2017 HSC Chemistry Exam Paper Solutions (with Option Topics)
Posted on November 1, 2017 by Michelle Wong
The HSC Chemistry exam was held today (1 November 2017). While the official exam paper has not yet been released by NESA, here are our draft suggested solutions and some explanations in case you wanted to know ASAP!
HSC Chemistry Section I A (Multiple Choice) Solutions
|1||A||While a measuring cylinder would be more practical, a burette would be more accurate.|
|2||D||Copper gives a blue-green flame.|
|3||A||2-chloro-1-fluorobutane: according to the frequently misinterpreted IUPAC organic naming rules as described in the “Blue Book”, the chloro prefix comes before fluoro (alphabetical order).|
|4||D||Condensers should never be stoppered to avoid pressure buildup. Sulfuric acid is necessary for the reaction but does not improve safety per se. The direction of water flow is correct (in the bottom, out the top for both reflux and distillation setups). A heating mantle would avoid the open flame of a Bunsen burner.|
|5||A||HSO4– is amphiprotic as it can donate a proton to form SO42-, and gain a proton to form H2SO4|
|6||B||Nuclear reactors produce neutron-rich radioisotopes. Both Co-60 and Np-239 are neutron rich (compare them to the masses on the periodic table, if the mass if higher than expected it is neutron-rich), but only Np-239 is transuranic.|
|7||C||Water and hydrocarbons are immiscible. Hexane does not decolourise bromine water int he absence of UV light.|
|8||C||Barium does not precipitate with chloride but lead does.|
|9||D||The black deposit is a result of incomplete combustion producing C(s).|
|10||B||The carbon rod of the dry cell and the PbO2 of the lead-acid cell are the cathodes, which have a positive charge.|
|11||D||S is oxidised as it goes from S (oxidation number of 0) to SO2 (oxidation number of +4).|
|12||D||Propene polymerises to give a polymer with no double bonds, and a methyl group on every second carbon.|
|13||B||Citric acid is triprotic, so it will require the most base for neutralisation.|
|14||A||HCl/Cl- cannot act as a buffer as HCl is strong, so the system does not form an equilibrium in solution.|
|15||B||This is tricky! Option B is the only option with an octet for each atom.|
|16||B||Adding more hydroxide would remove carbonic acid and shift the equilibrium to the right, removing CO2 and reducing pressure|
|17||C||1 mole of O3 weighs 3 x 16 = 48 g. 1 mole of any gas occupies 24.79 L at RTP. Hence the density is 48/24.79 = 1.936 g/L.|
|18||D||Y and Z are on the same side of the equation as their concentrations change in the same direction when the equilibrium shifts. The shift is equimolar for X, Y and Z (0.005 M each), so the mole ratio must be 1:1:1.|
|19||C||48% of 1.20 g is 0.576 g. Moles of sulfate = moles of barium sulfate, so 0.576 / 96.07 x 233.37 = 1.40 g.|
|20||C||Barium hydroxide is limiting (4 x 10-4 mol barium hydroxide vs 2 x 10-3 mol HCl). Therefore the moles of HCl left over is 2 x 10-3 – 8 x 10-4 = 1.2 x 10-3 mol. The total volume is 70 mL. Therefore H+ concentration is 1.714 x 10-3 M, so pH is 1.8.|
HSC Chemistry Section I B (Long Response) Solutions
|21(a)||2||Troposphere – causes respiratory irritation
Stratosphere – absorbs UV and turns it into heat
|21(b)||3||Boiling point – Ozone has a higher boiling point than oxygen
Density – Ozone has a higher density than oxygen
|21(c)||1||Cl•(g) + O3(g) → ClO•(g) + O2(g)|
|22(b)||2||From the graph, 0.58 absorbance means 3.45 ppm (indicate on the graph using dotted lines). Therefore the concentration of zinc is too high to be safe.|
|23(a)||2||The salt bridge is required to complete the circuit, to avoid a buildup of negative charge in the cathode compartment and hence allowing the cell to continue running.|
|23(b)||4||Zn(s) + 2Ag+(aq) → Zn2+(aq) + 2Ag(s)
n(Zn) = m / mm = 1 / 65.38 = 0.015295 mol
n(Ag) = 2 x n(Zn)
m(Ag) = n x mm = 3.30 g
Total m(Ag) = 10 + 3.3 = 13.3 g (3 s.f)
|24(a)||3||CH3COOH(aq) + NaOH(aq) → NaCH3COO(aq) +
n(CH3COOH) = c x V = 0.01255 mol = n(NaOH)
c(NaOH) = n / V = 0.01255 / 0.01930 = 0.6503 M (4 s.f.)
