ocr a f325 revision thread

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But I just looked in the book and they carried out a enthalpy of neutralisation calculation pg 156...using degrees C and not Kelvin :s-smilie:

Go in the internet and search SI Units. Degrees C, is not an SI Unit.

The Examples in the book are just Examples. All of these Papers I have done work on Kelvin.

If you convert to Degrees C AFTER you've done the calculation they will give you a mark but you will get it wrong if you do it before, as you will use the wrong number.

Reply 1581

chignesh10

Original post by Better

Oh I finish in around 1h45. But Funtry finished in 1h30...God I hate that guy hahahaha

I meant like the way you work through questions.

1 hour 30 minutes?! what is this madness lol.

I really don't know how ot describe my thought process as it changes for different types of questions but maybe if you could give me a specific question i could say. I am going to do a paper now but I will be back at around 5 and help you

Reply 1582

Better

Original post by chignesh10

Kk good luck. I'm going a bit slowly too, need to finish 2 more Maths Papers. Will have every C3 past paper done (all 22 of them) by tomorrow so yipeeee for me!!!!!

Then after that I can do Chem4 and Chem 5 Revision!

Good luck bro!!!

Reply 1583

Pride

Original post by Better

Read what I put above. Knowing that is the only way you could Logically Deduce, which Acid to choose in that Paper. Other wise you would have to go through a Process of Elimination which Funtry did. But that is not a Rigorous enough approach.

If you ask any 1st/2nd Year Chemistry Student or Teacher the justification of choosing the Acid they will say the same thing.

Assume for a Buffer the ratio is 1:1 and hence just choose the Acid with the closest pKa to the Ph they gave you.

They specifically stated in the Examiners Reports they will add more Stretch and Challenge Questions. So thanks, but no thanks mate.

lol, no need to be rude :smile:

But you would never just assume that the acid:conj. base ratio is 1:1, (or close to this) just because it's a buffer, in an exam. You wouldn't risk it.

So you would just have a look through them and think which one is it most likely to be. Trial and error tells you that the one closest to the pH is the most useful.

You can't just say well sometimes it can be 1:1 - that's not in the syllabus. Stretch and challenge sections of the textbook are just there for the high-level students to practice some higher-level thinking. Not so you can memorise some facts outside the syllabus, so that you can answer a question like this.

Reply 1584

Pride

Original post by Gulzar

But I just looked in the book and they carried out a enthalpy of neutralisation calculation pg 156...using degrees C and not Kelvin :s-smilie:

bearing in mind of course, if you're doing an enthalpy of neutralisation calculation, you're measuring the change in temperature, which is the same in degrees C and in K.

Reply 1585

Better

Original post by Pride

bearing in mind of course, if you're doing an enthalpy of neutralisation calculation, you're measuring the change in temperature, which is the same in degrees C and in K.

This for a Change in Temperature the Units are irrelevant i.e.

0-5 C = 273 - 278 K

Anyway time for me to work!

Reply 1586

ofudge

Hi guys,

has anyone done June 2010 paper?

Why for question 6Cii does the ligand substitution have to be 6NH3 and 4NH3 (as we've always been taught) being marked as only 1/2?

Reply 1587

MedMed12

Original post by ofudge

Hi guys,

has anyone done June 2010 paper?

Why for question 6Cii does the ligand substitution have to be 6NH3 and 4NH3 (as we've always been taught) being marked as only 1/2?

guidance says you get one mark because it wants 6, since its in excess :smile:

(edited 10 years ago)

Reply 1588

otrivine

I still dont get the question on the Jan 2013 paper which involves , how you get 0 degrees, that is weird :confused:

Reply 1589

emah123

Original post by Better

This for a Change in Temperature the Units are irrelevant i.e.

0-5 C = 273 - 278 K

Anyway time for me to work!

how many chem past papers have you?

Reply 1590

ofudge

Original post by MedMed12

guidance says you get one mark because it wants 6, since its in excess :smile:

Ok thank you, but isn't it still in excess with still just 4NH3?

