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Original post by C0balt
Concentrated H2SO4 reacts to produce HCl etc which can't be used to distinguish BaCl2 and MgCl2. You'd use H2SO4 conc to distinguish between the anion in two halides

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The thing is, the mark scheme allows non-concentrated H2SO4.
Is the underlying point here that too much H2SO4 will lead to too many spare H+ that can form HCl ...?

Original post by LibertyMan
The thing is, the mark scheme allows non-concentrated H2SO4.
Is the underlying point here that too much H2SO4 will lead to too many spare H+ that can form HCl ...?



H2SO4 is a strong oxidisng agent so it doesn't work with Br and I as they are oxidised but idk why it forms HCl, I just know the reaction as AS knowledge

Dilute sulfuric acid can be used to get precipitate of barium sulphate as barium sulphate is insoluble and magnesium sulphate is soluble so it works as a distinguisher

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Original post by SubwayLover1
What would you use?

:frown:


When concentrated sulfuric acid react with those two chloride you would see steamy white fine which is HCl

If you add dilute sulfuric acid then you'd get white precipitate of BaSO4 for BaCl

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Original post by AH5223
Thanks, i find them so confusing


Dude, don't forget elimination.
Original post by katemate98
Isn't it with the heterogeneous catalyst - adsorption, chemical reaction (bonds broken), desorption?

No, the heterogenous catalyst reacts them together to form less harmful products of CO2 and N2.


Original post by _kirstxn
theres a question in the textbook on p.231 "How are harmful Nox and CO formed in a car engine?" i feel like i know the basics of the information but im a bit 'iffy' on how to go about answering it

CO is formed via the incomplete combustion of a fuel in an engine due to insufficient supply of oxygen.
NOx is formed as a result of N2 from the air (important to say that) reacting with the oxygen in the engine.
I'm not sure if this has already been posted but this person is great at teaching. You can use this as revision if you wish. www.youtube.com/watch?v=jctsM6ISVDA&index=1&list=PL0yIQjy59R-CLL4AtNq_oi7HMB9wIb-Yt
[video="youtube;O2gb2wnKL0Y"]https://www.youtube.com/watch?v=O2gb2wnKL0Y[/video]

Here's a video that my friend uploaded about the CHEM1 exam that the AQA students sat.

I understand this thread is for helping students but I feel that its best to bring in a chilled atmosphere as most of you may be stressing at this time.
Original post by ilovereading
I never understand PPM questions, can anyone explain it to me?

Calculate the concentration of ozone in the sample of air in units of parts per million (ppm) by volume.

I have the volume of ozone, measured in m^3, present in the original sample of air which is 4.743x10-6

Thanks!

(AS Chemistry Edexcel)


Hey, I answered you on your thread earlier but I didn't quote so I don't know if you saw it. I'll just repeat my answer here too. :smile:

I think the equation is:

Concentration in ppm = (mass or volume of component/mass or volume of solution) X 1000000


I don't think there's much to ppm other than that?

You didn't give the volume of the original sample of air but I recognize that question, it's from the May 2012 Unit 2 paper which I did recently and it states that the sample of air was 100m3.

So. (4.743x10-6/100) x 106 = 0.04743 ppm
Holaaaa :smile:

I'm getting really confused about this multi-choice question, so here you go:

'The enthalpy change of neutralisation of an acid by an alkali is measured by adding 10cm3 of HCl to 10cm3 of NaOH. 10cm3 pipettes with an accuracy of +/-0.04cm3 are used to measure out both solutions. The overall % error in measuring the total volume of the reaction mixture is:

A- 0.04%
B- 0.08%
C- 0.4%
D-4.0%

I thought that the calculation would be (0.04/10)x2 ?
:s
Original post by Lainathiel
Holaaaa :smile:

I'm getting really confused about this multi-choice question, so here you go:

'The enthalpy change of neutralisation of an acid by an alkali is measured by adding 10cm3 of HCl to 10cm3 of NaOH. 10cm3 pipettes with an accuracy of +/-0.04cm3 are used to measure out both solutions. The overall % error in measuring the total volume of the reaction mixture is:

