1.B
2.C
3.D
4.B
5.D
6.A
7.C
8.D
9.A
10.A
11.A
12.C
13.C
14.B
15.B
16.C
17.D
18.C
19.B
20.D
1ai. Glucose (or beta glucose if you prefer)
1ii. 2CH3CH2OH(aq) + 2CO2(g). I antagonistically wrote enzymes over the reaction arrow too, which in hindsight was probably not a particularly bright idea.
1iii. C2H4(g) + H2O(g) ------H3PO4----> CH3CH2OH(l)
1bi) 3 fatty acids and 1 glycerol
ii) Stearic acid
iii) C18H36O2(l) + 26O2(g) ----> 18CO2(g) + 18 H2O(g)
2a) Tyrosine
b) The lower of the two dots in the middle.
c) Solvent G did not separate alanine and threonine and hence they produced only a single spot with the same Rf value, and three spots for Chromatogram I overall. Solvent F then differentiated between alanine and threonine so that they both produced individual spots with different Rf values. Whilst alanine and arginine had identical Rf values under Solvent F they had been separated by Solvent G and hence Chromatogram II had an additional spot and four spots overall.
3a) H2N-(CH2)6-NH2
b) Amide
bii) Amide/Peptide
c) Ceebs typing this up right now.
4a) Propan-1-ol and Propanoic acid
b) CH3CH2OH(l) + CH3CH2COOH (l) ----H2SO4(l)---> CH3CH2COOCH2CH2CH3 (l) + H2O(l)
c) Obtain a pure sample of propan-1-ol. Oxidise half of this sample using acidified dichromate to obtain propanoic acid. React propanoic acid with the remaining half of propan-1-ol in the presence of concentrated sulfuric acid to obtain propyl propanoate. Distil this product to obtain pure propyl propanaote.
d) Obtain an IR spectrum of the product. If it does not depict an O-H (acid) absorption band or O-H (alcohol) absorption band it can be said to be pure as any contamination from the reactants propanoic acid and propan-1-ol respectively would result in these absorption bands being present in the IR spectrum of the final product.
5a) 2
ii) 3
iii) 6
b) Oxygen and Hydrogen (Although VCAA may accept and even prefer O-H)
c) <Insert Propan-2-ol structure>
6a) You'll have to draw your own graph
ii) 36mg
b) UV-Visible Spectroscopy: The level of absorbance of certain wavelengths of light in the UV-Visible region can be used to assess the concentration of compounds which absorb in that specific wavelength. Throw in something about electrons moving to a higher energy level.
7a) Pb2+ aq) + 2I-(aq) ---> PbI2(s)
ii) To ensure that all water had been removed from the precipitate and was not contributing to the mass weighed.
iii) 0.0939g
iv) 0.331g
b) Nitrates are soluble in water with lead (II) cations and vice-versa and hence no precipitate was formed.
8a) NaOH(aq) + HCL(aq) ---> NaCL(aq) + H2O (l)
ii) 0.0400mol
iii) 0.00216mol
iv) 0.0270mol
v) 289g/L
b)Lower because a greater titre of HCl would be required resulting in the amount of NaOH in excess being calculated as greater than its actual value and the amount reacted with ammonium chloride to be less than its actual value. Hence the mass of ammonium chloride dissolved in solution will appear to be less than it really is.
DISCLAIMER: I didn't answer the questions exactly like this in the actual exam, unfortunately. A few of my response questions here are less detailed and would probably not obtain full marks.