AQA CHEM4 - 13th June 2012

Hi
Can someone help me with this question please.
It's from chem4 jan 2012.
Q4a.
25cm^3 sample of 0.0850 mol dm^3 HCL was placed in a beaker and 100cm^3 of distilled water were added calculate the ph of the new solution.
The answer I keep getting is 1.97 but the mark scheme says 1.77
Thanks

find the moles of hcl by c x v

divide n by new volume to get the concentration (100 + 25)

-log10 new concentration of H+

Reply 501

loz876

7

Original post by CollateralElement

Chem5 is a complete troll! I hate it. I'm looking forward to Chem4 if I had to compare against the two.... Did a paper this morning and minus the fact that I remembered most of the answers from the last time that I did it... I didn't find it all that bad... Who's ACTUALLY liking Chem5?

I actually prefer chem5 to chem4 :colondollar:

there's just less stuff to learn! as much as I love organic and mechanisms, I feel more confident for chem5 even though chem4 is a resit and chem5 is first time!

Reply 502

Doctor.

OP

12

Original post by popnit

If someone's free could they go through some questions on Jan 11 CHEM 4 with me please. Really stuck!

Sure what questions? I am just about to do it now :smile:

so when I get on my next break I should be able to be of some help I guess :h:

Original post by loz876

I actually prefer chem5 to chem4 :colondollar:

there's just less stuff to learn! as much as I love organic and mechanisms, I feel more confident for chem5 even though chem4 is a resit and chem5 is first time!

Too much memorising :tongue:

Reply 503

popnit

14

Original post by Doctor.

Sure what questions? I am just about to do it now :smile:

so when I get on my next break I should be able to be of some help I guess :h:

Thanks so much! PM when you're free pls!

Reply 504

xXACEXx

Original post by masterhr1

What happens when excess HCl is added to this?
Chap8 ExamStyle Q2b

This was posted from The Student Room's iPhone/iPad App

The NH2 at the bottom becomes NH3+ because of the acidic conditions

Reply 505

strawberries.

Hey guys, do we need to know the mechanism for the dehydration of an alcohol to an alkene? (elimination of H2O)
Jw, cos its in my study guide but I don't want to learn it if it's not required.

Thanks :smile:

Reply 506

NornIronKitKat

It was in unit 2 so I suppose they could ask us. It's not a difficult one, but probably more likely just to ask the name of the mechanism as part of a synthesis question.

Reply 507

loz876

7

Does anyone know if Chem4 and 5 are equally weighted? I know Chem2 was worth more than Chem1, but I dunno about A2 :s-smilie:

?

Reply 508

Doctor.

OP

12

Original post by strawberries.

Hey guys, do we need to know the mechanism for the dehydration of an alcohol to an alkene? (elimination of H2O)
Jw, cos its in my study guide but I don't want to learn it if it's not required.

Thanks :smile:

You'll have to know it to some extent, probably not know how ro do the mechanism. You will have to name It mist likely, for Organic Synthesis

Original post by loz876

Does anyone know if Chem4 and 5 are equally weighted? I know Chem2 was worth more than Chem1, but I dunno about A2 :s-smilie:

?

CHEM1- 100
CHEM2- 140
CHEM3- 60
CHEM4- 120
CHEM5- 120
CHEM6- 60

So yeah they're thw same

Reply 509

shuaib786

2

Does anyone know how to do question 6b from june 2011 paper???

Reply 510

Doctor.

OP

12

Original post by shuaib786

Does anyone know how to do question 6b from june 2011 paper???

Didn't understand why step 2 was the rate determining step, sorry not a clue!

Reply 511

shuaib786

2

Original post by Doctor.

Didn't understand why step 2 was the rate determining step, sorry not a clue!

I know. Hope we dont get something like that on Wednesday.

Reply 512

purple.piglet

1

Original post by shuaib786

Does anyone know how to do question 6b from june 2011 paper???

It's step 2 because they give you the rate equation on the previous page which indicates that the RDS and steps before include 2NO molecules and a H2 molecule :smile:

Reply 513

shuaib786

2

Original post by purple.piglet

It's step 2 because they give you the rate equation on the previous page which indicates that the RDS and steps before include 2NO molecules and a H2 molecule :smile:

but then why is not step 1???

Reply 514

Doctor.

OP

12

Original post by purple.piglet

It's step 2 because they give you the rate equation on the previous page which indicates that the RDS and steps before include 2NO molecules and a H2 molecule :smile:

Didn't think it had anything to do with that question :s-smilie:

Reply 515

purple.piglet

1

Original post by shuaib786

but then why is not step 1???

Because the rate eq. has [NO]^2 as well as

and as the H2 appears in step 2, the RDS is step 2. Because all steps up to and including the RDS itself must have the same number of moles of substance as specified in the rate eq.

Original post by Doctor.

Didn't think it had anything to do with that question :s-smilie:

Still part of Q6 so I suppose they're trying to catch us out with putting the question so far on that we forget to use it..! Otherwise, not really sure but it's all I can think of :colondollar:

Reply 516

Baraa3

Can someone please help me on the jan 2011 paper
Question 1aiv and 1av, I don't understand how to get the answers that are on the marksheme

Reply 517

Parle à ma main

1

Original post by Baraa3

Can someone please help me on the jan 2011 paper
Question 1aiv and 1av, I don't understand how to get the answers that are on the marksheme

iv- concentration is the number of moles per dm cubed, right? Which means if you increase the amount of volume, you're increasing the "dm cubed" part. That means that the moles are less squashed together, so the concentration decreases (think of juice- the more water you add, the less concentrated it becomes). Because the volume has increased 2x, the concentration has to decrease by 2 as well.

As for v, half the value of x (as they state the conc has halved and [x] means concentration), and times OH- by 3 because they say it has increased by 3, then just factor it into the equation. use the values given in (a)iii

Hope that made sense :o:

(edited 11 years ago)

Reply 518

Baraa3

Original post by Parle à ma main

iv- concentration is the number of moles per dm cubed, right? Which means if you increase the amount of volume, you're increasing the "dm cubed" part. That means that the moles are less squashed together, so the concentration decreases (think of juice- the more water you add, the less concentrated it becomes). Because the volume has increased 2x, the concentration has to decrease by 2 as well.

As for v, half the value of x (as they state the conc has halved and [x] means concentration), and times OH- by 3 because they say it has increased by 3, then just factor it into the equation. use the values given in (a)iii

Hope that made sense :o:

Thank you but I still don't understand how to plug it in to the equation could you show me how I've worked out both answers

This was posted from The Student Room's iPhone/iPad App

Reply 519

Baraa3

How worked out the answers

This was posted from The Student Room's iPhone/iPad App

AQA CHEM4 - 13th June 2012

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