NATIONAL SENIOR CERTIFICATE EXAMINATION … · 3.8.4 It is NOT possible for the engine noise to...

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NATIONAL SENIOR CERTIFICATE EXAMINATION

NOVEMBER PAPER 2012

PHYSICAL SCIENCES: PAPER I

MARKING GUIDELINES

Time: 3 hours 200 marks These marking guidelines are prepared for use by examiners and sub-examiners, all of whom are required to attend a standardisation meeting to ensure that the guidelines are consistently interpreted and applied in the marking of candidates' scripts. The IEB will not enter into any discussions or correspondence about any marking guidelines. It is acknowledged that there may be different views about some matters of emphasis or detail in the guidelines. It is also recognised that, without the benefit of attendance at a standardisation meeting, there may be different interpretations of the application of the marking guidelines.

NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER I – MARKING GUIDELINES of 12


QUESTION 1 MULTIPLE CHOICE 1.1 D (2) 1.2 B (2) 1.3 A (2) 1.4 C (2) 1.5 C (2) 1.6 A (2) 1.7 D (ALL Afrikaans candidates marked correct – error in question translation.) (2) 1.8 B (2) 1.9 C (2) 1.10 B (2) [20] QUESTION 2 – 7: Where candidates use equivalent wording to that given in the marking guidelines, their answers must be marked correct. QUESTION 2 TUGELA FALLS 2.1 Gravitational potential energy is the energy possessed by a body due to its position

above the Earth's surface. (or relative to a reference point). (2) 2.2 Ep = mgh Ep = 6 × 104 × 10 × 948 Ep = 5,69 × 108 J ANS (using 9,8) : 5,57 × 108J (3) 2.3 2.3.1 kinetic energy to electrical (potential) energy (mechanical energy to

electrical (potential) energy) (2)

2.3.2 EWP = =

t tp∆

85,69 10P

60×

=

=9,48 × 106 W Output = 80% of 9,48 × 106 = 7,58 × 106 W (7,58 MW) OR 8

pE in 5,69 10 J= ×

E electrical out 880 5,69 10100

= × × (method)

84,55 10 J= ×

WPt

=

Power out 84,55 10

60×

=

67,58 10 W= × (5)

conversion

P = Fv = 6 x 104 x 15,8 = 9,48 x 106 W



2.4 2.4.1 top of the waterfall (1) 2.4.2 Distance = 81 – 45 = 36 m (Reasonable accuracy of ± 1 m) (3) 2.4.3 Increasing (1)

2.4.4 (2) 2.4.5 10 m.s-1 down (or 9,8 m.s-1 down ) (2) 2.5 2.5.1 Vertical component of velocity vy = vsinθ = 8sin50 = 6,13 m.s-1 (method) vf

2 = vi 2 + 2a∆x 0 = 6,132 + 2(–10) ∆x (substitution signs for vi and a must differ) ∆x = 1,88 m (above top of falls)

Max 25

for not using vertical component of velocity)

ANS (using 9,8) : 1,92 m (5) 2.5.2 Time up Time down

vf = vi + a∆t ∆x = vi∆t + ½ a∆t2 0 = -6,13 + 10∆t (signs) (948 + 1,88) = 0 + ½ (10) ∆t2 ∆t = 0,613 s (up) ∆t = 13,9 s (down) [14,04s] c.o.e. from 2.5.1 Total time = 0,613 + 13,783 ANS (using 9,8) : 0,6255 s = 14,4 s (method) ANS (using 9,8) : 14,5 s Alternative: Time up and down ∆x = vi∆t + ½ a∆t2 948 = -6,13∆t + ½ (10)∆t2 c.o.e. from 2.5.1 ∆t = 14,4 s

Maximum 35

for not using vertical component of velocity.

ANS (using 9,8) : 14,5 s (5) OR v2 = u2 = 2as = 6,132 + 2(-10)(-948) = -137,83 m.s-1 BOTH formulae correct

One line correct Second line parallel to first (-1 if axes not labelled)

v (m.s-1)

t (s)

A B

0 0

t = (v – u) = -137,83 -6,13 = 14,4 s a -10

OR Ek bottom = Ep top ½ m (8sin50)2 = m(10)h h = 1,88 m



2.5.3 Horizontal component of velocity vy = vcosθ = 8cos50 = 5,14 m.s-1 ∆x = vi∆t + ½ a∆t2 (a = 0) OR x average velocity t∆ = × ∆

∆x = 5,14 × 14,4 OR i fv vx . t2+

∆ = ∆

∆x = 74 m c.o.e. from 2.5.2

Maximum 24

for not using horizontal component of velocity.

