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Multiple Choice:1. A (Fixed now!)
2. D
3. A
4. D (A is wrong because putting a stopper will cause pressure to build up -> explosion. B is most definitely wrong because concentrated H2SO4 is not safe, it only catalyses the reaction. C would not affect safety.)
5. A (amphiprotic: can gain or lose a H+, therefore HSO4-. B and D can only give away H+, C can only take H+)
6. B (collision of neutrons into the Np-237 nucleus to synthesise Np-239)
7. C (It should be noted that hexane does not react with bromine whereas hexane does. Because the Br-Br bond is what gives bromine water its colour, once the bond reacts with hexane then it is decolourised. It should also be noted that bromine being a non-polar substance dissolves well in hexane and hexane. What forms the immiscible layer is actually the water which is polar and does not dissolve in non-polar substances)
8. C (Check solubility rules. A is wrong because sulphate would precipitate with both barium and lead. B is wrong because nitrate is soluble for everything. D is wrong because carbonate would precipitate with both barium and lead)
9. D (incomplete combustion, B is wrong because the black thing left behind is carbon and in B only carbon monoxide is produced (which is a gas), C is wrong because hydrogen gas is not a product of combustion, hence D)
10. B (Zinc is the site of oxidation (high reactivity) whereas the carbon rod is the site of reduction. Pb is the site of oxidation, and PbO2 is the site of reduction.)
11. D (Work out the corresponding charge on each of the elements in both the reactant and product side of the equation. It is found that Cr went from +6 to +3 whilst S went from 0 to +2. Since oxidation is the losing of electrons, the element should be more positive. Therefore Sulfur is the one being oxidised.)
12. D
13. B (Citric can give off most amount of H+ ions per mole, meaning that more OH- ions are required to neutralise. Equivalence point is when moles of a standard solution (titrant) equal the moles of a solution of unknown concentration (analyte))
14. A
15. B (A, C and D are incorrect because N in the middle is bonded with 10 electrons (or it could just be incorrect representations that HSC markers made up))
16. B (A is incorrect because adding CO2 will increase the amount of gas molecules in a fixed volume despite the equilibrium will shift right. C is incorrect because decreasing volume = increase in pressure. D is incorrect because increase temperature will cause equilibrium to shift left, causing more CO2 gas to be produced)
17. C (molar mass / molar volume under room temperature and pressure = 1.936)
18. D (The sharp decline in concentration for all three substances means that there has been an increase in volume. Because X decreases after the disturbance whilst Y and Z increase, therefore Y and Z must be on the same side, having more moles of gas than X (LCP states that equilibrium will favour the side with most moles of gas when pressure decreases/volume increases). C would be a close answer in this case, but because Y and Z increase at the same rate, they must have the same molar ratio and this makes C a wrong answer.)
19. C (0.48 x 1.2 = 0.576 g, moles of SO4(2-) = 0.576/96.07 = 0.006 mol, 0.006 x (137.3 + 96.07) = 1.4g)
20. C (moles of Ba(OH)2 = 0.02 x 0.02 = 0.0004 moles, moles of HCl = 0.05 x 0.04 = 0.002 moles. When you write out the balanced chemical equation, the ratio of Ba(OH)2 to (HCl) is 1:2. Hence HCl is in excess by 0.002 - (0.0004 x 2) = 0.0012 moles. Concentration = 0.0012/0.07 = 0.017 mol/L. pH = -log(0.017) = 1.76 = 1.8 )
Short Answers: Q21. a) Ozone in the troposphere (lower atmosphere) has a negative effect and is harmful to human health by affecting lung functions, respiratory tracts and can potentially result in asthma especially when other air pollutants are present as well. Ozone in the stratosphere (upper atmosphere) is beneficial for humans because it is able to sacrificially absorb most of the UV radiation from the Sun, preventing humans from over-exposure and therefore decrease our chances of contracting skin cancer.
b) Oxygen properties: lower boiling point, odourless
Ozone properties: higher boiling point (higher dispersion force), pungent odour
Both are similar in that they all exist in a gaseous state.
c) Chlorine radical + Ozone (g) --> ClO radical + O2(g)
22. a) Beware when sketching the graph:
1. Line of best fit should go through as many points as it can, with equal number of data points below and above the line.
