o co h o · 2019. 8. 12. · question 31 answer is a question 32 question 33 answer is b question...

© The School For Excellence 2018 Trial Exam Revision Lectures – Chemistry – Solutions Page 1

QUESTION 1

Oil, natural gas and similar.

QUESTION 2

Fractional distillation. A physical process based on the different boiling points of the components of crude oil.

QUESTION 3

Plastics.

QUESTION 4

Carbon.

QUESTION 5

OHCOOOHHC 222115 1210152

QUESTION 6 (a) C4H8(l) + 6O2(g) 4CO2(g) + 4H2O(l) (note: although in a combustion reaction water

will be produced as a gas, it is most appropriate to use the standard state of the substance at room temperature and pressure in our equations)

(b) C2H5OH(l)+ C2H5COOH(l) C2H5-O-COC2H5 (l)+ H2O(l) (c) 2HCl(aq) + Mg(s) MgCl2(aq) + H2(g) (d) H2SO4(aq) + CuCO3(s) CuSO4(aq) + H2O(l) + CO2(g) (e) CO2(g) + Ca(OH)2(aq) CaCO3(s) + H2O(l) (f) CH3CH2OH(l) CH2CH2(g) + H2O(l)

QUESTION 7 Answer is D (States are missing! Add them in.)

conc. H2SO4

conc. H2SO4


QUESTION 8 (a) Gas

(b) Liquid

QUESTION 9

Octane has a higher boiling point than pentane. Both are non-polar molecules, so the main intermolecular force between the molecules in each case are dispersion forces (i.e. temporary dipoles). The strength of the dispersion forces depends on the number of electrons in the molecule. Octane (C8H18) has more electrons than pentane (C5H12), and consequently has stronger intermolecular forces. Stronger intermolecular forces require more energy to overcome, and thus octane has a higher boiling point. (Note – it is commonly, and incorrectly, stated that the stronger dispersion forces are due to higher molecular weight. At HSC level it is expected you understand that the fundamental reason is the number of electrons.)

QUESTION 10

C8H18 C2H4

Name of compound Octane Ethene

Name of series Alkane Alkene

QUESTION 11

QUESTION 12

Number of carbon atoms Alkane Name

2 Ethene

3 Prop-1-ene

4 But-1-ene

5 Pent-1-ene

6 Hex-1-ene

7 Hept-1-ene

8 Oct-1-ene


QUESTION 13

For a chemical reaction to occur, molecules need to collide with enough energy to overcome the activation energy (the “energy barrier” to the reaction). A catalyst, such as the zeolite in this case, lowers the activation energy for a reaction and so requires less heat energy input and a lower temperature.

QUESTION 14

Risk – hydrocarbons are flammable and ignition may cause burns. Precaution – keep away from sparks and sources of ignition such as naked flames. Risk – bromine water is corrosive and can harm skin and eyes. Precaution – wear safety glasses and wash hands if contact occurs. Use only the minimum amount necessary.

QUESTION 15

In a fume cupboard, carefully add 1 mL of cyclohexane to a labelled test tube. Carefully add 1 mL of cyclohexene to another labelled test tube. Add 1 mL of orange bromine water to each test tube, and place them both in a darkened cupboard for 10 minutes (alternatively – cover both test tubes in aluminium foil). After 10 minutes, observe any changed to the colour of the bromine water layer. Results – the bromine water layer in the cyclohexene test tube decolourised. The bromine water layer in the cyclohexane test tube did not change and remained orange.


QUESTION 16


QUESTION 17 (a) Hydrogenation.

(b) Hydrogen gas.

(c) pent-2-ene pentane

QUESTION 18 Answer is D QUESTION 19

(a) C2H4 (g) + H2 (g) C2H6 (g)

This is an example of an addition reaction. (b) Alkenes will rapidly decolourise bromine water. A simple test would be to add a few drops of bromine water to the mixture at the end of the reaction. If the colour of the bromine water disappears, that is, it changes from brown to colourless, then some ethene still remains.

QUESTION 20

Overall;

QUESTION 21 Answer is B QUESTION 22 (a) HDPE: this polymer is produced at relatively low pressures and low temperatures

(60ºC) using catalysts such as the Ziegler–Natta catalyst. LDPE: this polymer is produced at high pressure (100–300 MPa) and moderately high

temperatures (300ºC) using an initiator such as benzoyl peroxide. (b) HDPE consists mainly of linear chains which pack closely together forming extensive

dispersion forces. This gives strength to the polymer. Consequently, it can be used to make tough objects such as milk crates and agricultural pipes.

LDPE consists mainly of branched chains which are prevented from packing closely

together, resulting in weaker dispersion forces. This makes the plastic softer and transparent. It can be used to make plastic bags and food wraps.


QUESTION 23

PVC:

• Insulation of electrical cables – non-conductive, strong, resistant to chemicals

• Pipes for plumbing – chemically and biologically resistant, strong

• Packaging – strong and brittle to protect contents Polystyrene:

• Thermal insulation – expanded polystyrene is a good thermal insulator and has a low density

• Cases (CD cases, casing for electrical equipment) – crystal polystyrene is a stiff and strong protective material that can be clear if required (CD cases), or dyed if preferred (most electrical equipment)

QUESTION 24

(a) Chloroethene.

(b) Ethyl benzene.

QUESTION 25

QUESTION 26 Answer is D QUESTION 27 (a)


(b)

QUESTION 28


QUESTION 29

They are both covalent substances based on carbon.

QUESTION 30

Declining resource stock. Increasing demand for petroleum and plastics in developing countries and around the world.

QUESTION 31 Answer is A QUESTION 32

QUESTION 33 Answer is B QUESTION 34

(consider using a table for any questions that ask you to compare)

(a) Both ethylene and glucose are monomers that can undergo polymerisation reactions to form polymers. Ethylene will undergo addition polymerisation to form polyethylene. Polyethylene can be produced either by using a catalyst or an initiator. There is no byproduct and the reaction occurs at high temperatures. A new carbon-carbon covalent bond is formed.

Glucose undergoes condensation polymerisation. –OH groups on different monomers react to form an ether functional group plus a water molecule as a byproduct. Large polymers such as cellulose are formed in this way. An enzyme catalyst is required, and the reaction can proceed at low temperatures. A new carbon-oxygen covalent bond is produced.


(b) High density polyethylene (HDPE) is produced when ethylene is polymerised using a catalyst at low temperature and pressure. The polymer chains of HDPE are long and relatively unbranched and so can pack closely together with strong dispersion forces. As a result, HDPE is rigid and tough. It is used to make petrol tanks, pipes, buckets, bottles and toys. Low density polyethylene (LDPE) is produced when ethylene is polymerised using an organic peroxide initiator at high temperature and pressure. The polymer chains are shorter than those of HDPE and there is much more branching. The chains cannot pack closely together which weakens the dispersion forces. This results in a soft, flexible material which is used to make cling wrap and plastic bags.

Cellulose is a condensation polymer formed from thousands of glucose molecules. The polymer chains are unbranched. The –CH2OH groups alternate on opposite sides of the polymer chain, resulting in strong hydrogen bonding between linear layers. This gives the cellulose strength, and makes it insoluble in water. Consequently cellulose is found in the cell walls of plants and we use it in strong natural fibres such as cotton.



QUESTION 37 (a) Condensation. There should be (n-1) H2O moleules.

