Section I Part B

Question 16 (5 marks)

Factual material you need to know.
In describing the test be sure to use liquid compounds; remember that the simplest alkanes and alkenes (up to butane and butene) are gases at room temperature. This is a simple test so make sure you describe fully what you did and what you observed.
The equation given here is a simplified one that seems to be acceptable to the HSC examiners. The more correct equations are given on p 10 (though using 3-hexene instead of 2-hexene)
(p 9-10)

(a)

	C₈H₁₈	C₂H₄
Name of compound	octane	ethylene (ethene)
Name of series	alkanes	alkenes

(b)

Alkenes decolorise brown bromine water whereas alkanes do not. In our experiment drops of brown bromine water were added to 1 mL samples of the liquid compounds to be tested, hexane and 2-hexene, with the test tubes being gently shaken between additions. The 2-hexene, an alkene, decolorised the bromine water while the hexane, an alkane, did not.

Question 17 (5 marks)

You need to analyse this question carefully before starting to write.
      The normal treatment for municipal water supplies in Australia is clarification by flocculation and filtration followed by sterilisation by chlorination. Sterilisation is definitely to make the water safer to drink, but is clarification really to make the water safer to drink or is it more to make the water more appealing to look at and possibly to improve its taste? For this question if it is clarification you want to talk about you need to give it a safety slant. The process of flocculation can remove small amounts of harmful heavy metals if they are present and to some extent it can reduce the concentrations of viruses and parasites.
    So what features of catchments are these processes of clarification and sterilisation protecting people from? Farming activities particularly those involving sheep, cattle, pigs and poultry can lead to harmful bacteria, viruses and parasites getting into the water as rain washes across animal droppings (faeces). Forestry activities can lead to land clearing so that runoff from the area picks up considerable quantities of suspended solids that cause the water in the dam to become turbid. Roads through the catchment can have similar effects. Mining activities can also result in increasing sediment loads and to heavy metals getting into the water; mining can also alter pH necessitating pH adjustment in the water treatment plant. You could use any of these as your two features provided you geared your treatment processes to them.
      Of course the primary method of making water safe to drink is to keep virtually all human activity out of the catchment area, but that is probably not part of the question..
(p 289-91, 262-77)

The features of the catchment determine what undesirable substances get washed into the storage dam and so what treatment is needed to make the water safe to drink. Two features are:
(1) current (or previous) logging activities that have left parts of the catchment cleared of its original tree and scrub cover so that rain runoff washes considerable quantities of sediment into the dam, some of which stays in the water as suspended solids (very fine particles) and makes the water turbid. In such cases the water needs to be clarified by flocculation and filtration. In this process Fe³⁺ or Al³⁺ is added to the water, pH adjusted and the water gently stirred so that a precipitate of Fe(OH)₃ or Al(OH)₃ forms. As this precipitate coagulates into larger particles the fine suspended particles originally in the water adsorb onto it. The precipitate is removed with a coarse sand filter to produce very clear water. This process is intended primarily to improve the appearance of the water, to prevent annoying depositions in pipes and fittings and to remove any unpleasant taste the raw water had. However it can make the water safer to drink in that if the raw water had been contaminated with heavy metals this process would have precipitated much of them out as well, and it would also have removed (by adsorption) significant amounts of any viruses and parasites present.
(2) pastoral activities within the catchment such as grazing cattle or sheep. This results in the runoff water picking up bacteria, viruses and parasites as it passes over animal faeces and flows into the dam. Such microorganisms are harmful to people and so to make the water safe to drink they are removed by sterilising the water, generally with chlorine (though use of the more effective but more expensive ozone is gaining in popularity). This sterilisation is generally the last step in water treatment.