2. Descriptive title is a must!
3. Label all your axes, make sure your units are correct (Note: absorption does not have a unit)
4. Extrapolated lines must be shown in dashed lines
b) Find the point on the line of best fit that matches absorbance = 0.58, and if the corresponding zinc concentration value is less than 2.80 ppm, then the water is not overly polluted with heavy metals and therefore is safe for drinking.
23. a) Salt bridge is required for maintaining electrical neutrality in the galvanic cell. It delivers anion to the cathode to compensate for the zinc metal that has been oxidised into Zn2+ (aq) to ensure that the electrolyte is still neutral. In a similar way, the salt bridge delivers cations to the anions to compensate for the loss of Ag+ in the solution and react with NO3-(aq) to ensure that the electrolyte is neutral.
b) Zn(s) + 2AgNO3(aq) --> Zn(NO3)2(aq) + 2Ag(s)
moles of zinc solid lost = 1/63.55 = 0.0157 moles
moles of silver solid gained = 0.0157 x 2 = 0.0314 moles
mass of silver solid gained = 0.0314 x 107.9 = 3.388g
new mass of silver electrode = 10 + 3.388 = 13.388g = 13.4g (3.sig.fig)
24. a) NaOH(aq) + CH3COOH(aq) --> NaCH3COO(aq) + H2O(l)
moles of CH3COOH = 0.025 x 0.502 = 0.01255 moles
moles of NaOH = 0.01255 moles
Concentration of NaOH = 0.01255/0.0193 = 0.6503 mol/L
b) The pH level of the resulting salt solution is not 7, but was instead slightly below 7 because we are reacting a strong base with a weak acid. The final product NaCH3COO(aq) --> Na+(aq) + CH3COO-(aq). CH3COO-(aq) is a weak base, therefore the endpoint is slightly basic.
25.a) Cellulose --> (enzymatic hydrolysis, enzyme = cellulase) --> Dissarcharide Sugar (C12H22O11) --> (enzymatic hydrolysis) --> Glucose (C6H12O6) -->
(fermentation reaction, anaerobic condition, 25-30 degrees environment, yeast catalyst) --> ethanol (C2H5OH) --> (dehydration, concentrated sulphuric acid) --> Ethylene
Expand on the above process to obtain full marks
26. a) Release of sulphur dioxide ---->
1. form acidic rain (SO2 + H2O --> H2SO3), acidifying soils and lakes to make the conditions inhospitable for plantations and marine species, affecting wildlife habitats and marine ecosystem
2. toxic gas with pungent odour that seriously affect human respiratory system
3. Degradation of buildings and households
Expand on the above dot points to obtain full marks
b) The areas of high sulphur dioxide concentration coincides with metal smelters and coal-fired power plants. This is because most metal ores and fossil fuels contain Sulphur as their constituent. When smelted or combusted, the sulphur reacts with oxygen, forming sulphur dioxides that will gradually rise up in the atmosphere.
Smelting of copper ore: 2CuFeS2(s) + 4O2(g) → 2FeO(s) + Cu2S(s) + 3SO2(g)
During the combustion of sulfur-containing fossil fuel, sulphur is oxidised (S(s) + O2(g) → SO2(g)), resulting in sulphur dioxide.
27. Their boiling points are results of the different types of intermolecular forces within each compound. Acetic Acid (CH3COOH) is bonded together through hydrogen bonding between the hydrogen and the lone pair on the oxygen. Butan-1-ol is also bonded together through hydrogen bonding which involves the hydroxyl group in the compound. The reason why CH3COOH despite having less molar mass (i.e. less dispersion force) but still have similar boiling point is because two hydrogen bonds can be formed between two single CH3COOH compounds. Butyl acetate only contains dipole-dipole interaction, which is less powerful than hydrogen bonds and is easier to be overcome. Therefore CH3COOH has similar boiling points to the other two despite having a much higher molar mass (i.e. higher dispersion force).
28. Advantages of Ethanol:
1. Burns cleanly, releasing less CO2 per mole combusted (due to the addition of -OH in ethanols) (i.e. lower emissions)
2. Acts as a gas pipe anti-freeze
Disadvantages of Ethanol:
1. Does not supply as much energy as petroleum for each mole combusted
2. Over 15% ethanol in fuel can cause damage to automobile engine system
29. Water source Y would be more preferable.
Water source X is neutral in pH (pH = 7.3) and has low turbidity (2 NTU), which makes it more desirable than water source than Y.
