Chemistry 3/4 2013 Thread

[zvezda]

One mole of methane will release 900kJ, converting one mole of liquid water to steam requires 44 kJ.

So 900kJ/mol divided by 44kJ is 20.45 mol.

That means you can convert 20.45 mol of water to steam.

20.45 mol * 18.0 g/mol
=368 g.

And the solutions are here
http://www.cea.asn.au/sites/default/files/sample_paper_answers_2013.pdf

Has anybody noticed that for q7b these solutions contradict the vcaa assessor's report?
Here they say that "non-standard conditions" is an answer but the response that is given in these solutions is deemed an answer "that shows no understanding" from the vcaa 2009 report.

[Scooby]

Has anybody noticed that for q7b these solutions contradict the vcaa assessor's report?
Here they say that "non-standard conditions" is an answer but the response that is given in these solutions is deemed an answer "that shows no understanding" from the vcaa 2009 report.

What VCAA were looking for was an explanation of why the conditions were non-standard (so how exactly the non-standard conditions lead to a reduced reaction rate). It's fine to use it as an answer as long as you refer to something specific, rather than just saying "oh, yeah, conditions were non-standard, and that explains everything"

[zvezda]

What VCAA were looking for was an explanation of why the conditions were non-standard (so how exactly the non-standard conditions lead to a reduced reaction rate). It's fine to use it as an answer as long as you refer to something specific, rather than just saying "oh, yeah, conditions were non-standard, and that explains everything"

Ahh yeah i see the distinction. Lol.
Did you or anyone also notice how i they mentioned that the fact that the HCl was a limiting reagent also prevented a valid conclusion?
I can see how this would influence validity, but say that all variables were constant except concentration and limiting reagent in each flask, that shouldnt matter should it? Its only the rate of reaction thats being measured so one could just measure the rate up until a certain mass of CaCO3 has been lost?

[ineedhelp2]

Hello everyone,

I get extremely confused with electrolysis questions particularly the ones similar to Question 10 MCQ in 2011 Chemistry Exam 2.

Link: http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2011chem2-w.pdf

I now understand that the strongest oxidant has the highest E value and the strongest reductant has the lowest E value. But how would you go about doing this MCQ question?

Thankyou

[MonsieurHulot]

Has anybody noticed that for q7b these solutions contradict the vcaa assessor's report?
Here they say that "non-standard conditions" is an answer but the response that is given in these solutions is deemed an answer "that shows no understanding" from the vcaa 2009 report.

Also multiple choice Q21 is A according to the VCAA but B according to CEA. Which is it, anyone?

[jgoudie]

Well, this is a galvanic cell question so that might be where some of the confusion is coming from.  But anyway here goes.

From the first cell you have that the ammonia half cell is the cathode, thus this is reduction so where for the E1 must be larger than E3.  (as Ag(NH₃)₂⁺) is a stronger oxidant).

From the second cell you have that Ag(s) half cell is the cathode, thus E2, must be higher than E1.  (as Ag⁺ is the stronger oxidant).

This information together makes E2>E1>E3 which is C.

When given this type of information it is best to try and create your own little E.C. Series, for a galvanic cell the cathode/reduction/positive electrode's reaction will be higher in the E.C. than the negative.

Hope this is clear enough.

Hello everyone,


I get extremely confused with electrolysis questions particularly the ones similar to Question 10 MCQ in 2011 Chemistry Exam 2.

Link: http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2011chem2-w.pdf

I now understand that the strongest oxidant has the highest E value and the strongest reductant has the lowest E value. But how would you go about doing this MCQ question?

Thankyou

[lzxnl]

Another way you COULD do this question is note that the positive terminal will have the higher E nought value. Think about it; higher E nought means stronger oxidant means more readily reduced=>positive cathode. This would simplify your task.

[lmerrett]

Just wondering how you figure out the order of efficiency of energy sources ie. galvanic cells, fuel cells, electrolytic cells, nuclear fission, brown coal, black coal, photovoltaic cell, wind power, etc. there are so many!
Also how do you figure out the energy conversions and what types of energy would we need to know the conversions for? Thankyouuuuuuu

[barydos]

http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2008chem2-w.pdf
For question 3e)
What would be a good response to obtain the explanation mark?
Can I just say, acid IV is weak acid while acid I is a strong acid?

[presto]

Sorry, I'm not so sure if this post belongs here, but:
does anyone have any predictions on what the A+ cutoff will be for the chem exam this year?
Thanks!

