Physical Sciences Grade 12 FS March 2022 Memo

education Department of Education FREE STATE PROVINCE OL TEST MARKS: 100 TIME: 2 HOURS This paper consists of 11 pages and one information sheet. Copyright reserved Please turn over

t Grade 12 FS/March 2022 INSTRUCTIONS AND INFORMATION 1. Write your name and other information in the appropriate spaces on the ANSWER BOOK. 2. This question paper consists of SIX questions. Answer ALL questions in the ANSWER BOOK. 3. Start EACH question on a NEW page in the ANSWER BOOK. 4. Number the answers correctly according to the numbering system used in this question paper. 5. Leave one line between two sub-questions, for example between �STION L- 2.1 and QUESTION 2.2. 6. You may use a non-programmable pocket calculator. 7. You may use appropriate mathematical instrument �0 � 8. You are advised to use the attached DATA SH��- 9. Show ALL formulae and substitutions in �culations. 10. Round off your FINAL numerical ans��to a minimum of TWO decimal places where applicable. 11. Give brief motivations, discu�s, et cetera where required. Q 12. Write neatly and legibly\\ Copyright reserved Please turn over

FS/March 2022 Grade 12 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 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 When two objects collide during an ELASTIC COLLISION, A both momentum and kinetic energy are conserved. B both impulse and momentum are conserved. C only kinetic energy is conserved. D only momentum is conserved. (2) 1.2 A vector quantity with the same DIRECTION as the veloo ;n object � is the ... of the object. ...._\C, A rate of the change in momentum B momentum C impulse D inertia (2) 1.3 An airbag can protect the dr��f a vehicle from serious injuries during a collision. Which one of t (\JPtlowing best describes how that is possible? Net force on the driv Remain the same (2) Copyright reserved Please turn over

FS/March 2022 1.4 Two forces of 200 N and 100 N are simultaneously applied to a stationary box that has been placed on a flat surface. One of the forces is horizontal and the other one is applied at an angle of 30° to the horizontal as shown below. F1 = 200 N 30° 50 kg The normal force acting on the box is DECREASED by ... A increasing the angle at which the 200 N force is acting. B decreasing the angle at which the 200 N force is acting. C decreasing the magnitude of F1. D increasing the magnitude of F2. (2) 1.5 A ball is thrown vertically upward and returns to the thrower's hand. Taking upward as POSITIVE, which one of the following combinations best describes the velocity and acceleration of the ball when it is moving DOWNWARDS towards the thrower's hand? Ignore air resistance. Velocity Acceleration A + + B + - C - + D - - (2) 1.6 Consider the velocity versus time graph for an object moving VERTICALLY. Upward is taken as positive. Which one of the following statements is correct? A The object's speed is decreasing throughout the motion. B The object is travelling downwards throughout the motion. C The object is travelling with a constant velocity throughout the motion. D The object is travelling with a constant acceleration throughout the motion. (2) Copyright reserved Please turn over

FS/March 2022 1. 7 An object is dropped from a hot air balloon moving upward at a constant velocity. Which one of the following position versus time graphs best represents the motion of the object UNTIL IT HITS THE GROUND? A -- B -- E E C: 0 C: :2 0 (J) :;:::; ·;:;; 0 time (s) 0 0... 0... time (s) -- E C: - E C 0 time (s) D C: 0 ·;:;; :;:::; :E (J) 0 0 0... 0... time (s) (2) 1.8 The condensed structural formula off an organic compound is shown below. CH3 CH3 --CH--CH I I Br I CH3 Which one of the following is the correct IUPAC name of this compound? A 2-methyl-3-bromobutane B 2-bromo-3-methylbutane C 2-bromo-1, 1-dimethylpropane D 2-bromo-3,3-dimethylpropane (2) 1.9 Consider the functional group on the right. For which 0 one of the following homologous series is this the functional group? 11 I C-0-C- A Aldehydes / I B Alcohols C Ketones D Esters (2) Copyright reserved Please turn over