Water source Y is less hard (5 ppm) and contain much less phosphate (0.0001ppm) than source X, which makes it more desirable than X.
However, the water purification system does not have the facility to solve the issues of high calcium content (i.e. hard water) and high phosphate content (i.e. possible pollution by fertilisers and pesticides which can harm human health). However the water purification system can effectively address the issue of turbidity through sedimentation tank and sand filter, and neutralise the pH level of water source Y through pH control. Therefore water source Y is more desirable for usage.
30. Firstly provide an equation of Haber process: N2(g) + 3H2(g) (Equilibrium symbol) 2NH3(g)
1. High pressure is preferred for yield, because that forces equilibrium to shift to the right, favouring the side with less moles of gas. However this has to be compromised (200 atm) because high pressure within a sealed container is very dangerous.
2. Low temperature is preferred for yield, because that forces equilibrium to shift to the exothermic forward reaction, favouring the production of NH3. However this reduces the rate of production, therefore also has to be compromised at 450 degrees celsius.
3. Liquefaction of NH3 as it is produced to remove product, forcing equilibrium to shift to the product side, increasing yield.
4. Utilisation of a magnetite (Fe3O4) catalyst in order to reduce the activation energy required and improve the rate of production.
Industrial Chemistry: 31. Industrial Chemistry
a) ii) To carry our saponification:
1. Mix 250 ml of water with 100 g of NaOH in a beaker
2. Add coconut and olive oil into a container
3. Mix until thick
4. Let the mixture cure for a while before testing
Emulsion Testing:
1. Place 1g of the synthesised soap in one tube,. 1g of commercial detergent in another
2. Pour 20 ml of water in each test tube
3. Pour 10 ml of oil in each test tube
4. Stopper the test tubes and shake violently for a minute
5. See if both form stable emulsion (that is, water and oil separate).
b) i) When temperature increases, the equilibrium shifts to the reactant side, favouring the production of reactants, therefore decreasing the yield of products. The new curve should thus be below the original curve with the starting point still at the origin.
b) ii) Set up a MICE table in order to answer this question (Molar ratio, initial concentration, change in concentration, equilibrium constant)
c) i) X = NaHCO3, Y = CaO
c) ii) Brine is purified at the start of the Solvay Process by adding CO3(2-) and OH- to precipitate out the Ca2+, Mg2+ and Fe3+ impurities (e.g. 3NaOH(aq) + Fe(NO3)3 (aq) --> Fe(OH)3 + 3NaNO3(aq)).
c) iii) The brine is bubbled with ammonia to facilitate the production of HCO3-, which is then used to produce NaHCO3(s) - this is a very integral material that enables the production of sodium carbonate. At the same time, the carbonation of ammoniated brine also produces NH4+(aq) which would allow NH3 to be recycled, hence decreasing the overall expense of the Solvay process. Evidently, ammonia is a very important part of the process as it enables the most crucial reactant of sodium carbonate to be produced.
d) Speak about mercury cell, diaphragm cell and membrane cell in the order of their development.
In Mercury cell, the ineffective treatment of mercury is a concerning environmental issue. It is a neurotoxin, and if it leaches into waterways, it can disturb the marine ecosystem and cause unrecoverable mental damage to consumers of these marine species. The high amount of voltage and current required to carry out the process is a technology barrier that makes this form of sodium hydroxide production expensive.
Improvements in technology has led to the invention of diaphragm cell which utilises graphite anode and steel mesh cathode instead of mercury. Its low voltage requirement effectively addresses the technological issue that exists in the mercury cell. In terms of environmental issues, diaphragm cells has prevented the possibility of major heavy metal mercury leakage, however proposes another environmental problem associated with the use of asbestos in its diaphragms. The asbestos fibres can pollute the atmosphere, causing lung cancers and severe respiratory conditions in the long term, proposing health threat to humans. A technological issue that rises from this is the impurity of product (only 60% pure!).
Membrane cell is the most recent cell to be developed, effectively resolving the high power need through requiring even lower voltage supply. It utilises the new teflon technology to replace the asbestos diaphragm, hence addressing the issue of environmental pollution. The ion-selective membrane also resolves the issue of product impurity, making the final NaOH almost 99% pure.
Expand slightly along these lines in order to obtain full marks.
If there's any errors here, please point them out and I will change them.
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Topic: HSC Chemistry 2017 (Read 31386 times)