[ECheong]

Just wondering how you figure out the order of efficiency of energy sources ie. galvanic cells, fuel cells, electrolytic cells, nuclear fission, brown coal, black coal, photovoltaic cell, wind power, etc. there are so many!
Also how do you figure out the energy conversions and what types of energy would we need to know the conversions for? Thankyouuuuuuu

As a general rule of thumb, the less number of steps you need to get to the energy you 'want', the more efficient it is. This is because each conversion will inevitably lose to loss of energy (usually to heat). So, using this general rule of thumb.

Most efficient (in terms of electrical energy produced per mass, which is typically what we'll need to deal with)
Nuclear (think nuclear bomb, lots of energy from tiny source)
Galvanic cells = fuel cells (as far as I know since they're both essentially the same thing)
Black coal
Brown Coal (brown coal has more moisture in it than black coal, some energy is wasted heating the water into a gas during burning)
Least efficient

as for electrolytic cells, they work in reverse; electrical energy is converted into chemical energy

http://www.vcaa.vic.edu.au/Documents/exams/chemistry/2008chem2-w.pdf
For question 3e)
What would be a good response to obtain the explanation mark?
Can I just say, acid IV is weak acid while acid I is a strong acid?

I personally would just go and say that since for 0.10M of Acid(IV) gives a pH=2.1 we can say that it doesn't fully dissociate and hence is in equilibrium. In contrast, Acid(I) does fully dissociate (0.1M giving pH=1). Hence, when diluted, Acid(IV) can increase its percentage ionisation through LCP and Acid(I) cannot. In other words, Acid(IV) solution can offset the dilution and 'recover' a bit of the pH change whereas Acid(I), not being in equilibrium, cannot. Therefore the pH change will be greater for Acid(I) than Acid (IV)

[barydos]

As a general rule of thumb, the less number of steps you need to get to the energy you 'want', the more efficient it is. This is because each conversion will inevitably lose to loss of energy (usually to heat). So, using this general rule of thumb.

Most efficient (in terms of electrical energy produced per mass, which is typically what we'll need to deal with)
Nuclear (think nuclear bomb, lots of energy from tiny source)
Galvanic cells = fuel cells (as far as I know since they're both essentially the same thing)
Black coal
Brown Coal (brown coal has more moisture in it than black coal, some energy is wasted heating the water into a gas during burning)
Least efficient

as for electrolytic cells, they work in reverse; electrical energy is converted into chemical energy

I personally would just go and say that since for 0.10M of Acid(IV) gives a pH=2.1 we can say that it doesn't fully dissociate and hence is in equilibrium. In contrast, Acid(I) does fully dissociate (0.1M giving pH=1). Hence, when diluted, Acid(IV) can increase its percentage ionisation through LCP and Acid(I) cannot. In other words, Acid(IV) solution can offset the dilution and 'recover' a bit of the pH change whereas Acid(I), not being in equilibrium, cannot. Therefore the pH change will be greater for Acid(I) than Acid (IV)

Thanks ECheong, you're a beast at chem.

[neonperson]

For the VCAA sample exam, 
http://www.vcaa.vic.edu.au/Documents/exams/chemistry/chem-specs-samp-w.pdf

Question 5, to circle the methionine residue. It is the amino acid at the very end. However, the CEA answers do not have the COOH group circled. Why not? Isn't it also part of methionine?

[ch]

Can someone help explain?
fuel cell based on the oxidation of methane, the equation for the anode half reaction is
CH4(g) + 2H2O(l) → CO2(g) + 8H+(aq) + 8e–

Question 28
Assuming that all the energy of the oxidation reaction is converted to electricity, the amount of electric
charge, in coulomb, obtained from the oxidation of one mole of methane is closest to
A. 8 × 102
B. 1 × 103
C. 8 × 105
D. 1 × 106

[Scooby]

Can someone help explain?
fuel cell based on the oxidation of methane, the equation for the anode half reaction is
CH4(g) + 2H2O(l) → CO2(g) + 8H+(aq) + 8e–

Question 28
Assuming that all the energy of the oxidation reaction is converted to electricity, the amount of electric
charge, in coulomb, obtained from the oxidation of one mole of methane is closest to
A. 8 × 102
B. 1 × 103
C. 8 × 105
D. 1 × 106

Oxidation of one mole of methane produces eight moles of electrons

The charge on eight moles of electrons is given by Q = F x n(e^-), where the number of moles of electrons is eight

Therefore Q = 96500 x 8 = 772000 C, which to one significant figure is 8 x 10⁵

So it's C