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Suggested Solutions to the VCAA 2011 Chemistry Exam 1. Edward Cliff Multiple Choice Questions 1A – ethane – can only form dispersion forces. 2B – chlorines must be attached to 2 and 3 carbons. 3C – II and III are both 2,2,4-trimethylpentane 4C – 2,4-dimethylpentane only has 7 carbons. 5A – amide bonds are between amino acids and disulfide bonds are located between cysteine molecules. 6D – all amino acids are capable of acting as a base. 7C – substitution reaction (H is substituted for Br). Molecule X is 1,1,2trichloroethane. Always number to give lowest numbering. 8D – same principle as the acetyl salicylate ion; the ibuprofen lysine’s ability to form ion-dipole bonds makes it more soluble than ibuprofen. 9D – glycerol is a product and ethanol is a reactant (excess) in the conversion of canola oil to biodiesel. KOH is used to help formation of the biodiesel. 10B – 2 CO2 molecules are produced for every CH4 molecule. 11C – a weak monoprotic acid and a strong monoprotic acid of the same concentration will both neutralise the same quantity of base. 12C – methyl red turns yellow, which indicates pH > 6.3, and phenolphthalein stays colourless which indicates pH < 8.3. 13B – for every 2 mole of gypsum heated, 3 mole of water were given off. 14D – SO2 – use d=m/V, n=V/Vm and n=m/M 15A – 97.9g – use pV=nRT and n=m/M 16B – 81,018 gmol-1 - every glycosidic linkage (499 of them) was formed by condensation and hence represents loss of a water molecule. 17A – Z is more strongly adsorbed than W (spends more time adsorped to stationary phase) and has a lower Rf value (does not travel as far). 18B – atomic mass is key in determining how much IR energy is required to excite bonds. See page 90 of Heinemann Chemistry 2 for a more detailed explanation 19D – chromatography is used to separate the mixture and then mass spectrometry can be used to identify its components. 20B – Only I and III are correct. Copper undergoes reduction in the flame.

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Short Answer Questions 1 AND

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a) A b) H – polysaccharide c) D – saturated fatty acid. E – glycerol d) F – methyl ester e) B – glucose f) F – 0.32 g of bromine (Br2) = 0.002mol. Therefore the molecule reacts in a 1:2 ratio with bromine, which means it must have two double bonds. a) i) [CH3CH2]+ (Note: there must be a + sign) ii) 79Br is slightly more abundant than 81Br. The higher the peak, the more abundant the isotope as a greater percentage of molecular ions contain this isotope.

Br

b) i)

H

Br

C

C

H

H

H

Br

H

H

C

C

Br

H

H

ii) There are two chemical environments are two main peaks are present. There is a quartet present at approximately 5.8ppm on the NMR spectrum, and a doublet present at approximately 2.4ppm. The quartet splitting pattern is caused by the CH3 methyl group present in 1,1-dibromoethane (but represents the H adjacent to the bromines) and the doublet is caused by the single H adjacent to the bromines. 1,2-dibromoethane would produce only one peak as all four H atoms are in the same chemical environment. 3

a) 130ppm. Read off the graph. b) On the chromatogram of soft drink B, there is no visible peak at a RT value of 96 seconds, the characteristic retention time of caffeine, indicating that soft drink B does not contain caffeine. c) i) The open-ended peak indicates that peak area is greater than the scale can take into account; this means we cannot read the peak area off the graph and therefore cannot determine caffeine content ii) Dilution to reduce caffeine concentration e.g. 10 times

4

a) calculation questions – will type these up later tonight b) calculation questions – will type these up later tonight a) i) 2H2O (l) + TeO2 (s)  H2TeO4 (aq) + 2H+ (aq) + 2eii) and b) calculation questions – will type these up later tonight

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a) i) C – T – A – T - G ii) 12 hydrogen bonds b) phosphate, deoxyribose sugar and *adenine base a) i) NaOH (aq) H H

C

H

H H

H C O

C

C H

H

H

H

8)

H

C

ii) H iii) 3-methylbutan-1-ol iv) ethanol v) H2SO4 (l) vi) reaction II b) A mixture of (often organic) compounds is heated A fractionating column, containing glass beads, is used to separate the gases based on volatility As the gases move up the column and away from the heat source, the least volatile gases will recondense and move down the column, whilst the most volatile gas/es will continue to rise A condenser (e.g. running cold water) is used to condense the most volatile component and remove it from the mixture as distillate If the gas is not concentrated enough, we can repeat this to increase concentration. c) The infrared spectrum of the banana oil obtained after distillation has no broad peak at 3300 cm-1 which indicates an absence of the –OH (hydroxyl) functional group. This indicates that only the ester, 3-methylbutylethanoate, is present. a) Can’t see the bonds on the scanned paper. Please email answer to contact@connecteducation.com.au b) carboxyl -COOH and amine -NH2 c) Ensure all bonds are shown, including O-H. Glycine at pH=1 H

H

H

+

N

C

H

H

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O C O

H


VCAA Chemistry 2011 Exam 1 Suggest Solutions