. (2) 4.1.3 From the graph, determine the: (i) direction in which the ball is moving between points C and D (1) (ii) number of times the ball bounces (1) (iii) time at which the ball is at its maximum height Explain why the velocity at C is less than that at B. Copyright reserved (1) (2) Please turn over

Downloaded>fsom ‘Stanmorephysics. com FS/March 2023 Grade 12 4.2. Aballis thrown vertically upwards, from a balcony of a tall building, with a velocity of 12 m-s"'. On its way up, the ball passes a window which has a height of 1m. The balcony is 4 m above the ground. The velocity of the ball at the bottom of the window is 8,1 m-s"'. Ignore the effects of air resistance. Bt") A window jbalcony ground 4.2.1 Define the term free fall. (2) 4.2.2 Calculate the: (i) time taken for the ball to reach its maximum height (3) (ii) maximum height reached by the ball (3) (iii) | The time the ball takes to reach the top of the window (4) 4.2.3 Draw a velocity versus time graph for the motion of the ball from the moment that the ball is thrown upwards until it comes back to the position it was thrown from. Use the point from which the ball was thrown as reference. Clearly indicate the following on your graph: e The velocity with which the ball was thrown upwards. e The time taken by the ball to reach its maximum height. e The velocity with which the ball arrives on the ground. (3) [24] Copyright reserved Please turn over

Downloaded>fcom ‘Stanmorephysics. com FS/March 2023 Grade 12 QUESTION 5 A to F in the table below represent six organic compounds. A | 2,2,5-trimethylhex-3-yne B | CH3(CH2)3sCH3 H H QO | i I H H bs O H—C——C——C——H | | | tt L Cc H D | | Nou ~y H H—C——H H 7 | H f | ir Pi Er | E me 1 i, F H#—C-—— — ¢ a hy ty A | | | 4 ¢ 4 H H Te H | 5.1 Write down the: 5.1.1 Letters that represent TWO organic compounds that are isomers of each other (1) 5.1.2 Type of isomers (CHAIN, FUNCTIONAL or POSITIONAL) identified in QUESTION 5.1.1 (1) 5.1.3. GENERAL FORMULA of the homologous series to which compound D belongs (1) 5.1.4 NAME of the functional group of compound C (1) Copyright reserved Please turn over

Downloaded: fom ‘Stanmorephysics. com FS/March 2023 Grade 12 5.2 Write the IUPAC name of: 5.2.1. Compound B (1) 5.2.2 Compound F (3) 5.3 Write down the structural formula of compound A (3) [11] QUESTION 6 Compounds A, B and C, shown in the table below, are used to investigate a factor which influences the boiling point of organic compounds. COMPOUND A | CHsCH2CHO B | CHsCH2CH2CHO C | CHsCH2CH2CH2CHO 6.1 Define the term boiling point. (2) 6.2 | Which ONE of the compounds (A, B or C) has the highest boiling point? Explain (2) 6.3. For this investigation, write down the: 6.3.1 Independent variable (1) 6.3.2 Dependent variable (1) 6.4 Write down the names of the two types of van der Waals forces that occur between the molecules of compound A. (2) 6.5 How will the vapour pressure of 2-methylpropanal compare to that of compound B? Write down only HIGHER THAN, LOWER THAN or EQUAL TO. Fully explain the answer. (4) Copyright reserved Please turn over

Downloaded: fom ‘Stanmorephysics. com FS/March 2023 Grade 12 The boiling points of compounds D and E, shown in the table below are now compared. COMPOUND D- | CHsCH2COOH E CHsCH2CH2CH20H 6.6 — Write down the NAME of the functional group of: 6.6.1 D (1) 66.2 E (1) 6.7. The boiling point of compound D is HIGHER than that of compound E. Explain fully. (4) [18] GRAND TOTAL: 100 Copyright reserved Please turn over