|24(b)||2||NaCH3COO is not neutral as the acetate ion is basic and reacts with water to produce OH-:
CH3COO–(aq) + H2O(l) ⇌ CH3COOH(aq) + OH–(aq)
Hence the pH will be greater than 7.
|25||4||Hydrolysis: Water is added to break cellulose up into beta-glucose, using cellulase: H(C6H10O5)nOH(s) + (n – 1)H2O(l) → nC6H12O6(aq)
Fermentation: Glucose is fermented to produce ethanol and carbon dioxide, using yeast: C6H12O6(aq) → 2C2H5OH(aq) + 2CO2(g)
Distillation: Since fermentation can only produce ethanol concentrations up to 15%, the mixture needs to be distilled to produce higher concentrations before the next step.
Dehydration: Concentrated sulfuric acid catalyses the dehydration of ethanol to produce ethylene: C2H5OH(l) → C2H4(g) + H2O(l)
Addition polymerisation: Ethylene is reacted to produce polyethylene through a reaction such as free radical polymerisation (initiation, propagation, termination)
|26(a)||3||Sulfur dioxide causes acid rain (SO2(g) + H2O(l) → H2SO3(aq)) which has many detrimental effects:
Acid rain acidifies waterways, which causes an increase in toxic aluminium, which harms fish populations. Fish eggs will also not hatch.
Acid rain also damages forests, both directly (corrosive damage to leaves) and indirectly (leaches nutrients from soil)
|26(b)||4||Coal fired power stations produce SO2 from the combustion of sulfur impurities in coal: S(s) + O2(g) → SO2(g)
Metal smelters smelting sulfide ores also produce SO2 e.g. 2ZnS(s) + 3O2(g) → 2ZnO(s) + 2SO2(g)
Hence the concentration of SO2 is highest near these sources, and lower further away as winds and air currents dissipate the SO2. Most of the SO2 in the atmosphere is produced from industrial sources such as these, so the concentration in other areas are very low (< 0.2 x 1012 molecules per L).
|27||5||Acetic acid contains a COOH group in its structure, so it can form many strong hydrogen bonds.
Butan-1-ol contains an OH group in its structure, so it can form strong hydrogen bonds.
Butyl acetate contains an ester functional group, which causes its structure to be polar so it can form weaker dipole-dipole forces.
Hence due to these polar intermolecular forces, they have similarly strong intermolecular forces even though their dispersion force strength is different due to different molar masses (dispersion force strength is proportional to molar mass so acetic acid has the weakest dispersion forces, butan-1-ol is next weakest while butyl acetate has the strongest dispersion forces).
The same intermolecular force strength would require similar heat energy to break, hence the boiling points are similar.
Since ethanol is a liquid fuel, the same fuel infrastructure (current petrol stations, petrol tankers etc.) can be used to deliver the ethanol to consumers.
Ethanol can be produced from renewable sources like starch-based crops, unlike petrol which is derived from non-renewable petroleum which will likely diminish in the future.
Ethanol gives lower mileage, which means consumers will need to refuel more often
Blends containing over 15% ethanol can’t be used in current cars in many areas of the world including Australia, so expensive engine modifications would be required for greater use.
|28(b)||3||C2H5OH(l) + 3O2(g) → 2CO2(g) + 3H2O(l)
Therefore 1360 kJ per 2 mol CO2.
m(CO2) in 2 mol = n x mm = 2 x 44.01 = 88.02 g = 0.08802 kg
Therefore 1360 kJ / 0.08802 kg = 15450 kJ/kg (4 s.f.)
|29||4||Site Y would be preferred.
