physical sciences grade 11 term 2 resource pack

PHYSICAL SCIENCESGrade 11TERM 2RESOURCEPACK

Gr11_Physical_Science_Term2_Resource_Book.indb 1 2018/12/31 10:14:45 AM

Contents

Worksheets 3Topic 5: Geometrical Optics 4Topic 6: 2D and 3D Wavefronts 19Topic 7: Ideal Gases and Thermal Properties 27Topic 8: Quantitative Aspects of Chemical Change 36

Formal Experiment: Intermolecular Forces 43Technical Instructions 44Learner’s Worksheet 50Marking Guidelines 58

Assessments 63Topic 5: Geometrical Optics 64Topic 6: 2D and 3D Wavefronts 74Topic 7: Ideal Gases and Thermal Properties 83Topic 8: Quantitative Aspects of Chemical Change 91


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4 Grade 11 Physical Sciences

Topic 5: Geometrical Optics

WORKSHEET

MULTIPLE CHOICE1. Consider light travelling in air towards an air/water boundary, as shown in the diagram

below. Which of the following best describes the reflected ray 2?

1 2 Ray 2

Ray 3

air

water3

i i

i

A No light will be reflected, so ray 2 does not exist.

B Some light will be reflected and 1i = .2i

C All light will be reflected and 1i > .2i

D Some light will be reflected and 1i < .2i (2)

2. Consider the situation shown in Q1, above. Which of the following would best describe the situation with the refracted ray and the angle of refraction, 3i ?

A 1i = .3i

B 1i > .3i

C 1i < .3i

D No light will be refracted as it is all reflected. (2)

LONG QUESTIONS

3. Light is going from an insect into water, as shown in the diagram below.

air

water

3.1. Define the term ‘refraction’. (2)


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TOPIC 5: GEOMETRICAL OPTICS

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TOPIC 5

3.2. Why does the light ray change direction? (1)

3.3. In which direction does the light ray bend? (1)

3.4. Copy and complete the ray diagram to show how the ray of light reaches the fish’s eye. (3)

3.5. Where would the insect appear as seen by the fish? Show the position of the fish on your diagram. (2)

3.6. A friend says that light always travels in straight lines. How would you respond to her/him? (2)

4. Complete the following ray diagram to illustrate the path of the light.

Include the following labels in your sketch:incident ray, emergent ray, refracted ray, normal lines, angles of incidence and refraction and angle of deviation. (9)

5. Consider a ray of light entering a semi-circular glass block, as shown below.

5.1. Copy and complete the following ray diagram: (2)

Glassblock

O

5.2. Which medium has a higher optical density; the glass or the air? (1)

5.3. The ray does not bend when it enters the semi-circular block.

What does this tell you about the angle of incidence? Explain your answer. (3)What does this tell you about point O? (1)


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6. In the diagram below (not to scale), a ray of light, is travelling from glass towards the boundary with ice.

21˚

Ice n =1,31

Glass n =1,52

6.1. The angle on incidence is 21°. Copy and complete the ray diagram for the ray at the boundary of the glass and ice. (3)

6.2. Write down the equation for Snell’s Law. (1)

6.3. Calculate the critical angle for light going from glass into ice. (3)

6.4. How does the speed of light in the glass compare to speed of light in ice? (2)

7. In an experiment, light goes from air to diamond. The angle of incidence is 21° and the angle of refraction is 60°.

7.1. State Snell’s law. (2)

7.2. The refractive index for air is 1,00. Calculate the refractive index for this diamond. (4)

7.3. Calculate the speed of light in diamond. (3)

8. 8.1 What two criteria are needed for total internal reflection to occur? (2)

8.2. Explain how an optical fibre uses total internal reflection. (2)

8.3. Give two practical uses for optical fibres. (2)


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TOPIC 5

9. Consider a ray of light travelling through water into olive oil and then into air, as shown in the diagram below.

n =1,0

n =1,47

Air

Olive Oil

n =1,33Water

9.1. Redraw and complete the path of the ray through the three mediums. (4)

9.2. Explain why the ray changes direction as it goes from the water into the olive oil. (2)


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CONSOLIDATION EXERCISE

TOTAL: 54 MARKS

MULTIPLE CHOICE1. Light goes from air into a glass circular prism (as shown below), at 90° to the surface of

the prism. Which of the following best describes and explains the path of the ray inside the prism?

90˚

PATH OF THE RAY EXPLANATION

A Light bends towards the normal The glass is optically more dense than the air.

B Light bends away from the normal The glass is optically more dense than the air.

C Light does not bend The ray is travelling along the normal.

D Light does not bend The air and the glass have the same optical density.

(2)

2. A light ray travels from medium P to medium Q. Medium Q has a lower refractive index than medium P. The critical angle for this situation is 40°. Which one of the following sketches represents the correct path of the ray?

GRADE 11 TERM 2 WORKSHEET

QUESTION 9

CONSOLIDATION QUESTIONS

QUESTION 2

(2)


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TOPIC 5

LONG QUESTIONS3. The diagram below shows a ray box and a rectangular glass block placed on a sheet of

paper.

ray box

incident ray

glass block

emergent ray lateraldisplacement

The ray is laterally displaced when it leaves the block as shown in the diagram above.

3.1 Explain why the ray is laterally displaced. (1)

3.2 Explain why the emergent ray is parallel to the incident ray. (2)

3.3 Copy and complete the diagram to show the path of the ray of light. (3)

3.4 The block is placed in a medium of liquid carbon disulphide, which has a higher optical density than the block. Redraw the diagram showing the incident and refracted rays. (3)

4. During an experiment to verify Snell’s law, a ray of light was passed from one medium into another. The results are recorded in the table below.

MATERIAL 1

MATERIAL 2 n1 n2

nn

2

11i sin 1i 2i sin 2i

sinsin

1

2

ii

water air 1,33 1,0 25 34,2

75% sugar solution

water 1,47 1,33 25 27,8

diamond water 2,42 1,33 25 50,3

75% sugar solution

diamond 1,47 2,42 25 14,9

4.1 Calculate the ratio nn

2

1 for each of the four experiments. (4)

4.2 Is there a relationship between nn

2

1 and 2i ? State this relationship. (2)


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4.3 Draw the graph of nn

2

1 and 2i , with nn

2

1 on the x-axis. (4)

4.4 Complete the table by calculating sin 1i , sin 2i and sinsin

1

2

ii . (8)

4.5 Draw the graph of nn

2

1 and sinsin

1

2

ii with n

n2

1 on the x-axis. (4)

4.6 State the relationship between nn

2

1 and sinsin

1

2

ii . (2)

4.7 Compare this relationship to Snell’s law. (2)

5. The diagram below shows an optical fibre used for data transmission.

core cladding

5.1 The speed of light in a vacuum is 3,0 × 108 m.s-1. Calculate the speed of light in the fibre of refractive index 1,52. (3)

5.2 Calculate the minimum time taken for a pulse of light to travel along a straight optical fibre of length 3 000 m and refractive index 1,52. (3)

6. Consider a ray of light travelling through water into olive oil and then into air, as shown below.

n =1,47Olive Oil

n =1,33Water

6.1. Calculate the speed of light through olive oil. (3)

6.2. The angle of incidence in the water is 37°. Calculate the angle of refraction. (4)

6.3. What does the refractive index of the oil tell us about the optical density of the oil compared to the water? (2)


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TOPIC 5

MARKING GUIDELINES

MULTIPLE CHOICE1. B Some light will be refl ected and it will obey the law of refl ection . (2)

2. C Th e refracted ray will bend towards the normal. (2)

LONG QUESTIONS 3. 3.1 Th e change of direction of a light ray because its speed changes when it passes

from one medium into another. (2)3.2 Because it’s speed changes. (1)3.3 Towards the normal. (1)

3.4 for ray bending for bending towards normal for normal. (3)

3.5 See above; refracted ray appears to be straight ; similar length of incident and apparent rays . (2)

3.6 Light does not always travel in straight lines, it can be refracted or refl ected . (2)


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4.

GRADE 11 TERM 2 : WORKSHEET: MARKING GUIDELINES

QUESTION 3

QUESTION 4

Angle of incidence (1) Angle of refraction (2) Angle of deviation (3) Angle of incidence (4) Angle of refraction (5) Refracted ray Emergent ray Normal at incident ray Normal at emergent ray (9)

Angle of incidence (1) Angle of refraction (2) Angle of deviation (3) Angle of incidence (4) Angle of refraction (5) Refracted ray Emergent ray Normal at incident ray Normal at emergent ray (9)

5. Semi-circular glass block

5.1. (2)

Normal

Ray bends away from the normal

Glassblock

O

(2)

5.2. The glass (1)

5.3. 5.3.1 Angle of incidence = 0 (along normal) Because otherwise it would bend towards the normal. (3)

5.3.2 O is the centre of the diameter. (1)


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6.

21˚

21˚Ice n = 1,31

Glass n = 1,52

>

6.1. for refracted ray; ray bends away from normal; for indication of angle of refraction. (3)

6.2. sin sinn n1 1 2 2i i= (1)

6.3. sin sinn n1 1 2 2i i=

, ,sin sin1 52 1 31 901 ci =

,sin 0 861i =

, °59 51i = (3)

6.4. ,nn 0 86

2

1 = Light travels slower in glass (at 86% of the speed that it does in ice) (2)

7. 7.1 sin sinn n1 1 2 2i i= (2)7.2 sin sinn n1 1 2 2i i=

, sin sinn1 00 60 212{ { {=c c

,n 2 412 {= (4)

7.3 n vc=

,2 41{ ×v

3 108

={

., × m sv 1 24 108 1 {= - (3)

8. 8.1 The light needs to be in a more optically dense medium (higher n); the angle of incidence must be greater than the critical angle. (2)

8.2 The light enters the optical fibre. When the light hits the surface of the core, the angle of incidence is usually greater than the critical angle, thus all of the light is reflected back into the core. (2)

8.3 Telecommunications/transmissions of large amounts of data; endoscopes, where one set of fibres take light into the body and another set pick up the reflected light and take it back to the doctors. (2)


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9. 9.1

ray bends away from normal angle of refraction greater than angle of incidence in water.

ray bends towards normal

n =1,0

n = 1,47

Air

Olive Oil

n = 1,33Water

(4)

9.2 Going into the olive oil, the ray slowed down and bent towards the normal. (2)


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CONSOLIDATION EXERCISE (54 MARKS)MULTIPLE CHOICE1. C Ray does not bend as it travels along the normal. (2)

2. A Ray refracts away from normal going into a less optically dense medium, at less than the critical angle. (2)

LONG QUESTIONS

3. 3.1 The ray bends towards the normal as it enters the block. (1)

3.2 Because the sides of the block are parallel, the emergent ray will be parallel to the incident ray. The extent that it bends towards the normal when entering the block will be matched by the extent by which it bends away from the normal when exiting the block. (2)

3.3 The diagram, the ray, normal lines. (3)

GRADE 11 TERM 2 CONSOLIDATION QUESTIONS

Question 3 c


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QUESTION 3 d

3.4 Bends away from normal, emergent ray bends towards normal; rest (3)

4.

MATERIAL 1

MATERIAL 2 n1 n2 n

n2

11i sin 1i 2i sin 2i sin

sin1

2

ii

water air 1,33 1,0 1,33 25 0,422 34,2 0,56 1,33

75% sugar solution

water 1,47 1,33 1,1 25 0,422 27,8 0,47 1,1

diamond water 2,42 1,33 1,8 25 0,422 50,3 0,77 1,8

75% sugar solution

diamond 1,47 2,42 0,61 25 0,422 14,9 0,25 0,61

4.1 Above each. (4)

4.2 As nn

2

1 increases, so does 2i . (However these variables are not directly proportional to one another.) (2)


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4,3 for heading; for points; for line of best fit (not actually a straight line) (4)

0

10

20

30

40

50

60

0,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2,0

Angl

e of

Ref

ract

ion

(o )

Ratio of Refractive Indices (n1/n2)

Graph of the Angle of Refraction vs the Ratio of the Refractive Indices

0,0

0,2

0,4

0,6

0,8

1,0

1,2

1,4

1,6

1,8

2,0

0,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2,0

Ratio

of s

in θ

1/s

in θ

2


Graph of the Ratio of sin θ/sin θ vs the Ratio of Refractive Indices (n1/n2)

4.4. See above; for column sin 1i ; for column sin 2i ; for column

sinsin

1

2

ii . (8)

4.5 for heading; for points; for line of best fit. (4)

4.6 nn

2

1 and sinsin

1

2

ii are directly proportional (they are also equal) (2)

4.7 nn

2

1 = sinsin

1

2

ii is the same as Snell’s law sin sinn n1 1 2 2i i= (2)

0

10

20

30

40

50

60

0,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2,0

Angl

e of

Ref

ract

ion

(o )


Graph of the Angle of Refraction vs the Ratio of the Refractive Indices

0,0

0,2

0,4

0,6

0,8

1,0

1,2

1,4

1,6

1,8

2,0

0,0 0,2 0,4 0,6 0,8 1,0 1,2 1,4 1,6 1,8 2,0

Ratio

of s

in θ

1/s

in θ

2





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5. 5.1 n vc=

v nc=

,×

1 523 108

{={

., × m s1 97 108 1{= - (3)

5.2 v txD=

, ×1 97 10t 3000

8

{={

, s0 0000152 {= (3)

6. 6.1 n vc=

v nc=

,×

1 473 108

{{=

., × m s2 04 108 1= - (3)

6.2 sin sinn n1 1 2 2i i=

, ,sin sin1 33 37 1 47{ {i=c

,,

sinsin

1 471 33 37

i =c

,sin 0 544{i =

°33 {i = (4)

6.3 The oil is more optically dense than the water. (2)


Term 2 19

TOPIC 6: 2D AND 3D WAVEFRONTS

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TOPIC 6

TOPIC 6: 2D and 3D WavefrontsWORKSHEET

MULTIPLE CHOICE1. The ability of a wave to spread out as it passes as sharp edge is known as:

A Snell’s law.