Question 18 (5 marks)

The electron-dot structures in CCHSC are not strictly Lewis structures and while they would probably be acceptable as the answer for this question, it is safer to stick with the less clear though authentic Lewis structures given in the sample answer at right. Incidentally the structures in the Examiners Report with lines instead of pairs of dots or crosses are not Lewis structures either!
The difference in physical properties arises because O₂ is non-polar whereas O₃ is polar. This is not addressed in CCHSC and you may be puzzled as to why O₃ is polar, because the shared electrons are shared between atoms of the same element and so should be evenly distributed which would make the molecule non-polar. The polarity arises from the nature of the O–O single bond: it forms by a pair of electrons that had been owned exclusively by one O atom becoming shared between two O atoms (see the diagram at the bottom of p. 242). The donating O atom has effectively lost one electron (and so has become positively charged) while the receiving O atom has effectively gained one electron and so has become negatively charged. This would make the molecule polar:

and this structure can explain the physical properties. This is not the full story but it explains the polarity of the molecule and is adequate for HSC purposes.
It is not so easy to explain the different stability and reactivity in terms of polarity. The answer opposite is probably the best that can be expected at the HSC level. If saying that the double bond in O₂ is very stable and so less reactive than a single bond in O₃ seems inconsistent with what you know about alkenes and alkanes, namely that the double bond in alkenes is much more reactive than the single bonds in alkanes, bear this in mind: the reactivity of the double bond in alkenes arises because it can be opened out to form two new single bonds leaving a C–C single bond intact (for example the equation in Question 16 above). In O₂ for reaction to occur the double bond must be completely broken – it cannot open out to form new single bonds.
(p 244)

(a)

(b)

While O₂ and O₃ are both gases at room temperature, O₃ has a much higher boiling point than O₂.This is partly because it has a higher molecular weight (48 versus 32) but also because it is a polar molecule (central O atom positively charged) whereas O₂ is non-polar. O₃ is more soluble in water than O₂, again because it is polar (polar O₃ dissolves in polar water).
O₂ is more stable and less reactive than O₃. This is because the bond between the O atoms in O₂ is a strong double bond that has to be completely broken for O₂ to react with the other substances it reacts with such as NO, CO, SO₂, Mg, Fe or I^–, whereas in O₃ only a single bond has to be broken to transfer an O atom to the other reagent and that leaves behind a stable O₂ molecule (as when O₃ reacts rapidly with I^– solution to form IO^– or with gaseous NO to form NO₂).

Question 19 (5 marks)
There are 3 marks for drawing the graph, so make sure you use good techniques. Make the independent variable (concentration) the x axis, choose scales that use up most of the grid, mark the scales on the axes and label the axes. Plot each point as a fine dot and identify the points with a circle, square, triangle or cross (the sample graph in the Examiners' report is deficient in this last respect: make sure yours aren't!)	(a)
Read off the concentration as accurately as possible: 1.65 ( ± 0.05). Remember that this is a concentration in mg/L. You want the amount of Zn in one tablet, so you need to convert the concentration to an amount in 100 mL because that is the volume that one tablet was made up to. (p. 228-30)	(b)	From the graph concentration of Zn in the solution made from the tablets = 1.65 mg/L = 0.165 mg/100 mL Since the tablet was dissolved and the solution made to 100 mL, the amount of Zn in one tablet is 0.165 mg .
Question 20 (3 marks)
Know that all gases have the same molar volume. Look up its value in the Data Sheet at the end of the paper. (p 128-9)	(a)	One mole of any gas at 100 kPa and 25^oC has a volume of 24.79 L. Hence number of moles in the sample = 15.0 / 24.79 = 0.605 mol This means that 1.22 g is 0.605 mol Molar mass of this gas = 1.22 / 0.605 = 2.02 g/mol
If you do not know that hydrogen has a molar mass of 2 g/mol (atomic weight of 1), use the Periodic Table at the end of the paper to find what gas it could be.	(b)	The gas is hydrogen.
Question 21 (4 marks)
You need to know how CFCs got into the atmosphere, that they were harmful there and that there were international agreements to eliminate their use. Then you need to be able to interpret the graph in terms of increasing use and then cessation of use of CFCs. (p 250-8)	(a)	CFC-11 was widely used for a variety of purposes (particularly in refrigeration) in the second half of the 20th century. Some of it escaped into the atmosphere, and because it was not destroyed there, its concentration steadily increased as the graph shows from 1980 to 1990. By the 1980s it was recognised that CFCs were destroying ozone in the stratosphere (they slowly diffused there from the lower atmosphere), and starting in 1987 there was a series of international agreements that first restricted then finally eliminated the use of CFCs. The slowing down in the rate of increase that appears in the graph between 1990 and 1993 and then the slow decrease in CFC-11 concentration after 1993 reflects this reduced use and release of CFCs.
Here you need to explain why it is important to reduce the concentration of CFCs in the atmosphere and how monitoring their concentrations can show whether international programs to reduce their release to the atmosphere are working. (p. 250-8)	(b)	To check whether international agreements to reduce CFC-11 (and other CFC) concentrations in the atmosphere are working and that we are solving the ozone hole problem. CFC-11 (and other CFCs) are not destroyed in the lower atmosphere and so slowly diffuse into the stratosphere where they destroy ozone. We need ozone in the stratosphere because it filters out short wavelength u.v. radiation which is harmful to humans and other species. If the concentration of CFCs in the lower atmosphere falls, then less CFC will diffuse into the stratosphere and so there will be less destruction of ozone.