Downloaded>fsom ‘Stanmorephysics. com Grade 12 DATA FOR PHYSICAL SCIENCES GRADE 12 PAPER 1 (PHYSICS) FS/March 2023 GEGEWENS VIR FISIESE WTENSKAPPE GRAAD 12 VRAESTEL 1 (FISIKA) TABLE 1: PHYSICAL CONSTANTS / TABEL 1: FISIESE KONSTANTES Elektronmassa NAME / NAAM SYMBOL / SIMBOOL | VALUE / WAARDE Acceleration due to gravity 2 Swaartekgagversnelling g 9,8 ms Universal gravitational constant . Universele gravitasiekonstante GS 6,67 x 10°" N:m?-kg? Radius of the Earth 6 Radius van die Aarde Re 6,38 x 10°m Mass of the Earth 34 Massa van die Aarde Me 5,98 x 10 kg Speed of light in a vacuum 8 A Spoed van lig in 'n vakuum c 3,0 x 10° ms Planck’s constant ; Planck se konstante h 6,63 x 104 J-s Coulomb’s constant \ 7 Coulomb se konstante k 9,0 x 10° N-m2-C# Charge of electron 49 Lading op elektron e -1,6 x10" C Electron mass me 9.11x 10°" kg Copyright reserved Please turn over

Downloaded: fom ‘Stanmorephysics. com FS/March 2023 Grade 12 TABLE 1: FORMULAE/TABEL 2: FORMULES MOTION/BEWEGING vp = vj + At Ax = vjAt + 5 ane? or/of Ay = v;At + salt? vp =v? + 2aAx orlof v? =v? + 2ahy Ax = (2) At orfof Ay = (2) At FORCE/KRAG Fret = ma p=mv fy" = UsN fie = UN FretAt = Ap w=mg mM: mm: M M F=6—> orlof F =G—>+ g=G5 oof g=Ga5 WORK, ENERGY AND POWER/ARBEID, ENERGIE EN DRYWING W = FAxcos@ U=mgh orlof E, =mgh K= mv? orlof = E, = mv Whee =A4K or/of Whee = AEx AK = K,—K;, otlof AEy = Exp — Exi Wne = AK +AU or/of Wace = AE, +AEp |p At Pave = FVave | Poemiaaeta = F Vgemidaela WAVES, SOUND AND LIGHT/GOLWE, KLANK EN LIG = 1 v=fa T=- f £ ES hi f= ep, or/of f= fp E=hf otlof B= E = Wg + Excmax) orlof E =Wo+Kmayx where/waar E = hf and/en Wy = hf and/en Excmax) = mvp ax orlof Kmay = + mviax Copyright reserved Please turn over

Downloaded>fsom ‘Stanmorephysics. com Grade 12 ELECTROSTATICS/ELEKTROSTATIKA FS/March 2023 kQ1Q2 kQ Re r2 Boa Ww F V=— E=-— q q n=2 oflof n=2 e de ELECTRIC CIRCUITS/ELEKTRIESE STROOMBANE rak emf (e) =1(R +r) 1 emk(e) = 1(R +r) Ry =R,+Rp+ ... q = IAt W=Vq WwW P=— At W =VIAt P=VI W =I?RAt P=LPR w Vat = 2 R po R ALTERNATING CURRENT/WISSELSTROOM 1 — Imax / I — !maks rms — V2 wok — V2 — Vmax _— Vmaks Vins ~ V2 / Vg ~ v2 Pave =Vemslrms | Paemiddeld = Vvgklwak — 72 — 72 Fave _ TrmsR / Pgemiddeld ~ TygkR Vins Vogk Pave = R / Poemiddela =, Copyright reserved

Dowmloaded from Stanmorephysics. com education Department of Education FREE STATE PROVINCE CONTROL TEST / KONTROLE TOETS GRADE 12 / GRAAD 12 PHYSICAL SCIENCES FISIESE WETENSKAPPE MEMORUNDUM MARCH 2023 / MAART 2023 MARKS: 100 / PUNTE: 100 TIME: 2 HOURS / TYD: 2 UUR This memorandum consists of 9 pages. Hierdie memorandum bestaan uit 9 bladsye. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