B total internal reflection.

C diffraction.

D Huygens’ principle. (2)

2. The ability of a wave to spread out as it passes as sharp edge is best explained by:

A Snell’s law.

B Newton’s third law.

C diffraction.

D Huygens’ principle. (2)

LONG QUESTIONS

3. Give the name or term for the following:

3.1 The small circular wave that is created by a point on a wavefront. (1)

3.2 When two waves interact with each other in the same space at the same time. (1)

3.3 The bending of a wave as it passes an obstacle. (1)

4. For the diffraction pattern below, what would you expect to change if:

4.1. the wavelength gets larger? (1)

4.2. the wavelength gets smaller? (1)

4.3. a larger slit is used? (1)

4.4. a smaller slit is used? (1)

4.5. the frequency of the wave increases? (1)

4.6. the frequency of the wave decreases? (1)


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5. The following experiment is set up in a water ripple tank.

5.1. Copy and complete the diagram to show what happens to the wave as it passes through the slit. (4)

5.2. Name the phenomenon demonstrated in the experiment. (1)

5.3. Christiaan Huygens explained this effect by what is known as Huygens’ principle. 5.3.1 State Huygens’ principle. (2) 5.3.2 Explain how this principle explains this phenomenon. (3)

6. Two learners experimented to find out what effect the width of the slit has on the degree of diffraction. They set up the experiment as shown in the diagram below, using light with a wavelength of 520 nm. For each different slit width they measured the distance of the 1st dark band from the bright central band.

Light source

Single slit

Position of 1stdark band

The following results are obtained:

WIDTH SLIT (m) POSITION OF 1ST DARK LINE FROM THE CENTRE

5,3 x 10-8 19,6

4,9 x 10-8 21,2

4,2 x 10-8 24,8

6.1. State Huygens’ principle. (2)6.2. For this experiment name … 6.2.1 the dependent variable. 6.2.2 the independent variable. (2)6.3. Give the conclusion for this experiment. (2)


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[43 MARKS]

MULTIPLE CHOICE1. Which of the following will cause the maximum diffraction?

A Long wavelength, small gap

B Long wavelength, large gap

C Short wavelength, small gap

D Short wavelength, large gap (2)

2. Which property of sound enables it to undergo diffraction?

A That it is a longitudinal wave.

B That it is a wave.

C That is carries energy.

D That it carries sound. (2)

LONG QUESTIONS 3. Consider the diffraction patterns for green light, as shown below.

3.1. What gives rise to the alternating bright and dark areas? (4)

3.2. The two patterns came from green light of identical wavelength. What was done to change the degree of diffraction? (2)

3.3. Explain why there is a broad central band of light that is much wider than either the slit or the incident ray. (2)

3.4. What does this pattern tell you about the nature of the light? (2)

4. 4.1 Define diffraction. (2)

4.2 State Huygen’s principle. (2)

4.3 Use the answers to 4.1 and 4.2 to explain the pattern that is shown in the diagram below. (4)

wall


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5. The image below is typical of images found in several text books; source: http://labman.phys.utk.edu/phys136/modules/m9/diff.htm

In part A waves approach a barrier; Part B is the barrier.

5.1. What is in the barrier that enables the waves to pass through? (1)

5.2. How will the size of this feature affect the waves as they pass through? (2)

5.3. What phenomenon is observed in part C? (1)

5.4. D is a screen. Describe the pattern on the screen. (2)

5.5. Explain how the dark and bright areas on the screen arise. (3)

5.6. Part E is a graph of the brightness (intensity) for the wave on the screen. Using the principle of constructive and destructive interference, explain how variations of intensity comes about and how they are related to the bright and dark areas on the screen. (4)

6. Diffraction and interference are regarded as two ways of proving that something behaves as a wave.

6.1. What is meant by diffraction? (2)

6.2. How did Huygens use interference to explain diffraction? (2)

6.3. Give a practical example of your own where a particle does not undergo:

6.3.1 diffraction. (2)

6.3.2 interference. (2)


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TOPIC 6

MARKING GUIDELINES

MULTIPLE CHOICE1. C (2)

2. D (2)

LONG QUESTIONS3. 3.1 Wavelet – by definition (1)

3.2 Superposition or interference (1)

3.3 Diffraction (1)

4. These all use degree of diffraction widtham .

4.1. The pattern gets wider/shows more diffraction. (1)

4.2. The pattern is narrower/less diffracted. (1)

4.3. The pattern is narrower/less diffracted. (1)

4.4. The pattern gets wider/shows more diffraction. (1)

4.5. The pattern is narrower/less diffracted fv

m = so decreasesm . (1)

4.6. The pattern gets wider/shows more diffraction fv

m = so increasesm . (1)

5. 5.1 wall

for waves going through for waves bending at the top ; for waves bending at the bottom; for overall symmetry of diagram. (4)

5.2 Diffraction (1)

5.3 5.3.1 Every point on a wavefront acts as the source of secondary wavelets that spread out in the forward direction with the same speed as the wave. (2)

5.3.2 Every point on the wavefront is a wavelet. The wavelets destructively interfere with each other except on the wavefront. However, when the wavefront


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passes through a slit, the wavelets on each side are removed, which means that the wavelets spread out and cause interference. (3)

6. 6.1 The ability of a wave to spread out in wavefronts as they pass through a small aperture or around a sharp edge OR The bending of a wave around an obstacle or the corners of a narrow opening.

6.2 Every point on a wavefront acts as the source of secondary wavelets that spread out in the forward direction with the same speed as the wave. (2)

6.3 6.3.1 The position of the 1st dark band 6.3.2 The width of the slit (2)

6.4 The degree of diffraction is inversely proportional to the width of the slit. (2)


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[43 MARKS]

MULTIPLE CHOICE1. A The degree of diffraction is proportional to the wavelength and inversely

proportional to width. (2)

2. B Only waves, but all types of waves undergo diffraction. (2)

LONG QUESTIONS

3. 3.1 The bright bands come from constructive interference and the dark bands from destructive interference. (4)

3.2 The width of the slit was changed As the 2nd set of results shows more diffraction, the width was decreased. (2)

3.3 The broad central band comes from the wavefront spreading out as a result of the diffraction. (2)

3.4 Light has a wave nature. (2)

4. 4.1 Diffraction is the ability of a wave to spread out in wavefronts as they pass through a small aperture or around a sharp edge OR The bending of a wave around an obstacle or the corners of a narrow opening. (2)

4.2 Every point on a wavefront acts as the source of secondary wavelets that spread out in the forward direction with the same speed as the wave. (2)

4.3 Every point on the wavefront is a wavelet. The wavelets destructively interfere with each other except on the wavefront. However, when the wavefront passes an obstruction, the wavelets on the one side are removed , which means that the wavelets spread out on that side as there is no destructive interference. (4)

5. 5.1 A slit/gap (1)

5.2. Diffraction widtham A wider slit means less diffraction (or a narrower slit means

more diffraction.) (2)

5.3. Diffraction (1)

5.4. Alternating bright and dark bands (2)

5.5. The bright bands come from constructive interference and the dark bands from destructive interference. (3)

5.6. The bright band corresponds to a wave with high amplitude in part E because there has been lots of constructive interference resulting in the brighter band. The dark bands correspond to a part of the wave with zero amplitude due to the destructive interference, resulting in no light (dark.) (4)


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6. 6.1 The ability of a wave to spread out in wavefronts as they pass through a small aperture or around a sharp edge OR The bending of a wave around an obstacle or the corners of a narrow opening. (2)

6.2 Every point on the wavefront is a wavelet. The wavelets destructively interfere with each other except on the wavefront.) However, when the wavefront passes an obstruction, the wavelets on the one side are removed, which means that the wavelets spread out on that side as there is no destructive interference. (2)

6.3 Learners own

6.3.1 e.g.a ball does not diffract when it goes through a doorway etc. (2)

6.3.2 e.g. two balls collide, they bounce off each other, do not pass through & diffract. (2)


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TOPIC 7: IDEAL GASES AND THERMAL PROPERTIES

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TOPIC 7

TOPIC 7: Ideal Gases and Thermal PropertiesWORKSHEET

MULTIPLE CHOICE1. A gas is enclosed in a gas syringe. The pressure on the gas is increased, but the

temperature of the gas is kept constant. What will happen to the average kinetic energy of the particles?

A The average kinetic energy will increase.

B The average kinetic energy will decrease.

C The average kinetic energy will remain constant.

D Not enough information is given to answer this question. (2)

2. Two gas syringes, X and Y, each contains the same gas at STP. The volume of syringe X is 20 cm3 and that of syringe Y is 10 cm3 as shown below.

20cm

X Y

3 10cm 3

Assume ideal gas behaviour.

Which ONE of the following statements is CORRECT?

A The average kinetic energy of the molecules in X is less than that of the molecules in Y.

B The total kinetic energy of the molecules in X is less than that of the molecules in Y.

C The number of gas molecules in X is equal to the number of gas molecules in Y.

D There are more gas molecules in X than in Y. (2)

3. Which one of the following is not a property of an ideal gas?

A The force of attraction between the particles is zero.

B The molecules have elastic collisions with each other.

C The product of pressure and volume are constant for a sample of gas at constant temperature.

D The volume occupied by the gas is equal to the total volume occupied by the molecules. (2)


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LONG QUESTIONS4. An experiment is conducted to verify the relationship between the volume and pressure

of a gas at constant temperature. A 20 cm3 sample of gas is in a syringe, with a freely moving plunger, at 0°C and at 100 kPa.

4.1. What volume would the gas occupy if the pressure was decreased to 75 kPa? (3)

4.2. Name the law that you used in solving 4.1. (1)

4.3. If both the pressure and the kelvin temperature are doubled, how does the volume of the gas sample change? (2)

5. During an investigation into the relationship between the volume and the pressure of a given mass of gas at constant temperature (287 K), the following readings were obtained:

PRESSURE (kPa) VOLUME (cm3)

55 80

85 51,8

110 40

155 28,4

200 22

5.1. What is the relationship between the pressure and volume of a gas? (1)

5.2. Support your answer by using at least two sets of readings to show the relationship. (4)

5.3. Why can the volume not be read immediately after the pressure is changed? (2)

5.4. Draw a sketch graph of pressure against volume. (2)

5.5. Draw a sketch graph of pressure and the inverse of volume (1/V). (2)

6. The following relationship holds good for an ideal gas: pV T\

6.1. Write down what each of the symbols p, V and T represent (3)

6.2. Draw a sketch graph of pV and T. Put T on the vertical (y) axis. (3)

6.3. Under what conditions does the behaviour of a real gas deviate from that of an ideal gas? (2)

6.4. 32 g of a gas occupies a volume of 24,1 dm3 at 112,0 kPa and 51°C. Use this to calculate the molar mass of this gas. (5)


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MULTIPLE CHOICE1. Which of the following gases would have behaviour that deviates the most from an ideal

gas?

A helium

B hydrogen

C nitrogen

D chlorine (2)

2. In which ONE of the following graphs does the dotted line CORRECTLY represent the deviation of a real gas from ideal gas behaviour?

A. B.

-3

A

1/V (dm )

p (kPa)

B

p (kPa)

-31/V (dm )

C. D.

C

p (kPa)

-31/V (dm )

D

p (kPa)

-31/V (dm )

(2)

LONG QUESTIONS3. Consider an ideal gas.

3.1. State the ideal gas law in symbols. (2)

3.2. Use the ideal gas law to explain the warning that is given on aerosol cans “Do not heat or expose to high temperatures.” (3)

3.3. The pressure on a car tyre is 200 kPa at 27 °C. After a fast journey, the temperature increased to 57 °C. If the volume of the tyre remains constant, what is the pressure in the tyre at 57 °C? (3)


RESOURCE PACK


4. A gas syringe is connected to a pressure gauge by means of a short thick tube.

V = 100cm3

Pressure gauge

The initial volume all the enclosed air is 100 cm3, and the pressure is 100 kPa.

