2023 HSC Chemistry Solutions
Liquid hydrocarbon is considered hazardous waste which has to be disposed in a separate waste container for safe treatment.
This is atomic absorption spectroscopy where the sample is aspirated in a flame.
There are 5 carbon atoms so the prefix should be ‘pent’. The triple bond is the second bond (numbering from right to left) and the suffix for this bond is ‘yne’. Hence the correct answer is Pent-2-yne.
Ksp = [Na+][Cl–] = 6.13 × 6.13 = 37.6
Weak acids rarely dissociate into their ions, so the solution should mostly contain whole acid molecules. This rules out B and C which are fully dissociated. A concentrated solution will have more molecules, so the correct answer is D.
Recall that a lower pH corresponds to a more acidic solution. Hence we know the concentration of hydrogen ions is increasing. Also recall that pH follows a logarithmic scale, so a decrease in pH of 1 corresponds to a 10-fold decrease in [H+]. A change in pH from 8 to 5 therefore corresponds to a 10 × 10 × 10 = 1000 fold decrease in [H+].
There was a change of -0.3 mol of CO. Since the molar ratio of CO:H2 is 1:2, there will be a -0.6 mol change in H2, which gives 0.8 – 0.6 = 0.2 mol at equilibrium.
4 structural isomers of C3H6F2:
The equivalence point is at ~ pH = 5 which is acidic. Hence it results from a strong acid and a weak base.
Each compound involves different intermolecular forces. From weakest to strongest, dispersion forces (heptane) < dipole-dipole forces (heptan-2-one) < hydrogen bonding (heptan-1-ol) < more extensive hydrogen bonding due to COOH group (heptanoic acid).
The molar ratio of gaseous reactants to products is 4:1. Hence, according to Le Chatelier’s Principle, an increase in pressure will shift equilibrium to the right to favour the side with less gaseous moles. This will result in a higher yield.
13. (A): test 1 tells us the compound is not basic, which rules out magnesium carbonate (basic salt). Test 2 tells us that the anion forms a white precipitate with Ba2+. Hence it rules out lead acetate, as acetate does not precipitate with any cations. (A) is the correct option because Ag+ forms a brown precipitate with OH– and a white precipitate with Cl–.
From the table we can see that the heat of combustion increases approximately linearly between consecutive alcohol. Since propan-1-ol has a carbon chain in between ethanol and butan-1-ol, we can assume its heat of combustion is the average of 29.67 and 36.11, giving (B) as the correct answer.
Silver nitrate will react with the solution to form silver iodide and silver chloride precipitates. From the data sheet, the Ksp of silver iodide is 8.52 x 10-17, and silver chloride is 1.77 x 10-10. Hence AgI has a smaller Ksp and will precipitate out first, followed by AgCl.
From the graph, there is a sudden decrease in the rate of the forward reaction at time t. This immediate decrease is due to a decrease in temperature. Following this, the rate of the forward reaction increases again gradually, which indicates the forward reaction is exothermic (Le Chatelier’s Principle – decrease in temperature favours exothermic reaction). Thus the change in enthalpy is negative.
Options C and D are incorrect because those fragments cannot be derived from butan-2-one (draw out the structural formula of the molecule). CH3CO has a mm = 43 and is the fragment represented by the peak shown, hence it must be the charged fragment, since the mass spectrometer can only detect the charged particles. B is correct.
|E||2x||x||0.90 – 2x|
Short Answers Qs
Question 21 (2 marks)
Functional group isomer: butan-2-one and butanal (ketone and aldehyde)
Chain isomer: butanal and 2-methylpropanal
Question 22 (3 marks)
Explain that both HCl(aq) and NH4Cl(aq) are Arrhenius acids and Brønsted-Lowry acids since they both increase [H+] in aqueous solution, and both donate H+. Support explanations with equations (see sample answer).
HCl(aq) is an Arrhenius acid since it increases [H+] in aqueous solution.
HCl(aq) → H+(aq) + Cl–(aq)
HCl(aq) is a Brønsted-Lowry acid since it donates H+.
HCl(aq) + H2O(l) → Cl–(aq) + H3O+(aq)
NH4Cl(aq) is an Arrhenius acid since it increases [H+] in aqueous solution.
NH4Cl(aq) → NH4+ (aq) + Cl–(aq)
NH4+(aq) → NH3(aq) + H+(aq)
NH4Cl(aq) is a Brønsted-Lowry acid since it donates H+.