A- 0.04%
B- 0.08%
C- 0.4%
D-4.0%

I thought that the calculation would be (0.04/10)x2 ?
:s

When the values are added together the uncertainty is added together, so in this case 0.04+0.04 which is basically. 0.04*2=0.08
Divide by the total volume which gives 0.08/20=0.4%

(0.04/10)*2 is when you have two values which are multiplied. Volume can't be multiplied by volume though

When you might want to multiply is when you have a square for example, and you use a ruler with uncertainty 0.5cm and you measure a side 10cm. Thenyou will square the side. When you squared you multiply the percentage uncertainty by two, so 2(100*0.5/10)

Pay a visit to physics forums, and there's a sticky about uncertainty with all the rules, it explains better
Original post by LibertyMan
Dude, don't forget elimination.


You don't need elimination for OCR! Lucky us :biggrin:
Original post by kawehi
You don't need elimination for OCR! Lucky us :biggrin:


What

WHAT
Original post by LibertyMan
What

WHAT


Tbh it's the only good thing about OCR chemistry
Original post by kawehi
Tbh it's the only good thing about OCR chemistry


Is OCR chemistry focused more on content or problem solving? A lot of people disagree but I like OCR physics because there are less questions, marks and topics than other boards and more focus on harder questions where you have to apply skills to unfamiliar situations and get out alive. (Don't know what you think of that either)
Original post by LibertyMan
Is OCR chemistry focused more on content or problem solving? A lot of people disagree but I like OCR physics because there are less questions, marks and topics than other boards and more focus on harder questions where you have to apply skills to unfamiliar situations and get out alive. (Don't know what you think of that either)


I'm not really an expert on the topic, but I'm under the impression that OCR Chem has been more about content in the past. However, there seems to be a bit of a shift towards more applied topics in recent history..
I also *kind of* like OCR physics, for those same reasons! (I'm more of a biochemist, though :wink:) I've done a few AQA past paper questions, and I thought that all of the wording was much more similar to GCSE. Not saying that the questions are easy or anything, just a different structure..
Reply 3195
Original post by LibertyMan
The thing is, the mark scheme allows non-concentrated H2SO4.
Is the underlying point here that too much H2SO4 will lead to too many spare H+ that can form HCl ...?



if it comes up just say naoh because mg(oh)2 is insoluble and forms a white precipitate
Original post by thymolphthalein
Hey, I answered you on your thread earlier but I didn't quote so I don't know if you saw it. I'll just repeat my answer here too. :smile:

I think the equation is:

Concentration in ppm = (mass or volume of component/mass or volume of solution) X 1000000


I don't think there's much to ppm other than that?

You didn't give the volume of the original sample of air but I recognize that question, it's from the May 2012 Unit 2 paper which I did recently and it states that the sample of air was 100m3.

So. (4.743x10-6/100) x 106 = 0.04743 ppm


Thank you for answering my question! I switched to Biology soon after so I was meant to read, understand and reply when I came back to Chemistry (never happened). Again thanks >.<
"Write an equation for the reaction of iron with antimony sulfide to form antimony andiron(II) sulfide"
The question gives you Sb2S3 for antimony sulfide but how do you know what the formula is for iron(ll) sulfide???? :/
Original post by Signorina
"Write an equation for the reaction of iron with antimony sulfide to form antimony andiron(II) sulfide"
The question gives you Sb2S3 for antimony sulfide but how do you know what the formula is for iron(ll) sulfide???? :/


Well what group is S in? Group 6. They usually form 2- ions when reacted with metals. And it says iron (II) sulfide which means that the oxidation number for iron is +2. So it's 2 to -2 already and the "charge" balances so it's just FeS


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Original post by Signorina
"Write an equation for the reaction of iron with antimony sulfide to form antimony andiron(II) sulfide"
The question gives you Sb2S3 for antimony sulfide but how do you know what the formula is for iron(ll) sulfide???? :/


Use your knowledge of oxidation states - Iron(II)Sulfide: We're aware that Iron, in this instance, will have an oxidation state of +2, we're also aware of the fact that Sulfur will always have an oxidation state of -2 (Group 6), so in order to equate the oxidation state value to 0 we'll require one atom of Iron and one atom of Sulfur, thus creating Iron Sulfide: FeS!

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