FS/March 2022 Grade 12 1.10 The MELTING point of a compound is the ... A minimum temperature at which it boils. B maximum temperature at which it boils. C temperature at which its vapour pressure equals atmospheric pressure. D temperature at which the solid and liquid phases of a substance are at equilibrium. (2) [20] QUESTION 2 Two crates, A and 8, with masses of 8 kg and 5 kg respectively, are stationary on a rough, horizontal surface. The crates are connected by a light, inextensible string. When force F, with a magnitude of 80 N and making an angle of 30 ° with the horizontal, is applied to the 8 kg block, both blocks move to the right. F 8 A 5 kg 8 kg 2.1 State Newton's second law of motion in words. (2) 2.2 Draw a free-body diagram of ALL the forces acting on the 5 kg block (4) 2.3 The magnitudes of the frictional foroes acting on crates A and 8 are 7,68 N and 4,9 N respectively. Calculate the magnitude of the acceleration of block 8. (5) [11] Copyright reserved Please turn over

FS/March 2022 Grade 12 QUESTION 3 A truck of mass 2 000 kg is moving - Vi truck ___ eastward and collides with a car of mass 900 kg moving at a speed of 4,5 m-s- 1. After the collision, the -....E >, I I truck and the car are entangled ·u0 3,2 -----------�--------,,._..., I into a wreck which moves as ONE 0 2 / 3,5 time (s) UNIT with a speed of 3,2 m·s- 1. :/ The graph (not drawn to scale) - 4,5 ___ y represents the motion of the car vehicles just before and after the collision. 3.1 Are the vehicles moving in the SAME or OPPOSITE directions BEFORE the collision? (1) 3.2 State the principle of conservation of momentum in words. (2) 3.3 What do you understand by the term isolated system as used in physics? (2) Use the information in the graph to answer the following questions. 3.4 How long is the collision between the car and the truck? (1) 3.5 Use a calculation to show that the speed of the truck before the collision, v. in the graph, is equal to 6,665 m-s- 1. (4 ) 3.6 Calculate the average net force acting on the car during the collision. (4 ) 3.7 Determine, by means of calculations, what type of collision this is (elastic or inelastic) and give a reason as well for your choice. (5) 3.8 How does the magnitude of the average net force exerted by the truck on the car compare with the magnitude of the average net force exerted by the car on the truck? Choose from greater than, smaller than or equal to. Name a physics law or a principle to support your answer. (2) [21] Copyright reserved Please turn over

FS/March 2022 Grade 12 QUESTION 4 An object is projected vertically upward at a velocity of 10 m-s- 1 from the top of a building, which is 40 m high. Ignore air resistance. • 10 m-s- 1 40 m , Ground 4.1 Define the term projectile in words. (2) 4.2 Calculate the: 4.2.1 Maximum height the object reaches above the ground. (4) 4.2.2 Time it takes the object to hit the ground (from the instant it has been projected). (4) 4.3 Draw the following set of axes in your answer book. a Use it to draw a sketch graph of acceleration versus time to represent the motion of the object from the moment it is projected from the top of the building (at t = 0) until it strikes the GROUND. t1 and t2 represent the times when the object is at its HIGHEST position and when it strikes the GROUND respectively. GIVE AN INDICATION NEXT TO YOUR GRAPH WHICH DIRECTION YOU CONSIDER AS POSITIVE. (2) Copyright reserved Please turn over

FS/March 2022 Grade 12 The object is again projected vertically upward from the top of the building at 10 m-s- 1 as before. Someone in the building opens a window while the object is on its way up. On its way down, the object strikes the top of the window, at a height h, 3,4 s after it has been projected from the top of the building. Ground 4.4 Calculate the: 4.4.1 Magnitude of the velocity of the object when it strikes the top of the window. (3) 4.4.2 Height, h, above the ground. (5) [20] Copyright reserved Please turn over

