Chemistry 3/4 2013 Thread

[SocialRhubarb]

There's a C=C where there should be a C=O.

[clıppy]

There's a C=C where there should be a C=O.

Even after you said it, it was still hard to find.

[lzxnl]

This isn't the worst VCAA have done. Remember the giant robot in the Revs exam last year?
Or better still, giving integrals where the integrand isn't even defined for half of the integration domain?

I remember once VCAA gave the reaction of hydrogen and fluorine gas to form hydrogen fluoride as having an equilibrium constant of 313, but I did the calculations from some data I found on the internet, and the equilibrium constant was on the order of e^100 depending on which reaction you looked at (i.e. 0.5 H2 + 0.5 F2 => HF or H2 + F2 => 2HF). Come on.

VCAA 2009, question 5 MC

[clıppy]

This isn't the worst VCAA have done. Remember the giant robot in the Revs exam last year?
Or better still, giving integrals where the integrand isn't even defined for half of the integration domain?

I remember once VCAA gave the reaction of hydrogen and fluorine gas to form hydrogen fluoride as having an equilibrium constant of 313, but I did the calculations from some data I found on the internet, and the equilibrium constant was on the order of e^100 depending on which reaction you looked at (i.e. 0.5 H2 + 0.5 F2 => HF or H2 + F2 => 2HF). Come on.

VCAA 2009, question 5 MC

o.o
I am not looking forward to starting my first VCAA exam anytime soon

[lolipopper]

i got three questions:

Q1. out of propanoic acid and methyl propanoate, which reacts with a base?
 the way i saw it was that propanoic acid being an acid will, however doesn't methyl propanoate react with NaOH to reform the reactants methanol and propanoic acid as well?

Q2. are the Lisa Chem chemistry exams supposed to be much harder than the ones for VCAA?

Q3. im getting only 1 or 2 marks off for the insight chemistry exams, i am rank 1 and am aiming for 40 or above. Is it possible? (be honest)

[lzxnl]

i got three questions:

Q1. out of propanoic acid and methyl propanoate, which reacts with a base?
 the way i saw it was that propanoic acid being an acid will, however doesn't methyl propanoate react with NaOH to reform the reactants methanol and propanoic acid as well?

Q2. are the Lisa Chem chemistry exams supposed to be much harder than the ones for VCAA?

Q3. im getting only 1 or 2 marks off for the insight chemistry exams, i am rank 1 and am aiming for 40 or above. Is it possible? (be honest)

Q1. Methyl propanoate will react with sodium hydroxide solution as well...the base here is a necessary part of the reaction. I think the question is dodgy. And technically, you may not form propanoic acid as it's in basic solution; propanoate ion?

Q2. Comparing Lisa Chem exams to Checkpoints, they're roughly the same I would have thought. I personally haven't noticed many difficulty discrepancies in chemistry exams as there are in maths.

Q3. Yeah that's a pretty safe aim.

[Holmes]

Neap question

A colourless, odourless gas is evolved at the anode when a 0.60 M solution of potassium fluoride is electrolysed for 5.0 mins with a curren flow of 4.5 A. Determine the volume of gas evolved at SLC.

Firstly, the solutions indicate that one has to assume that the anode half equation is that of the oxidation of H20!!! How?!?

Secondly, I calculated the mol of electrons based on the potassium reduction equation, then MULTIPLIED this by 4 to get the mol of electrons for the 2H20 --> 02 + 4 H+ + 4e . The answers say that to get the correct answer, one has to DIVIDE the mol of electrons from the potassium reduction by 4.

Thanks so much guys

[jono88]

Is there a mistake in the 2013 Lisachem Trial Exam Unit 3/4?
Q1a) of Section B asks to write the empirical formula of caffeine from the diagram above.
Molecular forumla is C8H10N4O2
so shouldn't the empirical be C4H5N2O instead of C4H5N4O2 as stated in the solutions? O.O

[lolipopper]

Neap question

A colourless, odourless gas is evolved at the anode when a 0.60 M solution of potassium fluoride is electrolysed for 5.0 mins with a curren flow of 4.5 A. Determine the volume of gas evolved at SLC.

Firstly, the solutions indicate that one has to assume that the anode half equation is that of the oxidation of H20!!! How?!?

Secondly, I calculated the mol of electrons based on the potassium reduction equation, then MULTIPLIED this by 4 to get the mol of electrons for the 2H20 --> 02 + 4 H+ + 4e . The answers say that to get the correct answer, one has to DIVIDE the mol of electrons from the potassium reduction by 4.