ANS (using 9,8 m⋅s-2 in Question 2.5.2) : 74,52 m (4)

2.5.4 (2) [37] QUESTION 3 TOP-FUEL DRAGSTER 3.1 acceleration (1)

3.2 gradient (a) 94 242 0,5

−=

−

= 46,67 m.s-2 (4) (accuracy) 3.3 3.3.1 (uniform/constant acceleration) OR (uniformly/constantly increasing velocity) (forwards) (2) 3.3.2 (decreasing acceleration) OR (non-uniform acceleration) (forwards) (velocity increasing at a slower rate) (2) 3.4 Distance = area under graph = ½ × 2 × 94 = 94,0 m (4)

Alternative solutions

i fv vx . t2+

∆ = ∆ OR 2ix v t ½ a t∆ = ∆ + ∆ OR 2

fv 2

iv 2a x= + ∆

(0 94) .2

2+

= 2½ (46,7) 2= + 294 20 2 (46,7) x= + ∆ c.o.e 3.3

94,0m= 93,4 m= x∆ 94,6 m= 3.5 3.5.1 136 m.s-1 (1) 3.5.2 136 3,6 489,6× = km.h-1 (c.o.e. from Question 3.5.1) (1)

a (m.s-2)

t (s)

10

0

Line above time axis (positive) (–1 if axes not labelled) ( If line is below time axis (negative)) (Accept 9,8 for acceleration)

• Allow calculation using any TWO points from the graph within the region of constant acceleration, e.g. (0 ; 0) and (1,75 ; 82)

• Allow ± 2 m⋅s-1 (1 block) different when reading from graph.



3.6 Solution (using Newton's Second Law) vf

2 = vi 2 + 2a∆x Fnet = ma 0 = 1452 + 2a(150) = 1 000 × –70,083 a = –70,083 m.s–2 = –70 083 N

Magnitude of net force = 70 083 N (7,01 × 104 N) (6) Alternative solution (using Work-Energy Theorem) ∆Ek = ½ mvf

2 – ½ mvi2 ∆Ek = W = Fnet ∆x OR ∆Ek = Fnet∆x.cosθ

= [ ½ × 1 000 × 02] – ½ × 1 000 × 1452 10 512 500 = Fnet (150) = (–) 10 512 500 J 1,05 × 107 J Fnet = 70 083 N (7,01 × 104 N) Alternative solution (using Impulse)

i f(v v )x . t2+

∆ = ∆ netm(v u)F

t−

=∆

(145 0)150 . t

2+

= ∆ 1 000(0 145)

2,069−

=

t 2,069 s∆ = 70 082 N= − 3.7 Diagram Sound heard by observer at X (maximum of 3 marks) (2 marks)

• Shape of • Deeper note (lower wavefronts pitch) due to correct • apparent decrease

• Relative in frequency () difference in OR Less waves reach wavelengths observer per second. () correct OR Longer wavelength ()

• X and D labelled () ONE of these points only in correct places (5)

3.8 3.8.1 Any TWO of the following:

• It emits vast quantities of greenhouse gases (therefore contributing to climate change)

• After one race (of only 4 s) the whole engine has to be replaced. This is a waste of valuable resources.