FS/March 2022 Grade 12 QUESTION 5 The letters A to H in the table below represent eight organic compounds. H I HI-C-H H H H H H H I I I I I I H-C-C-C-C C C C-H A C 4 H 10 0 B I I I I I H H H H H H-C-H H-C-H I H-C-H H I H H H 0 H H H H H I I 11 I I I I H-C-C-C-C-H D H-C-C-C-C==C I I I I I I I \ H H H H H H H H E 3,4-dimethylpentan-2-ol F CH3(CH2 )2COOH G Ethyl ethanoat@ H P@ntan-3-on@ 5.1 Define the term saturated compound. (2 ) 5.2 Write down the following: 5.2.1 Letter that represents an UNSATURATED compound. (1) 5.2.2 IUPAC name of B (3 ) 5.2.3 Letter that represents a FUNCTIONAL ISOMER of compound F. (1) 5.2.4 NAME of the functional group of compound C (1) 5.2.5 General formula of the homologous series to which compound D belongs. (1) 5.3 Define the term homologous series. (2 ) 5.4 For compound E: 5.4.1 To which homologous series does it belong? (1) 5.4.2 Write down its CONDENSED STRUCTURAL FORMULA. (2) Copyright reserved Please turn over

FS/March 2022 Grade 12 5.4.3 Is it a primary, secondary, or tertiary compound? (1) 5.4.4 Explain your answer to question 5.4.3. (1) [16] QUESTION 6 A learner uses four organic compounds (A, B, C and D) to investigate the effect of the CHAIN LENGTH on BOILING POINT. The obtained results are shown in the table below. Condensed structural Compound Boiling point (° C) formula A CH3CH2CH2CH2CH2OH 138 B CH3CH2CH2OH 96 C CH3CH20H 77 D CH3OH 64 6.1 Define the term vapour pressure. (2) 6.2 Write down the INDEPENDENT variable for this investigation. (1) 6.3 State, with a reason, which ONE (A, B, C or D) of these compounds has the HIGHEST vapour pressure. (2) 6.4 Compound A is now compared to pentane. 6.4.1 Is the boiling point of A HIGHER THAN, LOWER THAN or EQUAL TO that of pentane? (1) 6.4.2 Refer to the TYPES of intermolecular forces to explain the answer to question 6.4.1. (4) 6.5 Write down the general conclusion that can be made about the boiling points of compounds A, B, C and D. (2) [12] GRAND TOTAL: 100 Copyright reserved Please turn over

Grade 12 FS/March 2022 DATA FOR PHYSICAL SCIENCES GRADE 12 CONTROL TEST - TERM 1 GEGEWENS VIR FIS/ESE WETENSKAPPE GRAAD 12 KONTROLETOETS - KWARTAAL 1 TABLE 1: PHYSICAL CONSTANTS/ TABEL 1: FIS/ESE KONSTANTES NAME/NAAM SYMBOUS/MBOOL VALUE/WAAROE Acceleration due to gravity Swaartekragversnelling g 9,8 m · s-2 TABLE 2: FORMULAE/ TABEL 2: FORMULES MOTION/ BEWEGING v1 = vi + af1t !ix= vif1t + !;,af1t 2 or/of 11y = vif1t + !;, af1t 2 2 2 v} = v?- + 2af1x or/of v} = vf + 2af1y (�+v_i·) 11t or/of 11y = (� !ix = - - +v ·) _i 11t 2 2 FORCE/ KRAG Fne t = ma p =mv Fne tf1t = 11p 11p = mv1 - mvi µs - µs - _ fs(max) / _ fs(maks) N N µk =- N WEIGHT AND ENERGY/ GEWIG EN ENERGIE w = mg or/of F9 = mg U = mgh or/of Ep = mgh K = �mv 2 2 or/of Ek = �mv 2 2 Copyright reserved

��t� education ���. r;J � ,�t. .. ... r � � ' ' Department of Education FREE STATE PROVINCE CONTROLTEST/KONTROLETOETS GRADE 12/ GRAAD 12 PHYSICAL SCIENCES FIS/ESE WETENSKAPPE MEMORANDUM MARCH 2022/ MAART 2022 MARKS: 100 / PUNTE: 100 This memorandum consists of eight pages. Hierdie memorandum bestaan uit agt bladsye. Copyright reserved / Kopiereg voorbehou Please turn over/ 8/aai asseblief om

