SC130 Examination revision set

Pure substances and Mixtures

1. Define and give examples which demonstrate the difference between the following;a) Mixture and pure substance; b) homogenous and heterogenous mixtures c) atoms and molecules d)

elements and compounds

2. a) Draw a diagram which illustrates the change in states of matter.b) Draw the arrangement of particles in each state and their relative energy levelsc) Explain at a molecular/particle level why does ice melt

3. How would you separate the following mixtures: List the equipment and processes useda) Dirt and pebblesb) Sugar from a soft drinkc) Water from a slushyd) Ball bearings, marbles, sand and salt in tub of water

4. Pictorially draw (using coloured dots as atoms) the followinga) Water at -10 oCb) Methane at 20oCc) A mixture of oxygen, nitrogen and carbon dioxide gas at room tempd) Elemental nitrogen at room temperature

Atoms and Compounds

5. Explain the difference between a physical/state change and a chemical reactionPhysical change is reversible and no new substance is formed

What are typical characteristics or indicators of a chemical change

colour change

gas bubbles

formation of new substance including a precipitate

Energy/temperature change (although physical changes may also involve temperature change)

irreservible

6. Draw and label the parts of typical atom SC130 Examination revision 2018S1

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7. Complete the table

Atomic Weight # of protons # of electrons # of neutrons Element12 6 6 6 Carbon107 47 47 60 Silver19 9 9 10 Fluorine40 (rounded) 20 20 20 Calcium

8. Explain the “Octet rule”Atoms will react to establish a stable arrangement of 8 outer valency electrons. An atoms valency is reflective of the number of electrons needed to form the full 8 outer shell

Atom Valency Atom ValencyBe +2 Al +3F -1 P -3K +1 I -1Si +/- 4 S -2

9. Draw the orbital arrangements for 3 of the above atoms

Remember shells of 2, 8, 8, 18 etc (reflects the periodic table arrangement)

10. Write the formula for the following:

Lithium Oxide Ammonium sulfiteAluminium Chloride AlCl3 Strontium

dichromateSr CrO4

Silicon sulfide SiS2 Lead carbonate PbCO3

Iron (III) Selenide Fe2Se3 Sulfuric Acid(Hydrogen sulfate)

H2SO4

Zinc chloride ZnCl2 Iron(II) permanganate

Fe(MnO4)2

Chromium Phosphide

CrP Calcium cyanide Ca(CN)2

Magnesium nitride Mg(NO3)2 Potassium nitrate KNO3

11. Write and balance the following equations

a) Zinc + hydrochloric acid -> Zinc Chloride + Hydrogen gasb) C3H8 (Propane) + oxygen -> Carbon Dioxide + Waterc) Aluminium chloride + Sodium hydroxide -> Aluminum Hydroxide + Sodium chloride

AlCl3 + 3 NaOH -> 3 NaCl + Al(OH)3

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d) Magnesium + Sulfuric acid -> Magnesium Sulfate + Hydrogen gase) Calcium Nitrate -> Calcium oxide + oxygen gas + NO2

2 Ca(NO3)2 -> 2 CaO + O2 + 4 NO2

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Forces and Energy

12. Explain the following terms or concepts a) Forceb) Newton third lawc) Energyd) Energy transformatione) Energy transferf) Frictiong) Newtonsh) Joules

13. List the types of energy and give an example

Energy type Kinetic or potential ExampleChemical Potential FoodKinetic energy Kinetic Moving ball/person/car etcElectrical (chemical) potential batteryThermal/heat Kinetic Burning woodGPE potential Anything with heightChemical potential PetrolLight Interesting (kinetic-ish) torchKineticElastic

KineticElastic potential (in strings)

tennis racket hitting a ball

14. List 5 energy transformations which occur within 10m of your current positionEg electrical->light in a light bulb Kinetic -> sound when tapping the keyboard

15. a) Create an energy pathway of at least 5 steps which ends with you tossing up a pen into the air.b) label each step as either a transfer of transformation of energy

light -> chemical (in plant) -> chemical (when eaten) -> kinetic (in muscle motion) -> kinetic (pen motion)transform transfer transform transfer