• It is responsible for noise pollution. • It consumes 23 litres of fuel in under 4 s – this fuel would need to be

produced using energy consuming processes. (Waste of energy) • In addition to greenhouse gases other poisonous gases would be

emitted in the exhaust fumes causing air pollution. • Fire hazard (2 × 2 = 4)

3.8.2 It is very dangerous …

• Detached retinas • Death • Deafness/ear pain • Fire hazard (2)

D X

Justification Any of these reasons one mark

• Allow semi-circles • Allow sinusoidal waves if

difference in wavelength clear



3.8.3 A shock wave is a (cone shaped) (high pressure) wave produced when the speed of the source is greater than or equal to the speed of the wave. (2)

3.8.4 It is NOT possible for the engine noise to cause a shock wave as the dragster

is travelling at (480/3,6) = 133 m.s-1 which is slower than the speed of sound. (3) (OR speed of sound is 340 × 3,6 = 1 224 km.h-1 which is faster than dragster) (NOTE: The rapidly expanding gases from the combustion of the fuel can

cause a shock wave.) [37] QUESTION 4 CAR CRASH 4.1 The (hardness/type of material of the) crash surface/barrier (barrier only gets 1) (2) 4.2 Any TWO of the following ...

Mass of car; speed of car; momentum of car; mass of barrier; angle of attack (barrier) (Allow 'car') (2 × 2 = 4)

4.3 Impulse is the product of the net force acting on a body and the time for which it

acts. (Product of net force and time gets one mark only) OR Impulse is the (instantaneous) change in momentum. (Instantaneous not essential) (2) 4.4 Neither are correct since the change in momentum is the same with both barriers. Impulse = F.∆t = area under graph Impulse for metal barrier = (½ × 0,04 × 12,6) = 0,252 kg.m.s-1 Impulse for foam rubber barrier = (½ × 0,12 × 4,2) = 0,252 kg.m.s-1 (6) 4.5 Plastic foam bumpers increase the time over which the momentum changes,

therefore decreasing the (net) force acting on the car, i.e. Fnet α 1/∆t. Smaller net force results in less damage to car. (3) OR Steel barrier has greater (max/average) force therefore more damage.

4.6 netm(v u)F

t−

=∆

(backwards = negative direction)

5

1 200 ( 8 20)t = 2,1 10− −

∆− ×

0,16 s= (5) Alternative Fnet = ma Both formulae correct -2,1 × 105 = 1 200 × a a = –175 m.s-2 vf = vi + a∆t -8 = 20 – 175∆t ∆t = 0,16 s



4.7 4.7.1 Work (done by a force) is defined as the product of the displacement and the component of the force in the direction of the displacement.

Work is the amount of energy transferred by the force. (2) Force times displacement (1) OR product of force and displacement (1) 4.7.2 W = F∆xcosθ = 1 500 × 180 × cos30 = 233 826 J or 2,34 × 105 J (4) 4.7.3 Power is the rate of doing work OR Work done per unit time OR Power is the rate at which energy is expended. (2) 4.7.4 P = F.v F = 1500cos30 = 1 299 N = 1 299 × 15 = 19 485 W or 1,95 × 104 W (4) Alternative ∆x = vi∆t + ½ a∆t2 (a = 0) 180 = 15∆t ∆t = 12 s

WP =

t∆

233 826

12=

4 19 486 W or 1,95 10 W= × 4.7.5 (a) Equal to. (1) (b) There has been no change to the component of the applied force in the

direction of the displacement and there has been no change to the distance over which it has acted therefore W = F∆xcosθ has not changed.

NOTE: The net force acting on the car has changed which will affect the change in kinetic energy of the car but NOT the total work done. (3)

OR (a) less than (b) Frictional forces have increased therefore loss in Ek OR Fnet opposite direction to motion therefore Wnet decreases

[38]



QUESTION 5 WATER WAVES 5.1 A wavefront is an imaginary line that connects points on a wave that are in phase. (2)

5.2 5.2.1 No. of waves arriving at A from 11,7X = 6,51,8

= (half number of waves)

No. of waves arriving at A from 14,4Y = 8,01,8

= (whole number of waves)

Therefore waves will arrive at A out of phase (crest meets trough) and will undergo destructive interference. (3)

OR Path difference between waves arriving at A from X and waves arriving at A

from Y = (14,4 – 11,7) = 2,7 m

No. of waves = 2,7 1,51,8

= =

Since the path difference corresponds to a half number of waves (1,5 waves) then they will arrive at A out of phase and will experience destructive interference.

