Section I  Part B (short answer and extended answer questions)
Answers are given in the right hand column. The left hand column contains some advice about how to go about answering the question – how to estimate the mark distribution and hence how to allocate time and amount of information to provide, how to make a good choice when a choice is required  and how to avoid common pitfalls and traps.
  
Question 16 (3 marks)
The verb here is the simple one, describe, so this is just recall of learnt information.
    There are two methods, nuclear reactors and particle accelerators. You need to mention both, preferably with an example form each, though for three marks not much detail is required. Give the equations if you can but they probably are not essential to get full marks.

 

(p 80-2)

 The first transuranic elements (elements with atomic numbers greater than that or uranium) were made by bombardment of heavy nuclei by neutrons in nuclear reactors, often just as a byproduct of reactor operation as for the very dangerous but well-known plutonium, Pu. It is made when the non-fissionable uranium-238 is hit by a neutron followed by the release of one electron to form neptunium which then emits another electron to form plutonium-239:


Other transuranic elements have been made in a similar way, usually by adding particular substances to the reactor.
     The other way in which transuranium elements have been made is by using cyclotrons or linear accelerators to bombard heavy nuclei with high-speed positive particles such as helium or carbon nuclei. For example californium, Cf, was made by bombarding U-238 with carbon nuclei:

Transuranic elements up to atomic number of about 118 have been made by one or other of these techniques. 
Question 17 (4 marks)
Make sure you give your answer to the correct number of decimal places. (a) [H+] = 0.2 mol/L
pH = –log [H+]  =  –(–0.699)  =  0.7 (rounded to one decimal place since there is only one significant figure in 0.2)
   
For the equation
H+(aq) + OH–(aq)  ®  H2O(l) is equally acceptable.
Show your working in sufficient detail for the examiner to be able to follow what you are doing. What is shown here is sufficient detail. Again watch number of decimal places. While too many decimal places in (a) probably would not be penalised (because there is only one mark for it, you could lose a mark here if you gave the answer as 0.854.

(p 134-5, 160-3 and CCPC 136, 217)
  

 (b) NaOH(aq) + HCl(aq)  ®  H2O(l) + NaCl(aq)
Moles of H+ initially  =  0.05 x 0.2  =  1.0 x 10–2
Moles of H+ removed by NaOH  =  0.02 x 0.01  =  2 x 10–4
Moles of H+ left  =  1.0 x 10–2 – 2 x 10–4  =  9.8 x 10–3 mol
This is now in 70 mL = 0.07 L so
[H+]   =  9.8 x 10–3/0.070   =  0.14 mol/L
pH = –log [H+]   =  –(–0.85)  =  0.9   (rounded to one decimal place since there is only one significant figure in 0.2 and 0.01.)
Question 18 (4 marks)
There are four significant figures in the mass of CO2 so the number of significant figures in the answer could be 4 (0.2422) if you used 44.01 as the molar mass, but only three if you used 44.0.
  		 (a) Mass of CO2 released  =  381.05 –370.39  =  10.66 g
Molar mass of CO2  =  44.0 g/mol
Moles of CO2 released  =  10.66/44  =  0.242 mol
Again show your working and again what is given here is quite adequate. Since I used three significant figures in the answer to (a), I can only get three in the answer to (b). Had you used 44.01 in (a) getting 0.2422 and used 180.2 (4 sig figs) for molar mass of glucose, you could have had 4 significant figures in your answer (21.82).
(p 33, 160)
  		 (b) C6H12O6(aq)  ®  2C2H5OH(aq) + 2CO2(g)
Moles of glucose that fermented  =  ½ x 0.242  =  0.121
Molar mass of glucose  =  6 x 12.0 + 12 x 1.01 + 6 x 16.0
                                   = 180 g/mol  
Mass of glucose used  =  0.121 x 180  =  21.8 g
Question 19 (7 marks)
To decide which metals to choose, use the table of standard electrode potentials at the back of the exam paper; the two with the greatest difference will be the correct pair.
Make sure your diagram includes a salt bridge; there would be no voltage without one. And make sure you label all the key features in your diagram.		 (a) Aluminium and silver
(b)
An equally acceptable equation is:
Al(s) + 3AgNO3(aq)  ®  
                        Al(NO3)3(aq) + 3Ag(s)
Be careful with signs in calculating the cell voltage. Maybe the examiners only wanted the magnitude of the voltage, but to be safe state which electrode (metal) is positive – the one where reduction is occurring.		 (c) Al(s) + 3Ag+(aq)  ®  Al3+(aq) + 3Ag(s)
This reaction is made up from the two half reactions:
Ag+(aq) + e–  ®  Ag(s) with voltage = Eo = 0.80 V
Al(s)  ®  Al3+(aq) + 3e– with voltage = –Eo = –(–1.68)  = +1.68 V
The voltage of the cell is the sum of the voltages of the two half reactions.
\ Cell voltage  =  0.80 + 1.68  =  2.48 V
with the silver metal positive with respect to the aluminium metal.
The first two steps would probably score both marks.
Actually even with 1.00 mol/L solutions and everything pure and clean you still will not get the calculated value because the standard condition is an activity or effective concentration of 1.00 mol/L which is only approximately equal to the actual concentration, but that is beyond expectations for HSC students.