(b) The biopolymer polyhydroxybutanoate (Biopol) is produced by the bacterium Ralstonia eutrophus. In recent years, the gene responsible for the production of the biopolymer has been transferred into the bacterium E. Coli. This has allowed for the more efficient production of biopol. Biolpol has thre advantage over petroleum-bsed polymers in that it is biodegradable, and can be broken down using alkaline solutions. It is also dervived from renewable resources. This means that biopol has a lower environmental impact. The physical properties of biopol are similar to the commonly used PET (polyethylene terephthalate which is produced from ethylene terephthalate, another derivative of ethylene), so it can be used to make plastic bottles and food wraps. For example, biopol has been used to make shampoo bottles in the past as it is water insoluble and non-toxic, however, this was abandoned due to the high cost of production. It is currently used in the medical industry as it is biocompatible. This has had a major impact on post surgical care as internal sutures no longer need to be removed. The high costs of producing polymers like biopol are still a significant disadvantage, although this is improving with time and new technologies. Therefore, despite being biodegradable, the environmental impact of biopol has been low due to its limited application in the production of disposable plastic items.

Assessment – biopol currently has had a significant impact in medical technologies, but its impact as a biodegradable polymer has been limited by cost. However, production costs are falling and demand for biodegradable plastics is increasing, so it is expected that in the future biopol will have a more significant impact.

QUESTION 38

Long-lasting, non-biodegradable polymers such as polystyrene greatly affect the environment, as microorganisms cannot break them down. Polystyrene is lightweight and can be made into a solidifying foam to be used in insulating cups. It can also be manufactured into a clear, hard and brittle plastic for drinking cups. These properties make polystyrene a useful polymer. However, due to its inability to biodegrade and most of its uses being short-lived, polystyrene debris builds up in the environment and fills land fill and, as the extract states, have a costly impact on waste management. Polylactic acid (PLA) is an alternative to traditional polymers. PLA is classed as a biopolymer as it is biodegradable. It is used for short-lived applications such as cold/warm drinking cups and plastic bags. As PLA is made from the waste products of corn crops, which are converted by bacteria into lactic acid and then reacted to form PLA, it is easily broken down by microorganisms once it has been disposed. Assessment - biopolymers such as PLA should be used instead of traditional polymers, such as polystyrene, as biopolymers break down in the environment and do not accumulate as debris.


QUESTION 41 Answer is B QUESTION 42 (a) Polyethene is produced by combining monomers which “add” across the double-bond

to form a polymer. i.e. a long chained molecule made from the monomer units

C C

H

H H

H

n C C H

H

H

H

H

Hn

(b) (i)

C C

H

H H

H

+ Br Br C CHH

Br

H

Br

H

1,2-dibromoethane

(ii)

C C

H

H H

H

+ OHH dilute H

2SO

4

C CH

H

H

O

H

H

H

ethanol used as a petrol extender and a fuel

QUESTION 43

Ethanol is widely used as a solvent due to its structure. Ethanol can act as a solvent for both polar and non-polar substances. The ethanol molecule has a polar hydroxyl end and a non-polar end (where the – CH

3 is). It is also a small molecule. The polar end dissolves polar

substances, such as water and sugars, by H–bonds or dipole-dipole interactions. The non-polar end can dissolve non-polar substances, such as hydrocarbons and esters, by dispersion forces.

QUESTION 44

Water and ethanol are both polar molecules as they contain a highly electronegative oxygen atom bonded to a hydrogen atom, but their structures have some important differences that change their solvent behaviour. Water is highly polar and can form hydrogen bonds, dipole-dipole interactions and ion-dipole interaction. This makes it an excellent solvent for polar molecules and ionic salts. Ethanol is also polar but has a non-polar alkyl chain as well. Ethanol can form hydrogen bond, dipole-dipole interactions and dispersion forces. Consequently, ethanol is not a good solvent for ionic salt, but it is a good solvent of polar and non-polar substances.

non-polar alkyl group

polar hydroxyl group


QUESTION 45 Answer is C

QUESTION 46 (a) (i)

400.079.24

92.9)( 2 COn

(ii)

400.0016.18

20.7)0( 2 Hn

(b) 0.100 mol hydrocarbon 0.400 mol C and 0.800 mol H 1 mol hydrocarbon 4 mol C and 8 mol H from (a) molecular formula = C4H8 (c) possible compounds for C4H8 but – 1 –ene but – 2 –ene cyclobutane 2 - methylpropane

QUESTION 47

(a) q = mc T 1.237g of ethanol burnt, q = 200 x 4.18 x 22J

when 1.00g of ethanol was burnt, q = 237.1

2218.4200 Jxx14.87kJ = 14.9kJ

(b) Molar mass of ethanol = [12 + (2x16.00)]g = 46.01

Hc = 46.01 x 14.9kJmol-1 = 684kJmol-1

(c) Complete combustion clear blue flame or no soot or no yellow flame.

(d) Heat from burning ethanol lost to surrounding air and the tin can, therefore less heat transferred to H2O. Approx. 50% heat lost to surroundings (1 mark)

QUESTION 48

(a) Sometimes cells cannot get enough oxygen to carry out aerobic respiration for energy, but they still need to obtain energy to stay alive. So they use an emergency system of reactions called anaerobic respiration. Anaerobic respiration is the use of a fuel, such as glucose, under conditions of low oxygen, to produce energy. Anaerobic respiration in yeast produced ethanol (fermentation). Anerobic respiration in mammals produces lactic acid (glycolysis).

(b) Fermentation of glucose to form ethanol must be carried out under anaerobic conditions (low oxygen) in order to avoid normal respiration that produces the by-products of carbon dioxide and water. If plenty of oxygen is present normal respiration will take place and ethanol will not be produced.


(c) C6H12O6 (aq) + 6O2(g) 6CO2 (g) + 6H2O (l)

QUESTION 49

Bubbling gas through limewater – changes from clear to milky.

CO2 (g) + Ca(OH)2 (aq) CaCO3 (s) + H2O (l)

QUESTION 50

(a) Aqueous: C6H12O6 (aq) 2CO2 (g) + 2C2H5OH (aq)

(b) Step 1 – filtration of the mixture to remove yeast.

Step 2 – Fractional distillation to remove the ethanol from the water (ethanol boils at 78°C, water boils at 100°C)

(Note – a Bunsen burner should not be used to distil ethanol, as ethanol is highly flammable. A heating mantle is a convenient substitute.)


QUESTION 51

(a) Mass CO2 = 381.05 g – 310.39 g = 10.66 g Moles CO2 = 10.66 g / 44 g/mol = 0.242 mol (b) C6H12O6 2C2H5OH + 2CO2 Moles glucose = 0.5 x moles of CO2 = 0.5 x 0.242 mol = 0.121 mol Mass glucose = 0.121 mol x 180 g/mol = 21.78 g

QUESTION 52 Answer is B QUESTION 53 Answer is B QUESTION 54

(a) C6H12O6(aq) yeast

2C2H5OH(aq) + 2CO2(g)

yeast essential, low O2, temp 30oC-37oC, sugar, sterile conditions, alcohol


Pure ethanol to ethyl butanoate.

• Esterification reaction. Ethanol is reacted with butanoic acid to form ethyl butanoate and water.

• Requires reflux and conc. sulphuric acid as a catalyst.

• After reflux the reaction vessel will contain ethyl butanoate, some ethanol, some butanoic acid, water, sulphuric acid.

• Purification of ester can be carried out, but this question does not really require this.

QUESTION 56 (a) Corn, wheat, sugarcane. (b) Biomass is composed mainly of cellulose. The action of cellulase enzymes will convert the cellulose to glucose (or acid digestion can also be used).

cellulose cellulase

C6H12O6(aq) OR

cellulose 42SOdiluteH

C6H12O6(aq)

The glucose can then be fermented by yeast to generate ethanol.

