Chemistry 3/4 2013 Thread

[lolipopper]

hi, i have a quick question.

Question 15.
When 0.10 g of magnesium is burnt in oxygen, 540 joule of energy is released. The closest
ΔH value for the combustion of magnesium will be, in kJ mol-1
A. 540
B. -130
C. -260
D. -540

For this I got the answer to be B however the answers say it is C. Their reasoning is that in the balanced equation for Mg combustion, the coefficient of Mg is 2 and thus the energy released must be multiplied by 2. How do you know you had to work out the ΔH for the reaction :
2Mg(s) + O2(g) -> 2MgO(s)
and not for
Mg(s) + 0.5O2(g) -> MgO(s).

And also when we usually work out the energy released by a specific amount of ethanol and use the VCAA Chemistry Data Book for the Molar Enthalpy, we dont multiply that by the coefficient so why in this case?

BTW this is from TSSM2010 U4.

[lzxnl]

hi, i have a quick question.

Question 15.
When 0.10 g of magnesium is burnt in oxygen, 540 joule of energy is released. The closest
ΔH value for the combustion of magnesium will be, in kJ mol-1
A. 540
B. -130
C. -260
D. -540

For this I got the answer to be B however the answers say it is C. Their reasoning is that in the balanced equation for Mg combustion, the coefficient of Mg is 2 and thus the energy released must be multiplied by 2. How do you know you had to work out the ΔH for the reaction :
2Mg(s) + O2(g) -> 2MgO(s)
and not for
Mg(s) + 0.5O2(g) -> MgO(s).

And also when we usually work out the energy released by a specific amount of ethanol and use the VCAA Chemistry Data Book for the Molar Enthalpy, we dont multiply that by the coefficient so why in this case?

BTW this is from TSSM2010 U4.

The answers are wrong then. The definition of the molar enthalpy of combustion is the enthalpy change of the combustion reaction of exactly ONE mole of whatever substance, and with all reactants and products in their standard states (so solid, gas, liquid, whatever phase they occur at 101.3 kPa and 298 K, and if solid, most stable allotrope, so for carbon the standard state would be graphite). So the molar combustion of magnesium would be the enthalpy change of Mg(s) + 1/2 O₂(g) => MgO(s)
The values in the VCAA chemistry data book agree with this convention. It's always for one mole of reactant in as much oxygen as needed.

[Limista]

how long should we be spending on multiple choice for the 2013 chem exam?

thanks 

[lzxnl]

I would generally not want to spend more than 35 minutes on the multiple choice section. It depends on the exam; I've done some that require 20 minutes and others that take up the full 35 minutes.

[Aurelian]

The answers are wrong then. The definition of the molar enthalpy of combustion is the enthalpy change of the combustion reaction of exactly ONE mole of whatever substance, and with all reactants and products in their standard states (so solid, gas, liquid, whatever phase they occur at 101.3 kPa and 298 K, and if solid, most stable allotrope, so for carbon the standard state would be graphite). So the molar combustion of magnesium would be the enthalpy change of Mg(s) + 1/2 O₂(g) => MgO(s)
The values in the VCAA chemistry data book agree with this convention. It's always for one mole of reactant in as much oxygen as needed.

To be fair the question says "calculate ∆H for the combustion of" not "calculate the molar enthalpy of combustion", so there's a level of ambiguity there. Poorly constructed question though...

[Jaswinder]

1) how does temperature affect the time taken for a sample to pass through a high performance liquid chromatography column?

2) how to draw four isomers of C4H9Br?

3) When asked to label the functional groups is Vitamin D and cholesterol, the answer suggests hydroxyl and alkene. If i label hydroxyl and alkyl, would that be correct too?

4) Write an equation for the conversion of propane to 1-propanol. With this what I wrote was C3H8 + H20 -> C3H7OH +H2, but in the answers they react propane first with chlorine and then with OH-. Why is that?

THanks

[Aurelian]

1) how does temperature affect the time taken for a sample to pass through a high performance liquid chromatography column?

2) how to draw four isomers of C4H9Br?

3) When asked to label the functional groups is Vitamin D and cholesterol, the answer suggests hydroxyl and alkene. If i label hydroxyl and alkyl, would that be correct too?

