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November 01, 2025, 11:41:12 am

Author Topic: Chemistry 3/4 2013 Thread  (Read 448755 times)  Share 

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jacobacla

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Re: Chemistry 3/4 2013 Thread
« Reply #960 on: June 16, 2013, 11:54:08 am »
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Would anyone happen to know where to find a copy of the STAV 2013 solutions? Thanks!  :)
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Edward21

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Re: Chemistry 3/4 2013 Thread
« Reply #961 on: June 17, 2013, 04:57:23 pm »
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Question in GC, what does the height of the peak tell you? I know the area under the peak is directly proportional to the amount of substance present. See I was doing the KBT 2010 midyear, and MC Q10 asks which is the most concentrated compound on a chromatogram, saying that the most concentrated is the one with the highest peak, not the one with the largest area under the peak which I picked... are amounts of substance and concentration now different things with peak area and peak height assigned to each of these???
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Patches

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Re: Chemistry 3/4 2013 Thread
« Reply #962 on: June 17, 2013, 05:40:15 pm »
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Isn't it impossible to compare concentrations of different substances in the same sample, because the detector will detect each substance to a different extent?

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Re: Chemistry 3/4 2013 Thread
« Reply #963 on: June 17, 2013, 06:18:21 pm »
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Were forensic applications removed from the course? I thought they were, since they removed tertiary structure of DNA...

Alwin

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Re: Chemistry 3/4 2013 Thread
« Reply #964 on: June 17, 2013, 07:07:34 pm »
+2
Question in GC, what does the height of the peak tell you? I know the area under the peak is directly proportional to the amount of substance present. See I was doing the KBT 2010 midyear, and MC Q10 asks which is the most concentrated compound on a chromatogram, saying that the most concentrated is the one with the highest peak, not the one with the largest area under the peak which I picked... are amounts of substance and concentration now different things with peak area and peak height assigned to each of these???

Hmmm, to me I believe that KBT may have generalised. It is true that (especially with a calibration curve for reference) the peak area is directly proportional to the concentration. However, on some less "strict" exams, and I'm not accusing KBT of anything, they use peak height in the same way as peak area, eg plot peak height vs concentration for a calibration curve. The generalisation is that the bigger peak height means bigger peak area, akin to similar triangles. They use this method since often measuring the 'base' of the triangle can be tricky, or all the peaks have the same base width but different heights - clearly highest peak means greatest area.

I may have gone rather overboard here, but I've calculated the relative peak areas of each


As you can see, the highest peak has the largest peak area :D
Hence, the answer is B as stated in the answers.


Were forensic applications removed from the course? I thought they were, since they removed tertiary structure of DNA...

Removed. Electrophoresis and enzymes as disease markers are no longer examinable, however still appear in the textbook since it hasn't changed as the study design has changed (no mid-years). Hmmm, some techniques such as PCR are not related to tertiary structure of DNA, but are still off the course.
« Last Edit: June 17, 2013, 07:12:51 pm by Alwin »
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Edward21

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Re: Chemistry 3/4 2013 Thread
« Reply #965 on: June 17, 2013, 10:04:26 pm »
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Hmmm, to me I believe that KBT may have generalised. It is true that (especially with a calibration curve for reference) the peak area is directly proportional to the concentration. However, on some less "strict" exams, and I'm not accusing KBT of anything, they use peak height in the same way as peak area, eg plot peak height vs concentration for a calibration curve. The generalisation is that the bigger peak height means bigger peak area, akin to similar triangles. They use this method since often measuring the 'base' of the triangle can be tricky, or all the peaks have the same base width but different heights - clearly highest peak means greatest area.

I may have gone rather overboard here, but I've calculated the relative peak areas of each

(Image removed from quote.)
As you can see, the highest peak has the largest peak area :D
Hence, the answer is B as stated in the answers.