( p 46-52, 66-70)
   

 (d) Ensured that the solutions were each 1.00 mol/L (since we have calculated the standard cell voltage which refers to that concentration) and that they were at a temperature of 25oC (because these are the conditions to which standard electrode potentials apply).
Cleaned the electrodes thoroughly before use.
Made sure that a good quality voltmeter was used (to ensure that negligible current flowed through the circuit since voltage decreases as current increases.
Ensure that there are no impurities in the solutions and that solutions are not too far from neutral because at high acidities other reactions can come into play and affect the voltage.
Question 20 (7 marks)

Analyse in this context means explain. The key reasons are (a) it is easily and cheaply obtained from oil and (b) it can be converted into a wide range of useful substance. For 7 marks you need to provide quite a lot of information (20 lines are provided so you should use most of them). You need to give at least two equations, preferably three or maybe four if they come easily to mind. The best two would be the conversion to ethanol and the formation of polyethylene.
     The next part of the question does not seem to have much to do with the first part!. In the middle of explaining the importance you have to deviate into a description of new materials. Frankly I do not consider the polymers I have described as new materials – they have been around for over sixty years! However if like me you do not know of any newer materials that are made from ethylene, then in an exam you provide the closest information you can. Hence what I have written. As for fuels that can be prepared from ethylene, I am at a complete loss to know what they are!. Ethanol can be prepared from ethylene and certainly there is a lot of talk around today about using ethanol as a fuel or fuel additive for motor cars. However that is ethanol prepared from plant material to minimise use of crude oil and hopefully to reduce greenhouse emissions. It would be quite senseless to make ethanol from ethylene to use as a car fuel as it would not alleviate either of these problems and it would be far more expensive than petrol. Hence my mention of ethanol as a convenience fuel – not a good example but the best I can come up with.
     As you may gather, I think there is a lot wrong with this question, but you as the candidate have to make the best of it and I hope my efforts give you some guidance.

(p 11-14, 18-21)

Ethylene is an important industrial starting material for the chemical industry because it can be used to make a wide range of useful products such as:
• ethanol which is used in industry both as a solvent and as the starting point for making many other chemicals such as acetic acid and ethyl acetate,
• ethylene oxide which is commonly used as a fumigant,    
• ethylene glycol which is used extensively as an anti-freeze in automobiles, and which is a starting material for the synthesis of poly(ethylene terephthalate) (PET or polyester), 
• polyethylene, an important polymer in widespread everyday use, and
• many monomers such as vinyl chloride and styrene from which other commonly used polymers are made (such as poly(vinyl chloride) and polystyrene).
    Its industrial importance also arises from the fact that it is easily and quite cheaply made from various fractions obtained from crude oil, either from steam cracking  or as a byproduct from catalytic cracking (in the production of petrol).
     Ethylene forms a wide variety of substances, because it contains a double bond across which many small molecules can be added.
     Ethanol is made by reacting ethylene with steam in the presence of concentrated sulfuric or phosphoric acid as catalyst:
CH2=CH2 + H2O  ®  CH3–CH2–OH
Ethylene oxide is made by reacting ethylene with oxygen in the presence of a catalyst:
CH2=CH2 + ½O2  ®   
This is converted to ethylene glycol by reaction with water, using acid as a catalyst:
+ H2O  ®  HOCH2–CH2OH 
Polyethylene is made by heating ethylene under pressure with a catalyst:
CH2=CH2 + CH2=CH2 + CH2=CH2 +  CH2=CH2 + ... ®
                                         –CH2–CH2–CH2–CH2–CH2–CH2–CH2–CH2–