Question 22 (5 marks)

You need to explain why using ethanol as a fuel appears carbon neutral and to explain why this is not correct. And as asked by the examiners, include relevant chemical equations. The relevant chemical equations are the ones for photosynthesis, fermentation and combustion. All three are required.
Note that you are asked to critically evaluate the extract. This means you must comment on the extent to which it is correct and in what ways it is incorrect.
(p 33-5)

The extract is partially but not completely true. At first sight ethanol looks like a 'carbon-neutral' fuel, meaning that its use as a fuel only releases into the atmosphere the amount of carbon dioxide that was removed from the atmosphere by its synthesis. This carbon neutrality arises as follows. Ethanol is made by fermentation of carbohydrates which were made by photosynthesis:
Photosynthesis:   6CO₂(g) + 6H₂O(l) ® C₆H₁₂O₆(aq) + 6O₂(g)
                                                                glucose
Fermentation:   C₆H₁₂O₆(aq)   ® 2C₂H₅OH(aq) + 2CO₂(g)
Overall result:    4CO₂(g) + 6H₂O(l)   ®   2C₂H₅OH(aq)) + 6O₂(g)
When ethanol is used as a fuel it undergoes combustion:
                   2C₂H₅OH(aq) + 6O₂(g)   ® 4CO₂(g) + 6H₂O(l)
This combustion reaction is the reverse of the overall synthesis reaction and so using ethanol appears carbon neutral.
     However this argument overlooks several energy inputs into this cycle. To bring about photosynthesis we need to grow crops; this requires the expenditure of much energy to cultivate and harvest the crops and to make the fertiliser needed to make them grow. This energy input is generally from fossil fuels and so there is considerable release of carbon dioxide involved. Then to ferment glucose or other sugars to ethanol there are further inputs of fossil fuel energy, first to heat the fermentation mixture, but more importantly to distill nearly pure ethanol (>95%) from the dilute aqueous solution that fermentation produces. The heat required is generally obtained from fossil fuels and so involves release of considerable amounts of carbon dioxide.
      Hence when all aspects are considered there are very significant releases of carbon dioxide to the atmosphere when ethanol is used as a fuel.
      Overall the extract presents a biased (and incorrect) picture of the true situation because it oversimplifies the energy inputs into the total process.

