Our Chemistry Team has worked hard to bring you the worked solutions to the 2016 Chemistry HSC Chemistry Paper's multiple choice section.
Do you ever have a hunch a multiple choice answer is right, but then can never prove it is? Don’t worry, Matrix is here to help! In this post, we give you the 2016 HSC Chemistry Paper worked solutions for multiple choice. It’s one thing to know what the answer is, it’s far more important that you know why it’s right.
Read on and learn why you got it right or wrong, so you can get it right every time!
|1||D||The compound is chloroethene, also known as vinyl chloride (the monomer from which PVC is made).|
|2||D||Sodium is not a heavy metal as it is one of the lightest metals (element 11, mass 22.99). It is naturally found in the environment in high concentrations without toxicity for many organisms e.g. in seawater.|
|3||C||You can determine the formula by drawing out the full structure and counting the atoms.|
|4||C||Polystyrene is very stiff, so it is used in screwdriver handles. Plastic bags are usually made of polyethylene.|
|5||B||Water and ethanol can participate in hydrogen bonding with each other. Hydrogen bonds form between a lone pair on a fluorine, oxygen or nitrogen atom, and a hydrogen atom covalently bonded to a fluorine, oxygen or nitrogen atom.
Note: Many schools will penalise you if you refer to hydrogen bonding as “bonding”, since it is considered an intermolecular force rather than a “true” chemical bond.
|6||A||Equimolar means containing the same number of moles.
Acetic acid is a monoprotic acid while barium hydroxide contains two hydroxide ions (proton acceptors) per unit, therefore they react in a 2:1 ratio and there will be excess barium hydroxide (a strong base) left over.
Acetic acid also reacts with sodium carbonate in a 2:1 ratio, but sodium carbonate is a weak base, so the resulting solution will contain the same concentration of sodium carbonate as the barium hydroxide solution, but will be less basic.
Sulfuric acid is a strong diprotic acid, so (C) and (D) would produce neutralised mixtures that would be neutral and weakly acidic, respectively.
|7||B||Lemon juice would be red, potato juice would be yellow. The other indicators would be the same colour for both solutions.|
|8||D||The aim is to test if the water is hard or soft. Hence it would be best to use soft water (low concentration of calcium and magnesium ions. e/g. distilled water) as a control.|
|9||D||Balance the equation so the sum of the top numbers (mass numbers) are the same on both sides, and the sum of the bottom numbers (atomic numbers) are the same on both sides.
X is 42He.
|10||C||A conjugate base contains one less proton.|
|11||D||The halogen substituents should be listed alphabetically. The numbers are assigned according to the first point of difference rule (not the lowest sum, although coincidentally this gives the same answer for this question).|
|12||C||0.01 M HCl has a pH of 2. To increase its pH by 2 units, it should be diluted 100-fold i.e. from 100 mL to 10000 mL.|
|13||A||Sodium has a bright yellow-orange flame test. Sulfate does not produce bubbles when reacted with acid (this would occur with carbonate). The precipitate is BaSO4.|
|14||B||Decreasing the volume causes the equilibrium to shift to the side with less gas moles i.e. towards N2O4 to minimise the disturbance (Le Chatelier’s principle).|
|15||A||W must be unsaturated, while X is saturated. Shorter alcohols are more soluble in water.|
|16||B||Aluminium and copper is the combination with largest potential difference (refer to the Table of Standard Reduction Potentials on the Data Sheet).
(A) = 0.74 V
(B) = 2.02 V
(C) = 0.63 V
(D) = 1.04 V
|17||D||This is an addition polymer as it has a carbon-only backbone. Breaking the polymer at every second C-C bond and re-forming the double bonds gives propene as the original monomer.|
|18||B||HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)
n(HCl) = c x V = 0.1 x 0.06 = 0.006 mol
n(NaOH) = c x V = 0.1 x 0.04 = 0.004 mol
NaOH is limiting, HCl is in excess. Since all of the other substances in the final solution are neutral, we only need to look at the concentration of HCl in the final solution to determine the pH.
n(HCl)excess = 0.006 – 0.004 = 0.002 mol
Total volume = 0.06 + 0.04 = 0.1 L
c(HCl) = 0.002 / 0.1 = 0.02 M = c(H+)
pH = -log10[H+] = 1.699 = 1.70 (2 s.f.)
Note: Only the digits after the decimal point are significant for pH. Since it is an exponential, the integer portion is used to determine the position of the decimal point, which is not considered significant.
|19||C||Ba(NO3)3(aq) + Na2SO4(aq) → BaSO4(s) + 2NaNO3(aq)
n(Na2SO4) = c x V = 0.2 x 0.2 = 0.04 mol
n(BaSO4) = 0.04 mol (1:1 mole ratio)
m(BaSO4) = n x MM = 0.04 x (137.3 + 32.07 + 16 x 4) = 9.3348 g = 9.33 g (3 s.f.)
|20||A||In atomic absorption spectroscopy, the intensity of light before absorption by the sample (i.e. at the lamp) is compared to the intensity after absorption (i.e. at the detector) at the same wavelength (in this case, 615.2 nm). The other options do not compare intensities at the same wavelength.|