NH4Cl(aq) → NH4+(aq) + Cl– (aq)
NH4+(aq) + H2O(l) → NH3(aq) + H3O+(aq)
Question 23 (3 marks)
For full marks explain why solution X increases significantly in pH, identify that solution Y is a buffer solution, and explain why pH change is minimised in a buffer solution (ideally using an example chemical equation).
Both solutions are neutral with pH = 7.00. Solution X is not a buffer solution, so when NaOH is added it dissociates completely into Na+ and OH–, hence causing the pH to increase from 7 to 12. Solution Y is very likely a buffer solution that contains similar moles of a weak acid and its conjugate base. Consider the buffer solution with the acid HA and its conjugate base A–:
HA (aq) + H2O (l) ⇌ A– (aq) + H3O+ (aq)
When NaOH is added, the OH– will react with the H3O+ to form H2O, minimising the increase in pH. This uses up H3O+, but equilibrium will shift towards the right (according to Le Chatelier’s Principle) to increase [H3O+] again. Hence the pH change is mostly buffered, and only increases very slightly from 7.00 to 7.02.
Question 24 (2 marks)
Write one chemical equation where (HC2O4–) accepts a proton, and one chemical equation where (HC2O4–) donates a proton (see sample answer).
HC2O4–(aq) + H2O(l) → H2C2O4(aq) + OH–(aq)
HC2O4–(aq) + H2O(l) → C2O42-(aq) + H3O+(aq)
Question 25 (5 marks)
A. Calculate the total heat energy produced from the combustion of octan-1-ol. Then calculate the moles of octan-1-ol combusted and the mass of octan-1-ol.
B. Identify and explain one key reason why biofuels are advantageous, e.g. renewable energy source.
Biofuels are renewable, because they can be produced from the fermentation of plant matter such as sugar cane. These crops can be grown relatively quickly, which makes them a more sustainable resource than non-renewable fossil fuels which are formed over millions of years.
Question 26 (5 marks)
A. For full marks mention that the concentration of NO2 decreases and explain why N2O4 is used up.
As NO2 is consumed by the reaction with water, the concentration of NO2 in the NO2/N2O4 mixture decreases. According to Le Chatelier’s Principle, the equilibrium in reactor 2 will shift to the left to produce more NO2, hence converting the N2O4 into NO2 and using up the N2O4.
B. For full marks mention how water may be recycled, and how NO can be re-used.
Firstly, the water that is produced from separator 1 can be recycled to react with NO2 in reactor 3. Secondly, the NO (g) released from separator 2 can be collected and re-used in reactor 2, for the reaction between NO and O2. This will save resources and lower the cost of the production process, whilst increasing efficiency.
Question 27 (4 marks)
Calculate the grams and moles of ethanol produced, which is equal to the moles of CO2(g) produced. Calculate the volume of CO2(g).
Question 28 (5 marks)
A. Describe the bromine test, and mention that the solution decolourises when bromine water is added.
The bromine test can be used to confirm that Q is an alkene. Adding the brown-coloured bromine water to Q will decolourise the brown due to an addition reaction that decreases [Br2].
B. Deduce molecular mass of R, then calculate the number of carbon atoms in the molecule. Hence, draw a dihalogenoalkane with chlorine atoms on adjacent carbon atoms (due to addition reaction).
R has a molecular mass of 112g/mol (peak at 114m/z due to Cl-37 isotope).
Question 29 (3 marks)
For full marks explain why hydrogen bonding makes alkanols soluble in water, and why an increase in chain length will decrease the solubility significantly.
Alkanols contain a hydroxyl group (-OH) which can form hydrogen bonds with polar water molecules, allowing them to dissolve in water. For smaller chain alkanols with a lower molar mass, the hydrogen bonding contributes more extensively to the intermolecular bonding. However, as the non-polar carbon chain increases in length, dispersion forces will predominate causing the alkanol to become increasingly more hydrophobic, causing the solubility to decrease exponentially. For longer chain alkanols, the addition of an extra carbon atom will have a comparatively smaller impact on the overall intermolecular forces. Hence the difference in solubility between consecutive alkanols becomes lesser, shown by the curve approaching 0 g/100mL.
Question 30 (4 marks)
For full marks mention: add acid to neutralise and confirm carbonate, add Ag– to test for Br–, two complete chemical equations.