Question 23 (4 marks)
Write the equilibrium reaction involved, state Le Chatelier's principle and explain how it affects the choice of pressure and temperature. You need to point out that Le Chatelier's principle is not the only consideration in choosing temperature – reaction rate is also a consideration. Note that you are not asked to discuss the other factors that affect yield such as use of a catalyst and removal of product as it forms. (p. 200-3)	The Haber process for synthesising ammonia uses the equilibrium reaction, N₂(g) + 3H₂(g) 2NH₃(g) (DH is negative) The yield of ammonia is optimised by using a high pressure (2.5 X 10⁴ kPa) and a moderate temperature (700 K). Le Chatelier's principle states that when an equilibrium reaction is disturbed, it will move in the direction that minimises the disturbance. In this reaction 4 moles of gas ® 2 moles of gas, so if the pressure is increased, the equilibrium moves to the right as a decrease in the number of moles of gas will lower the pressure. Hence a high pressure will improve the yield of ammonia. This reaction is exothermic (releases heat as it goes from left to right), so lowering the temperature will cause it to move to the right in order to release some heat and so partly counteract the lowering of the temperature. However equilibrium considerations are not the only ones needed in selecting the temperature. The rate of this reaction (like most reactions) decreases as temperature is decreased, so working at a low temperature will produce a good yield but a long time will be needed to get there. On the other hand a high temperature will get to equilibrium quickly but the yield will be very low. Hence a compromise is used: a moderate temperature is used because this gives a moderate yield moderately quickly.
Question 24 (5 marks)
Calculate the heat released per gram for each fuel and write it on the question paper (petrol 48, kerosene 48, hydrogen 143, ethanol 30) then choose hydrogen. No explanation is required.	(a)	hydrogen
You need to convert volume to mass (using density) then mass to moles and finally moles to quantity of heat. And show your working. How many significant figures should you use? There are probably three in the heat of combustion of petrol, two in the density, three in the molar mass and one or two in 80 L. Hence only two are justified in the answer, but to carry on into part (c) you should probably keep three.	(b)	Mass of petrol in the fuel tank = 0.69 X 80 = 55.2 kg = 55200 g Number of moles of petrol = 55200 / 114 = 484.2 mol Energy released = 484.2 X 5460 = 2.6 X 10⁶ kJ (2.64 is acceptable)
Again show your working and again be careful with significant figures. There are three in the heat of combustion of hydrogen but as explained above only two in the energy released so the answer should contain only two. Do not follow the examples in the examiners' report which completely ignore significant figures (giving six digits in the final answers!) (p. 36-7 and CCPC p. 276-81)	(c)	Number of moles of hydrogen required to supply this amount of energy = 2.6 X 10⁶ / 285 = 9.26 X 10³ mol Volume of this at 100 kPa and 25^oC = 9.26 X 10³ X 24.79 L = 2.3 X 10⁵ L
Question 25 ( 5 marks)
To identify the anode and cathode you need to look up the electrode potentials. Chlorine is higher than nickel (1.36 V versus –0.24 V). Therefore the chlorine half reaction, Cl₂(g) + 2e^– ® 2Cl^–(aq) goes as written. Hence reduction occurs at this electrode and that makes it the cathode. (p. 49)	(a)
Combine the chlorine half reaction (above) with the reverse of the nickel one, Ni(s) ® Ni²⁺(aq) + 2e^–noting that electrons balance, so just add.	(b)	Cl₂(g) + Ni(s) ® 2Cl^–(aq) + Ni²⁺(aq)
Either E_cell = electrode potential of the chlorine electrode – electrode potential of the nickel electrode that is E_cell = 1.36 – (–0.24) = 1.60 V or E_cell = voltage of the chlorine reduction half reaction + voltage of the nickel oxidation half reaction (which is minus the voltage for the reduction half reaction), that is E_cell = 1.36 + 0.24 = 1.60 V Be careful in your use of symbols E or e. Do not confuse electrode potentials (always for reduction half reactions) with half cell potentials (oxidation or reduction). To get the cell voltage we subtract electrode potentials or we add half cell potentials but of course we get the same answer regardless of method. (p. 64-70)	(c)	E_cell = 1.36 + 0.24 = 1.60 V with the chlorine electrode (Pt wire) being positive.
Actually if reaction proceeded for a considerable time a slight green colour would develop in the right hand beaker, because there is migration of ions through the salt bridge as current flows: Ni²⁺ ions migrate from the left hand beaker to the right hand one while chloride ions migrate from right to left. but that is beyond the requirements of this question. (p. 48-9)	(d)	The colour becomes more intense (darker) in the left hand beaker, because solid nickel is oxidised there to Ni²⁺which is green. This increases the concentration of Ni²⁺ in that solution which makes the colour more intense. There is no colour change in the right hand beaker, because the chloride ions generated there are colourless.