Reply 1591

georgiaaaxo

i was wondering if anyone could explain to me the enthalpy change of neutralisation and how it would work if you didn't have equal volumes of acid and base? so basically, if one was in excess - how do you change your calculation? what would m be in the equation for Q - the total volume of solution or just the solution that reacted? :s-smilie:

also, how do you scale the quantities to form 1 mole of water if there are different volumes therefore different amounts of acid and base? i just get confused if they're not equal :P
thankyou!

Reply 1592

MedMed12

Original post by ofudge

Ok thank you, but isn't it still in excess with still just 4NH3?

It is but because its 6 ligands my teacher told us to do the maximum :smile:

Reply 1593

MathsNerd1

Original post by MedMed12

...

I just got back from my shift which happened to be made longer to 10 hours and I didn't even have a break! -.-
It's safe to say I'm pretty exhausted but I shall push on with F325 and C3 :smile:

On a side note how many past papers do people feel I should be doing a night for this exam?

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Reply 1594

Lysipud

Got another one for you which I just can't seem to get my head around.

It's from Jan 13 question 6e
I don't understand the whole 3:1 ratio they have done in the mark scheme.

Reply 1595

bluedate

Original post by Lysipud

Got another one for you which I just can't seem to get my head around.

It's from Jan 13 question 6e
I don't understand the whole 3:1 ratio they have done in the mark scheme.

Cr^3+ (aq) +3e -----> Cr (s)

X (s) -----> X^2+ (aq) + 2e

In the question it said that the Cr metal gained in mass, so in the first equation I made it the product, while in the second equation the metal X is going from a solid to aqueous solution as it is losing mass. We know X is 2+ because in XSO4, sulfate is 2- (the sulfate cancels out in the equations btw)

Anyway, you can now make an overall equation by X 2 the first equation, and X 3 the second equation to get the electons same number, then combine them:

2Cr^3+ + 3X -----> 2Cr + 3X^2+

So we know the ratio of X:Cr is 3:2

We can then use the information we have to first find the no. moles of Cr
So mass/ mr is 1.456/52 = 0.028 mols
Use this to find X mols, so 0.028 X 3/2 (1.5) = 0.042mols

Now you have the number of moles and the mass from X from the question, you can now find the Mr
So m/n = 1.021/0.042 = 24.3
Look at the periodic table and you'll see that 24.3 is the Mr for Mg
X = Mg

I hope that makes sense

Reply 1596

nrp_95

Original post by Raj Kang

Can anyone help me with this question? June 2011 4c. Why is H squared when pyruvic acid is strong?

Pyruvic acid isn't strong, it's a carboxylic acid. So you have to use the Ka expression to get [H+] = square root of (Ka x HA)

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Reply 1597

Lysipud

Original post by bluedate

Thats great help thank you :smile:

Reply 1598

ankyloser

Please can someone help me with electrode potentials? I've got some chem factsheets and they contradict each other and the textbook.

It says on one sheet that reduction takes place at the cathode, and oxidation at the anode. But I thought it was opposite to electrolysis, so that the cathode is actually the positive electrode, like cation. So surely if reduction was happening at the cathode, it would be gaining electrons, and be becoming more negative? Or is it negative compared to its isolated solutions?

Any help would be brilliant, I've got myself into a right muddle :/

(edited 10 years ago)

Reply 1599

ankyloser

Hang on, so say you've got two half cells joined in a circuit:
Zn2+ +2e- --> Zn
Cu2+ +2e- --> Cu

The Zn is more reactive, is more likely to give up electrons, so the equilibrium is shifted to the left and it loses electrons. This means it is being oxidised, and is the more negative electrode as it has a greater build up of electrons on it, so is the anode.

The Cu on the other hand is less reactive, the equilibrium is in the forward direction, it gains electrons from the zinc and is reduced. Is has a smaller build up of electrons so is more positive, and is the cathode.

Is this all right?? So cathode IS reduction, and anode is oxidation?