4.1. What will happen to the reading on the gauge if the plunger is pushed down? (1)

4.2. Calculate the volume of the gas when the pressure is increased to 140 kPa. (3)

5. The diagram below shows three graphs which were obtained in an experiment to verify Boyles’ law for equal masses of oxygen, helium and neon gases.

3V (dm )

-11/p (kPa )

A

B

C

5.1. Write down the defining equation for Boyle’s law. (2)

5.2. Which of the graphs (A, B, or C) represent oxygen, which helium and which neon? Give precise reasons for your answers. (6)

5.3.

3V (dm )

-11/p (kPa )

B

A sketch graph of B only is given. Copy this into your answer sheet. If an equal mass of the same gas was investigated at a lower temperature, sketch, on the same set of axes as B, the graph that would be obtained. Label this graph D. Explain your reasoning. (4)


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6. A sketch graph of volume and temperature for an ideal gas is given below.

V

T

6.1. Redraw the graph and on the same axes show what the graph for a real gas would look like. (2)

6.2. Explain the deviation you have shown. (2)


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MARKING GUIDELINES

MULTIPLE CHOICE1. C From the definition of temperature, but showing the misconception that

pressure is a measure of average EK. (2)

2. D At equal pressure and at the same temperature, the number of molecules must be larger in the larger container. (2)

3. D Uses one of the misconceptions of the Ideal gas model, which is good for the learners to check their own understanding. (2)

LONG QUESTIONS4. ; ; ; ?p V p V100 20 751 1 2 2= = = = (3)

4.1. p V p V1 1 2 2=

, cmV75

100 2026 72

3

{{

{= =^ h

4.2. Boyle’s law (1)

4.3. V pnRT

2 = if both T and p double, then V remains the same (2)

5. 5.1 pV = k or pressure is inversely proportional to volume etc. (1)

5.2. PRESSURE

(kPa)VOLUME (cm3) pV

55 80 4400

85 51,8 4403

110 40 4400

155 28,4 4402

200 22 4400

All the pV values are essentially identical. for each pV value (2 max) and for the conclusion. (4)

5.3. Because the readings take a few seconds to settle. (2)

5.4. (2)

3V (dm )

p (kPa)


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5.5.

(2)

1/V (dm–3 )

p (kPa)

6. 6.1 p is pressure ; V is volume and T is absolute temperature . (3)

6.2.

graph shape

axes etc. (3)

T

pV

6.3. Real gas behaviour deviates at low temperature and high pressure . (2)

6.4.

3 3-

, ×

n RTpV

8 31 3241

, × × ×24 1 10 112 10{

=

=

=

{ {

.g mol

M nm

132

32 1

{

{

=

=

= -

(5)


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MULTIPLE CHOICE1. D H2 and He are small molecules with weak IMF; N2 has larger molecules, but

still has weak IMF; Cl2 has the biggest molecules and the strongest IMF; it has the highest boiling point. (2)

2. C Real gases deviate under high pressure (so only A or C); the P is greater than expected due to the size of the molecules. (2)

LONG QUESTIONS

3. 3.1. pV = nRT (2)

3.2. When temperature is high , the pressure inside the can will be high (P α T) and V is fixed . The pressure becomes so high that the can may explode. (3)

3.3. Tp

Tp

1

1

2

2=

×p 300200 3502 { {=

= 233 kPa (3)

4. 4.1 It will increase (1)

4.2. p V p V1 1 2 2=

×V 140100 100

2 {{=

, cm74 1 3{= (3)

5. 5.1 ( )pV k OR p V p V1 1 2 2{{= = (2)

5.2. All the gases have equal masses

n = m/M; and MHe = 4; MNe = 20; MO2 = 32

so pVHe > pVNe > pVO2

pV = nRT; ( )p nRTV or p nRT v1 1 1= =

This means the greater n is, the smaller the slope of the 1/p vs V graph

` A is O2 ; B is Ne ; C is He (6)


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5.3.

-11/p (kPa )

B

D

31/V (dm )

( )p nRT V1 1 {= so a lower T implies a greater slope of the graph. (4)

6. 6.1

V

T

graph shape

(2)

6.2. At very low temperatures, the volume of the gas is greater than expected due to the size of the molecules being significant. (2)


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TOPIC 8: Quantitative Aspects of Chemical Change WORKSHEET

MULTIPLE CHOICE1. How many moles of chloride ions are present in 111 g of calcium chloride?

A 0,5

B 2

C 1

D 1,47 (2)

2. What amount of oxygen gas (in moles) contains 1,8 x 1022 molecules?

A 0,03

B 33,34

C 1,2 # 1024

D 1,08 # 1046 (2)

3. Which of the following is true for a standard solution? In a standard solution the…

A concentration is always known.

B concentration is always 1 mol.dm-3.

C temperature is always 0 oC.

D pH is always 7. (2)

4.

PAGE 36

QUESTION 5

n = 0,2 mol C = 0,5 mol⋅dm-3

V = 0,3 dm3 V = 0,2 dm3

n = 0,2 mol c = 0,5 mol.dm-3

V = 0,3 dm3 V = 0,2 dm3

Beaker P contains 0,2 mol NaCl dissolved in 0,3 dm3 of water. Beaker Q contains 0,2 dm3 of NaCl solution with a concentration of 0,5 mol.dm-3. If the contents of beaker P are added to the contents of beaker Q, the concentration of the mixture will be …


Term 2 37

TOPIC 8: QUANTITATIVE ASPECTS OF CHEMICAL CHANGE

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TOPIC 8

A 1,17 mol.dm-3

B 0,67 mol.dm-3

C 0,60 mol.dm-3

D 0,58 mol.dm-3 (2)

LONG QUESTIONS

5. Coal burns in oxygen to produce energy, according to the following equation: C(s) + O2(g) → CO2(g)

5.1. If coal reacts with 9 000 000 000 m3 of oxygen, what volume of carbon dioxide is produced? (2)

5.2. In addition to your answer give an important science idea that you use to deduce the answer. (2)

6. The Haber process is used to produce ammonia according to the equation: N2(g) + 3H2(g) " 2NH3(g). At STP, 30 m3 of ammonia is produced.

6.1. What does “STP” stand for? (2)6.2. Give the values for STP. (2)6.3. What volume of nitrogen was reacted? (2)6.4. What volume of hydrogen was reacted? (2)

7. 20 cm3 of a 0,1 mol·dm-3 nitric acid solution neutralises 25 cm3 sodium carbonate solution according to the following balanced equation: 2HNO3(aq) + Na2CO3(aq) " 2NaNO3(aq) + H2O(ℓ) + CO2(g)

7.1. Write down the NAME of the salt formed in the reaction. (1)7.2. Calculate the mass of sodium carbonate used to prepare 250 cm3 of sodium

carbonate solution of this concentration. (5)

8. Calcium carbide reacts with water to produce ethyne and calcium hydroxide according to the following equation:

CaC2(s) + 2H2O(ℓ) "Ca(OH)2(aq) + C2H2(g)

8.1. 15,0 dm3 of ethyne gas is produced at STP. How many moles of gas are produced? (2)

8.2. 49,15 g of calcium carbide was used to produce the gas. What is the percentage purity of the reactant? (4)


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8.3. On Wednesday 12 August 2015 a series of explosions that killed over one hundred people and injured hundreds of others occurred at a container storage station at the Port of Tianjin in China. A fire had broken out at a warehouse, and the first people on the scene couldn’t keep the fire from spreading. Firefighters then arrived and tried to douse the fire with water as they were unaware that dangerous chemicals, including calcium carbide, were stored on the site, and so they set in motion a series of more violent chemical reactions. The first full explosion occurred soon after and registered as a magnitude 2,3 earthquake, generating shock-waves equivalent to 3 tons of TNT. Shortly after, a second more powerful explosion, equivalent to 21 tons TNT occurred, causing most of the damage and injuries with shock-waves felt many kilometres away. The resulting fireballs reached heights of hundreds of meters. Adapted from Wikipedia https://en.wikipedia.org/wiki/2015_Tianjin_explosions; accessed 22 Feb 2016

8.3.1 Why should “dry” firefighting methods have been used? (2)

8.3.2 Give the balanced reaction equation for C2H2 burning in oxygen to produce carbon dioxide and water. (3)

8.3.3 One ton of calcium carbide can produce 15 625 mol of C2H2. Calculate the mass of CO2 produced from one ton of calcium carbide. (3)


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CONSOLIDATION EXERCISE [33 MARKS]MULTIPLE CHOICE1. Which of these samples contains the same number of atoms as 1 gram of H2?

A. 22 g of carbon dioxide, CO2

B. 8 g of methane, CH4

C. 20 g of neon, Ne

D. 8 g of ozone, O3 (2)

LONG QUESTIONS2. 160 cm3 of a 1,5 mol.dm-3 nitric acid solution reacts with 15 g of calcium carbonate,

according to the following balanced reaction: 2HNO3 + CaCO3 Ca(NO3)2 + H2O + CO2

2.1. Identify the limiting reactant. (5)2.2. What mass of water is produced? (4)

3. It is found that 40 cm3 of a 0,5 mol.dm-3 sodium hydroxide solution is needed to neutralise 20 cm3 of the vinegar with a mass of 20,8 g. Vinegar is a solution of ethanoic acid in water. The balanced chemical equation for this reaction is:

NaOH + CH3COOH " CH3COONa + H2O

3.1. Calculate the number of moles of sodium hydroxide that reacted. (2)3.2. Calculate the mass of ethanoic acid present in the vinegar. (3)3.3. Calculate the percentage (by mass) of ethanoic acid present in the vinegar. (2)

4. Consider the following balanced chemical reaction: 2HNO3 (aq) + Ca(OH)2 (aq) Ca(NO3)2 (aq) + 2H2O (l)25,0 ml of the nitric acid of concentration of 0,15 mol.dm-3 reacts with the calcium hydroxide solution.4.1. How many moles of acid are used? (2)4.2. What mass of calcium hydroxide reacted with the nitric acid? (3)4.3. 13,6 ml of calcium hydroxide solution was used. What was the concentration of the

calcium hydroxide solution? (2)

5. A protein is found to consist of: 18.39% oxygen, 31,18% nitrogen, 41,38% carbon and 8,05% hydrogen.

5.1. Determine its empirical formula. (6)5.2. The molar mass of the protein is 174,0 g.mol-1. What is its molecular formula? (2)


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MARKING GUIDELINES

MULTIPLE CHOICE1. C M = 40 + 2(35,5) = 111; n M

m= (2)

2. A n =, ×

.N6 02 10

0 0323 = (2)

3. A by definition (2)

4. C ., ,

,,

, ,, mol dmC V

n CV0 2 0 3

0 20 5

0 2 0 10 6 3= = +

+=

+= -^ ^ ^ ^h h h h

(2)

LONG QUESTIONS5. 5.1 9 000 000 000 m

3 of CO2(g) is produced. (2)5.2 Science idea: The balanced equation for the reaction equation gives the mole

ratio of O2(g) : CO2(g) is 1 : 1, which is the ratio of the volumes of the gases. (2)

6. 6.1 Standard temperature and pressure. (2)6.2. 0 K and 101,3 kPa (2)6.3. 15 m3 (mole & volume ratio is 1:2) (2)6.4. 45 m3 (mole & volume ratio is 3:2) (2)

7. 7.1 sodium nitrate (1)7.2. , × , , moln CV 0 02 0 1 0 002HNO3 { {= = =

, mol in cmn 0 001 25from ratiosNa CO3

2 3 {= ^ h , mol in cmn 0 01 250Na CO

32 3 {=

, × × , gm nM 0 01 23 2 12 16 3 1 06 {= = + + =^ h (5)

8. ( )CaC s H O l Ca OH aq C H g22 2 2 2 2+ +^ ^ ^ ^h h h h

8.1.1 , ,,

, moln V22 4 22 4

15 00 67{= = = (-1 no unit) (2)

8.1.2 :CaC C H2 2 2

1 mol : 1 mol , moln 0 67CaC2 {= OR 49,15 g 0,768 mol , × , gm nM 0 67 64 42 88{ {= = = % purity = 0,67/0,768 = 87,24%

% ,,

× , %49 1542 88

100 87 24purity {= = (4)

8.2.1 Because CaC2 reacts with water to produce a flammable gas. (2)8.2.2 C H CO H O2 50 4 22 2 2 2 2+ + left right balancing (3)8.2.3 :C H CO2 2 2

1 mol : 2 mol 15 625 mol : 31 250 mol × gm nM 31250 44 1 375 000{ {= = = (i.e. 1,375 tonnes) (3)


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[33 MARKS]

MULTIPLE CHOICE1. C n 1

1 1H = = mol atoms; (A = 21 mol; B = 2

1 mol; D = 0,167 mol) (2)

LONG QUESTIONS

2. 2.1 2HNO3 + CaCO3 Ca(NO3)2 + H2O + CO2

n (HNO3) = CV

= (1,5)(0,16)

= 0,24 mol

n (CaCO3) = 40 12 4815

+ + = 100

15

= 0,15 mol

2:1 ratio. Therefore, HNO3 is the limiting reactant. (5)

2.2. 0,24 mol ÷ 2 = 0,12 mol

n (H2O) = m18 C.A.