C6H12O6(aq) yeast

2C2H5OH(aq) + 2CO2(g)

QUESTION 57

Small volumes of ethanol will form solutions with petrol. For example, 10–20% mixtures of ethanol in petrol are quite commonly used. Therefore, less petrol needs to be used to fuel a car, and the available petrol usage is extended in times of shortages.

QUESTION 58


QUESTION 59

Marking Criteria Marks

• Explains the problems and benefits arising from the use of ethanol.

• Makes use of data from the table.

• Provides a judgement of the potential for ethanol to be used as an alternative fuel.

4 – 5

• Describes the problems and benefits arising from the use of ethanol AND uses data from the table.

OR

• Provides a judgement of the potential of ethanol as an alternative fuel.

2 – 3

• Identifies one reason for the need for alternative fuels.

OR

• Identifies one aspect of data from the table with an advantage or disadvantage of ethanol.

1

A good answer to this question really requires a thorough assessment – not just stating what the positive and negative aspects to using ethanol as an alternative fuel, but most importantly, an overall JUDGEMENT (key word is “assess”) about its future use based on the points made. Writing down everything you know about ethanol does not guarantee a full score! QUESTION 60 Answer is A

QUESTION 61

Methanol, ethanol, propanol, butanol, etc.

QUESTION 62

The essential measurements required would be:

• Mass of water heated (alternatively the volume of water heated assuming 1 g/mL).

• Mass of spirit burner before heating.

• Mass of spirit burner after heating.

• Temperature of water before heating.

• Temperature of water after heating. The mass of the copper cup could also be measured, allowing the calculation of the heat absorbed by the cup.


QUESTION 63

(a)

(b)

QUESTION 64

Fuel Formula

Heat of combustion (kJ/mol)

Heat of Combustion (kJ/gram)

Methanol CH3OH 727 22.7

Ethane C2H6 1560 52.0

Ethanol C2H5OH 1366 29.7

Heat / gram = heat/mol divided by molecular mass.


QUESTION 65

(a) C4H9OH (l) + 6O2 (g) 4CO2 (g) + 5H2O (l) (b) The molar mass of 1-butanol is 74.12 g mol-1. Heat of combustion = 2676 / 74.12 = 36.10 kJ g-1

Fuel C is most likely to be 1-butanol. The values determined for fuels A and B are both greater than 36.10 kJ g-1. The methods used to measure heats of combustion in school laboratories usually result in significant heat losses. Therefore, the measured values are normally lower than actual values.

QUESTION 66

2676 kJ mol-1 / 74 g mol-1 = 36.16 kJ / g The answer is B.

QUESTION 67

(a) Selects appropriate X and Y scales Correctly labels Y axis Correctly determines molecular weights Correctly plots heat of combustion vs molecular weight and draws the line of best fit (b) (i) 1400 kJ mol-1

(ii) Incomplete combustion OR impurities in the fuel.

QUESTION 68

(a) Heat evolved/released when 1 mole of ethanol is completely combusted (excess O2)

(b) M = 46g, 46

15.1n

1627

156751518.4250

kJmoln

qH

JxxTmCq

(c) C2H5OH(l) + 3O2(g) 2CO2(g) + 3H2O(l) 1627 kJmolH

(d) Both the heat of combustion per gram and the heat of combustion per mole increase

as carbon chain increases. However the increase in heat of combustion per mole is almost linear, and the increase in heat of combustion per gram increases more slowly.


QUESTION 69

(a) Zinc is more reactive than copper and so will displace the copper ions in solution by undergoing oxidation. The less reactive copper will accept electrons from the zinc and form solid metal by undergoing reduction. Therefore, the red-brown deposit is copper metal. Since copper ions are blue in water, as they are converted into solid copper their concentration is reduced, causing the solution to become lighter in colour.

Zn (s) + Cu2+ (aq) Zn2+ (aq) + Cu (s) (b) n (Cu2+) initial = 0.05 x 0.250 = 0.0125 n (Cu) deposit = 0.325 / 63.55 = 0.00511 n (Cu2+) remaining = 0.0125 - 0.00511 = 0.007386 [Cu2+] = 0.007386 / 0.25 = 0.02955 molL-1 = 0.030 molL-1 (2 sig. fig.)


(a) +1

(b) +4

(c) +5

QUESTION 72

Reductant – CuS (S went from -2 to 0, so was oxidised) Oxidant – NO3- (N went from +5 to +4, so was reduced – note that some N atoms did not change their oxidation state)

CuS(s) + 4HNO3(aq) Cu(NO3)2(aq) + S(s) + 2NO2(g) + 2H2O(l)

QUESTION 73 Answer is B QUESTION 74 Answer is A QUESTION 75

Potassium nitrate (or sodium nitrate) – it does not form precipitates with other ions.

QUESTION 76

To allow current to flow through wire so it can be used to do some task.

QUESTION 77 Answer is B


QUESTION 78

Anions flow from Fe3+ I Fe2+ half cell to Pb I Pb2+ half cell.

QUESTION 79 Answer is D QUESTION 80 Answer is C QUESTION 81

(a)

(b) Ni(s) Ni2+

(aq) + 2e–

½ Cl2(g) + e- Cl

–

(aq) OR ½ Cl2(aq) + e- Cl

–

(aq)

Ni(s) + Cl2(g) Ni

2+

(aq) + 2Cl–

(aq) ORNi(s) + Cl2(aq) Ni

2+

(aq) + 2Cl–

(aq)


(c) Ni(s) Ni2+

(aq) + 2e–

Cl2(g) + 2e

–

2Cl–

(aq)

E–° Ni = 0.24 E

–° Ni = 0.24

E–° Cl

2(g) = 1.36 OR E

–° Cl

2(aq) = 1.40

E–° cell = 1.60 V E

–° cell = 1.64 V

(d) As the reaction proceeds, the [Ni2+

] at the anode increases.

Ni(s) Ni2+

(aq) + 2e–

Ni2+

ions are green and as the [Ni2+

] increases, the green colour will intensify at the anode (darker green).

QUESTION 82 (a) Lead

(b) 0.13 V + 0.80 V = 0.93 V (c) NaCl(aq) : not appropriate since the silver (and lead) can form precipitates. Glucose :

not appropriate since it does not allow the migration of ions(not ionic solution). NH4NO3(aq) : appropriate since all nitrates are soluble and it allows the effective migration of ions.

(d) While the voltmeter is in place, current cannot flow (voltmeters have a high resistance).

So when a conductive wire is placed there, silver will collect on the cathode (Ag+(aq) + e– Ag(s)) and the lead electrode will start to corrode away (Pb(s) Pb2+(aq) + 2e-). If we reconnect the voltmeter we will notice a lower voltage reading as the conditions have moved away from standard (the Pb2+ solution is >1M, the Ag+ solution


QUESTION 84

• Dry-cell battery continues to be an important source of low cost mobile power.

• It presents little danger to environment during use and disposal.

• It comes in a convenient range of sizes.

• In low power applications, the dry-cell continues to be a suitable technology.

QUESTION 85

(a) The anode is Pb(s) (anode is oxidation and oxidation is loss of electrons).

Overall reaction: Pb(s) + PbO2(s) + 2SO42-(aq) + 4H+(aq) 2PbSO4(s) + 2 H2O(l) (b) Because these batteries are rechargeable, due to the reversible reactions taking place,

they can last for several years. They discharge (galvanic cell) in order to produce energy to start the car, but once the engine is running energy is pushed back through the cell (electrolytic cell) to reverse the reactions and recharge the cell. This is a great benefit because it means that the battery does not need to be replaced very often, making it a reliable source of energy.