     Some new materials that have been made by using ethylene are the synthetic polymers, polyethylene, poly(vinyl chloride) or PVC, polystyrene and PET. These plastics are white, translucent or clear (unless deliberately coloured by addition of pigments), are relatively tough, generally flexible, good electrical insulators and reasonably stable under normal environmental conditions. They can all be easily moulded into various shapes and so are used for a variety of household and commercial products. Each polymer has particular properties that suit it to particular uses. For example polyethylene can be blown into film and so used as wrapping film, PVC is used for insulating electrical cables, PET is used for fibres (polyester) and drink bottles and polystyrene is used for making insulating foam and cups.
     The only fuel that can be made from ethylene is ethanol. This is sometimes used as a convenience fuel by campers and bush walkers, but made from ethylene it is of no help in overcoming our serious problems of running out of oil or global warming (greenhouse effect).
  
Question 21 (3 marks)
An alternative salt is sodium hydrogen sulfate, NaHSO4;
the HSO4– ion is a weak acid 		(a) ammonium chloride, NH4Cl
Assume that there is one mark for a verbal explanation and one for the equation.
   The equation for the ionisation of the hydrogen sulfate ion is:
HSO4–(aq) + H2O(l)    H3O+(aq) + SO42–(aq)
   		(b) A solution of ammonium chloride is acidic because the ammonium ion is acidic; that mean that in aqueous solution it partially ionises to form hydrogen ions and so makes the solution acidic:
NH4+(aq) + H2O(l)  H3O+(aq) + NH3(aq)
  
Question 22 (4 marks)
.The answer given here is really too long for 4 marks, though it is hard to treat the full ramifications of the question more briefly.
     The problem is that the opening statement is not strictly correct. There is no convincing evidence that global concentrations of sulfur dioxide have been increasing since the Industrial Revolution. This question repeats an error that is in the syllabus document. The difficulty is that pre-Industrial Revolution concentrations were so low and so variable (because of volcanic activity) and analytical techniques so primitive that we do not know what the concentrations really were. Undoubtedly there have been dramatic increases in localised industrialised regions, but globally it is not so clear cut, because SO2 and H2SO4 get washed out of the atmosphere by rain.

    The key points that you need to make are (1) that sulfur dioxide comes from burning coal and oil and from extracting metals (and give equations), (2) that it gets washed out of the atmosphere by rain (after being oxidised to sulfuric acid), (3) that there have been increases in the concentration of sulfur dioxide in industrialised areas but (4) that the evidence for significant increases globally is not at all strong. However you are required to discuss the evidence so a mere statement of these facts is not sufficient.

I have used the equation for silver because it is a simple one. Using the less common ore of copper Cu2S leads to a similar simple equation:
Cu2S(s) + O2(g)  ®  2Cu(s) + SO2(g)
However with PbS and ZnS the product of heating in air is PbO and ZnO.

(p 121-7)

 The evidence for this statement, particularly the bit about concentrations increasing since the Industrial Revolution is neither extensive nor convincing.
    There is strong evidence from a wide variety of chemical analytical methods that the the atmosphere contains small concentrations of sulfur dioxide gas and sulfuric acid aerosol particles, but there is virtually no evidence for the presence of sulfur trioxide; if formed it is converted very quickly to sulfuric acid:
     SO2(g) + ½O2(g)  ®  SO3(g)
.    SO3(g) + H2O(g)  ®  H2SO4(l)
     There is also strong evidence that the amount of sulfur dioxide released into the atmosphere has increased dramatically since the Industrial Revolution. This has come from the burning of coal and oil, both of which contain sulfur which during combustion gets converted to sulfur dioxide:
S(as element or in compounds) + O2(g)  ®  SO2(g)
and from the extraction of metals such as silver, lead, zinc and copper from sulfide ores. For example:
Ag2S(s) + O2(g)  ®  2Ag)(s) + SO2(g)
The quantities of coal and oil burnt and of mineral ores processed have been increasing greatly since the Industrial Revolution and this is the convincing evidence that the quantities of sulfur dioxide released have increased.
    There is also convincing evidence that in particular locations such as heavily industrialised areas and cities there have been significant increases in sulfur dioxide concentrations from time to time. In most industrial regions there were great increases up until about the 1950s or 60s but after those decades concentrations stabilised or decreased as a result of emission controls placed upon industry and transport. The evidence came both from direct chemical analysis and  from observed detrimental effects of atmospheric sulfur dioxide (or its oxidation product sulfuric acid) upon buildings, statues, plants, rubber and the pH of natural water bodies:
SO2(g) + ½O2(g) + H2O(l)  ®  H2SO4(aq)
    However there is very little evidence to show that globally the atmospheric concentrations of acidic oxides of sulfur have been increasing. This is because the concentration of sulfur dioxide in clean air is extremely low (of the order of 0.001 ppm) and until the last fifty years sufficiently sensitive and accurate analytical methods were not available for the accurate measurement of these low concentrations. In addition sulfur dioxide concentrations in the clean atmosphere have fluctuated quite markedly as a result of natural volcanic activity.
    The reason why there have been large releases of sulfur dioxide into the atmosphere without major increases in its global concentration is that sulfur dioxide and sulfuric acid are both soluble in water and so get washed out of the atmosphere by rain. This acid rain has detrimental effects upon buildings and water bodies, but it does prevent significant build-ups of sulfur dioxide concentrations in the atmosphere, except in locations close to industrial and minerals processing locations.
   
