link to exam here:
VCAA 2012 Chemistry Unit 3 Exam Discussion1.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
1ii. 2 CH3CH2OH and 2 CO2
1iii. C2H4 (g) + H2O (l) ------H+----> CH3CH2OH
1b) i ) 3 for product A and 1 for product B
ii) Stearic
iii) C18H36O2 (aq) + 26O2 (g) ----> 18CO2(g) + 18 H2O (l)
2a) Tyrosine
b) the lower dot on the 2nd column
c) the Rf values of threonine and alanine are equal in solvent G in chromatogram 1, therefore together they produce 1 spot, making 3 spots total.
the Rf values of threonine and alanine are different in solvent F in chromatogram 2, therefore they will produce different spots in chromatogram 2, making 4 spots total.
3a) H2N-(CH2)6-NH2
b) amide
bii) amide
c) the 2 monomers of nylon contain either 2 carboxyl groups with 0 amino groups or 0 carboxyl groups with 2 amino groups, whereas the monomers of protein contain 1 carboxyl and 1 amino group each. Therefore when condensation polymerisation occurs, the orientation of the amino and carboxyl groups will be different in nylon and protein.
4a) propan-1-ol and propanoic acid
b) CH3CH2OH(l) + CH3CH2COOH (aq) ----H2SO4---> CH3CH2COOCH2CH2CH3 (l) + H2O(l)
c) 1. begin with propanol, 2. oxidise some propanol into propanoic acid using dicromate as catalsyt and in a H+ medium, 3. react propanoic acid with propanol using H2SO4 as catalyst, to produce propyl propanote.
d) gas chromatography, if only one peak occurs, the sample is pure, if the retention time is the same as a known sample of propyl propanoate, it is propyl propanoate.
5a) 2
ii) 3
iii) 6
b) Oxygen and Hydrogen
c) CH3 - CH-OH - CH3
6a) draw graph
ii) 36mg
b) do it yourself lol
7a) PB2+ (aq) + 2I- (aq) ---> PbI2 (s)
ii) to remove any impurities and to remove all excess water
iii) 0.0939g
iv) 0.3313g
b) lead nitrate is highly soluble in water, therefore no precipitate can be extracted
8a) NaOH (aq) + HCL (aq) ---> NaCL(s) + H2O (l)
ii) 0.04mol
iii) 0.00216 mol
iv) 0.0270mol
v) 289g/L
b) lower because a higher amount of titre would be required to neutralise the solution, therefore the n(NaOH) !(not n(HCL) oops wasnt thinking when i wrote this)! that reacted with n(ammonia chloride) will be lower. thus leading to a lower calculated solubility