Another correct answer could be ‘an unstable ratio of protons to neutrons’, or variations on that theme.


• Uranium-238 has too many protons (more than 82) to be stable.

• Oxygen-18 has a proton: neutron that is too low (8:10).

• Carbon-14 has a proton: neutron that is too low (6:8).

• Therefore, A, C and D are wrong.

QUESTION 88

Polonium-210 is unstable due to the facts that:

• It has more than 82 protons. All nuclei with greater than 82 protons are unstable.

• The ratio of protons to neutrons in the nucleus is 84:126(1:1.5). A more stable ratio for very large nuclei is slightly above 1:1.5.

When polonium-210 undergoes alpha decay it becomes lead-206 which is stable. This nuclei has a proton to neutron ratio of 1:1.51. Polonium-210 will not undergo beta decay as this would result in an even lower proton to neutron ratio which would make it even less stable.


QUESTION 89

The ratio of neutron to proton in potassium-40 is 21:19 (1.1:1).

0140

20

40

19 CaK

The product of beta decay is calcium-20 as one neutron has been converted to a proton. This changes the neutron to proron ratio to 20:20 (1:1) which is a more stable ratio for small nuclei. QUESTION 90

Parent isotope - F189

Equation - 01

18

8

18

9 OF

QUESTION 91

(a) nPHeAl 1030

15

4

2

27

13

0130

14

30

15 SiP

(b) In phosphorus-30, the neutron: proton ratio is 1:1 (15:15), which is too low for a

nucleus with Z > 20. In positron decay, a proton becomes a neutron which makes the new ratio 16:14 (1.15:1) which is more stable.

QUESTION 92

(a) A nucleus will be unstable if the mass is too large (if Z>83), or if the neutron: proton ratio is too large or too small. For elements Z83). (also neutron: proton ratio is 1.6:1

so beta and positron decay less likely).

HeBhMt 42275

107

279

109

QUESTION 93

11C – too many protons/too few neutrons to be stable; positron emission or e– capture.

Actual decay mode: BeC

BC

11

5

0

1

11

6

11

5

0

1

11

6

14C – too few protons/too many neutrons to be stable; emission

0114

7

14

6 NC


QUESTION 94

Original activity = 100% (t = 0); 5.5 hours = 330 minutes = 3 half-lives (3 × 110) Activity after 3 half-lives = 1/2 × 1/2 × 1/2 × 100 = 12.5%.

QUESTION 95

PunPu 241941

0

239

94 2

01241

95

241

94 AmPu


(a) GapZn 67311

1

68

30 The charged particle is a proton

(b)

0168

31

68

32 GaGe The second particle is a positron

QUESTION 98

(a) X is a beta particle and Y is a neutron. (b) Neptunium is produced by neutron bombardment in a nuclear reactor which triggers

beta decay to produce the new element. Curium is produced in a particle accelerator such as a cyclotron in which the target nucleus is bombarded with positively charged nuclei.

QUESTION 99

Outline - carbon-11 is used in PET, a diagnostic technique that allows visualisation of blood flow and can detect blood clots and heart defects. The radioisotope carbon-11 is used to produce 11CO which is given to the patient as a gas. The 11CO binds to haemoglobin and travels around the blood stream. Carbon-11 decays according to;

01

11

5

11

6 BC

The positron produced collides with an electron, releasing gamma radiation which can be detected by the scanner. Justification - carbon-11 has a short half-life of only 20 minutes which reduces the harmful impact on the patient. Carbon is able to bond with O to produce a molecule that binds strongly to haemoglobin, so will successfully travel around the body. It is a positron emitter, and the resulting gamma rays can easily be detected outside the body as it is highly penetrating.

QUESTION 100 A good answer to this question will:

• Give two specific radioisotopes as examples, one used in medicine and one used in industry.

• Give a number of benefits and problems associated with the use of each isotope mentioned. These will be specific for the isotopes used (related to half-life, ionisation, penetration and effects on people/environment) and not vague generalisations.


• Refer to the chemical properties of the chemicals used. This is not just the radiation properties, but the way these isotopes react chemically with other substances to form compounds, ions, etc.

• Make some kind of overall value judgement (evaluation) on whether these radioisotopes have a positive or negative impact on society.


QUESTION 1

The indicator is not useful in distinguishing between neutral and basic substances as it appears yellow in pure water (neutral) as well as ammonia and oven cleaner, which are both basic.

QUESTION 2

(a) Answer is C – the colours described only overlap between 7.5 and 8.5.

(b) Answer is A – changes red to purple in basic solution (NaOH). Ammonia is the only basic solution in the options given, all others are acidic.

QUESTION 3

Identifies bromothymol blue as the best indicator. Provides a reason why bromothymol blue is the best indicator. Provides a reason why each of the other indicators are not suitable.

QUESTION 4 Answer is D QUESTION 5 Answer is C QUESTION 6 Answer is A

CO2 is quite common in the atmosphere and reacts with water in the atmosphere to form carbonic acid.

QUESTION 7

The oxides change from neutral to acidic to amphoteric for these oxides. CO is neutral and SiO2 is acidic. The remaining oxides (GeO2, SnO, PbO) are amphoteric.

QUESTION 8

The evidence of increasing concentrations of NO and NO2 in the atmosphere comes from

direct measurement, knowledge of sources and measurement of effects. Thus there is clear factual evidence to support this statement. Monitoring is necessary due to detrimental environmental and health effects.

Combustion of fossil fuels in motor vehicles and in power stations results in the formation of NO.

N2(g) + O

2(g) → 2NO(g)

In the lower atmosphere, in the presence of sunlight NO is oxidised to NO2

2NO(g) + O2(g) → 2NO

2(g)

Direct measurements are made by statutory bodies, such as NSW EPA, and also by scientists researching the atmosphere. Nitrogen oxides were found to be part of pollution generated over large cities from the onset of the industrial revolution and the use of the internal combustion engine. This was essentially unrestricted till the 1950s, when concerns grew from loss of life due to smog. Some reduction of NO was achieved by the introduction of catalytic converters in motor vehicles where 2CO(g) + 2NO(g) → N

2(g) + 2CO

2(g)


This removes NO from the exhaust gas. However the increased number and use of motor vehicles has offset further major reductions.

Measurement of effects of NOx highlights both issues of quantities produced and

detrimental effects which necessitates their monitoring.

One effect is production of photochemical smog. NOx is part of a complex set of chemical

species which have several reactions in the presence of sunlight resulting in photochemical smog.

Another effect is the direct toxicity of NOx to people, affecting especially the respiratory tract

and eyes. There is a need to monitor NOx in cities in order to warn people most susceptible

to acute effects. NO

x are also contributors to acid rain:

2NO(g) + HO( ) → HNO(aq) + HNO(aq)

2HNO2(g) + O

2(g) → 2HNO

3(aq)

The effects of acid rain are multiple:

• Lowering pH of natural waters affecting all biota in that water body.

• Acidic precipitation affecting vegetation by acid burning or defoliating plants

• Acidic runoff from soils leaching some nutrients, specifically Ca2+

, Mg2+

• Acidic leaching of toxic species, e.g. Al3+

Acidic attack on manmade structures, e.g. marble or limestone buildings.

QUESTION 9

Smelting and fuel combustion (eg coal) has increases since then to make more metals and energy available. This has caused more SO

2 going into the air making rain acidic.