5.2.2 No. of waves arriving at B from 9,0X = 5,01,8


No. of waves arriving at B from 7,2Y = 4,01,8


Therefore waves will arrive at B in phase (trough meets trough or crest meets crest) and will undergo constructive interference. (3)

OR Path difference between waves arriving at B from X and waves arriving at B

from Y = (9,0 – 7,2) = 1,8 m

No. of waves = 1,8 1,01,8

=

Since the path difference corresponds to a whole number of waves (1,0 wave) then they will arrive at B in phase and will experience constructive interference.

5.3 B (c.o.e. must agree with Question 5.2) (2) [10]



QUESTION 6 ELECTRIC CIRCUIT 6.1 Light emitting diode or An LED is a diode that emits light when it is forward biased. (2)

6.2 TVRI

=

T9R

0,2=

TR 45 = Ω (4)

6.3 eff 1 2

1 1 1R R R

= + 1 2eff

1 2

R RRR R

×=

+

eff

1 1 1R 270 30

= + eff(270 30)R(270 30)

×=

+

effR 27= Ω effR 27= Ω (3) 6.4 RX = (45 – 27) RX = 18 Ω (2) 6.5 P = I2R = 0,22 × 18 P = 0,72 W Alternatives

V = IR P = VI OR 2VP =

R

= 0,2 × 18 = 3,6 × 0,2 23,6

18=

= 3,6 V P = 0,72 W P = 0,72 W (4) 6.6 Ratio 30:270 1 : 9 IY = 1/10 of 0,2 A IY = 0,02 A Alternative (4) 6.7 It protects the LED by using up the 'extra volts' so that the LED does not get more

than 3 V as it would be damaged. (By limiting current through it so that the voltage across it is less than 3V.) (3)

V = IReff

= 0,2 × 27 = 5,4 V

VI = R

5,4270

=

IY = 0,02 A

conversion



6.8 6.8.1 Decrease (1) 6.8.2 Dimmer (1) 6.9 The LED is now in reverse bias therefore no current flows through this branch. The total resistance of the circuit now increases (as one of the branches has been

removed). The current through the ammeter decreases as shown by the formula, I = V/R (V is

constant therefore if R increases then I must decrease, I α 1/R). The bulb gets dimmer as shown by the formula P = I2R (R of bulb is constant

therefore as I decreases P will decrease, P α I2) (5) 6.10 Any TWO of the following

• Brighter • Use less energy/voltage/current/power … batteries last longer • Less fragile • More energy efficient therefore less wasteful of energy resources • The LEDs last longer therefore less likely to end up on landfill sites • Cheaper • Switching time faster • Do not release heat (2 × 2 = 4)

[33]



QUESTION 7 PHOTOELECTRIC EFFECT EXPERIMENT

7.1

Marking of graph • x-axis labelled with units • x-axis correct scale • Points correct • Straight line • Heading (Extrapolation not needed at this stage) (6)

7.2 Threshold frequency is the minimum frequency of light (electromagnetic radiation) at

which electrons will be emitted from a particular metal. (2) 7.3 + 5,6 × 1014 Hz (2)


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7.4 7.4.1 + 10,4 × 1014 Hz (1)

7.4.2 cf

λ =

8

14

3 1010,4 10

×=

× c.o.e. from 7.4.1

72,88 10 m−= × (288 nm) (3) 7.4.3 E = hf = 6,6 × 10-34 × 10,4 × 1014 c.o.e. from 7.4.1 = 6,86 × 10-19 J (2,41 eV)

Alternative

hcE = λ

34 8

7

6,6 10 3 102,88 10

−

−

× × ×=

×

= 6,86 × 10-19 J (2,41 eV) (3) 7.5 There would be NO CHANGE to the graph. Increasing the intensity would only

increase the number of electrons emitted OR it would have no effect on the maximum kinetic energy of the electrons as each photon would still have the same amount of energy(E = hf). (ONE photon transfers its energy to ONE electron.)