Question 26 ( 4 marks)

You need to be explain what a buffer is before you try to explain how it works. An equation is necessary. The example used here is the obvious choice. Another one that could have been used is the dihydrogen phosphate–hydrogen phosphate one:
H₂PO₄^–(aq) + H₂O(l)

H₃O⁺(aq) + HPO₄^2–(aq)
(p 166-8)

A buffer is a solution that maintains fairly constant pH even when acid or base is added to it. It does this because it contains comparable amounts of an acid and its conjugate base. A specific example in a natural system is the carbonic acid–hydrogen carbonate buffer. The equilibrium reaction involved is
H₂CO₃(aq) + H₂O(l)

H₃O⁺(aq) + HCO₃^–(aq)
At equilibrium this buffer produces a pH of about 7. If H₃O⁺ is added to this solution, the equilibrium moves to the left (Le Chatelier's principle) and so removes much of the added H₃O⁺ and so returns the pH to close to its original value. If hydroxide ion (or some other base) is added to the solution and removes H₃O⁺, the equilibrium moves to the right to produce H₃O⁺ ions and so again returns the pH to near its original value.

Question 27 (4 marks)

You need to know that the ammonium ion is a weak acid and that the ethanoate ion is a weak base. And also that the sodium, chloride and nitrate ions do not hydrolyse in aqueous solution.
(p 152-4)

(a)

Salt	Classification of solution
Ammonium chloride	acidic
Sodium ethanoate	basic
Sodium chloride	neutral
Ammonium nitrate	acidic

Use the equation you are required to write to explain how the hydrogen ion concentration of the solution is different from that of pure water.
(p. 152-4)

(b)

Ammonium chloride. The ammonium ion is a weak acid which means that in aqueous solution it ionises to produce hydrogen ions:
NH₄⁺(aq) + H₂O(l)

H₃O⁺(aq) + NH₃(aq)
This means that the solution contains a greater concentration of hydrogen ions than pure water does and so it is acidic.

You could have chosen ammonium nitrate; the argument would be exactly the same. Had you chosen sodium ethanoate the argument would have been:
The ethanoate ion is a weak base which means that in aqueous solution it undergoes the reaction,
CH₃COO^–(aq) + H₂O(l) CH₃COOH(aq) + OH^–(aq)
This means that there will be more OH^– ions (and so less H₃O⁺ ions) in the solution than in pure water and so it is basic (alkaline).

Question 28 (6 marks)

This is a standard volumetric calculation; the two key equations are 5.14 and 5.15 on p. 160. Show your working so if you make a careless mistake the examiner can give you some marks.
Watch significant figures. There are 4 in 1.314 g and in 250.0 mL so formally there should be 4 in the answer, 0.04958 mol/L and 0.09301 mol/L. However the accuracy of a volumetric flask in usually no better than 0.1 or 0.2% which would justify only three significant figures in the answer. Three or four would be acceptable, but fewer or more could lead to a loss of a mark.
Note that that HCl and Na₂CO₃ react in the ratio of 2:1 so take this into account in (c).
(p. 159-62)

(a)

Molar mass of sodium carbonate, Na₂CO₃ = 2 X 23.0 + 12.0 + 3 X 16.0
= 106.0 g/mol
Number of moles of Na₂CO₃ used = 1.314 / 106.0 = 0.01240 mol
This is made into 250 mL solution (= 0.250 L)
Molarity = 0.01240 / 0.250 = 0.0496 mol/L

(b)

2HCl(aq) + Na₂CO₃(aq) ® 2NaCl(aq) + H₂O(l) + CO₂(g)

(c)

Volume of Na₂CO₃ solution needed = 23.45 mL = 0.02345 L
Moles of Na₂CO₃ used = 0.02345 X 0.0496 = 0.001163 mol
Moles of HCl this reacted with = 2 X 0.001163 = 0.002326 mol
This was in 25.00 mL = 0.02500 L
Concentration of HCl = 0.002326 / 0/02500 = 0.0930 mol/L