CuS + O2 Cu + SO

2

SO2 + H

2O H

2SO

3

We see more of an effect of acid rain now. Metal and stone buildings are deteriorating quicker and forests are dying where more factories are.

QUESTION 10

Demonstrates an extensive knowledge of the industrial production of nitrogen oxides and sulfur dioxide and their impact on the environment. Includes examples and chemical equations. Provides an evaluation of these impacts.

QUESTION 11 Answer is D QUESTION 12 Answer is C

With an increase in pressure the equilibrium will shift to the side with less moles of gas (right) to minimise the disturbance; increasing the yield of phosgene. The reaction is exothermic (heat is a product), so reducing the temp forces the equilibrium to the right to


make more heat; increasing the yield of phosgene.


These reactions are not equilibrium reactions, so more sulfate does not change anything. Sodium sulfate is a neutral salt.

QUESTION 14

Concentrations vary at 4 to 8 minutes, 10 to 12 minutes and 14-16 minutes. At 4 to 8 minutes the system is moving to the right. There is no indication that Cl

2 or CO have been

initially removed or that COCl2 was initially added, hence the temperature has been

changed. The reaction is endothermic to the right, so this means heat has been added (temperature has been raised). The sharp drop in [CO] at 10 minutes means that CO has been removed. The system then moves to the right, seen as an increase in [CO] and [Cl

2]

and a decrease in [COCl2]. At 14 min all concentrations abruptly decrease, all by a 2/3 factor

hence the volume of the system has increased. As the right side is proportionally more affected than the left hand side, the system moves to the right to compensate, hence [CO], [Cl

2] increases and [COCl

2] decreases.

QUESTION 15 Answer is C QUESTION 16

(a) The addition of hydroxide ions will neutralise the carbonic acid and hydrogen ions in the soda water. This will shift equilibrium to the right to counteract the change. As a result the solubility of carbon dioxide in water will increase.

(b) This will increase the carbon dioxide solubility as equilibrium will shift to the right to

counteract the partial pressure increase. (c) No change; the addition of nitrogen does not affect the partial pressure of carbon

dioxide as these gases both fill the space above the liquid. Although the total gas pressure has increased, this has not happened due to volume reduction. The effective concentration of carbon dioxide above the water has not changed.

QUESTION 17 Answer is B QUESTION 18 Answer is B QUESTION 19 Answer is A

Find the moles of both gases (NO = 0.66 / 30 = 0.22 mol ; O3 = 0.72 / 48 = 0.15 mol). This means that ozone is the limiting reagent and only 0.15 mol of NO2 can be produced. Find volume of NO2 (0.15 X 22.71 = 0.34 L).


QUESTION 20 (a)

• Outlier plotted but not included in graph (line of best fit).

• Intersection point indicated.

• Lines connecting data points are straight.

• Points plotted correctly.

• Axes labelled.

• Linear scales used on axes. (b) 380 mL. HCl is the limiting reagent.

(c) Zn (s) + 2 HCl (aq) ZnCl2 (aq) + H2 (g) n(Zn) = 0.56/65.41 = 8.56 x 10-3 mol

n(H2) = 8.56 x 10-3 mol v (H2) = 8.56 x 10-3 x 24,79 = 0.212 L = 210 mL (2 sig. fig)


Moles of P2O5 (1.42 / 141.94 = 0.01 mol). Moles H3PO4 (2 X 0.01 = 0.02 mol). Moles NaOH (3 X 0.02 = 0.06 mol). Volume of NaOH (0.06 mol / 0.3 molL-1 = 0.2 L)

QUESTION 22 (a) Monoprotic

(b) Diprotic

(c) Monoprotic

(d) Diprotic

QUESTION 23

H2SO4(l) + H2O(l) H3O+(aq) + HSO4–(aq)

HSO4–(aq)+ H2O(l) H3O+(aq) + SO42– (aq)

QUESTION 24 Answer is B QUESTION 25 Answer is C QUESTION 26

(a) The H2SO4 as it is diprotic.

(b) The H2SO4 as it is strong and citric is weak (even though it is triprotic).


QUESTION 27

Hydrochloric acid is a strong acid and therefore fully ionises in aqueous solution. Ethanoic acid is a weak acid and does not fully ionise in aqueous solution. The total concentration of ethanoic acid must be higher than that of HCl, due to incomplete ionisation of CH

3COOH, to

give an equivalent [H+

] and therefore pH, as pH is equal to –log10

[H+

].

QUESTION 28 Answer is A

A monoprotic strong acid of this conc. would have pH = 1.0. Sulfuric is a strong diprotic acid so pH is less than 1.0.


(a) 10 mL diluted with 90 mL of water is a 10 x dilution. Therefore pH increases by 1.0. Initial pH is 2, so new pH is 3.

OR

c1V1 = c2V2

c2 = 0.01 mol L-1 x 10ml / 100mL

= 0.001 mol L-1

pH = -log10[H+]

= - log10(0.001)

= 3 (b) Both acids can be added to food to lower the pH, making it too acidic for decay-causing

bacteria to survive, and therefore preserving the food. The acids also have a sour taste which can be used to provide flavour to foods, such as sauces, soft drinks and confectionary.

(c) HCl has the lowest pH because it has the highest [H+]. This is due to the fact that, being

a strong acid, it ionises completely. Acetic acid has the lowest pH because it has the lowest [H+]. It is a weak acid and so does not ionise completely. Citric acid is also a weak acid, however, it ionises to a greater extent than acetic acid.

QUESTION 31 Answer is B QUESTION 32 Answer is C QUESTION 33 Answer is D


QUESTION 34

Arrhenius defines an acid as a substance that ionises in water to produce hydrogen ions

(H+

).He defines a base as a substance that ionises in water to produce hydroxide ions

(OH–

).Alternatively, Brönsted-Lowry defines the acid-base reaction as a transfer of a proton. They define acids as proton donors and bases as proton acceptors. The Arrhenius definition continues to be used despite the fact it only applies to aqueous solutions. This is becauseArrhenius’s theory is encompassed by the Brönsted-Lowry definition, most acid-basereactions occur in aqueous solutions, and the theory demonstrates the formation of salt and water during neutralisation reactions.


The only substance that can donate or accept a proton.


(a) H2O + HCl H3O+ + Cl- water as a base

H2O + NH3 OH- + NH4+ water as an acid

(b) HPO42- + OH- H2O + PO43- hydrogen phosphate as an acid

HPO42- + H3O+ H2O + H2PO4- hydrogen phosphate as a base

QUESTION 38

(a) Correct salt chosen eg. Ammonium chloride, NH4Cl. (1) (b) When ammonium chloride is dissolved in water it dissociates its ions, NH4+ and Cl- NH4+ + H2O H3O+ + NH3 Ammonium acts as an acid (proton donor) when dissolved in water. (2)

QUESTION 39 Answer is C QUESTION 40 Answer is D

This is the only combination of a weak acid and its conjugate base.

QUESTION 41


QUESTION 42

There is the same net ionic equation for neutralisation of strong acid and strong

base: H+

(aq) + OH –

(aq) → H2O

The heat of reaction is the same for these neutralisation reactions because the spectator ionsdo not influence the heat of reaction, and strong acids and bases completely dissociate

into their H+

/OH–

ions respectively.

QUESTION 43

Explain the amphiprotic nature of the hydrogen carbonate ion using correctly balanced equations. Equations must show an identified acid and base reacting with the hydrogen carbonate ion. Also describe other chemical properties that contribute to the effectiveness of the ion in neutralising both acidic and basic spills and include a judgment.