FS/March 2022 / VS/Maart 2022 Grade 12 / Graad 12 MEMORANDUM QUESTION 1 / VRAAG 1 1.1 A✓✓ 1.2 B ✓✓ 1.3 C ✓✓ 1.4 A✓✓ 1.5 D ✓✓ 1.6 D ✓✓ 1.7 A✓✓ 1.8 B ✓✓ 1.9 D ✓✓ 1.10 D ✓✓ [20] QUESTION 2 / VRAAG 2 2.1 When a net force is applied to an object, the object accelerates in the direction of the net force. Acceleration is directly proportional to the net force ✓and inversely proportional to the mass of the object. ✓ Wanneer 'n resu/terende/netto krag op 'n voorwerp inwerk, versnel die voorwerp in die rigting van die netto krag teen 'n versnelling direk eweredig aan die netto krag en omgekeerd eweredig aan die massa van die voorwerp. (2) Accepted labels 2.2 N Aanvaarbare b skrifte F9 / gravitational force / weight w ✓ F I ravitasiekra I ewi fk T FN / normal force / normal N ✓ FN I normaalkra I normaal Fr/tension T ✓ w Frls Frriction/f/friction / kinetic frictional force fk ✓ F wrywin/f/wrywing/kinetiese (4) 2.3 A: Fnet = ma ✓ F H + T + fk = ma } ---... ✓ both/beide 80cos 30° -T -7.68 ✓ = Ba T = 61,60 -Ba ................................. . ............. ( 1) B: Fnet = ma T + fk = ma T-4 9 ✓=Sa T = Sa + 4, 9 .................................................. (2) Solve for (1) and (2) a = 4,36 m · s- 2 ✓ (5) [11] Copyright reserved / Kopiereg voorbehou Please turn over/ 8/aai asseblief om

FS/March 2022 / VS/Maart 2022 Grade 12 / Graad 12 MEMORANDUM QUESTION 3 I VRAAG 3 3.1 Opposite (direction) ✓ Dieselfde (rigting) (1) 3.2 The total (linear) momentum of an isolated system remains constant (is conserved). Die totale lineere momentum in 'n geYso/eerde sisteem bly konstant (behoue). ✓ ✓ (2) 3.3 A system on which the net external force is zero. ✓ ✓ 'n Sisteem waarop die netto, eksterne krag nut is. (2) 3.4 3,5 - 2 = 1,5 s✓ (1) 3.5 :E p(before/voor) = :E p(after/na) } ✓ mtVit + mcVic = (mc+mt)Vt (2 000)Vit ✓ + 900 x (-4,5) ✓ = (2900)(3 ,2)✓ Vit =6,665 m·s- 1 (No mar k here. / Geen punt hier.) (4) 3.6 POSITIVE MARKING FROM 3.4. / POSIT/EWE NASIEN VANAF 3.4. Fnet�t = L'i p✓ Fnet�t = Lip ✓ Fnet( 1,5)✓ = 900(3,2 -(-4,5) ✓ Fnet(1,5) ✓ = 2 000(3,2 -6,665)✓ Fnet = 4 620 N Fnet = -4 620 N Fnet = 4 620 N; opposite to car's Fnet (on car/ op motor)= 4 620 N; direction of motion / teenoorqesteld aan opposite to car's direction of motion I motor se beweqinqsriqtinq ✓ teenoorqesteld aan motor se beweqinqsriqtinq ✓ (4) 3.7 �:K(before) = ½mv2 + ½mv2 -------­ ✓ Anyone (voor) = ½(2000)(6,665 )2 + ½(900)(-4,5)2 ✓ Enigeen = 53 534,725 J LK(after) = ½mv2 (na) = ½(2 900)(3,2)2 ✓ = 14 848 J LK(before/voor) # LK(after/na) ✓ OR/OF LK(before/voor) > LK(after/na) lnelastic/One/asties ✓ (5) 3.7 Equal to✓ Gelyk aan Newton's third law of motion✓ Newton se derde bewegingswet (2) [21] Copyright reserved / Kopiereg voorbehou Please turn over/ 8/aai asseblief om