®ewnioaded.fcom ‘Stanmorephysics. com FS/March 2023 Fisiese Wetenskappe Kontrole toets 1 Grade/Graad 12 Memo FS/Maart 2023 QUESTION 1/VRAAG 1 11. D¥vY 12 Cv¥v 13 AVY 14 Cv¥vY 15 Cv¥v 16 DYY 1.7 DvYY 18 Cv¥v 19 BYvY 1.10 Avv [20] QUESTION 2/VRAAG 2 2.1 Marking criteria/Nasienkriteria (If any of the underlined key words/phrases in the correct context is omitted deduct 1 mark/Indien enige van die onderstreepte sleutelwoorde/frases in die korrekte konteks uitgelaat is, trek 1 punt af.) When a net force acts on an object, the object will accelerate in the direction of the force. The acceleration is directly proportional to the force and inversely proportional to the mass of the object. vv Wanneer 'n netto krag op 'n voorwerp inwerk, sal die voorwerp in die rigting van die krag versnel. Die versnelling is direk eweredig aan die krag en omgekeerd eweredig aan die massa van die voorwerp. VV (2) 2.2 FY wy 120 N y (3) Accepted labels/Aanvaarde benoemings w_ | F,/Fw/weight/mg/gravitational force/ Fearth on 4kg block F/Fw/gewig/mg/gravitasie krag/ Faarde op 4kg blok T | FT/Ft/spanning/Fs FT/Ft/spanning/Fs F Fappliea/ Ftoegepas Notes/Aantekeninge e Mark awarded for label and arrow./Punt toegeken vir benoeming en pyitjie. e Do not penalise for length of arrows since drawing is not to scale./Moenie vir die lengte van die pyltjies penaliseer nie aangesien die tekening-nie volgens skaal is nie. e Any other additional force(s)/Enige ander addisionele krag(te): Max/Maks 2/5 e If everything correct, but no arrows/Indien alles korrek, maar geen pyltjies: Max/Maks 2/5 e If force(s) do not make contact with the dot//ndien krag(te) nie met die kolletjie kontak maak nie: Max/Maks 2/. Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

Oounicaded>.from ‘Stanmorephysics. com Fisiese Wetenskappe Kontrole toets 1. Grade/Graa 2.3 FS/March 2023 12 Memo FS/Maart 2023 UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF For the 8 kg block/Vir die 8 kg blok Fret = maw T-w=ma 120 - (8 x 9,8) = 8av a=5,2ms* ———+—» For the 4 kg block/Vir die 4 kg blok Fret = ma F-T-w=ma F -120 - (4x 9,8)v = 4(5,2)v F = 180 NV DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF For the 8 kg block/Vir die 8 kg blok Fret = mav -Ttw=ma -120 +(8 x 9,8) = 8av a=-5,2ms2 ————» For the 4 kg block/Vir die 4 kg blok Fret = ma -F +T+w=ma -F +120 + (4 x 9,8)v = 4(-5,2)v F = 180 NY QUESTION 3/VRAAG 3 3.1 3.2 Right as positive/Reg as positief Apa = m(vs— vi) ¥ = 0,75( -2,5 — 4) v = -4,875 kgm:s* (5) [10] «Ap = 4,875 kgm-s"' left/ in opposite direction. “/ links/ in teenoorgestelde rigting. Left as positive/Links as positief Apa = m(v¢ — Vi) v = 0,75[2,5 - (-4)] v = 4,875 kgm:s"t «Ap = 4,875 kgm-s"'v left/ in opposite direction. “/ links/in teenoorgestelde rigting. (4) Ape = -ApaY Ape = - (-4,875) = 4,875 kgm:s"! \ WA Aps = m(vF— vi) % 4,875 =1,25 (v4 — (-3)) Vv vi = 0,9 m-s Right/ in opposite direction /Regs/ in teenoorgestelde rigting Aps = -4,875 =1,25 (v;— 3) v7 m(vs — vi) ¥ vi = -0,9m-s"t vi = 0,9 m-s* Right/ in opposite direction /Regs/ in teenoorgestelde rigting 3.3 0 (zero/Nul) ¥ Copyright reserved/Kopiereg voorbehou (4) (1) Please turn over/Blaai asseblief om