Question 23 (6 marks)
Something you need to know.

(p 164-5)

 (a) Bromothymol blue, because it is the indicator that changes colour in the required pH range. From the graph its colour changes from yellow to blue over the pH range 6.0 to about 7.7. This means that a chart could be made showing colour versus pH in the pH range 6 to 8 against which a tested sample of pool water could be compared in order to estimate its pH. The other two indicators would have a fixed colour over the pH range 5.5 to 8.5 (yellow for methyl orange, colourless for phenolphthalein) and so would be of no use for pool water.
Explain how the equilibrium keeps moving to the right, mention Le Chatelier's principle and explain how increasing the concentration of OH– increases pH.

(p 117-20)

 (b) (i) to sanitise the pool, that is to remove bacteria and viruses etc to make the pool safe for people to swim in.
(ii) It would cause the pH of the pool to increase. As the HOCl is used up to destroy bacteria etc, the given equilibrium would move to the right – more OCl– would convert to HOCl (by Le Chatelier's principle) thus producing more OH–. Because of the equilibrium
2H2O  H3O+ + OH–
as [OH–] increases, [H3O+] decreases, which means that pH increases because pH = –log [H3O+].
Question 24 (5 marks)
Remember to put state symbols on your equation. (a) N2(g) + 3H2(g) 2NH3(g)
Note the key verb here, evaluate. This means that you must comment on how important the discovery was for the world.

(p 200-4)

 (b) The significance of Haber's discovery at that time was that it facilitated the expansion of agriculture and that it helped Germany's efforts in World War I.
     Early in the twentieth century the limited supply of Chile saltpetre, sodium nitrate, and the high demand for it as a nitrogenous fertiliser had been retarding the expansion of agriculture. When Bosch in 1914 was able  to make the Haber process work on an industrial scale, it became possible to make a synthetic fertiliser to supplement and later replace Chile saltpetre as the major nitrogenous fertiliser being used by agriculture. Haber's discovery allowed agriculture to expand and this was the major significance of his work. This was a very useful and valuable development for the whole world, both then and in subsequent times.
     In addition the Haber process facilitated the production of explosives, particularly in Germany where Haber and Bosch worked. Many explosives are made from nitric acid which in turn is made from ammonia. The Haber process gave Germany a distinct short-term advantage in World War I and probably helped extend the duration of the war. In the short term this was not particularly valuable for the world.
    Overall Haber's discovery was of great beneficial significance for the world, both then and now.
   
Question 25 ( 5 marks)
Plot the points as small dots then mark where they are with circles, squares or triangles. Then draw a line of best fit through the points. There are two choices here. You can decide that the points lie on a curve and so draw a smooth curve through them (green line) or you can decide that the points should like on a straight line because you know that absorbance should be proportional to concentration; in this case you would draw the blue straight line. (a)
The answers given are from the green curve. Perhaps you read off 31.5 for (ii) but bearing mind that the curve could be drawn with slightly different shapes, it is probably best to round off to 32.
     If you drew the straight line, then your answers would have been (i) 13 ppm (ii) 26.5 (rounded to 27) ppm
    Examiners reports from previous HSC exams have implied that validity means accuracy (though most chemists consider that it has a somewhat different meaning – the correctness of the arguments used to get the result from the measurements or of the conclusions drawn from the results). The answer given here is based upon validity = accuracy (as opposed to reliability which means reproducibility).
     Though probably beyond HSC requirements,  a further factor affecting the validity (accuracy) of these results is whether or not the line of best fit is a curve or a straight line (as expected from theory). This introduces an error of about ±1 ppm for result (i) and about ±5 ppm for result (ii) – using the results given above.