NOTE: Ek = hf – Wf therefore since hf and Wf have not changed then neither will Ek (3) OR Energy of photon only depends on frequency OR Increasing intensity does not affect frequency 7.6 7.6.1 Wf = 2,25 eV (see graph) (allow negative) (1) 7.6.2 Wf = 2,25 × 1,6 × 10-19 = 3,60 × 10-19 J (1) 7.6.3 The graph cuts the y-axis when the frequency is zero. If the frequency is zero then the energy (E) is zero since E = hf If the energy is zero then … E = Wf + ½ mv2 0 = Wf + ½ mv2 –Wf = ½ mv2 As shown the magnitude of the work function is equal to the kinetic energy

(½ mv2) of the electrons when the frequency is zero, which is the y-intercept. Alternative Photoelectric effect equation, E = Wf + ½ mv2 where E = hf and Ek = ½mv2 Therefore hf = Wf + Ek y-axis = Ek and x-axis = f Equation of a straight line, y = mx + c Rearranging photoelectric effect equation gives, Ek = hf – Wf Therefore when frequency (x-axis) = 0 Ek = - Wf (y-intercept) OR – Wf corresponds to c (the y-intercept) (3) [25] Total: 200 marks


NATIONAL SENIOR CERTIFICATE EXAMINATION NOVEMBER 2012

PHYSICAL SCIENCES: PAPER II

MARKING GUIDELINES

Time: 3 hours 200 marks These marking guidelines are prepared for use by examiners and sub-examiners, all of whom are required to attend a standardisation meeting to ensure that the guidelines are consistently interpreted and applied in the marking of candidates' scripts. The IEB will not enter into any discussions or correspondence about any marking guidelines. It is acknowledged that there may be different views about some matters of emphasis or detail in the guidelines. It is also recognised that, without the benefit of attendance at a standardisation meeting, there may be different interpretations of the application of the marking guidelines.

NATIONAL SENIOR CERTIFICATE: PHYSICAL SCIENCES: PAPER II – MARKING GUIDELINES Page 2 of 9


QUESTION 1 1.1 A (2) 1.6 D (2) 1.2 C (2) 1.7 D (2) 1.3 B (2) 1.8 B (2) 1.4 C (2) 1.9 A (2) 1.5 B or D (2) 1.10 C (2) [20] QUESTION 2 2.1 2.1.1 The ability of carbon atoms to form chains (or rings). (2) (Question removed from paper. 2 marks allocated to all learners) 2.1.2 A centre of reactivity in an organic molecule that determines that molecule's

chemistry. (2) A bond, atom or group of atoms that identifies to which Homologous Series

a compound belongs and is responsible for the chemical reactivity of that compound

2.1.3 Process whereby monomers are joined together to form a chain. (2) (increasing chain length /forming a larger chain) 2.2 2.2.1 Pent – 2 – ene (2-Pentene) (2) 2.2.2 2,4 – difluorohexane (absence of dashes – penalise once through question (2) 2.2.3 Propanoic acid (or methyl ethanoate/ethyl methanoate) (2) 2.3 2.3.1 A family/group of organic compounds identified by the same functional

group and obey the same general formula. (Will accept any other reasonable answer on discussion at memo meeting.) (2)

(increasing by –CH2 unit) 2.3.2 Butane • Weak London/van der Waals forces of attraction between

molecules. • Little energy to overcome forces. Butanol • Strong hydrogen bonding forces of attraction between

molecules. • Greater energy needed to overcome forces. • Identification of types of intermolecular forces in respective

molecules . • Comparison of strength of forces between molecules (4)

• State that more energy required to overcome forces 2.3.3 Butanol and isobutanol both have one – OH group, yet butanol has a longer

unbranched chain , thus greater surface area due to larger electron density. Stronger force of attraction between molecules. (3)

2.4 2.4.1 I – addition/hydrochlorination/hydrohalogenation II – substitution/hydrolysis III – esterification/elimination/condensation IV – elimination/dehydration (4)



2.4.2 (a) (-1 if H's not represented)

Butyl ethanoate (carry over error – 2 marks for name if wrong ester is drawn (4)

(b) Water (1) (c) • Act as catalyst (to speed up reaction). • To act as dehydrating agent (remove H2O). (2) (d) • Reagent flammable. • Warm water bath will heat mixture without coming into

contact with flame. (2) • Controls heat to prevent alcohol from vapourising too

quickly • Warm water bath distributes the heat evenly (any two from

above) also … ( safety factor plus any one reason above) 2.4.3 (a) Alkenes (1) (b) CH2 = CH – CH2 – CH3 / CH2CHCH2CH3 but-1-ene CH3 – CH = CH – CH3 / CH3CHCHCH3 but-2-ene CH2 = C (CH3) – CH3 / CH2C(CH3)CH3 (2) methyl propene (7) 2.4.4 (a) C4H8 + 6O2 → 4CO2 + 4H2O (2) (1 mark for O2 1 mark for balancing)