FS/March 2022 / VS/Maart 2022 Grade 12 / Graad 12 MEMORANDUM QUESTION 4 / VRAAG 4 4.1 An object which has been given an initial velocity and then it moves under the influence of the gravitational force only. ✓ ✓ 'n Voorwerp waaraan 'n beginsnelheid gegee is en wat dan slegs onder die invloed van die gravitasiekrag beweeg. (2) 4.2.1 U ositive / 0 ositief Down ositive / Af ositief Vt 2=Vi 2 + 2a�y ✓ Vf 2=Vi 2 + 2a�y ✓ 0 ✓= 10 2 + 2(-9,8) �y ✓ 0✓ = (-10)2 + 2(9,8) �y ✓ fly= 5,10m �Y= -5,10m Max height / Maks hoogte Max height / Maks hoogte = 40 + 5,10 =40 + 5,10 =45,10m ✓ =45,10m ✓ From maximum height From maximum height Vanaf maksimum hoogte Vanaf maksimum hoogte Vt 2=Vi 2 + 2a�y ✓ Vf 2=Vi 2 + 2afly ✓ (-10)2✓ =02 + 2(-9,8) �y✓ 102 ✓=02 + 2(9,8) �y✓ �Y=-5,10 m �y=5,10m Max height / Maks hoogte Max height / Maks hoogte =40 + 5,10 =40 + 5,10 =45,10m ✓ =45,10 m ✓ ✓ both/beide Vt = Vi+ a�t Vt = Vi+ a�t bothlbeide 0 = 10 + (-9,8)�t ✓ 0 = (-10) + (9,8)flt ✓ �t=1,0204 s �t =1,0204 s flY=Vi.1.t +�2 a�t2 �y=Vi�t +�2 a�t2 =(10)(1,0204) + =(-10)(1,0204) + (0,5)(-9,8)(1,0204)2 ✓ (0,5)(9,8)(1,0204)2 ✓ �Y=5,10m �Y =-5,10m Max height / Maks hoogte Max height / Maks hoogte = 40 + 5,10 = 40 + 5,10 =45,10m ✓ =45,10m ✓ (4) Copyright reserved / Kopiereg voorbehou Please turn over/ 8/aai asseblief om

COM FS/March 2022 / VS/Maart 2022 Grade 12 / Graad 12 MEMORANDUM 4.2.2 Up positive / Op positief Down positive / Af positief w2=vi2+ 2aAy v2 = vi2+ 2ady . = 10? + 2(-9,8)(-40) ¥ = (-10)? + 2(9,8)(40) v vi = -29,7321 ms" ve = +29,7321 m:s" Vi = vie at ¥ Vt = Vi+ aAt ¥ -29,7321 = 10 + (-9,8)At ¥ 29,7321 = -10 + (9,8) AtY At=4,05s ¥ At=4,05s Vv FROM MAXIMUM HEIGHT: POSITIVE MARKING FROM 4.2.1 FOR THE HEIGHT. VANAF MAKSIMUM HOOGTE: POSITIEWE NASIEN VANAF 4.2.1 VIR DIE HOOGTE. If At was NOT calculated in 4.2.1: | If At was NOT calculated in 4.2.1: Vi= vie adt vr= vi+ adt 0 = 10 + (-9,8)At ¥ 0 =-10 + (9,8)At ¥ At = T,0204 s At = T,0204 s Ay = wit +5 at? ¥ Ay = wit +5 adt? ¥ -45,10 = 0? + (0,5)(-9,8)At? ¥ 45,10 = 0? + (0,5)( 9,8)At? ¥ At = 3,0338 s At = 3,0338 s At = 3,0338 + 1,02 =4,05s v At = 3,0338 + 1,02 =4,05sv If At WAS calculated in 4.2.1: If At WAS calculated in 4.2.1: Ay = vit 6 at? ¥ Ay = vit +3 adt? -45,10 v= 02 + (0,5)(-9,8)At? ¥ 45,10 v= 02 + (0,5)(9,8)At?2 ¥ At = 3,0338's 7 At = 3,0338's 7 At = 3,0338 + 1,02 =4,05s ¥ At = 3,0338 + 1,02 = 4,05 s ¥ (4) 4.3 Up positive / Op positief | Down positive / Af positief a ' 1 a H ' 1 ' ' 1 ' : 9,8 ' ' } 1 1 ! 0 iti Yt >t 9 Te ¥t2 >t 9,8 H 1 ' \ in L 1 ' u L ' t } Horizontal line from t = 0 up to ta. v No mark on the value. Horisontale lyn vanaf t = 0 tot by te. Geen punt op die waarde nie. Above t-axis if down is positive. Below t-axis if up is positive. v If indication of direction is not given, the Bo t-as as af positief is. second mark is forfeited (Max 1/2). Onder t-as as op positief is. As rigtingaanduiding nié gegee is nie, word Copyright reserved / Kopiereg voorbehou die tweede punt verbeur (Maks. 1/2). (2) Please turn over / Blaai asseblief om