QUESTION 44 Answer is C QUESTION 45 Answer is A QUESTION 46 Answer is C

Both errors (rinsing the burette with water, not acid and rinsing the conical flask with base, not water) have the effect of requiring more acid to neutralise the base than should be the case. This results in a calculation of the base conc. that is too high.

QUESTION 47

(a)

(b)

(c)


QUESTION 48

QUESTION 49

(a)

1. Weigh the required mass of dried NaHCO3 directly into a clean dry beaker.

2. Dissolve in a small amount of distilled water.

3. Add to a clean volumetric flask washed with distilled water.

4. Wash the remnants from the beaker into the flask with a wash bottle containing distilled water.

5. Fill flask to the graduated mark with distilled water.

6. Stopper the flask and invert 10 times to mix thoroughly.


(b) n(NaHCO3) = 0.12 mol L-1 X 0.25 L = 0.03 mol

m(NaHCO3) = 0.03 mol X 84.008 = 2.5 g (to 2 significant figures)

QUESTION 50

(a) Ba(OH)2(aq) + 2HNO3(aq) Ba(NO3)2(aq) + 2H2O(l) (1) (b) n(Ba(OH)2) = 0.12 X 0.025 = 0.003 mol n((HNO3) needed to nuetralise = 0.003 X 2 (from equation) = 0.006 mol c(HNO3) = 0.006 / 0.01813 = 0.33 mol L-1 (3)

QUESTION 51

(a) A base is a proton acceptor (b) n(HCl) = 0.1000 x 0.0244 = 0.00244 mol n(base) = 0.00244 mol, because the acid/base mole ratio is 1:1 from the equation

HCl + NaX H2O + NaCl n(base) in 100mL = 0.00244 X 5 = 0.0122 mol M(base) = 1.00 g / 0.0122 mol = 82.0 g mol-1

QUESTION 52

(a)


(b)


(a) 2NH3 (g) + H2SO4(aq) (NH4)2 SO4(aq) (b) moles ammonia = 2.479 L / 24.79 L mol-1 = 0.10 mol moles sulfric acid required = 0.10 mol / 2 = 0.05 mol moles sulfuric present = 0.2 L x 0.5 M = 0.1 mol therefore sulfuric left over = 0.1 mol – 0.05 mol = 0.05 mol

(c) H2SO4 + 2NaOH Na2SO4 + 2H2O

moles NaOH required = 2 x 0.05 = 0.10 mol

therefore volume NaOH required = 0.1mol / 2.0 molL-1 = 0.05 L = 50 mL

QUESTION 55

(a) CaCO3 + 2HCl CaCl2 + H2O + CO2 (b) moles HCl = 0.600 M x 0.025 L = 0.015 mol (c) moles of NaOH used to neutralise excess acid = 0.100 M x 0.0142 L = 0.00142 mol

Therefore moles HCl used to neutralise CaCO3 = 0.015 – 0.00142 = 0.01358 mol Therefore moles of CaCO3 in original sample = 0.01358 mol / 2 = 0.00679 mol Therefore mass of CaCO3 = 0.00679 mol x 100.09 g/mol = 0.680 g (3 dec. places)


QUESTION 56 Answer is A

pH will change very little to begin with (buffer solutions absorb excess acid without changing pH very much). Eventually the buffer breaks down when too much acid is added and the pH drops.

QUESTION 57

(a)

(b)

(c)

(d)


The indicator the changes over the range of the titration equivalence point.


QUESTION 61 Answer is D


QUESTION 62

(a) Butyl propanoate and water.

(b)

(c) The reactants are volatile and yet need to be heated to reach their activation

energy/increase reaction rate. Refluxing cools the reactant gases and condenses them returning them to the reaction mixture for continued heating, whilst maintaining atmospheric pressure. The gases would otherwise escape before they reacted.

QUESTION 63

• Detailed description of necessary equipment and the refluxing process or suitable labelled diagram.

• Justification given for equipment, chemicals (including catalyst) and refluxing.

• Specifies chemicals for making the ester including the catalyst.

• Answer illustrated with correct equation.

QUESTION 64

Glucose is fermented to produce an ethanol mixture. Fermentation is carried out in large vats. Yeast is added to the glucose solution under anaerobic conditions as it provides the enzyme zymase which catalyses the process. Maximum yield is achieved at optimal conditions of temperature (around 35 ºC) and with inorganic nutrients such as ammonium phosphate to provide nitrogen and phosphorus. This produces approx. 10% ethanol mixture.

C6H12O6(aq) 2CO2(g) + 2CH3CH2OH(aq) The ethanol is purified by distillation to a 92% concentration. The process of distillation involves heating the mixture until the ethanol boils at 78ºC. The ethanol vapour is cooled and condenses and is separated. A diagram of fermentation is recommended. To produce the purity of ethanol needed in esterification, (98%), the alcohol is mixed with a dehydrating agent, which is later recovered and reused. Pure ethanol is mixed with pure butanoic acid and heated under reflux to produce the ester, ethyl butanoate. A diagram of esterification is recommended.


The mixture is heated in a heating mantle or water bath for a lengthy time so that the reactants reach their activation energy. To speed up the process, concentrated sulfuric acid is added which acts as a catalyst and assists in the dehydration process. Hydrogen is removed from the alkanol and hydroxyl from the acid so that they can link together forming the ester. Water is the other product

CH3CH2OH(l) + CH3CH2Ch2COOH(l) Ch3COOCH2CH2CH3(l) + H2O(l) The mixture is heated but this must occur without the loss of volatile reactants. The reaction is therefore carried out by refluxing where the volatile substances are condensed in an upright condenser and are returned to the reaction flask for further heating. The excess acid can be neutralised using sodium carbonate and the ester can be purified by distillation.

QUESTION 65 Answer is A QUESTION 66 Answer is A


QUESTION 1 (HSC 2001:25)

• Displays extensive knowledge of possible different reaction products, either due to identified varying reaction conditions or due to the presence of impurities using an appropriate example.

• Explains the need for monitoring these products to avoid problems. (5–6)

• Identifies at least one variation in reaction products using an appropriate example. AND

• Relates formation of this product to the need for monitoring to avoid problems. (3–4)

• Describes a specific chemical reaction. OR

• States purpose of monitoring. (1–2)

QUESTION 2 (HSC 2002:11) Answer is B QUESTION 3 (HSC 2002:12) Answer is D QUESTION 4

C4H9OH + 6O2 4CO2 + 5H2O

QUESTION 5

C4H9OH + 11/2O2 2CO2+ 5H2O + 2CO (several other options possible)

QUESTION 6 Water QUESTION 7

To monitor the level of unburnt VOC’s (fuel) and communicate with the fuel injection system to convert some pollutants to les harmful gases.

QUESTION 8 (HSC 2002:13) Answer is C QUESTION 9 (HSC 2001:24)

(a) Any relevant use stated (1) (b)

• States that the yield is increased.

• Explains the increase in terms of Le Chatelier’s principle and the removal of product. (2)

• States that the yield increased.

OR

• Indicates that equilibrium shifts right. (1)


(c)

• Indicates the conflicting effect of temperature on reaction rate and yield.

• Explains the effect of pressure on yield.

• Identifies that a compromise set of conditions must be used in equilibrium reactions. (3)

• Identifies that reaction has optimum conditions of temperature and pressure that affect reaction rate OR yield OR safety conditions.

