iGCSE 2012 Chemistry Discussion

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Perhaps I misunderstand the question but Carbonate ions cannot be identified in precipitation. They produce, apart from when with group 1 / Ammonium cations, precipitates as do hydroxides, phosphates and oxides. The only way to test for carbonate is to add HCl and collect the CO2 that is expelled.

Fine, fair enough.

Original post by Umackjiggles

Tell me, how can you tell the difference between magnesium hydroxide and magnesium carbonate when they both form a precipitate of the same colour and appearance? You can't you would have to add HCL and wait for CO2 to be released and test for that.

OK, you're technically correct, but do you really think the mark schemes will be prepared for hydroxide ions, carbonate ions, nitrate ions, and pretty much every negative ion possible except for sulphates and halides?

Original post by Umackjiggles

And btw nitrates can be identified in precipitation because they are all insoluble and thus, during precipitation, produce no precipitate if you are to assume the cation is not group 1 or ammonium. I can see that the question is vague and ambiguous and thus I think there will be multiple answers on th.e mark scheme as they are all equally valid.

Nitrates are all soluble and thus cannot be identified in precipitation. If they could, my answer would be downright wrong. Multiple answers, yes - but I still believe they will be nitrate and acetate :wink:

I could be wrong. I guess we'll have to wait until next year to find out (at which point I doubt anyone will bother :tongue:

Original post by sahajkaur

I think the question was asking which anion could not be shown to be present using precipitate reactions; thus Carbonate would be right.

Fine, fine, if you say so. Just be happy and forget about it, it's one mark. :smile:

Original post by sahajkaur

Being outnumbered doesn't equate being wrong so there's really no need to shoot down someone else's answer with such conviction unless you have the markscheme in front of you.

He was outnumbered on a provable fact and I'm sure he appreciated that. When I said "outnumbered", I meant to compare "most" (i.e. every insoluble compound with carbonate ions) to "group 1 metals" (i.e. every soluble compound with carbonate ions). There's no lack of conviction here - it is a fact that the number of insoluble carbonate compounds outnumbers the number of soluble ones. I don't need a mark scheme to know that, and I'm sure Umackjiggles got what I was saying.

Reply 501

member963009

Original post by Big-Daddy

Fine, fair enough.

Fine, fine, if you say so. Just be happy and forget about it, it's one mark. :smile:

I happily shrugged it off stating that maybe we're both right and the mark scheme will account for that. You're the one who's hell bent to prove that we're going to lose that mark :L

Reply 502

Big-Daddy

Original post by sahajkaur

I happily shrugged it off stating that maybe we're both right and the mark scheme will account for that. You're the one who's hell bent to prove that we're going to lose that mark :L

Hell bent? OK, OK, let's leave it there. :tongue:

Reply 503

RoyalBlue7

Original post by Umackjiggles

No, the rate of reaction stays the same however you produce half as much of the product because you have half as much (moles-wise) of the reactants (when liquid is not in excess).
EDIT : Sorry misunderstood your question, the rate of reaction would indeed change but this is not represented on the graph. See attachment.

hey I'm not asking if the ROR or the amount of product also halves, I'm asking about the time taken for the reaction to complete. That is the controversial page from the student book.

I think the reaction showing half the concentration should be half the initial ROR, produce half the mols of product AND BUT take half the time. Your attachment says "Reaction not still completed but will eventually end up with half...." My prob is why the hell does it take so long? Doesn't the reaction with the full concentration also reach the stage of half concentration????????????

Reply 504

member963009

Original post by StUdEnTIGCSE

If the ROR is half ofcourse it will take more time! :P Sorry I don't understand your question? If the ROR halves you can't expect the reaction to happen in the same amount of time....what? ROR directly influences time taken...hence it is rate of reaction.

Reply 505

Hazbo

for the suggest and ion that couldn't be tested by precipitation would OH- be one?

Reply 506

SupremeNunchucks

Original post by ak2109

1.2 tonnes

Dammn i got 22.3 :frown:

more marks lost there..

Reply 507

RoyalBlue7

Original post by sahajkaur

Well look at this model
Say you do a reaction with 100mol per unit volume and then 50 mol per unit volume (other conditions same) .
After some time 100 mol per volume acid MUST reach a state of 50 mol per unit volume and thus take then the same route as the 50 mol per unit volume experiment. Therefor time taken MUST be more than the 50 mol per unit volume even though initial ROR and Conc is more.
...
Its like 50 mol per unit volume takes 30 Secs

100 mole per unit volume takes 15 secs to reach 50 mol per unit volume state and then 30 sec to reach zero
:eek:

Reply 508

Big-Daddy

If the concentration is halved then there will be half the product formed, but the formation will take the same time (i.e. the rate of reaction will be half at every point).

Reply 509

member963009

Original post by Big-Daddy

If the concentration is halved then there will be half the product formed, but the formation will take the same time (i.e. the rate of reaction will be half at every point).

I'm fairly certain that if concentration halves, ROR halves, then surely it would take more time for the reaction to go to completion :s-smilie:

Reply 510

RoyalBlue7

Original post by Big-Daddy

If the concentration is halved then there will be half the product formed, but the formation will take the same time (i.e. the rate of reaction will be half at every point).

Hey compare it with the previous experiment, that uses twice the concentration. Compare and explain clearly the relative time it takes for completion.
This page from the student book almost gave me a laugh...

http://www.thestudentroom.co.uk/attachment.php?attachmentid=150253&d=1337686295

Reply 511

RoyalBlue7

Why has the reaction not yet finished, I mean why does it take more time?
The other experiment too should reach a stage similar to the beginning of the next experiment. Therefore the second exp should take less time
[ROR decreases with time as acid is not in excess]

Reply 512

Big-Daddy

Original post by sahajkaur

I'm fairly certain that if concentration halves, ROR halves, then surely it would take more time for the reaction to go to completion :s-smilie:

If rate didn't halve then the reaction would be complete in half the time as (the concentration having halved) there are only half the number of particles (assuming, as the question did and stated, that the volume stays the same) that can react will be present. So if the rate does halve, then the reaction will take the same time as the number of particles capable of reacting will have been halved.

I imagine that this is the case, but I could be wrong. If it is wrong, the solution would lie within simultaneous quadratic equations, but we don't want to go there just now do we? :eek:

Reply 513

member963009

Original post by Big-Daddy

I don't know :O I think it would be better for my health, blood pressure and state of mind if I left this forum right now because i am getting worked up about my A*s which doesn't help as its the worst subject tomorrow: Physics :frown:

Reply 514

RoyalBlue7

Hey the time taken must be less when half the number of particles are available.
When more particles are available its true that ROR (no of successful collisions per unit time increases) but as time progresses the ROR decreases as the number of particles decreases. At one point in time the number of particles will decrease and then be equal to the number of particles in the half the conc experiment. Then it will behave as if the conc was halved. So then the time taken it to COMPLETE is the same as for the exp with halve the number of particles.
Assumptions : Concentration of acid decreases as reaction proceeds, Volume of Acid is same, the acid is NOT in excess

Reply 515

RoyalBlue7

I drew the graph showing half the concentration levelling off at half the time or something because I had a good reason to. But the %^*%* examiner's is gonna cut marks.