Rough Solutions for Exam

[illuminati]

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

1. a) i. Glucose
        ii. 2CH3CH2OH(l) + 2CO2(g)
        iii. CH2CH2(g) + H2O(l) -----> CH3CH2OH (catalyst H3PO4, 300 degrees)
    b) i. 3, 1
        ii. stearic acid
        iii. CH3(CH2)16COOH (l) + 26O2(g) ------> 18CO2(g) + 18H2O(l)
2. a) tyrosine
    b) The lowest dot in the middle
    c) Two of the amino acids in the first chromatogram had the same retardation factor and hence the spots had doubled up, causing only 3 dots to appear on the first chromatogram. When run in different conditions, these two amino acids separated as they had a different retardation factor under these new condition and hence two dots appeared (multiple answers accepted, but you'd have to mention the same Rf values somewhere)

3. a) Semi structural: NH2(CH2)6NH2
    b) amide groups
    c) Protein: carboxylic acid and amine group on the same monomers. The carboxylic acid all face one direction when polymerised. Due to this alternating fashion, every single peptide bond will be the same in terms of orietnation.
        Nylon: two different monomers which contain either 2 amine groups or 2 carboxylic groups. The carboxylic groups are aligned opposite to each other on the monomer, causing the orientation to be alternating between bonds. There will be bonds where NH precedes the CO and the other bond where CO precedes NH.
       Hence, differences arise in their orientation. (different answers accepted, i'd say you need to find the difference between two monomers of protein and nylon for 1 mark)
    d) See the polymer below as an attachment.

4. a) propan-1-ol, propanoic acid
    b) CH3CH2CH2OH (aq) + CH3CH2COOH (aq) -----> CH3CH2COOCH2CH2CH3 (aq) + H2O (l) (Catalyst H+ or H2SO4)
    c) Oxidation of some propan-1-ol to propanoic acid (reagent dichromate or permaganate)
        React propan-1-ol with propanoic acid (H+ or H2SO4 catalyst)
        Separate the reaction mixture with fractional distillation (principles of fract. dist, use of unique boiling points)
        Each one of these points should be one mark.
    d) Chromatography, GC (Explanation: Use of unique retention times or retardation factor)

5. a) i. Two
        ii. Three
        iii. Six
    b) Oxygen and hydrogen, O-H
    c) Propan-2-ol

6. a) i. Draw da graph
        ii. 36mg
    b) IR, NMR, AAS, UV-visible, AES
        i. Infra-red, Radiofrequency, Visible, UV and Visible, Visible
        ii. Bonds in the molecule absorb energy that causes the bonds to change its dipole moment, qualitative (e.g. bending, twisting)
           Change in nuclear spin states, discrete energy absorption, qualitative
           Metals absorb a unique wavelength of light, quantative analysis can be performed based on the amount of light absorbed
           Solution absorbs wavelength of the UV-visible spectrum the more concentrated it is, quantative analysis of concentration based on amount of light absorbed
           Qualitative analysis. Uses the fact that a metal atom will absorb a discrete amount of energy and will release these quantas of energy as visible spectrum bands. Will not work with all metals, only those that emit visible light.
           (I'd say, 1 mark qualitative or quantative, one mark for the interaction)

7. a) i. Pb(CH3COO)2 (aq) + 2KI (aq) ----> PbI2 (s) + 2CH3COOK (aq)
           OR
           Pb2+(aq) + 2I- (aq) -----> PbI2 (s)
        ii. Ensuring that all water has been lost so that the results are accurate. (heating to constant mass)
        iii. 0.09390g
        iv. 0.331g
    b) Lead nitrate has such a high solubility that it would dissolve in solution and hence will not precipitate (other answers accepted)

8. a) i. HCl(aq) + NaOH (aq) ----> NaCl (aq) + H2O (l)
           OR
           H+ (aq) + OH- (aq) ------> H2O (l)
        ii. 0.00400 mol
        iii. 0.00216 mol
        iv. 0.0270 mol
        v. 289 g/L
    b) The solubility would be lower. If the burette was rinsed with water then the concentration of the HCl would be diluted. More HCl is required to neutralise the NaOH, which suggests that less NaOH reacted with the NH4Cl. Using mole ratios, NH4Cl would be lower than actual, and the concentration at saturation would be lower than actual. This in turn lowers the solubility. (1 mark for lower, 1 mark for explanation)

[pi]

You've still got it illuminati!

stickied

[ligands]

7ai) do you need a 2 infront of the ch3cook? and would kch3coo be accepted

[illuminati]

7ai) do you need a 2 infront of the ch3cook? and would kch3coo be accepted

Yep good pick up
And it should be accepted.