• Explains how temperature OR pressure affect yield OR reaction rate. (2)

• Identifies that reaction has optimum conditions of temperature and pressure that affect reaction rate OR yield OR safety conditions. (1)

QUESTION 10 (HSC 2002:14) Answer is C QUESTION 11 (HSC 2006:24) (a)

• Provides a balanced chemical equation for the Haber process.

• Includes states of matter.

• Uses equilibrium arrows in equation. (1) (b)

• Outlines a social or political issue at the time of development of the Haber process.

• Explains restrictions on supplies.

• Outlines the impact of the Haber process at the time of its discovery.

• Provides a judgement. (4)

• Outlines a social or political issue at the time of development of the Haber process.

• Outlines the impact of the Haber process at the time of its discovery.

• Explains restrictions on supplies (2–3).

• Identifies a use for ammonia.

OR

• Identifies a social or political issue at the time of development of the Haber process.

OR

• Identifies an advantage of the Haber process. (1)


QUESTION 12

Liquification removes Ammonia gas and thus lowering concentration of Ammonia. Due to Le Chatelier’s principle, the system will shift right to make up for the Ammonia removed. Liquification is effective because ammonia has a higher b.p than hydrogen and nitrogen. This means that Ammonia can be liquefied and hydrogen and nitrogen will remain as gases. Ammonia has a higher b.p due to its higher molecular mass and its ability to form Hydrogen bonds as well as dipole bond between molecules. Hydrogen and nitrogen are only able to form weak dispersion forces between like molecules.


(a) States that equilibrium has been reached. (1) (b) i.

• Correctly sketches all three trends.

• Identifies that equilibrium is re-established. (2)

• All three trends shown correctly. (1) (b) ii.

• Outlines how Le Chatelier’s principle is applied to trends in the graph.

• States effect in terms of relative number of moles of gas on either side. (2)

• States Le Chatelier’s principle.

OR

• States that equilibrium is disturbed. (1)

QUESTION 14 (HSC 2006:15) Answer is A QUESTION 15 (HSC 2001:12) Answer is D QUESTION 16 (HSC 2002:24)

• Describes how the use of AAS has changed analytical procedures and assesses the impact of AAS on scientific understanding of the effects of trace elements. (4)

• Describes how the use of AAS has changed analytical procedures.

AND/OR

• Describes how the use of AAS has enhanced scientific understanding of trace elements.

AND/OR • A good explanation of how AAS works.

AND/OR

• Describes the effects of trace elements. (2–3)

• Identifies some trace elements measured using AAS.

OR


• Describes AAS.

OR

• Indicates an advantage of AAS. (1)

QUESTION 17 (HSC 2010:20) Answer is B QUESTION 18 (HSC 2003:28)

(a)

• Plots data accurately.

• Draws a line of best fit. (2)

• Plots data accurately.

OR

• Draws line of best fit. (1) (b)

• Justifies the conclusion with reference to the data table.

e.g. The industrial plant because:

1. Highest concentrations are at sites 3 and 4 which are downstream of the sewage works and industrial plant.

2. The concentration near the plant is twice that of the unpolluted site 1. This indicates that the contamination must be moving downstream from the industrial plant. Although this is not conclusive, it is a fair inference.

• Identifies the correct source of cadmium pollution. (1)


(a)

• Correctly identifies the purpose of the light source AND the purpose of the flame. (2)

• Correctly identifies the purpose of the light source.

OR

• Correctly identifies the purpose of the flame. (1) (b)

• Describes how different parts of the procedure affect both validity AND reliability.

• Provides a judgement. (4)

• Describes how different parts of the procedure affect both validity AND reliability. e.g. Validity: An appropriate number of known solutions used to calibrate the machine

Variables controlled effectively on machine. Reliability: Sufficient repeats.

Anomalies identified (in this case measurement 4 appears anomalous and should not be included in average.)


OR

• Describes how different parts of the procedure affect validity OR reliability.

• Provides a judgement. (2–3)

• Makes a correct statement about validity OR reliability.

OR

• Describes how one part of the procedure increases or decreases either validity or reliability.

OR

• Identifies that absorbance = 0.64 is an outlier (1)

QUESTION 20 (HSC 2005:11) Answer is A QUESTION 21 (HSC 2006:14) Answer is C QUESTION 22 (HSC 2006:25) (a)

• Plots points correctly and draws a curve of best fit through the points. (2)

• Correctly plots points on the grid.

• Incorrectly plots points but correctly draws a curve of best fit through the points. (1) (b)

• Provides an accurate estimate for both samples based on the student’s graph.

• Provides a judgment for both estimates.

• Provides a reason for both judgments. (3)

Any TWO of:

• Provides an accurate estimate for either sample based on the student’s graph.

• Provides a judgment for either estimate.

• Provides a reason for the judgment. (2)

• Provides an accurate estimate for either sample based on the student’s graph.

OR

• Provides a reason why results obtained by AAS will be valid. (1)

QUESTION 23 (HSC 2010:10) Answer is A QUESTION 24 (2004:3) Answer is B


QUESTION 25 (HSC 2004:21) (a)

• Correctly distinguishes between the terms qualitative analysis and quantitative analysis. (2)

• Defines either qualitative analysis or quantitative analysis: Qualitative means comparative observation. No numbers. Quantitative means definite numerical results. (1) (b)

• Describes TWO factors that affect the concentrations of ions in natural bodies of water. (2)

• Describes a factor that affects the concentrations of ions in natural bodies of water. Predominant bedrock locally; rainfall; flush / flow rate of water body.

OR

• Identify two factors. (1) (c)

• Correctly identifies the reagent and observation for each of the specified anions. (3)

• Correctly identifies all 3 reagents OR 3 correct observations.

OR

• Correctly identifies 2 reagents and their observations. (2)

• Correctly identifies a reagent and observation for one of the anions.

OR

• Correctly identifies two reagents. (1)


(a)

• 0.052% (2)

• Calculates the mass of total dissolved solids. (1) (b)

• Provides characteristics and features of a test that could be used to identify lead ions in sample. ppt with HCl OR AAS. (2)

• Names a correct reagent for identifying lead ions. (1) (c)

• Provides a reason, with supporting argument, for monitoring a named ion. (2)

• States a reason for monitoring a named ion. (1)



(a)

• Carbon dioxide correctly identified. (1) (b)

• Provides specific reasons to show that if the tests are done out of order, they will not identify each of the three ions. (3)

OR

• Gives an example where two of the ions will not be distinguished if the tests are done out of order.

• Correctly identifies the anions detected by each of steps 1, 2 and 3. (2)

• Identifies the anion detected by step 1 or 2 or 3. (1)

QUESTION 28

Incubation causes water to evaporate from the sample. A desiccator that absorbs that water ensuring it does not become reabsorbed by the sample.

QUESTION 29 (HSC 2001:13) Answer is A QUESTION 30 (HSC 2007:27) (a)

• Recognises that AgNO3 is added in excess to the water sample.

• Writes correct balanced equation.

• AgCl precipitate filtered, dried and weighed.

• Mass of AgCl can be used to calculate mass of Cl ions

– Ag+ ions react with Cl – ions to form a precipitate (3)

• THREE of the above points. (2)

• ONE of the above points. (1)

(b)

• Calculates correct concentration of chloride ions in ppm to three significant figures. (3)

• Calculates correct mass of chloride in water sample. (2)

• Calculates correct formula mass for AgCl. (1)

(c)

• Relates a high concentration of Cl ions in water to a negative effect on society or the environment. (2)

• Identifies need to provide safe water for humans or to protect the habitat of other organisms. (1)


QUESTION 31 (a)

(b)

QUESTION 32 (HSC 2004:8) Answer is D QUESTION 33 (HSC 2003:4) Answer is B


QUESTION 34

UV

NO2 (g) NO (g) + O• (g)

O2 (g) + O• (g) O3 (g) Oxygen free radicals form when UV light breaks up the nitrogen dioxide molecule. The oxygen free radicals react with oxygen molecules to form ozone.


