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Predictions for Chemistry 2014 HSC? (5 Viewers)

strawberrye

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I looked at your paper. You posted your answer to question 4 as D. Shouldn't it be B? In fact, I'm certain that it should be B. D accounts for sulfate only, whereas the question is asking "What mass of barium sulfate precipitate would be expected to be formed.."
Nicely spotted:) You are right:) (When I checked on my multiple choice answers, I just realised I erroneously typed B when I wrote on my answer sheet as D), sorry about this, I will fix it up right away:)
 

GOsie

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Nicely spotted:) You are right:) (When I checked on my multiple choice answers, I just realised I erroneously typed B when I wrote on my answer sheet as D), sorry about this, I will fix it up right away:)
Also, I'm just wondering... wouldn't 11 be A or D not B or D?
Larger volume = lower pressure. More moles of gas on LHS therefore equilibrium position shifts to LHS which would have more moles of CO2
 

strawberrye

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Similarly, I think question 10 should be C, since C2H3F2Cl can have all three heteroatoms on one side, two F on one side and Cl on the other, or one F on one side and an F and Cl on the other?
Or did I miss something again?
Short falls of making and reviewing the same paper by one person is that you could review it 100 times and still miss out on some mistake, seriously need to make sure to find someone review my paper next time. Yes you are right, it should be C-this is a genuine mistake-will fix it right now.

Also, I'm just wondering... wouldn't 11 be A or D not B or D?
Larger volume = lower pressure. More moles of gas on LHS therefore equilibrium position shifts to LHS which would have more moles of CO2
For question 11, provided that I didn't change the last option to be increasing the temperature of the system, the only right answer for the online version of the paper should be B. The reason is volume A would suggest increase in carbon dioxide concentration over the long run, which is not supported by the grap, B-means there is a sudden increase in concentration of all species, but Le Chaterlier's principle suggest system shift to the right to decrease pressure, option C shifts equilibrium to right, and option D-increasing temperature for an exothermic reaction would shift equilibrium to the right because the system wants to get rid of some heat.
 

GOsie

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For question 11, provided that I didn't change the last option to be increasing the temperature of the system, the only right answer for the online version of the paper should be B. The reason is volume A would suggest increase in carbon dioxide concentration over the long run, which is not supported by the grap, B-means there is a sudden increase in concentration of all species, but Le Chaterlier's principle suggest system shift to the right to decrease pressure, option C shifts equilibrium to right, and option D-increasing temperature for an exothermic reaction would shift equilibrium to the right because the system wants to get rid of some heat.
Ok I think I get that thanks :)
 

strawberrye

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Ok I think I get that thanks :)
Should correct a part of my comment, the increasing temperature shift equilibrium to the left, think of heat as a product, so more heat, shift to the left to absorb some of the heat, hence D can't be a correct response.
 

GOsie

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Should correct a part of my comment, the increasing temperature shift equilibrium to the left, think of heat as a product, so more heat, shift to the left to absorb some of the heat, hence D can't be a correct response.
I thought that in equilibrium, pressure changes and temperature changes were all gradual. I thought the only thing that would be an instantaneous change would be to add one of the reactants/products or to remove one. I'm a little confused could you please clarify?
 

strawberrye

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I thought that in equilibrium, pressure changes and temperature changes were all gradual. I thought the only thing that would be an instantaneous change would be to add one of the reactants/products or to remove one. I'm a little confused could you please clarify?
Pressure change would be instantaneous because all the reactants are gases.
 

SuchSmallHands

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eurghhh something to do with radioisotopes?
Yeah I think we're due to be assessed on the medicine and industry section. Not as a seven marker but just in general, I searched past papers as far back as 2003 (first year of new syllabus) and in those 11 years they have never assessed the cell of your own choosing that you study. Seems about time for that too.
 

GOsie

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Yeah I think we're due to be assessed on the medicine and industry section. Not as a seven marker but just in general, I searched past papers as far back as 2003 (first year of new syllabus) and in those 11 years they have never assessed the cell of your own choosing that you study. Seems about time for that too.
In all my assessments this year I expected for radioisotopes and battery cells. I studied them super hard the night before and knew all my equations, was more prepared for these than I'd been for any test in my life. After the topic test and half yearly didn't have, I was certain the trial would.

History of Haber process, water treatment, ozone, cellulose, ethanol, polymerisation. Bluurghh
 

GOsie

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Shipwrecks:
i) Name of method for removing salt from an artifact recovered from a wreck (1 mark)
ii) Explain, with reference to two examples, chemical procedures used to clean and preserve artifacts from wrecks (5 marks)

Non-shipwrecks people:
Outline the difference in chemical structure between the Oxygen molecule (O2) and the Ozone molecule (O3), and with reference to these, outline two differences in their chemical properties (4 marks)
Bumping because neither of these questions received answers.
 

IR

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Bumping because neither of these questions received answers.
Ozone: Greater dispersion forces and a bent structure (allowing polarity). Polar structure = hydrogen bonds allowing stronger bonds = greater melting and boiling point. Coordinate covalent bond in ozone allows greater reactivity than the covalent bond in oxygen. Oxygen= linear structure= only dispersion forces and that too weaker due to lower molecular weight= weaker intermolecular forces= lower melting and boiling points.
 

GOsie

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Ozone: Greater dispersion forces and a bent structure (allowing polarity). Polar structure = hydrogen bonds allowing stronger bonds = greater melting and boiling point. Coordinate covalent bond in ozone allows greater reactivity than the covalent bond in oxygen. Oxygen= linear structure= only dispersion forces and that too weaker due to lower molecular weight= weaker intermolecular forces= lower melting and boiling points.
Mostly correct stuff but the setting out should be improved on. The question starts by asking about the differences in structure so that's what you should begin with.

Oxygen is a non-polar molecule with a linear structure, with one double O=O covalent bond.
Ozone is a polar molecule which gives it its bent structure. It contains a coordinate covalent bond (O=O-O you should draw the molecules).

Then it asks for two differences in their properties:

Ozone's higher molar mass (3 Oxygen molecules instead of 2), means it has increased dispersion forces acting on it which gives it higher boiling and melting points than Oxygen.
Since Ozone is polar, it also has strong dipole-dipole forces between molecules , which makes it more soluble in water (also a polar molecule) than oxygen which is non-polar.

Other properties you could include:
Ozone is more reactive and is a stronger oxidising agent, which is due to its unstable structure.
Ozone is more easily decomposed than oxygen which is very stable. This is also due to ozone's unstable structure.

Hope this helps :)
Someone please correct me if I made any mistakes.
 
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