-> GPE (pen gaining height) transform

16. a) What is the formula linking force and accelerationf = ma

b) Calculate the force needed to accelerate a 250 g cricket ball to 15m/s

f = ma m= 0.25kg

= 0.25 x 15 a = change in velocity = 0 to 15m/s

= 3.75 Joules (J)

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d) Calculate the acceleration of a 1.2kg netball if an 80N force was applied f = ma m= 1.2 kg 80 = 1.2 x a f = 80 N 80/1.2 = a a = 66.6 m/s2

18. A skater dropped into an empty old swimming pool

a) Discuss the energy changes involved and explain why they eventually stop at the bottom

GPE <-> KE + sound + heat. Slow down and stop due to friction

b) Was any energy lost to the system? What could be considered “lost” energy from the skaters perspective

The total energy is conserved but skater losses energy through sound & heat

19. Use the following information 20kg parcels, 80m distance between the truck and location of the parcel.

a) calculate the work required to carry one parcel to the truck (remember gravity)

w = f x d f = ma = 20kg x gravity(9.8)

= (20kg x 9.8) x 80

= 15680 J or 15.68 kJ

b) If approximately 19 kilojoules of energy was used in total how many parcels were carried? Just one

c) How many snickers bars (741 KJ /bar) would need to be consumed to replace this used energy?

19/741 = 0.025 of a bar ie. You would need to do approximately 39 trips to use up one snickers energy

Simple motion

20. A 50kg child bounced on a trampoline to a height of 4.5m. (really scary trampoline).

a) Calculate the child’s maximum GPE

GPE = mgh

= 50 x 9.8 x 4.5

= 2205 J

b) What would the max KE be? The same 2205 J

c) What would the GPE and KE be at the 3.6m height (ascending at the time)

Calculate the GPE at 3.6 m and substract this from 2205 to calculate KE

KE + GPE = Total energy. KE = 441 GPE = 1764

e) At what velocity would the child hit the trampoline meshSC130 Examination revision 2018S1

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KE= ½ mv2

2205= 0.5 x 50 x v2

2205= 25 v2

2205/25= v2

Square root of 88.2 = v

v = 9.4 m/s

f) If the child’s 83 kg mother matched her daughters jumping height would she have the same velocity as her daughter? (show calculations) yes as mass has no effect on velocity of an object. Gravity has

GPE = mgh = 3660 J = KE also

KE= ½ mv2

3660= 0.5 x 83 x v2

3660= 41.5 v2

3660/41.5 = v2

Square root of 88.2 = v

v = 9.4 m/s

21.

A 45kg boy running at 5.6m/s jumped onto a simple rope swing attached to a tree branch above a 4m riverbank. The rope was 5m long.

a) Calculate how high he would swing upward? KE would convert to GPE

KE= ½ mv2 GPE = 705.6 = mgh = 0.5 x 45 x 5.6 2 705.6 = 45 x 9.8 x h

= 705.6 J 705.6/(45 x 9.8) = h

h = 1.6 m

Total height = 1.6 + 4m = 5.6 m

b) if at the highest point he let go and dropped into the river. How much energy would be hit the water with GPE = KE GPE = mgh = 45 x 9.8 x 5.6 = 2469.6 J

c) How fast would he hit the water (Just the same as 20 f) see above)

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d) If his 90kg father ran at the swing slightly quicker, 6.0m/s, would he hit the water with the same velocity (show working) KE= ½ mv2 = 1620 J = GPE GPE = mgh -> height of 1.8 m total height is now 5.8 mPlace 5.8 in GPEKE =GPE (as per20f)

22.

A rev-head accelerated to a 5m/s/s from a standing start for a period of 9 sec.

a) How fast are they travelling in m/s and kph (pretty fast)Acceleration of 5m/s increase every second therefore 9 sec of acceleration 9 x 5m/s = 45m/s45m/s = 0.045km/ (1/3600 hour) = 162kph

b) The brakes were applied for 2.5 sec bringing the car to a speed of 12m/s, when a police car in view. Calculate the decceleration

a = change velocity/change in time

= (12 – 45)/2.5

= - 33/2.5

= - 13.2 m/s2

c) Car continued at this speed for a further 10 secs when the brakes where applied again and the car stopped completely at the traffic light. The total journey took 0.6 minutes = 36 sec

d) Draw a velocity time graph for the whole journeye) Comment upon the graph’s shapef) A mates car travelling at the same top speed deccelerated at the same rate but only for 2.0sec when

the police car appeared. Who they get pinged for speeding if it was a 60km/hr zone ?g) If the original car was 1.4 tonnes how much force was the engine applying to the wheels to reach the

top speed?