Thanks so much guys

hey Holmes:

being an aqueos solution the possible reactants are K+, F- and H2O. now you have to select the strongest oxidant (highest on right) and the strongest reductant (lowest on left). You will notice that H2O is the strongest reductant and strongest oxidant. thus the 2 equation for the electrolysis are:

At the anode : O2 + 4H+ + 4e- => 2H2O (reverse this reaction)
At the cathode : 2H20 + 2e- = H2 + 2OH-

thus considering the anode O2 will be the gas produced as it is both odorless and colorless.

Now for the second part.
1. calculate Q = (5 x 60)s x 4.5A = 1200C
2. calculate n(e-) = 1200C/96500 = 0.012435mol
3. calculate n(O2) = n(e-)/4 = 0.0031088mol (there are 4 mol of e- for every mol of O2. thus divide by 4 and not multiply)
4. now using n=cv calculate volume = 0.0031088/0.6 = 0.0051813L

hope it helps.

[lolipopper]

Is there a mistake in the 2013 Lisachem Trial Exam Unit 3/4?
Q1a) of Section B asks to write the empirical formula of caffeine from the diagram above.
Molecular forumla is C8H10N4O2
so shouldn't the empirical be C4H5N2O instead of C4H5N4O2 as stated in the solutions? O.O

i think you are correct because the proportion of atoms in the molecular formula should be a multiple of the proportion of atoms in the empirical formula. Although i haven't done that exam yet, im pretty sure they made that mistake.

[lzxnl]

Neap question

A colourless, odourless gas is evolved at the anode when a 0.60 M solution of potassium fluoride is electrolysed for 5.0 mins with a curren flow of 4.5 A. Determine the volume of gas evolved at SLC.

Firstly, the solutions indicate that one has to assume that the anode half equation is that of the oxidation of H20!!! How?!?

Secondly, I calculated the mol of electrons based on the potassium reduction equation, then MULTIPLIED this by 4 to get the mol of electrons for the 2H20 --> 02 + 4 H+ + 4e . The answers say that to get the correct answer, one has to DIVIDE the mol of electrons from the potassium reduction by 4.

Thanks so much guys

What do you think is meant to be oxidised at the anode? Fluoride ion? It's all the way at the top...the weakest reductant in the electrochemical series. Water will be much more readily oxidised than fluoride; strongest reductant always reacts with the strongest oxidant, regardless if it's galvanic or electrolytic.

Secondly, you do NOT reduce potassium. You just can't. It's a VERY, VERY weak oxidant and is far weaker than water. So essentially, you end up reducing and oxidising water in this cell.

Is there a mistake in the 2013 Lisachem Trial Exam Unit 3/4?
Q1a) of Section B asks to write the empirical formula of caffeine from the diagram above.
Molecular forumla is C8H10N4O2
so shouldn't the empirical be C4H5N2O instead of C4H5N4O2 as stated in the solutions? O.O

Mistakes happen everywhere in trial exams -.-

[Holmes]

being an aqueos solution the possible reactants are K+, F- and H2O. now you have to select the strongest oxidant (highest on right) and the strongest reductant (lowest on left).

Sorry for being pedantic, I'm sure it was just a mistyping thing, as the strongest oxidant would be the highest on the left, the strongest reductant the lowest on the right.

Now for the second part.
1. calculate Q = (5 x 60)s x 4.5A = 1200C
2. calculate n(e-) = 1200C/96500 = 0.012435mol
3. calculate n(O2) = n(e-)/4 = 0.0031088mol (there are 4 mol of e- for every mol of O2. thus divide by 4 and not multiply)
4. now using n=cv calculate volume = 0.0031088/0.6 = 0.0051813L

Also, sorry for not posting the answer in my post, the V(02) is 86 mL. I think the concentration was just a distractor in the question, because they used n=v/vm for the volume. (also, (5 x 60)s x 4.5A = 1200 1350C)

What do you think is meant to be oxidised at the anode? Fluoride ion? It's all the way at the top...the weakest reductant in the electrochemical series. Water will be much more readily oxidised than fluoride; strongest reductant always reacts with the strongest oxidant, regardless if it's galvanic or electrolytic.

Secondly, you do NOT reduce potassium. You just can't. It's a VERY, VERY weak oxidant and is far weaker than water. So essentially, you end up reducing and oxidising water in this cell.

Yeah, this makes sense. I thought that as the question didn't mention 'solution', there was only potassium fluoride in the cell. Is it the question's poor wording, or how am I supposed to recognise that there's also water in there?

[clıppy]

strongest reductant always reacts with the strongest oxidant, regardless if it's galvanic or electrolytic.

I was under the assumption that it changed depending on the type of cell and now I'm a bit lost. Can you clarify more on that?

[lzxnl]

I was under the assumption that it changed depending on the type of cell and now I'm a bit lost. Can you clarify more on that?

This NEVER changes. What changes is the polarity of the anode and cathode.

[clıppy]

This NEVER changes. What changes is the polarity of the anode and cathode.

Ahh okay, thanks nliu.