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AQA CHEM5 A2 Chemistry - 19th June 2013

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Original post by erniiee
It calls it green in the Nelson thornes textbook? :redface:


[cr(h20)6]3+ is green however due to variations ruby is also can be accepted in the mark scheme.
Original post by cheesypuff
[cr(h20)6]3+ is green however due to variations ruby is also can be accepted in the mark scheme.


It's actually ruby but seen as green.
Reply 342
Original post by cheesypuff
[cr(h20)6]3+ is green however due to variations ruby is also can be accepted in the mark scheme.


I think the colour it is when its pure is ruby, it just normally appears green for whatever reason (hydrolysis, impurities etc).
I have been self teaching chemistry since September; I'm doing the whole A-Level course in one year..
I'm confused about how unit 3 and unit 6 work. Is it one complete ISA that gets sent off? As in you do a practical followed by a written test?
It would be great if anyone can explain. Thanks.
Reply 344
Original post by Dedicatedstudent
Anyone got any tips for the transition metal ISA? Would be of great help seeing as I failed my last ISA, just can't understand why I can't do them! :angry:


It's not a bad paper. U need to be able to do calculations like finding the empirical formula and percentage yield, theoretical yield...etc

I would revise the colours table, which is obvious.
Reply 345
Think I've got Born-Haber, did 3 huge questions from the old spec and didn't have any problems so all's good :thumbsup:
Can anyone explain as I'm really confused, it's a bit vague sorry, came from a hand written notepad/questions from our teacher!

If you're given a rate equation, and it asks you, how do you know that this reaction takes place in more than two steps, more than 3 steps sort of thing, how do you know from the equation - what are you looking for?

Thank you
Original post by yoyo1994
It's not a bad paper. U need to be able to do calculations like finding the empirical formula and percentage yield, theoretical yield...etc

I would revise the colours table, which is obvious.

Ok thank you so much! :biggrin: I always think the paper is not too bad, then I get the result :/
Reply 348
has anyone done the chemistry empa yet? message me some tips please!
Just a little confused. We need to know Cr3+ as green when we're talking about the change in the oxidation state (i.e. +6 -> +3 is orange to green). So why is it that [Cr(H2O)6]3+ is ruby? I know this is mentioned many times, but what's the actual reason for these colour differences? Aren't they just the same? Or is the Cr3+ actually Cr(OH)63-? I don't want to learn two colours if it is that they are the exact same complex ion.
Reply 350
Original post by lukegreenhough
Just a little confused. We need to know Cr3+ as green when we're talking about the change in the oxidation state (i.e. +6 -> +3 is orange to green). So why is it that [Cr(H2O)6]3+ is ruby? I know this is mentioned many times, but what's the actual reason for these colour differences? Aren't they just the same? Or is the Cr3+ actually Cr(OH)63-? I don't want to learn two colours if it is that they are the exact same complex ion.


From chemguide:

Ligand exchange reactions involving chloride or sulphate ions
The hexaaquachromium(III) ion is a "difficult to describe" violet-blue-grey colour. However, when it is produced during a reaction in a test tube, it is often green.
We nearly always describe the green ion as being Cr3+(aq) - implying the hexaaquachromium(III) ion. That's actually an over-simplification.
What happens is that one or more of the ligand water molecules get replaced by a negative ion in the solution - typically sulphate or chloride.
Replacement of the water by sulphate ions
You can do this simply by warming some chromium(III) sulphate solution.
One of the water molecules is replaced by a sulphate ion. Notice the change in the charge on the ion. Two of the positive charges are cancelled by the presence of the two negative charges on the sulphate ion.
Replacement of the water by chloride ions
In the presence of chloride ions (for example with chromium(III) chloride), the most commonly observed colour is green. This happens when two of the water molecules are replaced by chloride ions to give the tetraaquadichlorochromium(III) ion - [Cr(H2O)4Cl2]+.
Once again, notice that replacing water molecules by chloride ions changes the charge on the ion.
Reply 351
Original post by lukegreenhough
Just a little confused. We need to know Cr3+ as green when we're talking about the change in the oxidation state (i.e. +6 -> +3 is orange to green). So why is it that [Cr(H2O)6]3+ is ruby? I know this is mentioned many times, but what's the actual reason for these colour differences? Aren't they just the same? Or is the Cr3+ actually Cr(OH)63-? I don't want to learn two colours if it is that they are the exact same complex ion.


The actual colour of PURE hexaaquachromium(III) is a ruby (which is often thought of a mixture of red and blue/purple, so is sometimes denoted as violet). This is the colour AQA expect you to remember as its in the Nelson thornes textbook..I haven't started past papers yet, but if it says otherwise in mark schemes e.g. violet, then use that.