Sources of contamination for X:
Sources of contamination for Y:
N2(g) + 3H2(g) ⇌ 2NH3(g) ΔH < 0
Removal of ammonia as it forms via liquefaction:
HSC Chemistry Section II (Option Topic) Solutions
Question 31: Industrial Chemistry
Question 32: Shipwrecks, Corrosion and Conservation
|32(a)(ii)||3||The expected result is that the rate of corrosion of the metals increases with increasing acid concentration. This will be indicated by increased mass loss of the nails immersed in high acid concentration. The presence of an acid will increase the rate of corrosion as the hydrogen ions can be reduced to hydrogen gas, therefore accelerating the oxidation of the metal as shown below:
Fe(s) → Fe2+(aq) + 2e–
2H+(aq) + 2e– → H2(g)
A higher acid concentration results in a higher cell potential for the oxidation of the metal, resulting in a faster rate of corrosion. Similarly, a more reactive metal will exhibit a faster rate of corrosion due to a higher oxidation potential.
|32(b)(i)||2||Increase the concentration of the electrolyte, increase surface area of electrodes|
|32(b)(ii)||4||The main reactions that would occur at the anode and cathode are as follows:
cathode: Cu2+(aq) + 2e– → Cu(s) E° = +0.34 V
anode: H2O(l) → ½ O2(g) + 2H+(aq) + 2e– E° = -1.23 V
Due to the reduction of copper(II) ions to metallic copper atoms at the cathode, the mass of the cathode increases. Additionally, as copper(II) ions give a characteristic blue colour, the loss of copper(II) ions from solution also decreases the intensity of the blue colour in the electrolyte solution. The oxidation of water at the anode results in the formation of oxygen gas, which can be observed as bubbles.
|32(c)(i)||2||Scientists predicted that the rate of corrosion would have been slow at this depth as the deep ocean would have a low concentration of oxygen, due to lack of photosynthesis and surface mixing, which should lead to less oxidation of metals. Additionally, the temperatures would also be at a minimum due to lack of sunlight penetration. Lower amounts of heat energy result in less kinetic energy, hence a lower number of successful collisions and hence lower rate.|
|32(c)(ii)||2||Wooden artefacts from long-submerged wrecks have a weak structure as the cell walls are broken down and their cellular contents are leached out and replaced with sea water (a concentrated aqueous ionic solution). The sea water is what maintains its original form. Immediately drying wooden artefacts will cause the wood to contract and collapse. Furthermore, the evaporation of water will cause salt crystals to form and expand, causing cracks that destroy the structure of the wood.|
|32(c)(iii)||3||Much of the chloride ions are present as insoluble hydroxyoxychlorides trapped within the metal oxide deposits. Cleaning can be achieved by electrolysis. The artefact is placed in a 0.5 mol L-1 NaOH electrolyte and connected as cathode. A stainless steel anode is used. When a low voltage is applied, the negatively charged chloride ions are repelled from the negatively charged cathode (i.e the artefact) and move toward the anode where they are oxidised.
|32(d)||7||Modern ocean-going vessels are constructed with steel. Steel is an alloy of iron with carbon, with varying amounts of other metals and silicon. Due to the impure nature of steel, along with the constant exposure to high electrolyte concentration, ocean-going vessels have a high tendency to undergo corrosion.
The corrosion occurs in 3 steps.
First, iron is oxidised and oxygen is reduced at 2 separate sites in the iron. The electrons provided from oxidation flow through the iron to an impurity like carbon, and reduce oxygen in the thin film of moisture on the iron surface. Iron acts as the anode and an impurity in the iron will act as the cathode.
Oxidation: Fe(s) → Fe2+(aq) + 2e–
Reduction: O2(g) + 2H2O(l) + 4e– → 4OH– (aq)
Then Fe2+ and OH– will combine and form a precipitate: 2Fe2+(aq) + 4OH–(aq) → 2Fe(OH)2(s)
Iron(II) is then oxidised to form rust: 4Fe(OH)2(s) + O2(g)→ 2Fe2O3·H2O(s) + 2H2O(l). The overall reaction is 4Fe(s) + 3O2(g) + 2H2O(l) → 2Fe2O3·H2O(s).