(b) R

mnM

= MR = 56 g.mol–1

1456

n = Carry over

n = 0,25 mol mole ratio of C4H8 : CO2 = 1 : 4 ∴ 1,00 mol of CO2 produced

R

mnM

=

. Rm n M= 1,00 44= × 244 g of COm = (4) (if only the answer is put down showing no working = 1 mark only) [48]



QUESTION 3 3.1 3.1.1 A molecular fragment with an unpaired electron. (2) 3.1.2 Propagation Termination (2) 3.1.3 (3) 3.1.4 (3) 3.1.5 Any 2 possible practical and actual uses of PVC. ( each) (4)

• Pipes • Floor tiles • Garden furniture • Bottle top lids

[14] QUESTION 4 4.1 A prediction of an outcome of an experiment. (2) A prediction of the relationship between two variables 4.2 • As time of reaction progresses, the rate of loss of mass will decrease. • As time increases, the rate of the reaction decreases. • (Accept any hypothesis that is relevant to investigation) (1) (if in a question or a personalised format = 0 marks) 4.3 4.3.1 Independent variable – the variable that is changed or controlled by the

experimenter. Dependent variable – is the outcome variable that is produced as a result of

the independent variable/the variable that depends on the change in the independent variable. (4)

4.3.2 Leonard – as the loss of mass is measured against the constant time

intervals which are controlled by the experimenter. (2) 4.4 To prevent any loss of mass out of the flask due to splashing (spitting). (2) Only allow for CO2 to escape 4.5 Yes. Changing the timekeeper can change reaction/response times on the

stopwatch. (2) Keeping the variable of time controlled, keep it a fair test



4.6

0

Description Mark allocated Heading Heading must be specific Axes labeled with units (1) Scale correct on both axes Half page minimum Points plotted correctly • All points plotted correctly (2)

• 1 or 2 points plotted incorrectly (1) • More than 2 points plotted incorrectly (0)

Line of best fit Correct shape between t = 0 and t = 60 (1) Correct shape between t = 60 and t = 240 (1)

Graph drawn on wrong axes = 4 marks (max) (7) 4.7 4.7.1 Increasing loss of mass due to:

• HCl being at a high concentration • temperature in flask increases (exothermic reaction) • the rate of reaction increases (3)

4.7.2 Reactants are being used up, the rate of reaction decreases. (2) 4.7.3 CaCO3 being used up, thus no more product formed. (2) Reaction has run to completion reaction is over



4.8 4.8.1 The variable that does not change during the experiment. (2) 4.8.2 2: mass of CaCO3/concentration of HCℓ 4: temperature/concentration of HCℓ (2) 4.8.3

1 mark for correct slope 1 mark for correct finish (6) [37] QUESTION 5 5.1 5.1.1 0,4mol ( no marks for a mass) (2) (- 1 for no units) 5.1.2 0,5 0,4− 0,1mol= (2) (carry over)

5.2 nCV

= nCV

=

0,10,4

= 0, 2

0,4=

30, 25 mol dm−= 30,5mol dm−= (3) no conversion of volume = 1 mark max but check units penalise only once for incorrect units (-1) if expressed unit as mol.cm-3

5.3 Kc

[ ][ ]

2 42

2

N O

NO=

if use round brackets ( -1)

( )2

0,50,25

= ( carry over)

Kc

8=

(3)



5.4 5.4.1 The gas mixture will become lighter/less red-brown. (becomes colourless = 1 mark) (2) 5.4.2 • Decreasing volume will increase the pressure.

• Reaction will tend to favour reaction that reduced pressure, i.e. less number of moles.