(p 222-3)
  

(b) (i) 12 ppm
(ii) 32 ppm
Result (i) is reasonably valid (valid meaning accurate) in that it comes from an interpolation of a graph that can be drawn with reasonable accuracy (though some variation in its shape would alter the value a little bit). Result (ii) is less valid in that it comes from an extrapolation of the graph and we cannot be 100% sure that the curve follows the shape we have given it beyond the last experimental point. However the extrapolation is not very great, so  result (ii) still has fair validity (accuracy). The accuracy of result (ii) could have been greatly improved by using a standard solution that contained 35 or 40 ppm copper so that the answer could have been obtained from an interpolation.
Question 26 ( 4 marks)
(a) Carbon dioxide, CO2
You need to know the solubilities of carbonates, sulfates and chlorides to answer this question.

(p 206-11)

 (b) Carbonate forms a precipitate with both barium ions and with silver ions so it would interfere with the tests for sulfate and chloride. Hence it has to be removed (and tested for) first. Dilute acid reacts with carbonate to produce carbon dioxide gas and so removes it from the solution as well as showing that it was present.
   Silver ions produce a precipitate with sulfate (if its concentration is high enough) as well as with chloride, so adding silver ions before adding barium ions and getting a precipitate would not establish that chloride was in the solution. Barium ions form a precipitate with sulfate but not with chloride, so testing with barium ions before using silver ions (and removing all the precipitated barium sulfate) means that only chloride (if present) would be left in the solution. Adding silver ions at that stage and getting a precipitate would establish that chloride was present in the solution.
Question 27 (4 marks)
(a) 4.8 x 10–4 mol
Show your working (b) Moles of Ca2+ that this EDTA reacted with  =  4.8 x 10–4
This is in a volume of 50.0 mL = 0.050 L
Concentration of Ca2+ ions in the water  = 4.8 x 10–4/0.050
                                                           = 9.6 x 10–3 mol/L
Concentration in mg/L  =  9.6 x 10–3 x 40.1 x 103
                                  =  385 mg/L
 
The titration gave a value that was higher than the AAS value. You should know that AAS is very good at measuring one specific ion regardless of what other ions are present. The information given tells you that EDTA reacts with other metal ions as well as calcium, so this provides the clue as to the explanation of the discrepancy.

(p 225-31)
  

(c) EDTA reacts with other metal ions commonly present in water, such as magnesium, so the titration would measure the total concentration of all such metal ions. AAS, because it uses a specific lamp for each metal ion, would measure calcium ion concentration alone (and not include any magnesium ion present). The AAS measurement shows that 16% of the nominal calcium determined by titration is some other metal ion, probably magnesium.
Question 28 ( 4 marks)
Do not rush into an account of the chemistry of the Ozone hole. Read the question carefully and answer it as asked. It does not ask for any chemistry at all! Just explain how the contour map was obtained and describe how the ozone hole has changed. Note that there are two aspects of the changes – the seasonal changes through any given year and the changes that have occurred over many years.

(p 254-60)

 Such contour maps are produced by total ozone mapping spectrophotometers (TOMS) mounted in satellites that orbit the earth. These instruments use uv absorption by ozone molecules to measure ozone concentration in the atmosphere as a function of altitude and geographic location; the source of the uv radiation for the measurement is the sun. These data can be used to construct contour maps such as the one given which show the total amount of ozone in the atmosphere per unit area of earth surface as a function of geographic position.
     During the last twenty years or so atmospheric ozone concentrations over the Antarctic have fluctuated from season to season in each year as well as from one year to the next. Ozone concentrations start to fall in early Spring (September) and reach minimum values in about October to November, then they start rising again in December and January, reaching steady state values until the next Spring. This fall in ozone levels over the Antarctic in Spring is called the Ozone Hole.
     The amount by which ozone concentration falls during Spring, the 'depth' of the ozone hole, varies from year to year as does the extent of the hole, that is the surface area of the Earth over which there is a significant depletion of ozone. The hole was first observed in the early 1980s. It increased in depth and area over the next twenty years or so, but is now showing signs of stabilising. It is expected that the hole will become smaller and shallower in the next few decades as the concentrations of the chemicals that have been causing it (CFCs and related substances) gradually fall in response to international agreements to restrict the use of such harmful chemicals.