`0,12 = m18 `m = 2,16 g (4)

3. 3.1 n(NaOH) = CV = (0,5)(0,04) = 0,02 mol (2)

3.2. n(CH3COOH) = n(NaOH) = 0,02 mol ` m(CH3COOH) = nMr = (0,02)(60) = 1,2 g (3)

3.3. % mass of CH3COOH = 1,2 / 20,8 x 100 = 5,77 % (2)

4. 4.1 , , , moln C V 0 15 0 025 0 00375a a a { {= = =^ ^h h (2)

4.2. HNO3 : Ca(OH)2 2 mol : 1 mol0,00375 : 0,001875 mol

, , gm nM 0 001875 74 0 14{ {= = =^ ^h h (3)

4.3. .,,

, mol dmC Vn

0 01360 001875

0 14bb

b 3

{{= = = - (2)


5. 5.1 ,

, moln 1618 39

1 15O {= = ,

, moln 1432 18

1 30N {= = ,

, moln 1241 38

3 45C {= = ,

, moln 18 05

8 05H {= =

Ratio O : N : C : H

1,15 : 2,30 : 3,45 : 8,05

1 : 2 : 3 : 7

So empirical formula is C3H7ON2 (6)

5.2. .( ) , ; ×M C H ON g molM M87 0 174 2 87actual3 7 21{= = =-

Molecular formula is C6H14O2N6 (2)


FORMAL EXPERIMENT


FORMAL EXPERIMENTGRADE 11 TERM 2: CHEMISTRY

Intermolecular Forces

62 marks

This section provides guidance and assessment of the learner’s knowledge and understanding when carrying out a virtual experiment using the NECT

video of the same name.

If your class is carrying out the experiment using laboratory apparatus and taking down their own results, you must set up your classroom appropriately and give the learners the relevant instructions. You may find it useful to refer to the Technical Instructions which precede the Learner’s Instructions

while preparing for this experiment.

If the learners are proceeding with the virtual experiment, then continue with the NECT programme by using the information, handouts and marking

guidelines contained in this section of this Resource Book.


INTERMOLECULAR FORCES

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TECHNICAL INSTRUCTIONS

AIM: TO VERIFY INTERMOLECULAR FORCES AND THE EFFECTS OF INTERMOLECULAR FORCES ON EVAPORATION, SURFACE TENSION, SOLUBILITY IN WATER AND BOILING POINTS.

BACKGROUND INFORMATION

The liquids which we will work with are ethanol, water, acetone (nail polish remover), methylated spirits, glycerine and cooking oil. Here is table showing the types of intermolecular forces between their molecules.

SUBSTANCE TYPE OF MOLECULEVAN DER WAALS

FORCESHYDROGEN BONDS

Ethanol Polar dipole-dipole

Water Polar dipole-dipole

AcetonePolar (can also

dissolve in non-polar substances)

London forces anddipole-dipole forces

Methylated spirits Polar dipole-dipole

Glycerine Polar dipole-dipole

Cooking oil Non-polar London forces

Chloroform Slightly polar dipole-dipole

APPARATUSPart 1: Evaporation rate: ethanol, water, acetone, methylated spirits

4 evaporating dishes 4 mass meters (electronic mass meters work best) 4 × measuring cylinders or syringes (to measure 20ml)

Part 2: Surface tension: water, cooking oil, glycerine, acetone, methylated spirits 5 × small glass measuring cylinders 5 × stirring rods 1 microscope glass sheet 5 × measuring cylinders or syringes (to measure 50ml)

Part 3: Solubility: Solvents: water, ethanol, chloroform Solutes: sodium chloride, iodine, potassium permanganate 9 x 100 ml beakers 3 spatulas 3 stirring rods


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3 × 10 ml test tubes 3 × 50 ml syringes (or 100 ml measuring cylinders) Masking tape (to label the beakers) A fine-line permanent marker

Part 4: Boiling points: glycerine, acetone, methylated spirits 3 × boiling (test) tubes 3 × 30 ml syringes 1 large beaker Hot plate or burner with tripod and gauze mat Access to water

INVESTIGATING THE EFFECTS OF INTERMOLECULAR FORCESThe following four experiments investigate the effect of various physical properties (evaporation, surface tension, solubility, boiling point and capillarity) of substances and determine how the intermolecular forces between the molecules relate to these properties. Each experiment looks at a different property.

PART 1: EVAPORATION RATE

AIM: TO VERIFY EVAPORATION AND TO DETERMINE THE RELATIONSHIP BETWEEN EVAPORATION RATE AND INTERMOLECULAR FORCES.Substances: ethanol, water, acetone, methylated spirits

METHOD1. Place an evaporating dish onto each of four electronic mass meters placed in the same

warm spot in the laboratory.

2. Zero each balance (so that it will only read the mass of the substance placed in the dish).

3. Measure 20 ml of each substance into each of the evaporating dishes.

4. Measure the mass of 20 ml of each substance.

5. After 6 minutes, measure the mass of each substance.

PART 2: SURFACE TENSION

AIM:To verify surface tension and to determine the relationship between surface tension and intermolecular forces

Substances: water, cooking oil, glycerin, acetone, methylated spirits



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METHOD1. Place about 50 ml of each substance into separate small measuring cylinders.

2. Observe the shape of the meniscus. (This is the level of the liquid). Note what happens at the edges where the liquid touches the glass. (You can place a few drops of food colouring in each substance to help you see the meniscus more clearly.)

3. Now place a drop of the substance on a small piece of glass. Observe the shape of the drop.

PART 3: SOLUBILITY

AIM: TO VERIFY SOLUBILITY AND TO DETERMINE THE RELATIONSHIP BETWEEN SOLUBILITY AND INTERMOLECULAR FORCES.

PRECAUTIONChloroform is a volatile solvent. Its fumes cause drowsiness. It is unsafe to allow these fumes to be inhaled by anyone for even short periods of time. Work with chloroform in a well-ventilated space e.g. outdoors, or in a protected space such as a fume cupboard.

Water

Water

Water

Ethanol

Ethanol

Ethanol

Chloroform

Chloroform

Chloroform

Sodiumchloride

Potassiumpermanganate

Iodine

METHOD:1. Place 50 ml of each solvent given into three separate beakers.

2. Arrange the beakers as shown in the diagram.

3. Into the first set of beakers add about 2 g of sodium chloride.

4. Into the second set of beakers add two or three crystals of potassium permanganate.

5. Into the third set of beakers add one or two pieces (or crystals) of iodine.

6. Stir the contents of each beaker four times (using a clean stirring rod each time).

7. Wait about a minute then observe what happens to the contents of each beaker.


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PART 4: BOILING POINT

AIM: TO VERIFY BOILING POINT AND TO DETERMINE THE RELATIONSHIP BETWEEN BOILING POINT AND INTERMOLECULAR FORCES

WARNINGMethylated spirits and acetone are highly fl ammable. Th ey will easily catch fi re if left near an open fl ame. For this reason, they must be heated in a water bath. Th is experiment MUST be performed in a well-ventilated room. Th e fumes from methylated spirits and acetone are toxic.

NOTE1. Cooking oil produces smoke from its surface long before it boils. It is therefore diffi cult

to establish the boiling point of cooking oil. We have therefore left cooking oil out of this experiment.

2. Since we are using a water bath to heat the substances, we can measure (or compare) the temperature at which water boils. Th erefore, we are testing three liquids: acetone, ethanol and glycerine.

3. Th e water bath can be heated using a hot plate, or by using a burner, tripod and gauze mat.

4. If you have access to a digital thermometer, place the probe in the water bath so that you can read the approximate temperature at which each substance (A, B and/or C) begins to boil.

METHODTh e boiling points of these liquids are easy to look up on the internet or other media, so we are turning this investigation around a little, and identifying the liquids according to their boiling points. You will start with 20 ml of three liquids in separate test tubes labelled A, B and C.

A B C

Test tubes



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1. Half-fill the beaker with water and place it on the hot plate (or gauze mat).

2. Place the three test tubes in the beaker.

3. Observe the order in which each substance begins to boil.

4. If a substance does not boil by the time the water boils, record its boiling point as “above the boiling point of water”.

In the investigation shown in the video the liquids were as follows:

A. acetone

B. ethanol

C. glycerine


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NAME: GRADE:

Formal ExperimentINTERMOLECULAR FORCES


BACKGROUND INFORMATIONThe liquids which we will work with are ethanol, water, acetone (nail polish remover), methylated spirits, glycerine and cooking oil. Here is table showing the types of intermolecular forces between their molecules.












APPARATUS:Part 1: Evaporation rate: ethanol, water, acetone, methylated spirits

4 evaporating dishes 4 mass meters (electronic mass meters work best) 4 × measuring cylinders or syringes (to measure 20ml)

Part 2: Surface tension: water, cooking oil, glycerine, acetone, methylated spirits 5 × small glass measuring cylinders 5 × stirring rods 1 microscope glass sheet 5 × measuring cylinders or syringes (to measure 50ml)



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Part 3: Solubility: Solvents: water, ethanol, chloroform Solutes: sodium chloride, iodine, potassium permanganate 9 × 100 ml beakers 3 spatulas 3 stirring rods 3 × 10 ml test tubes 3 × 50 ml syringes (or 100 ml measuring cylinders) Masking tape (to label the beakers) A fine-line permanent marker

Part 4: Boiling points: glycerine, acetone, methylated spirits 3 × boiling (test) tubes 3 × 30 ml syringes 1 large beaker Hot plate or burner with tripod and gauze mat Access to water

INVESTIGATING THE EFFECTS OF INTERMOLECULAR FORCESThe following four experiments investigate the effect of various physical properties (evaporation, surface tension, solubility, boiling point and capillarity) of substances and determine how the intermolecular forces between the molecules relate to these properties. Each experiment looks at a different property.


AIM: TO VERIFY EVAPORATION AND TO DETERMINE THE RELATIONSHIP BETWEEN EVAPORATION RATE AND INTERMOLECULAR FORCES.Substances: ethanol, water, acetone, methylated spirits

METHOD1. Place an evaporating dish onto each of four electronic mass meters placed in the same

warm spot in the laboratory.

2. Zero each balance (so that it will only read the mass of the substance placed in the dish).

3. Measure 20 ml of each substance into each of the evaporating dishes.

4. Measure the mass of 20 ml of each substance.

5. After 6 minutes, measure the mass of each substance.


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RESULTS

SUBSTANCE

MASS BEFORE

(G)

MASS AFTER 6 MINUTES

(G)

CHANGE IN MASS

(G)

%CHANGE IN MASS

(%)

RANKING OF EVAPORATION

RATE (1 = FASTEST;4 = SLOWEST)

Ethanol

Water

Acetone

Methylated spirits

(12)

CONCLUSION (4)


AIM: TO VERIFY SURFACE TENSION AND TO DETERMINE THE RELATIONSHIP BETWEEN SURFACE TENSION AND INTERMOLECULAR FORCESSubstances: water, cooking oil, glycerin, acetone, methylated spirits

METHOD 1. Place about 50 ml of each substance into separate small measuring cylinders.

2. Observe the shape of the meniscus. (This is the level of the liquid). Note what happens at the edges where the liquid touches the glass. (You can place a few drops of food colouring in each substance to help you see the meniscus more clearly.)

3. Now place a drop of the substance on a small piece of glass. Observe the shape of the drop.



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RESULTSFor each substance draw the shape of the meniscus. (5)

10

20

30

40

50

10

20

30

40

50

10

20

30

40

50

10

20

30

40

50

10

20

30

40

50

Water Cooking oil Glycerine Acetone Methylated spirits

For each substance draw the shape of the droplet, and the side view of the droplet’s shape.

SHAPE

WATER COOKING OIL GLYCERINE ACETONE METHYLATED SPIRITS

SIDE VIEW (10)

Background information: Glass is a polar substance.

2.1 Which substance is most strongly attracted to glass? Justify your answer with reference to the observations. (4)

2.2 Rank the substances in increasing order of surface tension. Justify your answer. (4)


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PART 3: SOLUBILITY

AIM: TO VERIFY SOLUBILITY AND TO DETERMINE THE RELATIONSHIP BETWEEN SOLUBILITY AND INTERMOLECULAR FORCES.

PRECAUTIONChloroform is a volatile solvent. Its fumes cause drowsiness. It is unsafe to allow these fumes to be inhaled by anyone for even short periods of time. Work with chloroform in a well-ventilated space e.g. outdoors, or in a protected space such as a fume cupboard.

Water

Water

Water

Ethanol

Ethanol

Ethanol

Chloroform

Chloroform

Chloroform

Sodiumchloride

Potassiumpermanganate

Iodine

METHOD:1. Place 50 ml of each solvent given into three separate beakers.

