VCE Chemistry Question Thread

[keltingmeith]

Hey
I'm a year 9 student (not VCE) but could someone simply yet in detail explain to me about chemical bond, nuclear decay and reaction rates?
Thankyou so much

I can, but those are all fairly lengthy topics. What do you want to know in particular about them?

Can someone please explain to me why the melting point of propane is lower then that of methane and ethane?
cheers

Reasons beyond VCE's understanding - it has to do with the degrees of freedom of movement in the molecules. See, methane has very few degrees of movement, ethane slightly more, and propane a LOT more. Because of this, methane and ethane can much more easily pack and form a crystal structure, and so more energy is required to break that crystal structure and melt the compound.

[Individu]

Hey guys, got a good question for all of us (which I personally struggled to figure out but I feel that others could benefit as well):

The concentration of Fe2+ in solution can be determined by titration with an acidified aqueous solution of potassium dichromate. The unbalanced equation for this reaction is: __Cr2O7 2-  +  __Fe 2+  +  __H+  --->  __Cr 3+  +  __Fe 3+  +  __H2O. When correctly balanced, the number of mole of Fe 2+ reacting with 1 mole of Cr2O7 2- is:
A) 1 mole
B) 3 mole
C) 6 mole
D) 14 mole

Now, I worked out the half-equations which are:

• Cr6+  +  3e- --->  Cr3+ and
• 3Fe2+  ---> 3Fe3+  +  3e-

The following is how I approached the question, and I was wondering if this is the correct method of working out this type of problem:

We can deduce from the information given and these half-equations that the no. of mole of Fe2+ reacting with Cr2O7 2- is 6 mole (answer C). This is because 3 mole of Fe2+ would be required to react with 1 mole of Cr atoms, as seen in the half-equations. However, since there are 2x the amount, in mol, of Cr atoms in a mole of Cr2O7 2-, this would mean that 1 mole of Cr2O7 2- would yield 2 mole of Cr atoms and thus we would require 6 mole of Fe2+.

Again, my question is: is this method of using the half-equations the proper approach to answering this type of question?

[HopefulLawStudent]

Could someone please explain how to do question 5 from the 2009 chemistry exam? I don't get it.

[Callum@1373]

Could someone please explain how to do question 5 from the 2009 chemistry exam? I don't get it.

The molecule with the longest retention time would adsorb to the stationary phase the best. As the stationary phase is non-polar, the answer is the most non polar molecule, which is option A

[HopefulLawStudent]

So it adsorbs more when the stationary phase and the mobile phase are both either polar/non-polar? This is probably a really dumb question but... How do we know that A's non-polar?

[Callum@1373]

So it adsorbs more when the stationary phase and the mobile phase are both either polar/non-polar? This is probably a really dumb question but... How do we know that A's non-polar?

No, this question is talking about a non-polar stationary phase. It would therefore interact more with the stationary phase if (a) it was a non-polar molecule and (b) the mobile phase was polar.

A is non-polar because it has a long hydrocarbon chain, whereas B,C and D have functional groups which will form hydrogen bonds therefore polar molecules.

[HopefulLawStudent]

Would it work the other way round too? i.e. If we had a polar molecule and non-polar mobile phase, would there be more interaction than if we had a non-polar mobile phase and non-polar molecule?

Ohhhhhh. I did not know that. We only briefly skimmed over polar/non-polar molecules and that section of organic chem.

PS: Thank you for all of your help. I really appreciate it.

[blacksanta62]

Hey need some help with this neap exam question

Triiodine ions can be produced in acidic solutions as follows:

IO_3^2- (aq) + I^-(aq) -----------> I_3^-(aq)

Write the balanced redox equation for the above reaction.


Edit: worked it out. Allow the triiodine ion to have an oxidation number of -1/3. From there you can see that the I from IO_3^2- (aq) reduces in its oxidation number (from +5 -------> -1/3) while the I^-(aq) increases in it's oxidation number (from -1 -----> -1/3). Work out the rest from there Thought I would do this since it wasn't so straight forward

[helloworld]

Does anyone have answers to thushans notes for chem on here?

[Adequace]

Can someone confirm that the relative formula mass of Fe2(CO3)3 is 291.714? The book has a different answer..

[Sine]

Can someone confirm that the relative formula mass of Fe2(CO3)3 is 291.714? The book has a different answer..

291.6gmol-1
292gmol-1 (3 sig figs)

Fe2 = 55.8 x 2
C3 = 12.0 x 3
O9 = 16.0 x 9

This is when using the vce databooklet.

[Adequace]

291.6gmol-1
292gmol-1 (3 sig figs)

Fe2 = 55.8 x 2
C3 = 12.0 x 3
O9 = 16.0 x 9

This is when using the vce databooklet.

Thanks, the book had something around ~297 lol.

[keltingmeith]

291.6gmol-1
292gmol-1 (3 sig figs)

Fe2 = 55.8 x 2
C3 = 12.0 x 3
O9 = 16.0 x 9

This is when using the vce databooklet.

Nitpick: you should leave that to one decimal place, because you're adding the numbers together, not multiplying/dividing.

[Sine]

Nitpick: you should leave that to one decimal place, because you're adding the numbers together, not multiplying/dividing.

oh forgot about that good pickup.

[Shax]

Guys, the textbook has a compound named as "2-Methylpropene".
I think it is redundant to put the 2 there since it doesn't add anything. Methylpropene is just enough as the numbering always starts from the double bond and the methyl group cannot be attached anywhere else (if it was one or three, it would be called butene).
I have heard it said that vcaa assessors consider putting a number that is not required in the IUPAC name as a mistake. What do you think guys??