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f) Deceleration was 13.2m/s2 for 2 sec = loss of 26.4 m/s Original velocity 45m/s – 26.4m/s = final velocity 18.6 m/s

= 0.018km/(1/3600hr) = 64.8 kph

g) f = ma = 1400kg x 13.2m/s2 = 19335 N

23.

Draw a circuit with the following

a) 1.5 V battery with 3 light bulbs (A,B,C) and a switch all in seriesb) Same components but 2 bulbs (D,E) are on one strand of a parallel circuit and the other bulb (F) on the

other strand.

c) Which bulb(s) would have the highest current flowing through it from either circuit a) or b) F as 1.5 V is over a strand of wire with only 1 bulb as resistance

d) Which bulb(s) has the lowest current A,B,C 3 bulbs of resistance in the wiree) In circuit a) the current was measured at 3.4amps. Calculate the resistance of each bulb.

V = IR 3 bulbs = 0.44 ohms 1.5 = 3.4 x R 1 bulb = 0.44/3 = 0.146 ohms1.5/3.4 = R R = 0.44 Ohms

f) In circuit b) the total resistance was measured at 2.5 ohms calculate the overall current V = IR 1.5 = I x 2.5 1.5/2.5 = I I = 0.6 amps

g) How much current will b flowing through lamp D and how much through lamp F

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The 0.6 amps will be split 1:2 ratio (3 parts in total) = 1/3 : 2/3 = 0.2 amps : 0.4 amps

Lower resistance strand (one bulb F) will have 0.4 amps current (the 2/3 in the ratio) and bulbs D and E will have 0.2 amps 1/3 in the ratio.

24. Draw the following optical devices (use scientific diagrammatic symbols)

a) concave mirror b) plane mirror c) broad concave lens d) convex lens e) triangular prism

f) thin convex lens g) convex mirror

25. Draw the ray diagrams for all of the above. Eg. I have drawn the shading incorrectly on the mirror… sorry

26.

Complete the ray diagram for the rectangular prism

Explain why it changes direction

On entry of the prism and high side of the ray hits the prism first, slows down and therefore bending upwards

On exit the high side of the ray hits the air first, speeds up and therefore bends downwards

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27.

The following data is the forearm length measured in different aged children

Person 1 Person 2 Person 3Child age Forearm length

(m)Forearm length

(m)Forearm length

(m)17 0.65 0.68 0.6515 0.50 1.52 0.5112 0.33 0.41 0.328 0.28 0.26 2.325 0.23 0.27 0.26

a) Which results could be considered errors or outliers b) What would be the one of the statistical operations you would undertake when analyzing this data

Average (mean) the 3 samplesc) Between which age periods (eg 12-15 years) showed the least amount of growth?

5 – 8 yrsMeasured in actual amount (m) and % age? 8 average = 0.27m ; 5 average = 0.25Growth = difference/original x100

= 0.02/0.25 x 100= 8%

d) What would be a reasonable hypothesis for these results The forearm lengthens with age with an increasing percentage of growth.

e) Was there any patterns or explanations for this growth Growth spurt at puberty

28. Read the webpages and watch the videos on Science and faith and SCI EXP

a) Science can document miracles but has limits in explaining them. What are the general limits of the scientific method.

Science is limited to the observable and physical data through experimentation using an analytical method. It can infer the ‘how’ of physical observations but not the ‘why’ behind it

It is limited and cannot make moral judgements, aesthetic judgements, tell you how to use Scientific knowledge or draw conclusion about supernatural

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b) What is Dr Gerald Schroders (video 3 faith and Science) main parallels in regard to the big bang and Genesis

1 This is the nature of God and aligns with the Genesis story of creation

c) Watch the God in the lab video and read 5 reasons science and faith are compatible: What are some of the key reasons these Scientist believe there faith has been strengthened by science inquiry.

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