ewnioaded.fcom ‘Stanmorephysics. com FS/March 2023 Fisiese Wetenskappe Kontrole toets 1 Grade/Graad 12 Memo FS/Maart 2023 3.4 POSITIVE MARKING FROM 3.1 AND 3.2/ POSITIEWE NASIEN VAN 3.1 EN 3.2 Object A/Voorwerp A OR/OF | Object A/Voorwerp A FretAt = Apy FrecAt = Apy Fret (0,2) = —4,875V FretAt = ma(Vag — Vai) Fret = -24,375 N Fyee(0,2) = 0,75(—2,5 — 4)¥ Magnitude/Grootte = 24,375 NV Fret = -24,375 N Magnitude/Grootte = 24,375 NV Object B/Voorwerp B OR/OF | Object B/Voorwerp B FnecAt = Apv FrecAt = ApyY Fret (0,2) = 4,875¥ FrecAt = Mg (Vpe — Vai) Fret = 24,375 N Fret (0,2) = 1,25[0,9 — (-3)]” Magnitude/Grootte = 24,375 NV Fret = 24,375 N Magnitude/Grootte = 24,375 NV (3) 3.5 POSITIVE MARKING FROM 3.1 AND 3.2/ POSITIEWE NASIEN VAN 3.1 EN 3.2 IK; = 5myv} + Mav} v = +(0,75)(4)? +2(1,25)8)°¥ = 11,625 J UK = =mgv3 += mgv3 if = 7MaVvg +> mavg = 5 (0,75)(2,5)? + 5 (1,25) (0,9)? = 4,369 J Inelastic /Onelasties ¥ 2K; # XK, orlof EKy < UK,“ (5) [17] QUESTION 4/VRAAG 4 4.1.1. An object which has been given an initial velocity and then.it moves under the influence of the gravitational force only. 'n Voorwerp wat 'n aanvanklike snelheid gekry het en daarna beweeg dit slegs onder die invloed van gravitasiekrag. (2) 4.1.2 It is moving with constant acceleration. Vv“ Dit beweeg met konstante versnelling. (2) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

ounicaded>from ‘Stanmorephysics. com Fisiese Wetenskappe Kontrole toets 1. Grade/Graa 4.1.3 4.1.4 4.2.1 4.2.2 (i) (ii) (ili) 1.2 (s)¥ 12 Memo Upwards/opwaarts ¥ Once/ one/1 v / Een keer/ een/1 FS/March 2023 FS/Maart 2023 (1) (1) (1) Energy is lost“ in deforming the ball “during the bounce. Energie gaan verlore v_met die vervorming van die bal” tydens die botsing. (2) Motion under the influence of gravity/weight/gravitational force only. vv Beweging slegs onder die invloed van gravitasie/gewig/swaartekrag. (2 or/of 0). OR/OF Motion during which the only force acting on an object is the gravitational force. Beweging waar die enigste krag wat op die liggaam inwerk, gravitasie/ gewig/swaartekrag is. (2 or/of 0). (2) (i) VF = vi + aAtyY Vr = vi + aAtY 0 = 12 +(-9,8)Atv 0 = -12 +(9,8)Atv At = 1,22 sv At = 1,22 sv (3) ve = v7 + 2adyv 0 = 122 + 2(-9,8)Ayv Ay = 7,35 mv v7 = v7 + 2abyY 0 = -122 + 2(9,8)Ayv Ay = 7,35 mv (iii) UPWARDS AS POSITIVE/OPWAARTS AS POSITIEF vp = vi + 2ady 0 = 8,12 + 2(-9,8)(1) v vr = 6,78 m-s" Vi = vi + aAtv 6,78 = 8,1 +(-9,8)Atv’ At = 0.134 sv Ay = vit +5ahtev 1 = 8,1t +4(—9,8)At7v At = 0.134 sv 1= CHS) atv At =0.134 sv Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

ewnioaded.fcom ‘Stanmorephysics. com FS/March 2023 Fisiese Wetenskappe Kontrole toets 1 Grade/Graad 12 Memo FS/Maart 2023 DOWNWARDS AS POSITIVE/AFWAARTS AS POSITIEF v7 = v? + 2aAy 0 = -8,17 + 2(9,8)(1) W vi = 6,78 mst v= vit atv. Ay = vit +5adev -6,78 = 8,1 +(9,8)Atv’ oe 1 ay At = 0.134 sv’ 1= 8,1t +5 (9,8)de At = 0.134 sv _ vitvE Ay = (“£) atv -1= —8,1+(—6,78) (ay At =0.134 sv (4) 4.2.3, POSITIVE MARKING FROM QUESTION 4.2.2/POSITIEWE NASIEN VANAF VRAAG 4.2.2 UPWARDS AS POSITIVE / DOWNWARDS AS POSITIVE / OPWAARTS AS POSITIEF AFWAARTS AS POSITIEF v(m-s") 4 v (m-s*) 12 12 0 > 0 > t (s) 1,22 t (s) -12 -12 Criteria for graph/Kriteria vir grafiek Straight line starting at v = 12 m-s"' with a negative final velocity or straight line | “ starting at v = -12 m:s" with a positive final velocity./Reguitlyn wat begin by v=12m-s" met negatiewe eindsnelheid of reguitlyn wat begin by -12 m-s" met positiewe eindsnelheid. Straight line cuts time axis calculated in Question 4.2.2/Reguiltlyn sny tyd-as by | “ die tyd bereken in Vraag 4.2.2. Correct final velocity of 12 m-s"' or -12 m:s"'/Korrekte eindsnelheid van v 12 m:s" of -12 ms". (3) [24] Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