As the info from Chemguide says, in reality, you will rarely see that ruby colour due to replacement of some of the water ligands. The colour you see instead is green. So if you're asked the colour of [Cr(H2O)6]3+ the technically correct answer is ruby, although I think green is accepted as that is the generally seen colour.

(after reading a little further, I've found out that the extreme colour change is due to the slow rate at which Chromium actually substitutes ligands, and the fact that those particular ligand exchanges result in sensitive changes in the energies absorbed which correspond to those shades in the visible spectrum, but you definitely don't need to know this for the exam :lol:)

Interestingly, from a google search of chromium(III) nitrate, I found this picture from this website http://fphoto.photoshelter.com/image/I0000hF8WVHhI_6s which shows the violet colour of chromium(III) nitrate, to the left.

Original post by erniiee
The actual colour of PURE hexaaquachromium(III) is a ruby (which is often thought of a mixture of red and blue/purple, so is sometimes denoted as violet). This is the colour AQA expect you to remember as its in the Nelson thornes textbook..I haven't started past papers yet, but if it says otherwise in mark schemes e.g. violet, then use that.

As the info from Chemguide says, in reality, you will rarely see that ruby colour due to replacement of some of the water ligands. The colour you see instead is green. So if you're asked the colour of [Cr(H2O)6]3+ the technically correct answer is ruby, although I think green is accepted as that is the generally seen colour.

(after reading a little further, I've found out that the extreme colour change is due to the slow rate at which Chromium actually substitutes ligands, and the fact that those particular ligand exchanges result in sensitive changes in the energies absorbed which correspond to those shades in the visible spectrum, but you definitely don't need to know this for the exam :lol:)

Interestingly, from a google search of chromium(III) nitrate, I found this picture from this website http://fphoto.photoshelter.com/image/I0000hF8WVHhI_6s which shows the violet colour of chromium(III) nitrate, to the left.





Ohhhhhhhh okay. Thanks for the help :smile: wish the textbook would inform us of this! I suppose I will just wait until I start doing the question papers.
Reply 353
Original post by lukegreenhough
Ohhhhhhhh okay. Thanks for the help :smile: wish the textbook would inform us of this! I suppose I will just wait until I start doing the question papers.


No problem! Yea always best to just use whatever is accepted in the mark scheme. Another discrepancy is the hexaaquairon(III) complex..I have no clue what the accept on MSs for that!

Haha, what a coincidence - your avatar next to your name has green and red coloured bubbles..how relevant to our Chromium troubles!
Even weirder that I was just about to ask about hexaaquairon(III). Just looked through papers and they haven't asked for that colour. I guess that the examiners will avoid questions where the colour of the complex is so varied.
Reply 355
Original post by lukegreenhough
Even weirder that I was just about to ask about hexaaquairon(III). Just looked through papers and they haven't asked for that colour. I guess that the examiners will avoid questions where the colour of the complex is so varied.


Haha fair enough, stops me from worrying..I really need to revise electrochemistry :eek:
Original post by erniiee
Haha fair enough, stops me from worrying..I really need to revise electrochemistry :eek:


Hello have I done this right?

Use the following data to calculate the entropy change when one mole of ammonia is formed from its elements:
S (JK-1mol-1): N2(g) = 191.6, H2(g) = 130.6, NH3(g) = 192.3

Given that the enthalpy of formation of ammonia is -46 kJmol-1, determine the range of temperatures for which this reaction is feasible.


I got temp must be lesser than or equal to 463 k. BUT mark scheme is saying Reaction feasible below 462 K.

What happened? If an exam, would I still get the mark? :smile: THANKS!
Reply 357
for electrochemical cells, redox etc this is true isnt it....

the more positive state is the oxidised state
the more negative state is the reduced state

?
Reply 358
Original post by laurawoods
Hello have I done this right?

Use the following data to calculate the entropy change when one mole of ammonia is formed from its elements:
S (JK-1mol-1): N2(g) = 191.6, H2(g) = 130.6, NH3(g) = 192.3

Given that the enthalpy of formation of ammonia is -46 kJmol-1, determine the range of temperatures for which this reaction is feasible.


I got temp must be lesser than or equal to 463 k. BUT mark scheme is saying Reaction feasible below 462 K.

What happened? If an exam, would I still get the mark? :smile: THANKS!


Haven't tried the calculation, but I suspect you've made a rounding error somewhere which is why your answer is (only slightly) off. Try it again without rounding anywhere!
Original post by erniiee
Haven't tried the calculation, but I suspect you've made a rounding error somewhere which is why your answer is (only slightly) off. Try it again without rounding anywhere!


there was no rounding involved !

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