VCE Chemistry Question Thread

[lArcdeTriomphe]

No, you do not have to understand the process of fractional distillation - it was part of the previous study design, but has now been removed.

Having said that, knowing its principles won't necessarily hurt.

[Kel9901]

Anyone know whether the process of fractional distillation is required to understand in 3/4 chem?

You need to be able to state 'fractional distillation' (e.g. a question like 'what is a method that can be used to separate the ester from the reactants and water in an esterification reaction?), but you don't need to know the principles

[FarAwaySS2]

I need help with this question.

What would be the sulfate ion concentration in a solution prepared by mixing 50.0 mL of 0.100 M sodium sulfate and 25.00 mL of 0.250 M aluminium sulfate solutions?

The answer is 0.317 M. I keep getting it wrong.

Help would be much appreciated!

[lzxnl]

I need help with this question.

What would be the sulfate ion concentration in a solution prepared by mixing 50.0 mL of 0.100 M sodium sulfate and 25.00 mL of 0.250 M aluminium sulfate solutions?

The answer is 0.317 M. I keep getting it wrong.

Help would be much appreciated!

50.0 mL of 0.100 M sodium sulfate contains 0.05*0.1 mol sodium sulfate, or 0.005 mol sodium sulfate, which is 0.005 mol sulfate ions (molecular formula of sodium sulfate contains only 1 sulfate ion). 25.00 mL of 0.250 M aluminium sulfate is 0.025*0.25 = 0.00625 mol aluminium sulfate. However, if you look at the molecular formula of aluminium sulfate, one mole of aluminium sulfate contains three sulfate ions. So, there are really 0.00625*3 = 0.01875 mol sulfate ions in your second solution.

Combine these and you should get 0.02375 mol sulfate ions in a total volume of 75 mL. Divide and you'll find the concentration is around 0.317 M.

[FarAwaySS2]

Thank you so much! I think I kept getting it wrong cos I kept messing around with the mols.

50.0 mL of 0.100 M sodium sulfate contains 0.05*0.1 mol sodium sulfate, or 0.005 mol sodium sulfate, which is 0.005 mol sulfate ions (molecular formula of sodium sulfate contains only 1 sulfate ion). 25.00 mL of 0.250 M aluminium sulfate is 0.025*0.25 = 0.00625 mol aluminium sulfate. However, if you look at the molecular formula of aluminium sulfate, one mole of aluminium sulfate contains three sulfate ions. So, there are really 0.00625*3 = 0.01875 mol sulfate ions in your second solution.

Combine these and you should get 0.02375 mol sulfate ions in a total volume of 75 mL. Divide and you'll find the concentration is around 0.317 M.

[paper-back]

In chromatography which of the following would provide the lowest Rf value?
(A) Substance - polar, Stationary phase - non polar, Mobile phase - polar
(B) Substance - polar, Stationary phase - polar, Mobile phase - non polar

The answer says (B), why isn't it A?

[eagles]

Like attracts like.

In option (A) the polar substance will be more attracted to the polar mobile phase than the non-polar stationary phase. Hence, the substance will have more of a tendency to form bonds with the mobile phase and then travel a further distance up.

In option (B) the polar substance will be more attracted to the polar stationary phase than the non-polar mobile phase. Hence, the substance will have more of a tendency to form bonds with the stationary phase than travel a further distance up.

Rf value is calculated by the the total distance travelled from the origin, divided by the total distance of the solvent front.

Hence, option (B) will have a lower Rf value because it travels less. 

[paper-back]

Oh shoot, it's talking about R_f value not R_t
Thank you eagles

[paper-back]

Do we need to cover both Low resolution and high resolution NMR?

[zsteve]

With NMR, does it work like I think it does?
[Proton NMR]
We know that the majority of H nuclei will just be protons, and a minority will be deuterium.
Therefore, the majority of H nuclei will have an overall spin, hence overall magnetic field, and are therefore able to absorb energy to promote spin of a nucleon, and then to release that amount when the nucleon defaults back to its original aligned spin.

We are told that "samples may be dissolved in a solvent which will not give a signal, such as D2O".
D2O is heavy water, i.e. the hydrogens in it are all deuterium.
But why does D2O not give out a signal?
Deuterium has two nucleons, therefore there is a possibility if the states are (u, d) that there will not be any net spin. My textbook says

" In many nuclei, the orientations of the spins of all the nucleons (protons and neutrons) are paired and so cancel out"

Does this mean that all nuclei with an even nucleon count have paired spins? Or could a deuterium nucleus have both nucleons as up/down (hence getting a net spin)?
OR could a low energy (2x low energy nucleons) deuterium nucleus absorb one quantum of energy, promote one nucleon to a higher energy state, and then stop responding to EMR (as it is now stable, with no net spin)?

[13-C NMR]
Ok, and this one. We know that 12-C is the most common isotope, and that 13-C is less common.
From what I can understand, the fact that 13-C NMR can work is because of the large number of molecules in the sample. With billions of sample molecules in the sample, there will be many occurrences of 13-C in different parts of the molecule. 12-C atoms will not have net spin, and hence will not absorb EMR. However, 13-C will always have a net spin, so it will be continually absorbing/releasing EMR.
In short, if we could guarantee that all instances of carbon atoms in a sample were 12-C, 13-C NMR wouldn't work.

Have I got all this correct?

[Zues]

how do i calculate ppm? if its ppm = .../... how do i get each ...

hope that makes sense

[soNasty]

ppm = mg/L

[Zues]

and also a question, they say mols dont change (which i can see dont). But when accounting for dilutions in say titrations you often get the moles and multiply by the dulution factor e.g. a 25ml sample diluted to 500mls, so you get the mols and x 500/25 = a different number of mols.

Or was it concentration in the place of mols?


also with titrations
1) the known standard is in the burette. we know its concentration, the volume dispensed however is the value of the titre (which is the volume of the known?) Essentially tells us how much of this we need to meet mole rations
2) a known volume of on of the solutions is pipetted into a flask, no concentration is known

the species under investigation is in step 2 and to find its concentration we use the mols in 1) and mol ratios?
is that correct

[soNasty]

and also a question, they say mols dont change (which i can see dont). But when accounting for dilutions in say titrations you often get the moles and multiply by the dulution factor e.g. a 25ml sample diluted to 500mls, so you get the mols and x 500/25 = a different number of mols.

Or was it concentration in the place of mols?


also with titrations
1) the known standard is in the burette. we know its concentration, the volume dispensed however is the value of the titre (which is the volume of the known?) Essentially tells us how much of this we need to meet mole rations
2) a known volume of on of the solutions is pipetted into a flask, no concentration is known

the species under investigation is in step 2 and to find its concentration we use the mols in 1) and mol ratios?
is that correct

usually they say that they take a 'x ml pipette from the diluted sample'. That's when you would multiply by the ratio by like total diluted volume/x
the titre is the volume of substance required in order to reach the equivalence point (equal stochiometric ratios)

[Zues]

is the titre the volume of the thing inside the burette?

also with the mols this question says calculate the amount of mol present in the 25ml sample. So in the beginnging of the question they has a 25ml sample diluted to 250ml and 20 ml aliquot

ive got the mol of the aliquot but why do i need to multiply by 500/20 to find mol in 25ml sample. wouldnt it be 250/25 since the 25ml sample is diluted. i.e mol in 20ml x 250/25?

again this brings me to my point, the moles do change? but i have been explained before that moles do not change, if i do multiply by 250/20 or 25 it does actually change??