2. Arrange the beakers as shown in the diagram.

3. Into the first set of beakers add about 2 g of sodium chloride.

4. Into the second set of beakers add two or three crystals of potassium permanganate.

5. Into the third set of beakers add one or two pieces (or crystals) of iodine.

6. Stir the contents of each beaker four times (using a clean stirring rod each time).

7. Wait about a minute then observe what happens to the contents of each beaker.



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RESULTS:Record your results in the table below. If you observe only a small amount of the solid dissolving, then write that very little solid dissolved. If the entire solid dissolves, then write that all the solid dissolved. (9)

SUBSTANCE WATER ETHANOL CHLOROFORM

Sodium chloride

Potassium permanganate

Iodine

3.1 Classify the solutes as ionic or molecular, and polar or non-polar. (2)

SUBSTANCE IONIC OR MOLECULAR POLAR OR NON-POLAR

Sodium chloride


Iodine

CONCLUSION:


AIM: TO VERIFY BOILING POINT AND TO DETERMINE THE RELATIONSHIP BETWEEN BOILING POINT AND INTERMOLECULAR FORCES

WARNINGMethylated spirits and acetone are highly flammable. They will easily catch fire if left near an open flame. For this reason, they must be heated in a water bath. This experiment MUST be performed in a well-ventilated room. The fumes from methylated spirits and acetone are toxic.


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NOTE1. Cooking oil produces smoke from its surface long before it boils. It is therefore diffi cult

to establish the boiling point of cooking oil. We have therefore left cooking oil out of this experiment.

2. Since we are using a water bath to heat the substances, we can measure (or compare) the temperature at which water boils. Th erefore, we are testing three liquids: acetone, ethanol and glycerine.

3. Th e water bath can be heated using a hot plate, or by using a burner, tripod and gauze mat.

4. If you have access to a digital thermometer, place the probe in the water bath so that you can read the approximate temperature at which each substance (A, B and/or C) begins to boil.

METHODTh e boiling points of these liquids are easy to look up on the internet or other media, so we are turning this investigation around a little, and identifying the liquids according to their boiling points. You will start with 20 ml of three liquids in separate test tubes labelled A, B and C.

A B C

Test tubes

1. Half-fi ll the beaker with water and place it on the hot plate (or gauze mat).

2. Place the three test tubes in the beaker.

3. Observe the order in which each substance begins to boil.

4. If a substance does not boil by the time the water boils, record its boiling point as “above the boiling point of water”.

RESULTS4.1 Write down the boiling points of these liquids.

Glycerin ………….. Acetone ………………. Ethanol ……………. (1)



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4.2 In which order did the liquids boil? Write down the labels A, B or C.

1st ………………. 2nd ………………… 3rd …………….. (2)

4.3 Identity the substances.

A …………………………………………………………….…

B ………………………………………………………………..

C ……………………………………………………………….. (2)

4.4 How are the intermolecular forces inside a substance related to its boiling point? Justify your answer.

(4)


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Formal ExperimentMARKING GUIDELINES

62 MARKS


BACKGROUND INFORMATIONThe liquids which we will work with are ethanol, water, acetone (nail polish remover), methylated spirits, glycerine and cooking oil. Here is table showing the types of intermolecular forces between their molecules.














Term 2 59

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AIM: TO VERIFY EVAPORATION AND TO DETERMINE THE RELATIONSHIP BETWEEN EVAPORATION RATE AND INTERMOLECULAR FORCES

RESULTS

SUBSTANCE

MASS BEFORE

(G)

MASS AFTER 6 MINUTES

(G)

CHANGE IN MASS

(G)

%CHANGE IN MASS

(%)

RANKING OF EVAPORATION

RATE (1 = FASTEST;4 = SLOWEST)

Ethanol 15,4 14,2 1,2 7,8 2

Water 19,3 19,0 0,3 1,6 4

Acetone 15,1 11,6 3,5 23,2 1

Methylated spirits

16,2 15,0 1,2 7,4 3

(12)

CONCLUSION (4)• The intermolecular forces between the molecules of acetone are the weakest.

• At the same temperature more acetone molecules were able to break free (evaporate) from the surface of the liquid.

• The other three substances (liquids) each have strong hydrogen bonds between their molecules, so it takes more energy for molecules to break free of the liquid surface.


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AIM: TO VERIFY SURFACE TENSION AND TO DETERMINE THE RELATIONSHIP BETWEEN SURFACE TENSION AND INTERMOLECULAR FORCES

RESULTSFor each substance draw the shape of the meniscus. (5)

10

20

30

40

50

10

20

30

40

50

10

20

30

40

50

10

20

30

40

50

10

20

30

40

50

Water Cooking oil Glycerine Acetone Methylated spirits

For each substance draw the shape of the droplet, and the side view of the droplet’s shape.

SHAPE [One each]

Water Cooking Oil Glycerine Acetone Methylated spirits

SIDE VIEW [ One each] (10)

Background information: Glass is a polar substance.

2.1 Which substance is most strongly attracted to glass? Justify your answer with reference to the observations. (4)

Water is most strongly attracted to glass because it has the most concave menis-cus and one of the flattest water droplets. Because water is a polar solvent and glass is also a polar substance, they are strongly attracted to each other.

2.2 Rank the substances in increasing order of surface tension. Justify your answer. (4)



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Cooking oil AcetoneGlycerineMethylated spiritsWaterCooking oil is a non-polar substance with the weakest intermolecular (or cohesive) forces between its molecules. [The learners could discuss the shape of cooking oils droplets to prove that oil is non-polar].

PART 3: SOLUBILITY

AIM: TO VERIFY SOLUBILITY AND TO DETERMINE THE RELATIOSHIP BETWEEN SOLUBILITY AND INTERMOLECULAR FORCES.

RESULTS:[One each; Some learners may record colour and appearance of mixture – that’s fine. They MUST state whether the solute dissolved or not as shown in these answers to gain the mark.] (9)

SUBSTANCE WATER ETHANOL CHLOROFORM

Sodium chloride All the solid dissolved

Very little solid dissolved (OR also

accept)No solid dissolved

No solid dissolved


All the solid dissolvedVery little solid

dissolvedNo solid dissolved

Iodine No solid dissolvedSome solid dissolved

to form a yellow brown solution

Some solid dissolved to form a purple (pink)

solution

3.1 Classify the solutes as ionic or molecular, and polar or non-polar. (2)

SUBSTANCEIONIC OR MOLECULAR

[ ALL CORRECT]POLAR OR NON-POLAR

[ ALL CORRECT]

Sodium chloride ionic polar

Potassium permanganate ionic polar

Iodine molecular non-polar


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CONCLUSION (3)Ionic solutes dissolve in water because these are both polar substances.

Ethanol can act as a polar and a non-polar solvent because it dissolves both polar and non-polar substances.

Chloroform acts as a non-polar solvent even though it has a slightly polar molecule.


AIM: TO VERIFY BOILING POINT AND TO DETERMINE THE RELATIONSHIP BETWEEN BOILING POINT AND INTERMOLECULAR FORCES.

RESULTS4.1 Write down the boiling points of these liquids.

Glycerine: 290 °C Acetone: 57 °C Ethanol: 79 °C (1)

4.2 In which order did the liquids boil? Write down the labels A, B or C.

1st A 2nd B 3rd C (2)[Any two in correct order]

4.3 Identity the substances.

A Acetone

B Ethanol

C Glycerine

[Any two in correct order c.o.e. from 4.2] (2)

4.4 How are the intermolecular forces inside a substance related to its boiling point? Justify your answer.

The stronger the intermolecular forces the higher the boiling point of the liquid.

Acetone has the weakest intermolecular forces and it has the lowest boiling point. (4)


ASSESSMENTS


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Topic 5: Geometrical OpticsQUESTIONS

MULTIPLE CHOICE1. Study the ray diagram below. Material 1 has a lower refractive index than Material 2.

Material 1ray 1

ray 2

ray 3

Material 2

Which of the following options is correct?

Rays which will be observed

A Ray 2 only

B Ray 1 and ray 2

C Ray 3 only

D Ray 3 and ray 1

(2)2. Which of the following correctly describes the conditions required for total

internal reflection to take place?

A The ray must go from a medium with a higher to a medium with a lower refractive index, and the angle of incidence must be greater than the critical angle.

B The ray must go from a medium with a lower to a medium with a higher refractive index, and the angle of incidence must be greater than the critical angle.

C The ray must go from a medium with a higher to a medium with a lower refractive index, and the angle of incidence must be less than the critical angle.

D The ray must go from a medium with a lower to a medium with a higher refractive index, and the angle of incidence must be less than the critical angle. (2)


Term 2 65


3. Light travels faster in pure water than it does in a 30% sugar solution. Which of the following deductions is/are correct? i) 30% sugar water has a higher refractive index than pure water. ii) Light travelling from 30% sugar water into pure water will usually bend

towards the normal. iii) The 30% sugar water is optically less dense than pure water.

A i) onlyB i) and ii) onlyC i) and iii) onlyD i), ii) and iii) (2)

4. A light wave travels obliquely from air into a glass block and its speed changes as it bends towards the normal.

Which ONE of the combinations below correctly describes the changes in the FREQUENCY of the wave and the REFRACTIVE INDEX of the block compared to that of air?

FREQUENCYREFRACTIVE INDEX of the block

compared to air

A Remains the same Larger

B Remains the same Smaller

C Increases Larger

D Decreases Larger (2)

5. Which one of the following statements is always true?A Light always travels at a speed of 3 × 108 m.s-1.B All waves travel at a speed of 3 × 108 m.s-1.C All colours of light travel at the same speed in a vacuum.D Light can travel at a speed of higher than 3 × 108 m.s-1 in a medium with a

very low refractive index. (2)

6. Which of the following statements correctly describes the conditions required for refraction of a light ray to be observed?

i) The incident ray must travel along the normal.ii) The light must go from one medium to another, with different optical densities.

iii) The angle of incidence must be greater than 0° but less than 90°.A i) onlyB i) and ii) onlyC ii) onlyD ii) and iii) only (2)


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LONG QUESTIONS7. Learners investigate how the path of a light ray incident on an air-glass boundary

changes as it enters the glass. Their results are shown in the table below.

angle Î (°) angle r̂ (°) sin i sin r

15 10 0,259 0,174

25 16 0,423 0,276

45 28 0,707 0,469

55 33 0,819 0,545

60 35 0,866 0,574

70 39 0,940 0,629

7.1 For this investigation, write down the: i) dependent variable. (1) ii) independent variable. (1) iii) fixed (control) variable. (1)7.2 Draw a graph of the sin r against sin i. (6)7.3 Calculate the slope of the graph. (2)7.4 Re-arrange Snell’s Law to express the slope of the graph in terms of ni and nr. (2)7.5 From your answers above, and knowing that the speed of light in air is

3 × 108 m.s−1, calculate the speed of the light through the glass. (4)

8. Refer to the diagram of a periscope shown below.

Light Ray

Periscope

Eye


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8.1 State the law of reflection. (2)8.2 Why does the light ray, as drawn, not refract as it enters the prism? (2)8.3 Which has a higher refractive index, the prism or the air? Explain why this is

necessary. (2)8.4 Prisms can make use of a phenomenon called ‘total internal reflection’.

8.4.1 State two conditions necessary for total internal reflection to occur. (2)8.4.2 Give two other examples of where total internal reflection is used. (2)

9. 9.1 A marble is placed at the bottom a polystyrene cup which is full of water. It is viewed from above as shown in the diagram below.

9.1.1 Define refraction. (2)9.1.2 Although the opaque side of the cup is directly between the marble

and the observer’s eye, the marble can still be seen. Use a ray diagram to show how this is possible. (3)

9.1.3 If there was no water in the cup, could the marble be seen from the same position? Draw a diagram to explain your answer. (4)

9.2 Consider the diagram below of a ray of light entering a glass prism.

air

glass

air

9.2.1 Redraw the diagram and complete the path of the ray through the prism and out into the air. (2)

9.2.2 On your diagram, label the following: incident ray, emergent ray, refracted ray, normal lines. (4)


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10. Fibre optics uses the principle of total internal reflection. A diagram of a fibre optic cable is shown below.

Glass cladding

Glass corePrimary buffer

Secondary buffer

Source: errantscience.com

10.1 Define the term critical angle. (2)10.2 Calculate the speed of light in glass given that the refractive index of glass is

1,9. (3)10.3 Explain why it is important for the glass cladding around the glass core to

have a lower refractive index than the core. (2)10.4 What is the purpose of the secondary buffer? (2)10.5 Give two advantages of using optical fibre over conventional copper cable to

transmit information. (2)10.6 Give one disadvantage of using optical fibre. (1)

11. The diagram below shows a ray box and a rectangular glass block placed on a sheet of paper.

The ray is laterally displaced when it leaves the block as shown below.

ray box

incident ray

glass block

emergent raylateral displacement

11.1 Explain why the emergent ray is parallel to the incident ray. (2)11.2 Copy and complete the ray diagram, including the normal lines. (2)


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11.3 Using the diagram as shown above, measure the angle of incidence and the angle of refraction. Use these values to complete the table below.