com FS/March 2022/ VS/Maart 2022 Grade 12 / Gr aad 12 MEMORANDUM 4.4.1 U ositive / 0 ositief Down ositive / Af ositief Vt = Vi + ab.t ✓ Vt = Vi + ab.t ✓ = 10 + (-9,8)(3,4) ✓ = -10 + (9,8)(3,4) ✓ = -23,32 m·s-1 Vt = 23,32 m·s·1 Vt = 23,32 m-s- 1 downward/afwaarts ✓ Vt = 23,32 m-s· 1 downward/afwaarts✓ (3) 4.4.2 U ositive / 0 ositief Down ositive / Af ositief b.y = Vib.t +� ab.t2 ✓ b.y = Vib.t +� ab.t2 ✓ 2 2 = (10)(3,4) ✓+ (0,5)( -9,8)(3,4)2 ✓ = (-10)(3,4) ✓ + (0,5)(9,8)(3,4)2 ✓ = -22,644 m = 22,644 m h = 40 - 22,644 ✓ h = 40 -- 22 644 ✓ = 17,36 m ✓ (17,356) = 17,36 m ✓ (17,356) POSITIVE MARKING FROM 4.2.2. AND 4.4.1. POSIT/EWE NASIEN VANAF 4.4.2 EN 4.4.1. b.y = Vib.t +�2 ab.t2 ✓ b.y = Vib.t +� ab.t2 ✓ 2 = (-23,32)(0,65) ✓+ (0,5)(-9,8)(0,65)2 ✓ = (23 ,32)(0,65) ✓+ (0,5)(9,8)(0,65)2 ✓ = -17,23 m ✓ = 17,23 m ✓ h = 17,23 m ✓ h = 17,23 m ✓ Vt 2 = Vi 2 + 2ab.y ✓ Vt 2 = Vi 2 + 2ab.y ✓ (-29,7321)2 ✓= (-23,32)2 + 2(-9,8)b.y ✓ 29,73212 ✓= 23,322 + 2(9,8) b.y ✓ If ground striking velocity was not used in 4.2.2, that calculation must be shown here to earn one of the substitution marks. As grondtref-snelheid nie in 4.2.2 gebruik is nie, moet daardie berekening hier gewys word om een van die substitusiepunte te kry. b.y = 17,36 m ✓ (17,35 m) b.y = 17,36 m ✓ (17,35 m) h = 17,36 m ✓ 17,35 m h = 17,36 m ✓ 17,35 m (Vt+ V·) (Vt+ V·) liy = t lit✓ liy = I lit✓ 2 2 (-29,7321) + (-23,32) (29,7321 + 23,32) = ✓(0,65)✓ = ✓(0,65)✓ 2 2 = -17,24 m ✓ = 17,24 m ✓ h = 17,24 m ✓ h = 17,24 m ✓ (5) 6t in 4.2.2 minus 3,4 s. [20) Copyright reserved/ Kopiereg voorbehou Please turn over/ 8/aai asseblief om

