HSC 2012-2015 Chemistry Marathon (archive) (1 Viewer)

Ekman · Oct 22, 2015

re: HSC Chemistry Marathon Archive

rand_althor said:
Not sure if this question has popped up before:

Compare the chemical processes involved in the production of polyethylene and a named biopolymer (not cellulose) and evaluate the effect on society and the environment of both types of polymers. (7 marks)

This exact question popped up in my trials. Where did you get it from?

Ekman · Oct 22, 2015

re: HSC Chemistry Marathon Archive

Also can someone please re-iterate the rules about sig figs? If the data gives 0.001molL-1, does that mean it has 1 sig fig?

rand_althor · Oct 22, 2015

re: HSC Chemistry Marathon Archive

Ekman said:
This exact question popped up in my trials. Where did you get it from?

It was in my trials too.

Mr_Kap said:
I really dont know how to get 7 marks. Like i can only find differences between them not similarities.

Can you put in dot points what you need to mention for full marks.

The marking criteria (6-7 marks):
- Describes ONE or TWO similarities between the chemical processes involved in the production of polyethylene and the named biopolymer
- Describes at least TWO differences between the chemical processes involved in the production of polyethylene and the named biopolymer
- Describes an impact of the use of each polymer on society
- Describes an impact of the use of each polymer on the environment
- Make a clear judgement about the use of each type of polymer on society and the environment
- Answer is cohesive, comprehensive and demonstrates a thorough understanding of the science of the chemical processes involved

Mr_Kap · Oct 22, 2015

re: HSC Chemistry Marathon Archive

Ekman said:
Also can someone please re-iterate the rules about sig figs? If the data gives 0.001molL-1, does that mean it has 1 sig fig?

shit. i forgot sig figures rules.Someone help!!!

and Is 1.00 mol/L 1 sig fig or 3 sig fig???

Crisium · Oct 22, 2015

re: HSC Chemistry Marathon Archive

Ekman said:
Also can someone please re-iterate the rules about sig figs? If the data gives 0.001molL-1, does that mean it has 1 sig fig?

I brought it up with Fizzy at one point in time

You round your answer off to the lowest amount of significant figures in the question (e.g. There's 90 kg, 555 N so you choose the number of sig figs in the 90 kg) BUT that value actually has to be used in the actual calculation - My school got sneaky with the half-yearlies where they included some BS value that wasn't included in the calculation and heaps of people rounded it off to the amount of significant figures it had (since it had the lowest)

Mr_Kap · Oct 22, 2015

re: HSC Chemistry Marathon Archive

what about 0.001 mol/L. How many sig figures is that..

and what about 1.00 mol/L how many sig figures is that

Crisium · Oct 22, 2015

re: HSC Chemistry Marathon Archive

Mr_Kap said:
what about 0.001 mol/L. How many sig figures is that..

and what about 1.00 mol/L how many sig figures is that

LOL

Oh you're talking about the general sig fig rules?

You only begin counting sig figs after the zero's so it's 1 sig fig in the first thing

For the second thing the .00 tells us something interesting after the 1 so it's 3 sig figs

InteGrand · Oct 22, 2015

re: HSC Chemistry Marathon Archive

Ekman said:
Also can someone please re-iterate the rules about sig figs? If the data gives 0.001molL-1, does that mean it has 1 sig fig?

Yeah that's 1 sig. fig.

Crisium · Oct 22, 2015

re: HSC Chemistry Marathon Archive

http://www.purplemath.com/modules/rounding2.htm

Have a quick skim over the examples

Good luck for chem everyone

I'm gonna head off and get some pracs memorised

InteGrand · Oct 22, 2015

re: HSC Chemistry Marathon Archive

$Note however that the real rule for taking logs to the base 10 with sig. figs is to apply the typical sig. fig. to the \textsl{mantissa} of the answer (i.e. the part after the decimal point of the logarithm's value). So say we're given a concentration of $\text{H}^+$ of $0.050$ mol/L (i.e. to 2 sig. fig.). This means $\text{pH}=-\log_{10}(0.050)=1.301029996....$, and the proper way to round this is to say $\text{pH}\approx 1.30$ (i.e. report the fractional part (the part after the decimal point) to the no. of sig. figs given in the data).$

$The reason for this is that the only uncertainty comes about from the fractional part. Note that any piece of data $d>0$ given to us can be written (uniquely) in the form $d=a\times 10^k$, where $k$ is an integer, and $a$ is a real number satisfying $0<a<10$. There is no uncertainty in the integer $k$, all the uncertainty in the data $d$ is captured in $a$. If we now take a log base 10, we have $\log_{10} d = \log_{10}\left( a\times 10^k \right) = \log_{10}a + \log_{10} 10^k = \underbrace{k}_{\text{no uncertainty}} + \underbrace{\log_{10}a}_{\text{all uncertainty is here}}$, and so we just round $\log_{10}a$ to the required no. of sig. figs, as it is \emph{this} that has the uncertainty.$

$But note that as $0<a<10$, we have $0< \log_{10}a < 1\Rightarrow \log_{10}a$ is just the fractional part of $\log_{10} d$ !! In other words, apply the rule for rounding \emph{only} to the fractional part, i.e. just round the part after the decimal point to the appropriate sig. figs amount. This explains why for logs (to base 10), only round the mantissa.$

InteGrand · Oct 22, 2015

re: HSC Chemistry Marathon Archive

$In summary, if our datum $d$ has $N$ significant figures, then $\log_{10} d$ should be reported to $N$ decimal \textsl{places} (the integer part of the log indicates the order of magnitude, and is not a significant figure, so we only round the part after the decimal point).$

See: http://www.laney.edu/wp/cheli-fossum/files/2011/01/Significant-Figure-Rules-for-los.pdf

Loudvicuna · Oct 22, 2015

re: HSC Chemistry Marathon Archive

Screen Shot 2015-10-02 at 9.32.47 PM.png

Anyone?

kawaiipotato · Nov 1, 2015

re: HSC Chemistry Marathon Archive

Loudvicuna said:
View attachment 32586

Anyone?

Le Chatelier's Principle (LCP) states that a disturbance experienced by an equilibrium will cause the equilibrium to shift, in order to minimise the disturbance. In step 1, the mole ratio of gases of reactants to products is 9:10 and thus, to maximise the yield, a low pressure is used. With a lower pressure, the equilibrium will favour the products as it has more gaseous moles and shift to produce more nitric oxide used in step 2. Since it is also the most highly exothermic reaction in the process, a relatively higher temperature can be used (to maximise reaction rate) as the equilibrium will not shift backwards too substantially. In step 2, the use of a high pressure causes the equilibrium to shift forward, to the less gaseous moles and producing more nitrogen dioxide. The use of lower pressure (lower than step 1 but still moderate) is used to balance out the reaction rate and the yield produced (since equilbrium will shift forward). In step 3, the use of high pressure is also used, as the mole ratio of raectants to products is 3 to 1. This will allow the equilibrium to shift forward to produce more of the required nitric acid. The use of a moderate temperature is also used to balance out reaction rate and yield.
The optimising of step 1 helps optimise step 2 due to the increased concentration of nitric oxide and also helps optimise step 3 due to increase in nitrogen dioxide.

anomalousdecay · Nov 1, 2015

Re: HSC Chemistry Marathon Archive

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HSC 2012-2015 Chemistry Marathon (archive) (1 Viewer)

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