Angle of incidence (θ1) Sin θ1 Angle of refraction (θ2 ) Sin θ2

(4)11.4 Air has a refractive index of 1,0.

5.4.1 State Snell’s law. (2)5.4.2 Calculate the refractive index of the block. (3)

12. The diagram below shows a light ray travelling through an optical fibre. The fibre is made of two different types of glass, the one being the glass core and the other the glass cladding.

CoreCore

Cladding

12.1 State two conditions needed for total internal reflection to occur. (4)12.2 Which part of the optical fibre will have a higher refractive index? (1)12.3 The speed of light in the cladding is 2,7 × 108 m.s−1. Calculate the refractive

index of the cladding. (3)12.4 The critical angle for the boundary of the two surfaces is 27°. Calcuate the

refactive index of the core. (4)


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MARKING GUIDELINES

MULTIPLE CHOICE

1. D As the ray is going into a more optically dense medium, the refraction will be towards the normal. However, there still will be some reflection, (ray 1), even it is limited. [CL3] (2)

2. A These are the standard conditions for total internal reflection. [CL2] (2)

3. A Only (i) is true. Light travels slower in the sugar solution, therefore the sugar solution is more optically dense than water, A light ray passing from sugar solution into water would bend away from the normal. [CL3] (2)

4. A Although the speed of the light decreases in the block, the frequency of the light does not change, but the wavelength changes. [CL2] (2)

5. C A is not true because light changes speed in different mediums; B is not true because only electromagnetic waves travel at the speed of light; D is not true as the speed of light is the maximum speed that light can travel at. [CL2] (2)

6. D A light ray travelling along the normal will not bend when it passes from one medium to another, so no refraction will be observed. The speed of the light ray will decrease when it enters the more optically dense medium. [CL2] (2)

LONG QUESTIONS7. 7.1 i) sin of the angle of refraction (or the angle of refraction) [CL2] (1) ii) sin of the angle of incidence (or the angle of incidence) [CL2] (1) iii) the mediums [CL2] (1)

7.2

0.70.60.50.40.30.20.1

00 0.2 0.4 0.6 0.8 1

Graph of sin r against sin i for glass

sin

r (an

gle

of re

fract

ion)

sin i (angle of incidence)

Heading ; axes labelled , plotting of points line of best fit [CL2] (6)

7.3 ,,

,slope xy

0 92 00 6 0

0 65{ {DD

= =-

-=

^

^ h

h [CL3] (2)


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7.4 sin sinn ni ri ri i=

sin sinnn

r ir

i { {i i= [CL3] (2)

7.5 ,nn 0 65

r

i {= As vin air = .× m s3 108 1- , ni = 1.

. ,n 0 651 1 5r {= =

.,

× ×; m sn vc v n

c1 5

3 10 2 108

8 1{= == = - [CL3] (4)

8.1 The angle of incidence equals the angle of reflection. [CL1] (2)8.2 Because the light ray enters along the normal. [CL3] (2)8.3 The prism – total internal reflection is used in the prism, so it must have a

higher n than the air. [CL3] (2)8.4.1 Light goes from more optically dense to less optically dense material

(or from higher to lower refractive index); the angle of incidence is greater than the critical angle. [CL1] (2)

8.4.2 In optical fibres; in binoculars, telescopes & microscopes (any 2 reasonable). [CL2] (2)

9. 9.1.1 The change of direction of a light ray because its speed changes when it passes from one medium into another. [CL1] (2)

9.1.2

Ray must bend away from normal Ray must reach eye Normal must be shown [CL2] (3)

9.1.3

No , because the ray from the object straight to the eye would be

through the side of the cup, which is opaque. If there is no water, there would be no refraction. [CL3] (4)


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9.2

normal normal

incident rayemergent ray

refracted ray

9.2.1 refracted ray ; emergent ray , as per diagram above. [CL2] (2)9.2.2 Labels as shown in the diagram above. [CL2] (4)

10. 10.1 The angle of incidence which will give an angle of refraction of 90° and the refracted ray is parallel to the boundary between the media. [CL1] (2)

10.2 .,

× , × m sn vc

1 93 10 1 58 10

88 1

{{ {= == - [CL2] (3)

10.3 In order to have total internal reflection , the light must be in a medium with a higher refractive index. [CL3] (2)

10.4 To protect the fibre and to isolate it. [CL1] (2)10.5 Optical fibres are less expensive to make , can carry much larger amounts

of information and are less likely to be stolen. (Any two). [CL1] (2)10.6 Breaks are difficult to repair. [CL2] (1)

11. 11.1 Because the sides of the block are parallel , the emergent ray will be parallel to the incident ray; the extent that it bends towards the normal when entering the block will be matched by the extent by which it bends away from the normal when exiting the block. [CL3] (2)

11.2 Complete the ray diagram , including the normal lines. [CL2] (2)

ray box

incident ray

glass block

emergent raylateral displacement


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11.3Angle of incidence (θ1) Sin θ1 Angle of refraction (θ2) Sin θ2

46 0,72 22 0,375

(allow for a few degrees of error in the measurement) [CL2] (4)

11.4 11.4.1 sin sinn n1 1 2 2{{i i= [CL1] (2)

11.4.2 sin sinn n1 1 2 2i i= , ( , )n1 0 72 0 3752{ {=^ h

,n 1 922 {= [CL2] (3)

12. 12.1 Light goes from more optically dense to less optically dense material (or from higher to lower refractive index); the angle of incidence must be greater than the critical angle. [CL1] (4)

12.2 The core. [CL2] (1)

12.3 n ==, ×, ×

vc

2 7 103 0 10

8

8

{{

[CL2] (3)

,1 1{=

12.4 sin sinn n1 1 2 2i i= , ×sin sinn 27 1 1 90× °1 { { {= [CL3] (4)

,,

,n 0 451 1

2 451 {= =


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Topic 6: 2D and 3D WavefrontsQUESTIONS

MULTIPLE CHOICE1. Which one of the following statements is true for all transverse waves?

A They are electromagnetic in nature.B They transfer energy.C They have a speed of 3 × 108 m.s−1.D They all undergo the same degree of diffraction. (2)

2. Two types of monochromatic light are shone through the same slit. Both undergo diffraction, but light 1 has more diffraction than light 2. Which of the following could explain this observation?A Light 1 is more intense than light 2.B Light 2 is more intense than light 1.C Light 1 has a longer wavelength than light 2.D Light 2 has a longer wavelength than light 1. (2)

3. Which of the following phenomena can occur with a longitudinal wave like sound?A Interference only.B Interference and diffraction only.C Interference and refraction only.D Interference, diffraction and refraction. (2)

4. Sound waves bend readily around buildings whereas light waves only bend very slightly around buildings.Which ONE of the following statements BEST explains this observation?A Sound waves have much longer wavelengths than light waves.B Sound waves have much shorter wavelengths than light waves.C Sound waves have higher frequencies compared to light waves.D Sound waves have greater amplitudes compared to light waves. (2)

5. The amplitudes of two pulses combine when they meet in a medium. What is the term used to describe this phenomenon?A reflectionB refractionC superposition (2)D diffraction


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ASSESSM

ENTS

6. It often happens that through an open door you can hear someone talking in another room, although you cannot see them. This is due to: A reflection.B refraction.C superposition.D diffraction. (2)

LONG QUESTIONS7.1 All types of waves undergo diffraction.

7.1.1 Name two types of waves. (2)7.1.2 Define the term diffraction. (2)7.1.3 State the conditions needed for diffraction to be observed with a single slit. (2)

7.2 Light of a wavelength of 400 nm passes through a single slit of width 8 × 10-6 m. A diffraction pattern is seen on a screen, as shown in the diagram below.

X

7.2.1 Describe the pattern observed on the screen. (2)7.2.2 Describe how each of the following changes will affect the width or

broadness of the portion X in the diagram above:7.2.2.1 The wavelength of the light is increased to 420 nm. (1)7.2.2.2 The width of the slit is changed to 1 × 10-5 m. (1)

8.

silt screen

8.1 Define the term diffraction in words. (2) In the diagram shown above, wave fronts of light are seen approaching a narrow

opening. The light has a wavelength of 650 nm. The screen, placed some distance from the slit as shown, has a clear diffraction pattern on it.


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8.2 Express 650 nm in m, using scientific notation. (1)8.3 Describe the pattern seen on the screen. (2)8.4 Name the two important principles that are used to explain the diffraction

pattern. (2)8.5 The width of the slit is decreased slightly. Explain how this change will affect:

2.5.1 The brightness of the diffraction pattern. (1)2.5.2 The width of the diffraction pattern. (1)

8.6 Light of wavelength 600 nm is used. How will this change the diffraction pattern that is observed? (1)

9. Sound waves from a siren pass the edge of a wall as shown in the diagram below.

wall

9.1 Redraw and complete the diagram to show what happens to the sound wave as it passes the wall. (4)

9.2 Name this phenomenon. (1) 9.3 Huygen’s principle is often used to explain this phenomenon.

9.3.1 State Huygen’s principle. (2)9.3.2 In your own words, and using a simple diagram, explain how

Huygen’s principle describes how this phenomenon occurs. (3)

10. Two pulses, as shown below, meet at a single point.

10.1 What do we call the meeting of two pulses at a single point? (1)10.1.1 What type of interference will happen in this case? (1)10.1.2 Redraw the diagram, showing one of the original pulses and the final

pulse. (3)


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ASSESSM

ENTS

10.2 Now two pulses, which are totally out of phase, as shown below, meet at point.

10.2.1 What name is given to this interference? (1)10.2.2 Redraw the diagram to show the original pulses and the resultant pulse.

Label the resultant clearly. (3)10.3 You are given the diffraction pattern for red light below:

A B C D

10.3.1 Which of the points A, B C, D would correspond to the phenomenon described in 10.1.1?

10.3.2 Which of the points A, B C, D would correspond to the phenomenon described in 10.2.1? (4)

11. The diagram below shows wave fronts approaching a single slit in a ripple tank.

slit

11.1.1 Describe how you would devise an experiment to test the effect of the wavelength of the waves on the degree of diffraction. (3)

11.1.2 State the relationship you would expect between the degree of diffraction and the wavelength. (1)

11.1.3 Draw a sketch graph for this relationship. (2)


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11.2 11.2.1 Describe how you would devise an experiment to test the effect of the width of the gap on the degree of diffraction. (2)

11.2.2 State the relationship you would expect between the degree of diffraction and the width. (1)

11.2.3 Draw a sketch graph of ‘degree of diffraction’ against width. (2)11.2.4 How would you get a straight line graph for this experiment? (2)

12. Beams of red, green and blue light are each shone through a narrow single slit. 12.1 Use a diagram to explain what the diffraction pattern of red light would

look like. Indicate the bright and dark bands on your sketch. (3)12.2 Explain, using Huygen’s principle, how diffraction occurs. (3)12.3 Explain, in terms of interference, how the light and dark bands form. (2)12.4 Compare the diffraction patterns for the red, green and blue light. (2)


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ENTS

MARKING GUIDELINES

MULTIPLE CHOICE

1. B There are non-electromagnetic transverse waves (such as a transverse wave set up in a rope.) However, learners often forget that waves carry energy. [CL1] (2)

2. C Degree of diffraction is greater for light 1 therefore it has a greater wavelength. [CL2] (2)

3. D Waves will show interference, diffraction and refraction. [CL3] (2)

4. A Sound waves diffract around a wall or building because they have much longer wavelengths than light waves. [CL3] (2)

5. C This is by definition. [CL3] (2)

6. D The bending of a wave going through a gap (the door) is diffraction. [CL1] (2)

LONG QUESTIONS7. 7.1 7.1.1 Transverse waves and longitudinal waves [CL1](2)

7.1.2 The ability of a wave to spread out in wavefronts as they pass through a small aperture or around a sharp edge. OR The bending of a wave around an obstacle or the corners of a narrow opening. [CL1](2)

7.1.3 The slit must be small relative to the wavelength , and the screen must be relatively far away. [CL2] (2)

7.2 7.2.1 There will be a large central bright band , with smaller bright bands on either side alternating with dark areas. [CL3] (2)

7.2.2 X will increase with increasing wavelength. [CL2] (1) 7.2.3 The width of the slit is increased, so the width of X will decrease.