ewnioaded.fcom ‘Stanmorephysics. com FS/March 2023 Fisiese Wetenskappe Kontrole toets 1 Grade/Graad 12 Memo FS/Maart 2023 QUESTION 5/VRAAG 5 5.1.1 B,E v (NB: both compounds must be correct to award one mark./ beide verbindings moet korrek wees om een punt toe te ken). (1) 5.1.2 Chain isomers/Kettingisomere “ (1) 5.1.3 CpH2n+1 COOHY (RCOOH) (1) 5.1.4 Ketonev/Ketoon (1) 5.2.1 Pentanev/Pentaan (1) 5.2.2 2,2-dimethylv hex-3v -enev /2, 2-dimetiel ~heks-3 v-eenv (3) 5.3 i i | H— y a H —c C co F eH "| | He TH H Marking guideline/Nasienriglyn Functional group (triple bond) in correct v position/Funksionele groep (drievoudige binding) in korrekte posisie The three methyl groups in correct positions/Die v drie metielgroepe in korrekte posisies Whole structure correct/Hele struktuur korrek v (3) [11] QUESTION 6/VRAAG 6 6.1. NB: Deduct 1 mark for omission of any of the underlined words in the correct context./Trek 1 punt af vir weglating van enige van die onderstreepte woorde in die korrekte konteks. The temperature at which the vapour pressure of a liquid equals atmospheric pressure. Vv Die temperatuur waarteen die dampdruk van 'n vioeistof gelyk_is aan die atmosferiese druk. VV (2) 6.2 C,” Has the longest chain length/Het die langste kettinglengte. v (2) 6.3.1 Chain length/Kettinglengtev (1) 6.3.2 Boiling point (bp)/Kookpunt (kp) “ (1) Copyright reserved/Kopiereg voorbehou Please turn over/Blaai asseblief om

ewnioaded.fcom ‘Stanmorephysics. com FS/March 2023 Fisiese Wetenskappe Kontrole toets 1 Grade/Graad 12 Memo FS/Maart 2023 6.4 6.5 6.6.1 6.6.2 6.7 Copyright reserved/Kopiereg voorbehou London forces v (and) Hydrogen bondsw/Londenkragte ~ (en) waterstofbindings. (2) Higher/Hoér. v e The compounds have the same functional group and the same molecular mass. 2-methylpropanal has branching and hence a smaller surface area. v 2-methylpropanal has weaker intermolecular forces” e Less energy is required to overcome (weaker) intermolecular forces. “ e Die verbindings het dieselfde funksionele groep en dieselfde molekulére massa. e 2-metielpropanaal het ’n vertakking en dus 'n kleiner opperviakte. v~ 2-metielpropanaal het swakker intermolekulére kragte “” Minder energie is nodig om (swakker) intermolekulére kragte te oorkom.v (4) Carboxyl group/Karboksielgroep. ¥ (1) Hydroxyl group/Hidroksielgroep. ~ (1) Both compounds D and E have hydrogen bonding between molecules. “ Compound E has one site for hydrogen bonding,” while compound D has two sites for hydrogen bonding (can form dimers). More energy is needed to overcome intermolecular forces in compound D. ” Beide verbindings D en E het waterstofbinding tussen molekules. Vv Verbinding E het een plek/posisie vir ‘n waterstofbinding, Vv terwyl verbinding D twee plekke/posisies het vir waterstofbindingsv (kan dimere vorm). Meer energie is nodig om die intermolekulére kragte in verbinding D te oorkom.v (4) [18] GRAND TOTAL: 150 GROOTTOTAAL: 150 Please turn over/Blaai asseblief om