VCE Chemistry 2010 Unit 3 Exam Suggested SolutionsNote, this is an unofficial release, to be used as a guide for how you performed,
not the actual marking scheme. Released as a VN community resource.
MCQQuestion 1 - C
Question 2 - D
Question 3 - A
Question 4 - C
Question 5 - C
Question 6 - A
Question 7 - A
Question 8 - B
Question 9 - B [Note, this is at the VCE level though, but determining the concentration of red dye at 500nm has a fairly large amount of uncertainty involved. In actual practice, I'd expect the chemist to run the test at 640nm first [red dye neglegible, blue dye near max absorption] to determine the concentration of the blue dye, then make up the standards containing the determined concentration of blue dye and varying concentration of red dye.]
Question 10 - C
Question 11 - A [The only isomer is 1,1,2-trichloroethene]
Question 12 - D
Question 13 - B [Note, congratulations to VCAA for applying some decent chemistry, option D gives the correct isomer]
Question 14 - D
Question 15 - B [Note, imagine as a cyclic chain = 51 peptides, take away two peptides to make two strands yields 49]
Question 16 - C
Question 17 - D
Question 18 - B
Question 19 - C
Question 20 - B
Short AnswerQuestion 1a.  +8H^+(aq) + 5e^- \to Mn^{2+}(aq) + 4H_2O(l))
b.  = 22.0 mL)
c.  = 8.80\times 10^{-4} mol)
d. _{aliquot} = 4.40\times 10^{-3} mol)
,
_{250mL} = 5.50\times 10^{-2})
e. _{total} = 3.07 g)
,
(3 sig figs)
Question 2a. C=O, carbonyl group [NOT carboxyl, ketone or aldehyde]
b. 3
c. 3
d.i. 2
d.ii. H H
| |
-C-C-H
| |
H H
e. ethyl methanoate HCOOCH
2CH[/sub]3[/sub]
H
|
O=C H H
| | |
O-C-C-H
| |
H H
Question 3a.  \to CH_3CH_2OH^+(g) + e^-)
or
 \to NO_2^+(g) + e^-)
[Note: or any correct split of the molecule. Most of you would've played the safe card and drew the parent molecular ion and stayed away from cleaving bonds.]
b. A. Impossible to have peaks at 45, 31, 29, 27, etc
c. NO
+ [Note: the positive charge is essential]
Question 4a. A, D [Note: quite sure about D, technically the O
- will react, and P=O will be of an equivalent state as the O
-, so if anything is going to react, it won't be the OH. But I have no idea how much VCAA would simplify organic chemistry, they may be targeting the point that a phosphoester bond is formed, thus you need an OH group.]
b.i. Hydrogen bonding
b.ii. C, F
Question 5a.i. Different volatile chemicals separate in the chromatograph, and produce distinct peaks. The peak corresponding to ethanol is separate from other volatile chemicals, and the measurement of this peak size will give a relative amount of ethanol.
a.ii. The peak at 0.9 minutes corresponds to ethanol.
b. 0.22% m/v
Question 6a. An enzyme's function is an effect of its structure. Its structure is usually optimized for binding to a particular substrate in a particular manner, thus only catalysing one particular reaction specific to this structure.
b. 55
oC, pH 5
c. Changing the pH can affect the tertiary structure (particularly ionic attractions between acidic and alkaline groups), affecting its structure and thus its activity.
d. 
,
 = 0.1 mg \times 300 \frac{mmol}{mg} \times \frac{1 mol}{1000 mmol} \times 180 \frac{g}{mol} = 5.4 g)
Question 7a.i. 3
a.ii.  = 3\times \frac{10000}{833} = 36.0 mol)
,
 = 1.15\times 10^3 g)
,
 = <br />1.46 L)
b.i.  = 6.00\times 10^{-3} mol)
,
 = 1.00\times 10^{-3} mol)
, thus 6 double bonds [7 if you include the C=O]
b.ii. C
22H
32O
2Question 8a. Ether
b.i. Carbohydrates, specifically a hexose monosaccharide.
b.ii. -OH
c.i. Oxidation
c.ii. Acidified Permanganate (H
+/MnO
4-) or Acidified Dichromate (H
+/Cr
2O
72-)
d. Ethanoic Acid or Acetic Anhydride, both are possible [Note, sulfuric acid is used as catalyst for both mechanisms]
e. The carboxyl group up the top loses H+, becomes COO-, otherwise the structure is the same. [tex]C_6H_4(COO^-)(OOCCH_3)[Note: cbf drawing]
Question 9a.i. Methanol or Ethanol
a.ii. Since the first fraction is collected at 97.2 degrees, this must be the lightest alkanol. Methanol/Ethanol boil before this temperature, thus if they were present, the first fraction would be collected at 64.5/78.3 degrees.
b. Alkanols are volatile and combustible. Alkanol vapours are in fact explosive. Using a naked flame (bunsen burner) is a serious hazard in dealing with these chemicals.
c. Propan-1-ol is overall polar, and the presence of OH also allows for hydrogen bonding to occur, the intermolecular attractions (dipole-dipole and H-bonding) is strong, thus it will have a high boiling point. Butane is overall non-polar, and the only intermolecular attraction is dispersion forces (weak), thus it will have a low boiling point. The difference between the strength of intermolecular forces is very large, polar forces are much stronger than dispersion forces, thus explaining the very large difference (~240K) between the BP of the two chemicals.
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