[CL2] (1)

8. 8.1 The ability of a wave to spread out in wavefronts as they pass through a small aperture or around a sharp edge. OR The bending of a wave around an obstacle or the corners of a narrow opening. [CL1] (2)

8.2 650 × 10-9 m OR 65,0 × 10-8 m OR 6,50 × 10-7 m [CL2] (1)8.3 There will be a large central bright band , with smaller bright bands on

either side alternating with dark areas. [CL3] (2)8.4 Huygen’s principle and the principle of superposition (or interference)

[CL2] (2)8.5 2.5.1 It will be less bright. [CL3] (1) 2.5.2 It will be less wide. [CL2] (1)8.6 The wavelength is increased, so the diffraction will be more noticeable, with

the bright bands wider. [CL2] (1)


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9. 9.13.1

for diagram. for showing increasing bending and for 3 or more waves aft er the wall [CL3] (4)

9.2 Diff raction [CL1] (1)9.3 9.3.1 Every point on a wavefront acts as the source of secondary wavelets

that spread out in the forward direction with the same speed as the wave. [CL1] (2)

9.3.2

for diagram

Every point on the wavefront is a wavelet. Th e wavelets destructively interfere with each other except on the wavefront. However, when the wavefront passes an obstruction (the wall) the wavelets on the one side are removed, which means that the wavelets spread out without any interference, causing the circular spreading seen in the diagram. (source for diagram: https://www.ux1.eiu.edu/~cfadd/1160/Ch25WO/Huygn.html) [CL4] (3)


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ASSESSM

ENTS

10. 10.1 Superposition (or interference) [CL1] (1)10.1.1 Constructive interference [CL1] (1)10.1.2 drawing ; new pulse has same phase ; double the amplitude

of original pulse [CL2] (3)

10.2 10.2.2 for original pulse (either) and for the horizontal zero line. [CL2] (3)

10.3 10.3.1 A , B and C are constructive interference.

10.3.2 D is destructive interference. [CL4] (4)

11. 11.1.1 Change the wavelength of the waves by changing the frequency of the waves and keep everything else constant. Measure the degree of diffraction in each case. [CL3] (3)

11.1.2 Degree of diffraction is directly proportional to the wavelength. [CL2] (1)11.1.3

Deg

ree

of d

iffra

ctio

n

Wavelength

straight line going through the origin [CL2] (2)11.2 11.2.1 Change the width of the gap. Measure the degree of diffraction in

each case. [CL3] (2)11.2.2 Degree of diffraction is inversely proportional to the width. [CL2] (1)


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11.2.3 hyperbolic curve

Deg

ree

of d

iffra

ctio

n

Width [CL3] (2)11.2.4 Plot degree of diffraction against the inverse of width. [CL4] (2)

12. 12.1

Image source: https://i.stack.imgur.com/3Jsx4.jpg.

Bright and dark bands alternate, the bright central band for diagram [CL3] (3)12.2 Every point on the wavefront is a wavelet. The wavelets destructively

interfere with each other except on the wavefront. However, when the wavefront passes through a slit, the wavelets on each side are remove, which means that the wavelets spread out and cause interference. [CL4] (3)

12.3 The bright bands are constructive interference and the dark bands are regions of destructive interference. [CL3] (2)

12.4 Red will show the most diffraction i.e. the widest central band and the most spread bands, blue will show the least and green in the middle. [CL2] (2)


Term 2 83


ASSESSM

ENTS

Topic 7: Ideal Gases and Thermal PropertiesQUESTIONS

MULTIPLE CHOICE1. Which of the following statements regarding an ideal gas is NOT TRUE?

A The forces of attraction between the gas molecules are negligibly small.B All collisions between the gas molecules are perfectly elastic.C The gas molecules collide with the sides of the container and thus exert pressure.D The volume of the individual gas molecules, collectively, is significant. (2)

2. Which graph best represents the relationship between the pressure p and the Celsius temperature T of a gas sample at constant volume? (2)

p (kPa)

T (˚C)A

p (kPa)

BT (˚C)

p (kPa)

C T (˚C)

p (kPa)

D T (˚C)

3. When the volume of a given mass of gas is halved at a given temperature, the pressure is doubled. Which of the following best explains this fact?A The force of attraction between the molecules is doubled.B The kinetic energy of the molecules is doubled.C The number of collisions per unit time between the molecules and the walls

of the container is doubled.D The momentum of the molecules is doubled. (2)


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4. The diagram shows two containers A and B connected by a tube with a stop-cock (tap.) The volume of container A is V and it contains a gas at a pressure p. Container B is evacuated. The stop-cock (tap) is opened and the pressure inside A decreases to ¾ p. The volume of the tube and stop-cock is negligible. If the temperature of the gas remains constant, what is the volume of B?A 1 3 VB 2 3 VC 3 4 VD 4 3 V

BA

(2)

5. Which of the following statements is FALSE? As the temperature of the gas in the container increases, …A the molecules move more quickly.B the molecules expand.C the molecules collide more often.D the molecules move further apart. (2)

6. Which of the following will double the volume of a given mass of gas?A Pressure is halved at constant temperature.B Temperature in celsius is halved at constant pressure.C Pressure is doubled and kelvin temperature is halved.D The pressure is doubled at constant temperature. (2)

LONG QUESTIONS7. An agricultural company wants to measure the average volume of a very small

type of seed. They use the apparatus sketched below to do this, with a constant temperature being maintained at all times.

20 cm

X Air tight piston Y

pressure gauge

Seed chamber

Funnel

tap

3

7.1 With the tap open and the piston at X, the pressure gauge indicates that the atmospheric pressure is 1,02 × 105 Pa. The tap is closed and the piston is slowly pushed to Y. The gauge now reads 1,62 × 105 Pa. The internal volume of the piston is 20 cm3. Show that the volume of the air in the rest of the apparatus is 34 cm3. (6)

7.2 The tap is now opened and the piston returned to X. A sample of seeds is poured in through the funnel and the tap is closed. The piston is slowly pushed to Y and the pressure gauge reads 1,95 × 105 Pa. 7.2.1 Calculate the volume of air in the apparatus. (3)7.2.2 Calculate the volume of seeds present. (1)


Term 2 85


ASSESSM

ENTS

8. 8.1 At sea level a small, soft plastic milk bottle is sealed up with 2 dm3 of air inside it at 25 °C. The bottle is now submerged below the water, to a pressure of 160 kPa at 15 °C. What volume does the air inside the bottle now occupy? (3)

8.2. During the testing on Earth, with an air pressure of 100 kPa, it was found that the tyre pressure in a moon buggy operated best at 280 kPa.2.2.1 To what pressure would the tyres have to be inflated on the Moon so

as to be ready to use? (1)2.2.2 Explain your answer given above. (2)

8.3 Calculate the pressure of 80 g of butane gas (C4H10) enclosed in a 8 dm3 cylinder at 18 °C. (4)

9. The sketch below is of a gas syringe. The total capacity of the syringe is 150 cm3 and the plunger is free to move. 120 cm3 of nitrogen, initially at 298 K is trapped in the syringe at atmospheric pressure, then the end of the syringe is slowly immersed in hot water at 77 °C.

120 cm3

9.1 Explain why the pressure in the syringe remains at atmospheric pressure, even when the end of the syringe is immersed in hot water. (2)

9.2 Calculate the volume of the gas at 77 °C. (4)9.3 Hence, will the plunger remain in the syringe? (1)9.4 Calculate the mass of the nitrogen gas initially in the syringe. (5)

10. You are given a sample of oxygen gas.10.1 Under which conditions of temperature and pressure will this oxygen sample

behave like an ideal gas? (2)10.2 The temperature of the gas increases from 0 °C to 273 °C. The initial volume

of the sample was 5 dm3. What will the final volume of the sample be? (3)10.3 A cylinder with a volume of 30 dm3 contains oxygen gas at a pressure of 41,6

kPa and a temperature of 27 °C. Calculate the mass of oxygen in the cylinder. (5)


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11. A pump is used to inflate a rubber balloon.

V= 120 ml

450 mmballoon

valve

air

The cylinder of the pump is 450 mm long when the piston is fully out, as shown above, and the volume of the cylinder is 120 ml. At the outlet of the cylinder is a valve which allows air to pass through it when the pressure in the cylinder exceeds the pressure on the other side by more than 50 kPa. On the reverse stroke, the cylinder fills with air at atmospheric pressure.

Initially the balloon is deflated. The internal volume of the balloon can rise to 240 ml before the rubber of the balloon starts to stretch. Atmospheric pressure is 100 kPa. Assume that the air temperature remains constant at 25 °C in all processes.11.1 With the balloon detached, how far must the piston be pushed down the

cylinder before the valve opens? (5)11.2 If the piston is pushed fully down, what volume will the air that escapes into

the atmosphere occupy? Explain your answer. (4)11.3 When the balloon is attached, how many strokes of the pump are needed

before the rubber of the balloon starts to stretch? (2)11.4 After the balloon begins to stretch, for each subsequent stroke, the piston has

to be pushed further down than the previous stroke before the valve opens. Explain why this is so. (4)

12. A small group of learners investigate the relationship between pressure and volume of an enclosed mass of gas at 298 K. They record the volume of the gas for different pressures in the table below:

Pressure (kPa) Volume (cm3) 1/V (cm-3)

80 43 0,02

160 27 0,04

200 22 (a)

240 18 (b)

12.1 Write down the name of the law being investigated. (1)12.2 Two 1/V values, (a) and (b) have been left out of the table. Calculate these

values. (2)12.3 Draw a graph of pressure against 1/V on graph paper. (4)12.4 Use the graph to determine the volume of the gas at 170 kPa. (2)12.5 Calculate the slope of your graph. (2)12.6 If the experiment was repeated at a higher temperature of 350 K, slightly

different results would be expected. On the same set of axes as used earlier, draw a sketch of the graph you would expect and label this graph. (2)


Term 2 87


ASSESSM

ENTS

MARKING GUIDELINES

MULTIPLE CHOICE

1. D The other distractors are statements of the ideal gas assumptions, this negates an assumption. [CL1] (2)

2. B The equation is pV = nRT i.e. p = mT, where T is in kelvin and m = (nR/V). However, for a celcius temperature, the graph becomes p = mT + c; where c is (nR/V) + 273. [CL3] (2)

3. C This comes from the definition of pressure. Since T is constant, both A and D are excluded. If a gas deviates from ideal gas behaviour and the force of attraction increases, then the pressure would decrease. [CL2] (2)

4. A This is a fairly tricky calculation. butp V nRT V RT0 0A A B= = ; Once the tap is opened; nRT = 4

3 pA(VA + VB) So p V V p V4

3A A B A A+ =^ h since nRT is equal before and after the tap was

opened. p V p V p V4

343

A A A B A A+ =

p V p V43

41

A B A A=

V V3 1B A=

V V31

B A= [CL4] (2)

5. B This question is designed to clarify the difference between the volume the gas occupies and the negligible size of the particles. [CL1] (2)

6. A The correct answer is simply Boyle’s law. The other options are designed to test the understanding of kelvin temperature and the correct use of the gas laws. [CL2] (2)

LONG QUESTIONS7. 7.1 p V p V1 1 2 2=

, × , × ( )V V1 02 10 20 1 62 105 5{ { {+ =^ h

, × , × , ×V V20 4 10 1 02 10 1 62 105 5 5{+ =^ ^h h

, × , ×V0 6 10 20 4 105 5{=^ h

cmV 34 3{= [CL3] (6)7.2 7.2.1 p V p V1 1 2 2= , × , × ( )V1 02 10 54 1 95 105 5{ {=^ h , × , × V55 08 10 1 95 105 5= ^ h , cmV 28 24 3{= [CL2] (3) 7.2.2 Vseed = 34,0-28,24 = 5,75 cm3 [CL3] (1)


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8. ; ; ; ; ; ?dm kPa K kPa KV p T p T V2 101 298 160 28831 1 1 2 2 2= = = = = =

Note that p1 has to be understood from ‘at sea level’.

8.1

× ×

, dm

Tp V

Tp V

V

V298

101 2288

160

1 22 3

1

1 1

2

2 2

2

2

{ {

{

=

=

=

[CL2] (3)

8.2 8.2.1 On the Moon the tyres would have to be inflated to a pressure of 180 kPa. [CL3] (1)

8.2.2 The difference in pressure is 180 kPa , but on the Moon there is no atmospheric pressure i.e. 0 kPa, so the pressure in the tyre would only be 180 kPa. [CL4] (2)

8.3 ? × ; ;m Kp V T8 10 273 183 3= = = +-^ h n needs to be calculated

n Mm=

48 10

80=+^ h

, mol1 4 {=

p VnRT=

=×

, ,8 10

1 4 8 31 2913 {{

-

^ ^h h

Pa416 933 {= [CL3] (4)

9. ; ; ; ?K KV T T V120 298 3501 1 2 2= = = =9.1 Because the plunger is free to move , it will move so that the force on each

side of it is equal , i.e. each side is at atmospheric pressure. [CL3] (2)

9.2 V TV T

21

1 2=

298

120 350{

{ {=

^

^

h

h

, cm140 9 3{= [CL2] (4)9.3 The plunger will remain in the syringe. [CL2] (1)9.4 pV nRT=

n pVRT=

=× ×

,101 10 120 10

8 31 2983 6{ {

{-

^

^ ^

^h

h h

h

, moln 0 0489=

× , × , gm n M 0 0489 28 0 14{ {= == [CL3] (5)


Term 2 89


ASSESSM

ENTS

10. ; ; ; ?K KV T T V5 298 5961 1 2 2= = = =

10.1 At high temperature and low pressure [CL2] (2)

10.2 V5

10 5962 {

{=^ h

dm20 3{= [CL2] (3)

10.3 pV nRT=

n pVRT= [CL3] (5)

, × ×

,41 6 10 30 10

8 31 3003 3{ {

{= -^

^ ^

^h

h h

h

, moln 0 05008=

× , × , gm n M 0 05006 32 0 16{ {= ==

11. ; ; ?kPa kPaV p p V120 100 1501 1 2 2= = = =

11.1 V150

100 1202 {

{=

^ h

= 80 ml 80 ml is 2/3 of the volume , so the plunger will be 1/3 down i.e. 150 mm.