FS/March 2022 / VS/Maart 2022 Grade 12 / Graad 12 MEMORANDUM QUESTION 5 / VRAAG 5 5.1 Compounds in which there are no multiple bonds between C atoms in their hydrocarbon chains. ✓✓ Verbindings waarin daar geen meervoudige bindings tussen C-atome in hul koolwaterstofkettings is nie. (2) 5.2.1 D ✓ (1) 5.2.2 4-ethyl-2,5-dimethylheptane 4-etiel-2,5-dimetielheptaan (3) Marking criteria I Nasienriglyne: • Correct stem, i.e. heptane✓ Korrekte stam, d.i. heptaan • Substituents (ethyl & methyl) correctly identified. ✓ Substituente/sykettings (etiel & metiel) korrek geidentifiseer. • IUPAC name completely correct including numbering, sequence, hyphens, and commas✓ IUPAC-naam hee/temal korrek ins/uitende nommers, vo/gorde, ko eltekens en kommas. 5.2.3 G ✓ (1) 5.2.4 Carbonyl (group) ✓ Karboniel(groep) (1) 5.2.5 CnH2n ✓ (1) 5.3 A series of organic compounds that can be described by the same general formula OR A series of organic compounds in which one member differs from the next with a CH2 group.✓✓ 'n Reeks organiese verbindings wat deur diese/fde a/gemene formule beskryf kan word OF 'n Reeks organiese verbindings waarin die een lid van die volgende verskil met 'n CH2-groep. (2) 5.4.1 Alcohols✓ Alkohole (1) CH3 CH3 5.4.2 CH3-CH-CH-CH-CH3 I ✓✓ OR/OF I CH3CHCHCHCH3 I OH I CH3 I I OH CH3 OR/OF CH3CH(OH )CH( CH3)CH (CH3)2 (2) 5.4.3 Secondary ✓ Sekonder (1) 5.4.4 The carbon atom bonded to the hydroxyl/OH group is bonded to two other carbons atoms. ✓ Die koo/stofatoom wat aan die hidroksie//OH-groep verbind is, is oak aan twee ander koolstofatome verbind. (1) [16] Copyright reserved / Kopiereg voorbehou Please turn over/ 8/aai asseblief om

FS/March 2022 / VS/Maart 2022 Grade 12 / Graad 12 MEMORANDUM QUESTION 6 / VRAAG 6 6.1 The pressure exerted by a vapour at equilibrium✓with its liquid in a closed system✓ Die druk uitgeoefen deur 'n damp in ewewig met sy vloeistof in 'n ges/ote sisteem. (2) 6.2 Chain length/length of carbon chain✓ Kettinglengte/lengte van koolstofketting (1 ) 6.3 D ✓; lowest boiling point. D; laagste kookpunt (2) 6.4.1 Higher than✓ Hoer as (1) 6.4.2 Marking criteria • Compare structures. ✓ • Compare strength of intermolecular forces. ✓ • Compare the energy required to overcome intermolecular forces. ✓ Nasienriglyne • Vergelyk strukture. ✓ • Vergelyk sterkte van intermolekulere kragte. • Vergelyk energie benodig om intermolekulere kragte te oorkom. ✓ • Structure/Struktuur: Between the molecules of A (in addition to London forces) hydrogen forces are present. ✓ Between pentane molecules London forces ✓ are present. Tussen die molekule van A (bykomend tot Londonkragte) is waterstofbindings. Tussen pentaanmolekule is Londonkragte. • Intermolecular forces / lntermolekulere kragte Stronger intermolecular forces are present in compound A than in pentane. ✓ Sterker intermolekulere kragte is teenwoording in verbinding A as in pentaan. (Or opposite arguments / Of teenoorgestelde argumente) • Energy/Energie: More energy is needed to overcome the intermolecular forces in A. ✓ Meer energie is nodig om die intermolekulere kragte in A te oorkom. (4) (Or opposite arguments / Of teenoorgestelde argumente) 6.5 Boiling point increases✓with an increase in the chain length / size of the molecule.✓ Kookpunt neem toe met 'n toename in die kettinglengte I grootte van die molekuul. (2) [12] GRAND TOTAL/ GROOTTOTAAL: 100 Copyright reserved / Kopiereg voorbehou