• Compares the environmental effects of ozone in the upper and lower atmosphere, relating these to concentrations. (4)

• Compares the environmental effects of ozone in the upper and lower atmosphere. (3)

• Identifies the concentration of ozone (in the upper atmosphere) and one related environmental effect.

OR

• Identifies the concentration of ozone (in the lower atmosphere) and one related environmental effect.

OR

• Identifies two environmental effects of ozone. (2)

• Identifies that the upper atmosphere has a higher concentration of ozone than the lower atmosphere.

OR

• Identifies an environmental effect of ozone 1.

QUESTION 36

QUESTION 37 (HSC 2004:12) Answer is C QUESTION 38 (HSC 2010:16) Answer is C QUESTION 39 (HSC 2003:9) Answer is A QUESTION 40 (HSC 2004:4) Answer is B QUESTION 41 (HSC 2006:5) Answer is B QUESTION 42 (HSC 2010:4) Answer is B QUESTION 43 Answer is C & B


QUESTION 44

CFCs do/did not cause any problems in the troposphere. They are inert, non-toxic and insoluble in water. They are relatively dense gaseous molecules and only slowly diffuse through the lower atmosphere into the stratosphere. Their stability in the troposphere has meant that they are not destroyed and persist for many decades after release. The problems with their use come when the molecules reach the stratosphere (approximately 15 km above the Earth’s surface). Here, the CFCs are broken down by the high energy UV light and chlorine free radicals are produced.

CCl3F → Cl• + CCl2F•

These react with ozone and remove it from the stratosphere. This depletion of ozone has resulted in increased levels of high energy UV radiation reaching the Earth’s surface.

Cl• + O3 → ClO• + O2

ClO• + O• → Cl• + O2

QUESTION 45

CFCs are inert, non-toxic and insoluble in water. They are relatively dense gaseous molecules and only slowly diffuse through the lower atmosphere into the stratosphere. Their stability in the troposphere has meant that they are not destroyed and persist for many decades after release. The steps to alleviate the problems associated with CFCs appear to be effective, as the levels of ozone in the stratosphere have stabilised in recent years but have not yet started to increase. It may still be several decades before the problems associated with the use of CFCs and halons have been eliminated, as one chlorine free radical can cause the breakdown of 1000 ozone molecules (in a chain reaction, which regenerates the chlorine free radical) and CFCs released decades ago will still be diffusing up into the stratosphere. Hence it is unlikely that the ozone concentration in the stratosphere will change significantly in the next few years.


• Provides points for and/or against the uses of CFCs.

• Describes mechanism of ozone degradation by CFCs.

• Provides points for and against chemicals used as alternatives to CFCs, using data in the table and explains why they can be used as replacements.

• Provides a judgement about suitability of replacements. (6–7)

• Provides points for and/or against the uses of CFCs.

• Explains why other chemicals are used as replacements for CFCs, using data in the table.

• Provides a judgement about the suitability of replacements.

OR

• Describes mechanism of ozone degradation by CFCs. (4–5)

• Describes the uses of CFCs and the problems associated with their use.


AND/OR

• Describes alternatives to CFCs, using data in the table. (2–3)

• Identifies that CFCs destroy ozone in the stratosphere.

OR

• Identifies what either HCFC or HFC stands for.

OR

• Identifies one use of CFCs.

OR

• Identifies one type of replacement chemical not in the table. (1)

QUESTION 47 (HSC 2003:10) Answer is A QUESTION 48 (HSC 2006:28)

• Identifies an instrument that could have been used to obtain the information.

• Provides the characteristics and features of how data can be obtained using the instrument identified.

• Outlines how ozone concentrations above Antarctica have changed over time. (3–4)

• Outlines how the information could have been obtained.

OR

• Outlines how ozone concentrations have changed over time.

OR

• Outlines how CFCs can decrease the concentration of ozone. (2)

• Identifies an instrument used to obtain information about ozone levels.

OR

• Identifies that ozone concentrations above Antarctica have decreased.

OR

• Identifies that the rate of depletion is less due to phasing out of CFCs.

OR

• Identifies that ozone concentrations above Antarctica are less than other places on earth.

OR

• Identifies that ozone hole is getting bigger. (1)

QUESTION 49 46 ppm of CaCO3



• Describes test with expected distinguishing results. (2)

• Describes test.

OR

• Gives results without test (1)

(b) 6 X 10-4 mol L-1 = 6 X 10-4 X 84 = 0.0504 g/L = 50.4 mg/L 1000 ml / 150mL = 6.67 50.4/ 6.67 = 7.56 mg / 150mL


• 4.8 X 10-4 moles

OR

• Correct working with a substitution error. Marks (1)

(b)

• Calculates correct concentration of calcium. (2)

• 384 mg/L (1)

(c)

• Identifies Mg2+ or another cation capable of reacting with EDTA was also present 1.

QUESTION 52 (HSC 2005:15) Answer is A QUESTION 53 (HSC 2010:25)

Dissolved O2 is the concentration of oxygen gas dissolved in a water sample. BOD is a value obtained by calculating the change in dissolved O2 concentrations after 5 days. BOD is directly related to the amount of bacteria consuming O2 during the decay of organic matter. Consequently the relationship between dissolved O2 and BOD is necessary as BOD cannot be calculated without being able to measure the dissolved [O2]. It is important to measure both dissolved O2 and BOD as the dissolved [O2] values allow for BOD value to be obtained. The BOD value is then used as a measure of microbial activity in water, which determines the quality of water.



• Provides a thorough description of the process of eutrophication.

• States two or more tests used to monitor eutrophication.

• Provides an assessment.

OR

• Applications of tests.

OR

• Explains one or both tests in detail. (4)


AND

• States two tests used to monitor eutrophication.

OR

• Explains one test in detail.

OR

• Basic description AND states two tests and an assessment.

• Explains why one named test is useful for monitoring eutrophication. (3)


OR

• States two tests used to monitor eutrophication.

OR

• Explains why one named test is useful for monitoring eutrophication. (2)

• Provides a basic description of the process of eutrophication. OR

• States one appropriate test used to monitor eutrophication. (1)

QUESTION 55 (HSC 2001:11) Answer is B QUESTION 56 (HSC 2001:14) Answer is B QUESTION 57 (HSC 2006:13) Answer is D



• Provides characteristics and features of two possible sources of contamination.

• Describes at least three methods that could be used to purify the water.

• Provides a judgement about the effectiveness of each of the methods used or a holistic judgement with supporting statement. (6–7)

• Provides characteristics and features of two possible sources of contamination.

• Describes two methods that could be used to purify the water.

• Provides a judgement in general terms about the effectiveness of at least one method used.(4–5)

• Outlines at least one possible source of contamination.

• Describes at least one method that could be used to purify the water. (2–3)

• Identifies one possible source of contamination.

OR

• Identifies one method that could be used to purify the water. (1)

QUESTION 59

wwionConcentrat /%2010020

4

QUESTION 60

%10%100500

x

1.0500

x

gx 50

QUESTION 61

%15100170

x

mlx 5.25

QUESTION 62

200x g

o co h o · 2019. 8. 12. · question 31 answer is a question 32 question 33 answer is b question...

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