[CL4] (5)11.2 120 ml – Once the pressure inside the pump is greater than 150 kPa, the

valve opens. As the plunger pushes down the pressure is maintained at >150 kPa. But the 80 ml of air expands again to 120 ml once it has escaped at the atmospheric pressure (=100 kPa). [CL3] (4)

11.3 2 strokes = 240 ml [CL2] (2)11.4 Once the balloon begins to stretch the pexterior > 100 kPa. Therefore pinside

must be greater than 150 kPa before the valve opens, thus the piston needs to be pushed further down. [CL3] (4)


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12. 12.1 Boyle’s law [CL1] (1)12.2 (a) 0,045 (b) 0,055 [CL2] (2)12.3

0.070.060.050.040.030.020.01

00 50 100 150 200 250 300

6.7

Graph to show the relationship between inverse of volume and pressure

Inve

rse

of v

olum

e (c

m )-3

Pressure (kPa )

heading; axes; plotting points; line of best fit [CL2] (4)12.4 1/V = 0,042 (allow for some error) [CL2] (2) V = 23,8 cm3 12.5 Slope = (0,05-0) /(200-0) = 0,0025 cm−3.kPa−1 [CL3] (2)12.6 The slope is smaller, as slope is 1/nRT. ( for plotting) [CL3] (2)


Term 2 91


ASSESSM

ENTS

Topic 8: Quantitative Aspects of Chemical ChangeQUESTIONS

MULTIPLE CHOICE1. The molar mass of sodium sulfate is … A 70 g.mol-1

B 98 g.mol-1

C 119 g.mol-1

D 142 g.mol-1 (2)

2. During a reaction 0,02 moles of magnesium were ignited in excess oxygen at standard temperature and pressure. The reaction that occurred is shown below:

2Mg(s) + O2(g) 2MgO(s) The volume of O2 that reacted with the magnesium was … A 0,320 dm3

B 0,160 dm3

C 0,224 dm3

D 0,224 cm3 (2)

3. Consider the reaction:C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g)

If the rate of appearance of CO2 is 0,8 mol∙s-1, the rate of disappearance of O2 is: A 0,8 mol∙s-1

B 0,48 mol∙s-1 C 0,27 mol∙s-1 D 1,33 mol∙s-1 (2)

4. Which of the following solutions contains the greatest number of dissolved ions? A 50 cm3 of 0,1 mol∙dm-3 LiF B 100 cm3 of 0,2 mol∙dm-3 KCl C 100 cm3 of 0,1 mol∙dm-3 MgCl2

D 50 cm3 of 0,2 mol∙dm-3 Na2O (2)


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5. Water is formed when oxygen reacts with hydrogen according to the following unbalanced reaction: H2 + O2 " H2O

What mass of oxygen is required to react completely with 2 g of hydrogen? A 12 g B 16 g C 96 g D 144 g (2)

6. Consider beakers A and B below:

200 cm3

0.25 mol.dmNaOH(aq)

-3150 cm3

0.1 mol.dmNaCl(aq)

-3

Beaker A Beaker BBeaker A200 cm3

0,25 mol.dm−3

NaOH(aq)

Beaker B150 cm3

0,1 mol.dm−3

NaCl(aq)

20 cm3 of the NaOH(aq) solution in beaker A is added to the NaCℓ(aq) solution in beaker B. Which one of the following represents the correct calculation for the new concentration of Na+(aq) ions in beaker B?

A ,, ,

0 170 015 0 005+

B ,, .

0 170 015 0 05+

C ,, ,

0 150 015 0 05+

D ,, ,

0 150 015 0 005+

(2)

LONG QUESTIONS7. Iron (Fe) reacts with sulfur (S) to form iron sulfide (FeS) according to the

following balanced equation:Fe(s) + S(s) → FeS(s)

7.1 Define the term limiting reactant. (2)7.2 Calculate which of the two substances will be used up completely if 20 g of

Fe and 10 g of S are mixed and heated. (5)7.3 How many grams of the other substance are in excess? (2)


Term 2 93


ASSESSM

ENTS

7.4 Magnesium burns in air to form magnesium oxide according to the following balanced equation:

2Mg(s) + O2(g) " 2MgO(s) If the percentage yield of this reaction is only 80%, calculate the mass of

magnesium that needs to be burned to produce 30 g of magnesium oxide. (6)

8. A standard solution of ethanol (CH3CH2OH) is made up so that it will have a volume of 0,25 dm3 and a concentration of 0,5 mol.dm-3.

The standard solution is made up using distilled water.

8.1 Calculate the mass of CH3CH2OH required to make up the standard solution. (4)

8.2 27 g of propane is burnt in air. The balanced reaction equation is shown below:C3H8 + 5O2 " 3CO2 + 4H2O

8.2.1 27,0 g of propane is burnt. What mass of CO2 will be produced? (5)8.2.2 What would the volume of this gas be at STP? (2)

9. A group of learners react ethanoic acid and potassium hydroxide. They repeat the experiment, a titration, a number of times until they obtain three results that have a high degree of precision. The concentration of the standard KOH solution they use is 0,1 mol.dm-3. They calculate that the average volume of KOH required to neutralise 25,0 cm3 of vinegar solution is 44,1 cm3.

CH3COOH(aq) + KOH(aq) " CH3COOK(aq) + H2O(l)

9.1 9.1.1 Explain how the learners will determine whether their results “have a high degree of precision”. (2)

9.1.2 Calculate the number of moles of ethanoic acid in the vinegar solution. (4)9.1.3 Hence, calculate the mass of ethanoic acid in the vinegar solution. (3)9.1.4 If 5 g of white spirit vinegar was used to make up the 25 cm³ vinegar

solution used in the titration, calculate the percentage (by mass) of ethanoic acid in white spirit vinegar. (2)

9.2 During the titrations the learners used a burette to measure out the volume of KOH required to neutralise the vinegar solution. Describe two precautions that the learners should take when using the burette to ensure that the measurements that they take with the burette are as accurate as possible. (2)


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10. Magnesium burns in air to form magnesium oxide according to the following balanced equation: 2Mg(s) + O2(g) 2MgO(s)

2,4 g of magnesium combines with 0,8 g of oxygen, which is at STP.

10.1 What does STP stand for? (1)10.2 Calculate how many moles of magnesium and

how many moles of oxygen are present. (3)

10.3 Which of the two substances will be used up completely? (2)

10.4 Determine how many moles of the other substance is in excess. (2)

10.5 What mass of product will form? (3)

11. 11.1 20 g of impure iron reacts with excess sulfuric acid, releasing 5 dm³ of hydrogen gas. The balanced equation is shown below:

Fe (s) + H2SO4 (aq) FeSO4 (aq) + H2 (g)11.1.1 Calculate the number of moles of hydrogen gas released. (2)11.1.2 Write down the number of moles of pure Fe which reacted. (1)11.1.3 Calculate the percentage purity of the iron. (3)

11.2 Consider the balanced equation shown below. The yield for the production of water is 68,7%.

2 HNO3 (aq) + NO (g) 3 NO2 (g) + H2O (g)11.2.1 If 44,1 g of HNO3 reacts completely with nitrogen monoxide, calculate

the theoretical mass of water which is produced. (4)11.2.2 Considering the yield for this reaction (68,7%), now calculate the

actual mass of water which is produced. (2)

12. A standard solution of Mg(OH)2 is made up so that it will have a volume of 0,25 dm³ and a concentration of 0,5 mol.dm-3.

The standard solution is made up using distilled water. 12.1 Name the solute used to make this solution. (2)12.2 Calculate the mass of solid Mg(OH)2 required to make up the standard

solution. (6)12.3 If the solution needed to be diluted to a concentration of 0,2 mol.dm-3, how

much additional distilled water would need to be added? (4)


Term 2 95


ASSESSM

ENTS

MARKING GUIDELINES

MULTIPLE CHOICE QUESTIONS

1. D Na2SO4 MR = 2(23) + 32 + 4(16)=142 g.mol-1. [CL2] (2)

2. C 0,02 mol Mg reacts with 0,01 mol O2 [CL2] (2)

3. D 5 mol O2 gas produce 3 mol CO2. Therefore O2 is used at a rate of (5 × 0,8)/3 = 1,33 mol.s−1. [CL2] (2)

4. B This is comparison (2 # 50 × 0,1) vs (2 × 100 × 0,2) vs (3 × 100 × 0,1) vs (3 × 50 × 0,2). There is no need to convert units to do the comparison. [CL3] (2)

5. B The reaction needs to be balanced 2H2 + O2 2H2O; 1 mol of H2 reacts with 0,5 mol O2 i.e. 16 g [CL3] (2)

6. A Only 20 cm³ of beaker A (= 0,02 × 0,25 mol) is transferred. Moles in B was 0,015; total volume is (0,015 + 0,02 = 0,17 dm3). [CL4] (2)

LONG QUESTIONS7. 7.1 The reactant that will be used up totally in a chemical reaction

[CL1] (2)7.2 , , ; ,mol moln n55 8

20 0 358 3210 0 3125Fe S{ { {= = = =

From equation : : .isn n 1 1Fe S { Thus S runs out first; S is the limiting reagent [CL3] (5)7.3 , , , ,m nM 0 358 0 3125 55 8 2 56{ {= - == ^ ^h h [CL2] (2)

7.4 Actual yield = 30 g; theoretical yield = , , g0 830 37 5{ {=

,

,, moln

24 3 1637 5

0 93MgO {{=

+=

: : ,hence moln n n2 2 0 93Mg MgO Mg{ {= =

, × , , gm nM 0 93 24 3 22 61 {= == [CL4] (6)

8. 8.1 n = CV = 0,5 (0,25) = 0,125 mol m = nM = 0,125 (46 ) = 5,75 g [CL2] (4)8.2.1 n = m/M = 27/(44 ) = 0,81 ratio 1: 3 0,81: 2,45 m =nM = 2,45(44 ) = 108 g [CL2] (5)8.2.2 V = n(22,4 ) = 55,0 dm³ [CL2] (2)


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9. 9.1.1 If their results are very similar (within a range of 0,5 ml ) they will have a high degree of precision. [CL2] (2)

9.1.2 n(KOH) = CV = 0,1 × 0,044 = 0,0044 mol KOH `n(acid) = 0,0044 mol CH3COOH [CL4] (4)9.1.3 m = n × M = 0,0044 × 60 = 0,264 g [CL4] (3)

9.1.4 ,

× , %50 264

100 5 28= method ans [CL4] (2)

9.2 • Rinse burette out with water and KOH solution.• When filling burette let KOH run through tap and ensure no air bubbles.• When taking readings ensure eye is parallel with the bottom of the

meniscus to avoid the error of parallax.• Estimate to one decimal place (any two). [CL3] (2)

10. 10.1 Standard temperature and pressure [CL1] (1)

10.2 ,,

, ;,

,mol moln n24 32 4

0 099 320 8

0 025Mg O2{ { {= = = = [CL2] (3)

10.3 nMg:nO2 = 2:1 so 0,025 mol O2 will react with 0,05 mol Mg ; hence O2 will run out first. [CL3] (2)

10.4 excess Mg = 0,099-0,05 = 0,049 mol [CL2] (2)

10.5 , × × , , gm nM 0 05 2 24 3 16 4 03{ { {= + == ^ h [CL2] (3)

11. 11.1 11.1.1 n(H2) = V/Vm = 5/22,4 = 0,223 mol [CL2] (2)11.1.2 mol Fe = 0,223 mol (1:1) [CL2] (1)11.1.3 m(Fe) = nM = 0,223 × 56 = 12,5 g

` %purity = 12,5/20 × 100 = 62,5% [CL3] (3)

11.2.1 n(HNO3) = m/M = 44,1/63 = 0,7 mol ` n(H2O) = 0,7/2 = 0,35 mol (2:1) m(H2O) = nM = 0,35 × 18 = 6,3 g H2O (theoretical mass) [CL3] (4)11.2.2 % yield = actual yield / theoretical yield × 100 ` actual yield = 68,7/100 × 6,3 = 4,33 g H2O [CL3] (2)

12. 12.1 magnesium hydroxide ID correct name [CL1] (2)

12.2 , × , ( ) ,both subs moln CV 0 5 0 25 0 125{ {= ==

, × , , gm nM 0 125 58 3 7 29{ { { {= == [CL2] (6)

12.3 ,,

,coe

dmV Cn

0 20 125

0 625 3{ {= ==^ h

, , ,V dm0 625 0 25 0 375added3{ {= - = [CL4] (4)


physical sciences grade 11 term 2 resource pack

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