4) Write an equation for the conversion of propane to 1-propanol. With this what I wrote was C3H8 + H20 -> C3H7OH +H2, but in the answers they react propane first with chlorine and then with OH-. Why is that?

THanks

1) Higher temperature results in lower retention time.

2) Difficult to explain this in words... Just draw out as many skeletal structures of the isomers as you can see by moving around atoms.

3) Personally I don't really think "alkene" is a functional group at all, which I believe is what VCAA thinks too. So, just saying "hydroxyl" would be sufficient.

4) Addition of water relates to the conversions of alkenes to alcohols, not alkanes. Alkanes need to be first substituted with a halogen, such as Cl, by reacting that alkane with the halogen in its elemental form while exposed to UV light. The chlorine substituent is more reactive (a better "leaving group") than the hydrogen it replaces; as a result it's a lot easier to subsequently substitute that chlorine for a hydroxy substituent to finally give us an alcohol.

[lolipopper]

Question: what is the reasoning behind the fact that the Equilibrium constant doesn't change due to a change in concentration or amount, but does due to temperature change?

how long should we be spending on multiple choice for the 2013 chem exam?

thanks 

i usually tend to spend 10-15 minutes on them and then set aside 5 minutes for going over them.

[Stevensmay]

2) how to draw four isomers of C4H9Br?

Make sure to actually include the Hydrogen's and not just their bonds like I did.

https://www.dropbox.com/s/cfji5i1b19bjvug/2013-10-19%2012.58.52.jpg

[Jaswinder]

1) Higher temperature results in lower retention time.

3) Personally I don't really think "alkene" is a functional group at all, which I believe is what VCAA thinks too. So, just saying "hydroxyl" would be sufficient.

1) how does it lower the retention time?

3) This question was from a VCAA paper where it asked us to label 2 functional groups in is Vitamin D and cholesterol. would alkyl be right or does it have to be alkene?

https://www.dropbox.com/s/cfji5i1b19bjvug/2013-10-19%2012.58.52.jpg

Yeap I drew the same ones, however with the isomer in the bottom right corner, i put the Br in the methyl group would that still be right? :/

thanks guys

[Stevensmay]

Yeap I drew the same ones, however with the isomer in the bottom right corner, i put the Br in the methyl group would that still be right? :/

Yeah it is just a different way of drawing the one I have.

As we heat things up, the atoms have more kinetic energy, so they move around faster. It makes sense then that they would also move faster within the tube and thus exit it faster, resulting in a lower retention time.

[achre]

Can someone explain to me why C is the right answer? I just guessed B, but C seems wrong because if fragment A was 'larger' than B, it would be more negatively charged and move across the plate faster, right?

[Alwin]

Can someone explain to me why C is the right answer? I just guessed B, but C seems wrong because if fragment A was 'larger' than B, it would be more negatively charged and move across the plate faster, right?

1. electrophoresis sadly is no longer on the chem study design for 2013

2. electrophoresis works on a similarish (note the italics) to mass spec. A larger fragment does not mean that is has a greater charge. A proton can have a +1 charge while glycine in acidic environment also has +1 charge

3. Since you 'guessed' the ans, I won't go into lots of specifics of electrophoresis (see point 1) and leave it at: a larger fragment will take longer to move because it has a greater mass/charge ratio. Much like a very fat cat moves much slower towards a milk than a fast skinny one

hope it kind of makes sense =]

[simba]

Hey there!
I was just wondering what the main topics we covered this year that won't be assessed on the exam are. I've heard that we don't have to revise the industrial processes, but I'm not sure if that's true and if theres anything else we don't need to revise...
Any info would be appreciated

[achre]

Hey there!
I was just wondering what the main topics we covered this year that won't be assessed on the exam are. I've heard that we don't have to revise the industrial processes, but I'm not sure if that's true and if theres anything else we don't need to revise...
Any info would be appreciated

ahat produced a nice summary document of the new SD here. If you covered stuff that's not on the SD anymore (like fractional distillation or forensics), obviously that won't be on the exam. Industrial processes are exam assessible, but the specifics for a single chemical's production is not exam assessible. (Sulfuric acid, nitric acid, etc.)