• Reaction will thus favour forward reaction producing more N2O4. • More product/N2O4(colourless) will reduce the intensity of the

mixture colour. (4) (If answer to Question 5.4.1 is 'darker', explanation in Question 5.4.2 must

be consistent with that )(max 3 marks) [16] QUESTION 6 6.1 6.1.1 (a) A cell whereby chemical energy is converted into electrical energy. (2) (b) • Condition of electrolyte concentration of 1 mol⋅dm-3. • Temperature of 25 °C. (2) (if pressure mentioned , -1 mark)

6.1.2 ncV

=

n cV= 1.0, 275= 0, 275= mol

r

mnM

= r 3M (AgNO )

rm n.M= 108 14 48= + + 0, 275 170= × 1170 g mol−= m 46,75 g= OR

r

mcM V

=

rm c M V= 1 170 0,275= × × 46,75 g= (4) 6.1.3 θE cell = θE cathode – θE anode = 0,80 – (0,34) = 0,46 V Nothing – light bulb will not light up (3)

6.1.4 • To join/link 12

cells together to form a circuit.

• To maintain neutrality in the half cells. (2) 6.1.5 Oxidation: Cu → Cu2+ + 2e− Reduction : Ag+ + e− → Ag Cu + 2Ag+ → Cu2+ + 2Ag (–1 for any errors) (3)



6.2 3Cr 3e Cr+ −+ → Thus 0,01 mol of Cr deposited will require 0,03 mol of electrons

METHOD 1 Total charge required to deposit Cr = 0,03 × 96 500 = 2 895 C Thus Q = I t∆

t∆ = QI

= 28950,8

= 3 618,75 s (6)

[22] QUESTION 7 7.1 A concentrated solution of sodium chloride (2) Saline solution (salt solution = 1 mark NaCl = 1 mark) Seawater = 0 marks 7.2 • it is an ion selective membrane (if they say 'only permits ...', can get 2 marks

as implies ion selective) • Na+ ions can now pass through it. (3)

7.3 Na e Na+ −+ ( )2 22H O 2e H g 2OH− −+ + • H2O is a stronger oxidising agent than Na+ ions. • Thus 2H O will be reduced in preference to Na+. (3) (can get 1 mark if just puts down the correct H2O equation) 7.4 Chlorine NaOH • Water purification

• Bleach • Production of PVC (impact)

• Soap + detergents • Drain cleaner • Production of paper (impact) (4)

[12] QUESTION 8 8.1 It is able to be recharged. (2) 8.2 Oxidation: Pb + SO4

2– → PbSO4 Reduction: PbO2 + SO4

2– + 4H+ → PbSO4 + 2H2O Pb + PbO2 + 2SO4

2¯ + 4H+ → 2PbSO4 + 2H2O (2H2SO4) (–1 for any error) (3) 8.3 • Conc. of H2SO4 will decrease. • H2SO4 is being used up in the reaction to react with Pb to produce PbSO4. (3)

METHOD 2 OR Ne = n × NA = 0,03 × 6,02 × 1023 = 1,806 × 1022 Q = 1,806 × 1022 × 1,6 × 10–19 = 2 889,60 C

Thus t∆ = QI

= 2889,600,8

= 3 612 s


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8.4 • Conc. of H2SO4 will increase. • Pb will be produced and plate at the negative electrode. • PbO2 is produced at the positive electrode. Any 2 (2) 8.5 Eθ cell = Eθ cathode – Eθ anode = 1,69 – (–0,36) = 2,05V Total = 6 × 2,05 = 12,3V (3) 8.6 Closer the plates – the less the internal resistance of the cell. (2) 8.7 8.7.1 The ability of a charged battery to deliver a specific amount of electrical

charge./The total charge a battery can deliver. (2)

8.7.2 WVQ

=

W VQ= and Q I t= ∆ = 22,5 x ((60 x 60) x 2) 12.= 162 000 162 000= C W 1 944 000 J= ( )61,94 10 J× (4) 8.7.3 Total energy 1 944 000= J and rate of transfer 21= J per second

Thus time taken 1 944 00021

=

925 710= s t∆ 25= hours 43 minutes (4) 8.8 Negative effects

• Toxicity of Pb poisons environment. • Non-biodegradable casings (plastic). • Sulphuric acid hazardous Minimise • Recycle the Pb and the plastic casings. (6)

[31] Total: 200 marks

Can also use W = V. I . Δt

NATIONAL SENIOR CERTIFICATE EXAMINATION … · 3.8.4 It is NOT possible for the engine noise to...

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