[Tonychet2]

7ai) do you need a 2 infront of the ch3cook? and would kch3coo be accepted

Yep good pick up
And it should be accepted.

i wrote the ionic equations for 7a that still full marks right?

[jadams]

Q5aiii) Bit obscurely worded: "In the 1H NMR spectrum, the signal at 3.6 ppm is split into a septet (7 peaks). What is the number of equivalent protons that are bonded to the adjacent carbon atom(s)?"

I gave both perspectives, but by equivalent protons does it mean other 1H protons, or 1H protons in the same environment?

I gave 0 in same environment, and 6 adjacent which cause splitting.

[illuminati]

7ai) do you need a 2 infront of the ch3cook? and would kch3coo be accepted

Yep good pick up
And it should be accepted.

i wrote the ionic equations for 7a that still full marks right?

Should be fine
The equation I wrote is just safer...

[illuminati]

Q5aiii) Bit obscurely worded: "In the 1H NMR spectrum, the signal at 3.6 ppm is split into a septet (7 peaks). What is the number of equivalent protons that are bonded to the adjacent carbon atom(s)?"

I gave both perspectives, but by equivalent protons does it mean other 1H protons, or 1H protons in the same environment?

I gave 0 in same environment, and 6 adjacent which cause splitting.

Interesting pickup
That answer you gave will get full marks
I still think 6 is just fine, vcaa examiners try to reward
but then again vcaa is dodgy so... we'll see.

[HERculina]

What about OH- + HCL ---> CL- + H2O?

[jadams]

Q5aiii) Bit obscurely worded: "In the 1H NMR spectrum, the signal at 3.6 ppm is split into a septet (7 peaks). What is the number of equivalent protons that are bonded to the adjacent carbon atom(s)?"

I gave both perspectives, but by equivalent protons does it mean other 1H protons, or 1H protons in the same environment?

I gave 0 in same environment, and 6 adjacent which cause splitting.

Interesting pickup
That answer you gave will get full marks
I still think 6 is just fine, vcaa examiners try to reward
but then again vcaa is dodgy so... we'll see.

Yeah I'm a bit worried that vcaa examiners may resent marking my exam...I crossed out a whole question, re wrote my answer in 2 dot points in a different colour pen, legible if an effort is made to read, and wrote Refer to answer in blue pen.

Otherwise, I think I've dropped 2, possibly 3 marks

[illuminati]

What about OH- + HCL ---> CL- + H2O?

Eurgh, that is a very halfway between molecular equation and ionic.
I don't know, but i'd say if you're counting go worse case scenario and say you dropped that mark.

[Nobby]

What would you give my (ever so wobbly) answer for SA. 7.b)

There is not nearly enough lead to produce a ratio of 60.0g lead nitrate to 100mL water (and hence form precipitate). Thus, no lead nitrate precipitates.

[illuminati]

What would you give my (ever so wobbly) answer for SA. 7.b)

There is not nearly enough lead to produce a ratio of 60.0g lead nitrate to 100mL water (and hence form precipitate). Thus, no lead nitrate precipitates.

So wobbly. honestly hard to tell.
But based on current knowledge it is a possibility.

[Tonychet2]

What would you give my (ever so wobbly) answer for SA. 7.b)

There is not nearly enough lead to produce a ratio of 60.0g lead nitrate to 100mL water (and hence form precipitate). Thus, no lead nitrate precipitates.

sorry not trying to be harsh or mean but 0/1

[illuminati]

What would you give my (ever so wobbly) answer for SA. 7.b)

There is not nearly enough lead to produce a ratio of 60.0g lead nitrate to 100mL water (and hence form precipitate). Thus, no lead nitrate precipitates.

sorry not trying to be harsh or mean but 0/1

I wouldn't be so sure.
It's because you can argue that they have given you the solubility of the lead nitrate. Since you know that lead nitrate has a high solubility in water, you know that after the solution has become saturated you'd have a solid in there. If you had written that the solution was not saturated and therefore no excess salt crystals would be in solution that could have gotten one mark.
What he's written is just a more simplified version.