VCE Chemistry Question Thread

[lzxnl]

In the context of IR Spectroscopy, we can use a spring analogy for the bond between two atoms. A higher frequency of radiation means the spring vibrates a lot/quickly, while a lower frequency of radiation means that it does not vibrate as much.

If you have say, a C-H bond, the lower mass of the hydrogen atom means the spring is able to vibrate much more because its so light. Think about it -- if we have the much heavier O attached to our carbon, then its going to vibrate much more slowly (ie. lower frequency of radiation) simply because it's harder for it to vibrate as much.

Furthermore, just like with springs, the stiffness matters. A stiffer spring can contract much faster than one that isn't as stiff. Hence stronger bonds will absorb more energy than weaker bonds, which is why for example C=O absorbs at a higher frequency than C-O.

Indeed. There is a result from quantum mechanics that if you take a quantum mechanical spring, the energy gap between consecutive energy levels is Planck's constant h times the frequency of oscillation. In other words, as the energy of a photon is Planck's constant times its frequency, the frequency of light absorbed is equal to the frequency of oscillation, which means if you increase the frequency of oscillation, you increase the frequency, and hence the energy, of the light absorbed. The frequency of oscillation of a spring is given by if I remember correctly, where m is the (reduced) mass of the system and k is the spring constant, essentially how stiff the spring is. You can see from this equation the importance of the mass of the atoms and the bond strength.

Of course, you don't need this for VCE

[psyxwar]

Indeed. There is a result from quantum mechanics that if you take a quantum mechanical spring, the energy gap between consecutive energy levels is Planck's constant h times the frequency of oscillation. In other words, as the energy of a photon is Planck's constant times its frequency, the frequency of light absorbed is equal to the frequency of oscillation, which means if you increase the frequency of oscillation, you increase the frequency, and hence the energy, of the light absorbed. The frequency of oscillation of a spring is given by if I remember correctly, where m is the (reduced) mass of the system and k is the spring constant, essentially how stiff the spring is. You can see from this equation the importance of the mass of the atoms and the bond strength.

Of course, you don't need this for VCE

when I saw you replied I thought you were gonna rip my layman explanation to shreds HAHA

[lzxnl]

when I saw you replied I thought you were gonna rip my layman explanation to shreds HAHA

Well it wasn't wrong as such; it was just a little vague so I elaborated

Solving the Schrodinger equation for a quantum mechanical spring is so annoying. I still don't fully understand the solution process hahahaha

[Zues]

wouldnt the number of moles in the 20ml be same as the 25ml and the 250 ml? because moles are always same (youve just added solvent). But i dont get it, if the moles STAY THE SAME why does the mole value change in different quantities ? plus if i have a 20ml sample and im trying to find the moles in 25ml, wouldn't x 25/20 work?

[lArcdeTriomphe]

I think you are getting confused between two types of scenarios (I'm not sure which scenario you are referring to because you haven't attached the actual question).

Scenario 1: you have a solution, and you've diluted it. In this case, yes the number of moles in 20ml would be the same as 25ml or 250ml.

Scenario 2 (which I believe is what the question is most likely asking): you have an extemely concentrated solution, you've diluted it AND then you take an aliquot from this diluted solution. e.g. I have 10.00ml of conc HCl. I dilute it to 100.0 mL, and then I pipette out 20.00mL (i.e. you've taken 1/5 of the diluted solution). In this case, to find out the n(HCl) in conc 10.00mL, you need to do the following:
n(HCl) in conc 10.00mL = n(HCl) in diluted 100.0mL = n(HCl) in diluted 20.00mL x (100.0/20.00)

so, i'm assuming in your question, you had 25ml of concentrated bleach - it was then diluted to 250.0 ml (in a volumetric flask) and then you pipetted out 20.00ml. you therefore need to multiple n(ClO-) in 20.00ml by 250.0/20.00 to find n(ClO-) in diluted 250.0ml solution, which then equals n(ClO-) in concentrated 25.00ml solution

[paper-back]

In question 4a VCAA CHEMISTRY EXAM 1 2011 why do they halve the number of mol of MgNH4PO4.6H2O to get to number of mol of P2O5?
Why don't you double it as wouldn't you have double the amount of potassium in the P2O5 solution?