Removed. Electrophoresis and enzymes as disease markers are no longer examinable, however still appear in the textbook since it hasn't changed as the study design has changed (no mid-years). Hmmm, some techniques such as PCR are not related to tertiary structure of DNA, but are still off the course.
Oh wow, I see. If the triangles were close, would I use bh/2 like in maths to find the area? I have a feeling I may be screwed over by VCAA on something like this at the end of the year D:
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Ancora_Imparo

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Re: Chemistry 3/4 2013 Thread
« Reply #966 on: June 17, 2013, 10:08:07 pm »
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VCAA always makes things really clear from the outset. If it's a dodgy multiple choice, they may scrap the question altogether (this has happened numerous times in the past). If it's short-answer, as long as you correctly justify your answer, you'll be fine. And you probably wouldn't ever have to find areas of triangles...
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Edward21

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Re: Chemistry 3/4 2013 Thread
« Reply #967 on: June 17, 2013, 11:01:19 pm »
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VCAA always makes things really clear from the outset. If it's a dodgy multiple choice, they may scrap the question altogether (this has happened numerous times in the past). If it's short-answer, as long as you correctly justify your answer, you'll be fine. And you probably wouldn't ever have to find areas of triangles...
I hope so.. :) Btw, I love your forum name! I may even put it down as my Year 12 Yearbook quote  :D
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Re: Chemistry 3/4 2013 Thread
« Reply #968 on: June 18, 2013, 06:17:56 pm »
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just a few questions:

what is the oxidation number of vanadium in VO2NO3? ans: +4

Hydrochloric acid reacted with 10g of calcium carbonate to produce 2L of carbon dioxide gas at STP. This data indicates: what i think = d, ans :A
a)the calcium carbonate was in excess
b) the hydrochloric acid was in excess
c) the reaction did not go to completion
d) there were equal moles of reactants

A sample was prepared by diluting 20ml of a tomato sauce sample with water in a 250ml volumetric flask. The dilute solution was then analysed using AAS. The absorbance was found to be 0.116 (concentration = ~220mgL-1). The mass of sodium ions in the original 20ml tomato sauce sample is therefore:- ans : 58mg
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lzxnl

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Re: Chemistry 3/4 2013 Thread
« Reply #969 on: June 18, 2013, 07:21:00 pm »
+1
This compound is ionic; you have a nitrate ion NO3-, and VO2+
I'm fairly sure the answer should be +5 not +4

For that question, 22.4 L/mol, so if 2 L you have around 0.089 moles of carbon dioxide formed.
Calcium carbonate has a molar mass of roughly 100 g/mol so 10 g is 0.1 moles
n(CaCO3 reacted) = n(CO2) formed so the fact that less CO2 is formed than amount of CaCO3 initially suggests that not all of the CaCO3 reacted, i.e. excess.

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Re: Chemistry 3/4 2013 Thread
« Reply #970 on: June 18, 2013, 10:38:16 pm »
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oh i see thanks nliu. also the last one?

A sample was prepared by diluting 20ml of a tomato sauce sample with water in a 250ml volumetric flask. The dilute solution was then analysed using AAS. The absorbance was found to be 0.116 (concentration = ~220mgL-1). The mass of sodium ions in the original 20ml tomato sauce sample is therefore? ans : 58mg
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Edward21

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Re: Chemistry 3/4 2013 Thread
« Reply #971 on: June 18, 2013, 11:02:00 pm »
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oh i see thanks nliu. also the last one?

A sample was prepared by diluting 20ml of a tomato sauce sample with water in a 250ml volumetric flask. The dilute solution was then analysed using AAS. The absorbance was found to be 0.116 (concentration = ~220mgL-1). The mass of sodium ions in the original 20ml tomato sauce sample is therefore? ans : 58mg
The amount of Na+ is the same in the 20mL original, as in the diluted solution. Therefore is we know the concentration in the diluted solution, calculating as follows the amount per 250mL gives you the answer. Ie. 220mg/L well we have 0.250L (250mL of this) so 220*0.250= 55mg in the 250mL and therefore in the 20mL undiluted tomato sauce. If you read this off a calibration curve, you may have some slightly off results in reading the concentration values, but the 55mg is close enough to the 58mg? yes  ;)
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Re: Chemistry 3/4 2013 Thread
« Reply #972 on: June 19, 2013, 08:13:55 pm »
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This compound is ionic; you have a nitrate ion NO3-, and VO2+
I'm fairly sure the answer should be +5 not +4

But if the ionic compound is made up of NO3- and VO2+

If we consider the VO2+ by itself, O is -2 and the overall molecule is +2
Shouldn't that mean V is +4?

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Re: Chemistry 3/4 2013 Thread
« Reply #973 on: June 19, 2013, 08:25:19 pm »
+1
VO2+
Oxygen is -2
V is x
x-2*2=1
x=5
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Re: Chemistry 3/4 2013 Thread
« Reply #974 on: June 19, 2013, 08:46:41 pm »
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VO2+
Oxygen is -2
V is x
x-2*2=1
x=5

I thought it was VO with the 2+ on top

Sorry lol