Firstly add a few drops of 1.0 mol L-1 HCl to the water sample and observe for any effervescence. If bubbling occurs, this confirms the presence of CO32- ions, as carbonate will react with HCl in a neutralisation reaction to produce CO2:
2HCl (aq) + CO32- (aq) ⇌ 2Cl–(aq) + H2O(l) + CO2 (g)
Then add an excess of HCl slowly until no more bubbling is observed. This means that all of the
CO32- is consumed. To test for the presence of Br–, add a few drops of AgNO3 solution and observe. If a creamy precipitate forms, this means Br– is also present in the water sample.
Br– (aq) + Ag+ (aq) → AgBr (s)
Question 31 (7 marks)
Plot curve of concentration vs absorbance, and use it to find the concentration of [Cu(C3H6O3)2]2+ in the solution after equilibrium. Then use the ICE table to calculate the concentrations of the reactants and products, and put them into the equation for equilibrium constant.
Interpolate from the curve to get A = 0.66, [[Cu(C3H6O3)2]2+] = 0.046 mol L-1
Question 32 (5 marks)
For full marks show all correct working out and obtain the right answer to 4 significant figures.
This represents the excess NaOH remaining after reaction with ammonia.
n(NaOH) initial =C x V = 0.05 x 1.124 = 0.0562 moles
n(NaOH) reacted with ammonia = initial – excess = 0.0562 – 0.0303 = 0.0259 moles
Since ammonium and hydroxide react in a 1:1 ratio,
n(NH4+) = 0.0259 moles
m(NH4+) = n x mm = 0.0259 x (14.01 + 1.008 x 4) = 0.4672g
Question 33 (6 marks)
A. Between 6 to 8 minutes, the amount of all species in the reaction remains constant which means the system is at equilibrium (rate of forward and reverse reactions are equal).
B. Identifies that the reverse reaction is favoured and explains 2 possible reasons why.
At 8 minutes, the concentrations of the reactants A2 and B2 gradually increase, whilst the concentration of AB2 gradually decreases. Thus any change would have favoured the reverse reaction. One possible explanation is an increase in temperature, which would favour the reverse endothermic reaction in order to use up the heat applied (Le Chatelier’s Principle). Alternatively, a decrease in pressure would favour the side with more gaseous moles, which is the reverse reaction, in order to increase the pressure again.
Question 34 (5 marks)
For full marks, all correct working out including relevant equations.
Question 35 (6 marks)
- Gives correct working out with correct answer to 4 significant figures.
Question 36 (9 marks)
For full marks, draw the correct structure for each compound and name the correct functional groups, with full justification and reference to given data.
A (hex-3-ene) – C=C double bond
B (hexan-3-ol) – -OH hydroxyl group
C (hexan-3-one) – C=O ketone group
C (hexan-3-one) – C=O ketone group
- Benzene ring or C=C double bond (13C NMR)
- Assuming only hydrogens and carbons (and alkene), 6 carbons and 12 hydrogens (mass = 84.156g mol-1)
- O-H (IR)
- C-H (IR)
- 12 H (1H NMR)
- Not a carboxylic acid (missing 9.0 – 13.0 shift) (1H NMR)
If B can be oxidised to C (and C is not a carboxylic acid), B is likely a secondary alcohol (supported by infrared data), and C is likely a ketone. In dilute H2SO4, alkenes can undergo an addition reaction with water to form alcohol, with the product being a secondary alcohol in this case. Hence, A is likely an alkene.
Since A has a molar mass of 84.156g mol-1 and only has 3 signals on the 13C NMR spectrum, it is likely a 6-carbon symmetrical, and hence C is a 6-carbon ketone. The only hexanone isomer that matches the 1H NMR spectrum is hexan-3-one. Therefore, B is hexan-3-ol, and A is hex-3-ene.
Question 37 (5 marks)
A. For full marks include correct equation for Q and correct justification.
Hence Q = Keq, so the system is at equilibrium.
Adding CO2 will cause equilibrium to shift to the left, to decrease the concentration of CO2 (Le Chatelier’s Principle). Hence let the additional [CO2] be and the subsequent change in [CO2] be
|2CO (g)||CO2 (g)||C (s)|
|I||1.10 × 10-2||1.21 × 10-3 + y||–|
|E||1.10 × 10-2+2x||1.21 × 10-3 +y – x||–|