[sin0001]

In question 4a VCAA CHEMISTRY EXAM 1 2011 why do they halve the number of mol of MgNH4PO4.6H2O to get to number of mol of P2O5?
Why don't you double it as wouldn't you have double the amount of potassium in the P2O5 solution?

The amount/mol of 'P' remains the same as P2O5 is precipitated into MgNH4PO4.6H2O
So n(P) = 2 x n(P2O5) and n(P) also = n(MgNH4PO4.6H2O); note that n(P) in both equations is the same, so we can merge those two equations together: n(P) =  2 x n(P2O5) = n(MgNH4PO4.6H2O), so n(P2O5) = 1/2 x n(MgNH4PO4.6H2O)

[SammyBoy]

Can someone help me out with this question

A 0.500g sample of sodium sulfate (Na2SO4) and a 0.500g sample of aluminum sulfate (Al2(SO4)3) were dissolved in a volume of water and excess barium chloride was added to precipitate barium sulfate. What was the total mass of barium sulfate produced?

Thanks.

[Kel9901]

Can someone help me out with this question

A 0.500g sample of sodium sulfate (Na2SO4) and a 0.500g sample of aluminum sulfate (Al2(SO4)3) were dissolved in a volume of water and excess barium chloride was added to precipitate barium sulfate. What was the total mass of barium sulfate produced?

Thanks.

You're looking for the total sulfate content of the two salts, so the mass of barium sulfate precipitate can be calculated.
sodium sulfate:
n(Na2SO4)=m/Mr=3.52*10^-3
n(SO4 2-)=n(Na2SO4)=3.52*10^-3

aluminium sulfate:
n(Al2(SO4)3)=m/Mr=1.46*10^-3
n(SO4 2-)=3*n(Al2(SO4)3)=4.38*10^-3

n(SO4 2-) total=3.52*10^-3+4.38*10^-3=7.90*10^-3

n(BaSO4)=n(SO4 2-)=7.90*10^-3
m(BaSO4)=n*Mr=1.84 g

[zsteve]

*Bump previous question Could someone answer this one?

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?

[alchemy]

What colour do you expect a flame to be if potassium chloride is dropped in it?
One of the first questions in the textbook, but how am I supposed to know off the top of my head?

[doher109]

What colour do you expect a flame to be if potassium chloride is dropped in it?
One of the first questions in the textbook, but how am I supposed to know off the top of my head?

This is not required knowledge. You are required to understand the benefits of flame tests and they limited nature of the information they provide. If a question was to appear on an exam there would be a table provided with several metals and the colour they produce.

[RazzMeTazz]

With subshells I know they go in order of: s,p,d,f but what comes after these?
Are there any others?
Thanks

[paper-back]

The amount/mol of 'P' remains the same as P2O5 is precipitated into MgNH4PO4.6H2O
So n(P) = 2 x n(P2O5) and n(P) also = n(MgNH4PO4.6H2O); note that n(P) in both equations is the same, so
we can merge those two equations together: n(P) =  2 x n(P2O5) = n(MgNH4PO4.6H2O), so n(P2O5) = 1/2 x n(MgNH4PO4.6H2O)

Sorry I'm still a little confused, can you eleborate a bit?

[RazzMeTazz]

Sorry I'm still a little confused, can you eleborate a bit?

Sorry to intrude but I can hopefully try and explain:

In MgNH4PO4.6H2O there is only one phosphorus atom but in P2O5 there are two phosphorus atoms.
So if you had say 2 molecules of MgNH4PO4.6H2O you would have a total of 2 phosphorus atoms. If you then wanted to convert this back into P2O5 you would essentially only be able to get one molecule of P2O5 (As one molecule contains 2 atoms of Phosphorus and you only had two atoms of phosphorus available to you from the start.)

From that ^^ we can see that the amount of P2O5 that you have is always 1/2 of that of the amount of MgNH4PO4.6H2O  that you have.

I used very small scale quantities for this explanation but think of this in a larger scale, such as number of moles and it explains why the number of mol of P2O5 is half that of the number of mol of MgNH4PO4.6H2O,

That's why  n(P2O5) = 1/2 x n(MgNH4PO4.6H2O)

Hope that helps!