It is our understanding of the corrosion process that has led to the development of effective protection methods, which has been aided by the works of Volta and Davy.
Volta demonstrated that metal wires in the solution generated electric current. He constructed the first electric battery by connecting two different metals in an electrolyte solution. He later stacked cells to form what became known as the Voltaic pile. This was the first source of a direct electric current. Davy used the electricity generated by the Voltaic pile to decompose compounds into their constituent elements, and thus developed electrolysis.
These findings provided the basis of our understanding of electron transfer reactions, which contributed to our current understanding of the corrosion process and assisted in the development of a number of protection techniques that reduce the extent of corrosion. These include creating physical barriers and using cathodic protection methods. The use of paints, lacquers and polymer-based coatings prevent the steel from coming into contact with seawater (the electrolyte) and hence prevent corrosion.
Cathodic protection prevents corrosion of steel structures by making the steel structure the cathode in an electrochemical cell, thus ensuring the iron in the steel is protected from oxidation. There are two types of cathodic protection: the sacrificial anode technique which sets up a galvanic cell by attaching a metal that is more active (a stronger reactant) than Fe to the steel hull. This active metal forms the anode. An impressed current system makes the steel hull the cathode (negative terminal) of an electrolytic cell with an inert anode attached to the positive terminal. An external power source applies a continuous low voltage. Thus the corrosion protection methods used today have been largely due to the discoveries of Volta and Davy.
Question 34: Chemistry of Art
|34(a)(i)||2||Flammable reagents were cleared from the laboratory before the experiment due to
the need for a naked flame.A platinum wire loop was cleaned thoroughly by dipping into concentrated HCl. The cleanliness of the wire was checked by placing the wire in the flame and checking for a burst of colour.The clean wire loop was then dipped into an aqueous solution of calcium nitrate
and placed in a blue Bunsen flame. The colour of the flame was recorded.
|34(a)(ii)||3||Calcium = orange-red (brick red) flame
According to the Bohr model, an atom contains electrons which can only occupy
|34(b)(i)||2||Pigment: insoluble, provides colour and opacity
Medium: binds pigment to surface
|34(b)(ii)||4||Red ochre: iron(III) oxide Fe2O3
The colour of red ochre relates to the transition metal it contains (iron), as well as the rest of the compound.
Transition metal compounds are coloured because they have an incomplete d subshell. In compounds, the energies of the d orbitals split, with an energy gap in the visible light range. Electrons can transition from the lower to higher d orbitals with the absorption of a visible wavelength, and the unabsorbed wavelengths will be observed as colour.
The energy gap depends on a number of factors, including the identity of the metal, the
|34(c)(i)||3||Hund’s rule states that all degenerate orbitals in a subshell must be occupied before the
electrons pair up. In the oxygen atom, this means two of the p orbitals should contain one electron each, so there will be two unpaired electrons in the diagram.
|34(c)(ii)||2||As seen from the low 1st, 2nd and 3rd ionisation energies in the graph, the first three electrons in Sc are easily removed. However, the 4th electron requires a very high (>7000 kJ mol-1) ionisation energy, so Sc will predominantly have an oxidation state of 3+..|
|34(c)(iii)||2||Vanadium(III) compounds are coloured because V3+ has electrons in the 4s and 3d subshells, which are very close in energy, with energy gaps in the visible light range. Electrons can transition from the lower to higher energy orbitals with the absorption of a visible wavelength, and the unabsorbed wavelengths will be observed as colour.
However, scandium(III) compounds are not coloured as Sc3+ does not have any electrons in the 4s or 3d subshells, so transitions will not absorb in the visible range and produce colour.
|34(d)||7||In a complex ion, the transition metal ion covalently bonds to ligands. The ligands act as Lewis bases and donate lone pairs to the empty orbitals in the valence shell of the transition metal ion, which acts as a Lewis acid. This forms coordinate covalent bonds.
e.g. pentaamminechlorocobalt(III) ion: CoCl(NH3)5]2+
Co3+ has the electron configuration 1s2 2s2 2p6 3s2 3p6 3d6
[include orbital diagram and Lewis structures]
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