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A-level Chemistry Revision Squad!

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Reply 560
Avatar for voltz
voltz
6
The H-S-H has 3 bonded pairs and 1 lone pair so its a pyramidal strucutre, which would have a bond angle of 107 degrees?
Original post by voltz
The H-S-H has 3 bonded pairs and 1 lone pair so its a pyramidal strucutre, which would have a bond angle of 107 degrees?


Cools
The answer is between 95-107.5 degrees
So you'd look at the shape. Gotcha


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These are the files that were within the attachment size limits for pdf. I think there was only a couple of files I couldnt upload so if you really need them let me know how I can share them.
F321_MS_June15.pdf200.0KB
F321-01 QP Jun15.pdf472.1KB
F325-01 QP Jun15.pdf529.7KB
F324_MS_June15.pdf622.8KB
F324-01 QP Jun15.pdf544.2KB
F322-01 QP Jun15.pdf498.1KB
Reply 563
Avatar for OGGUS
OGGUS
21
Original post by IamJuggernaut
These are the files that were within the attachment size limits for pdf. I think there was only a couple of files I couldnt upload so if you really need them let me know how I can share them.

Thanks man
Reply 564
Avatar for Dinaa
Dinaa
17
http://qualifications.pearson.com/content/dam/pdf/A%20Level/Chemistry/2013/Exam%20materials/6CH04_01_que_20130114.pdfquestion 8 help please. Finding order wrt to Br-
Original post by Dinaa


The paper doesn't come up
Can you link the paper again?


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Reply 566
Avatar for Dinaa
Dinaa
17
Original post by Bloom77
The paper doesn't come up
Can you link the paper again?


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http://qualifications.pearson.com/content/dam/pdf/A%20Level/Chemistry/2013/Exam%20materials/6CH04_01_que_20130114.pdf
January 2013 unit 4 question 8- edexcel. :redface:
Reply 567
Avatar for voltz
voltz
6
Yeah, including that it has 3 if Im correct, 2 to the hydrogens and 1 to the B, and then a lone pair
Original post by Dinaa


Is the answer d) alizRin yellow R
Because you are adding NaOH into ethanoic acid and NaOH is a strong base so the pH would go very high and by looking at the data booklet, the highest pH indication is D) alizarin yellow R


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Original post by Bloom77
Is the answer d) alizRin yellow R
Because you are adding NaOH into ethanoic acid and NaOH is a strong base so the pH would go very high and by looking at the data booklet, the highest pH indication is D) alizarin yellow R


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I would disagree since we aren't looking at the end point but in fact the equivalence point. The equivalence point for the titration between a weak acid and strong base is 9 so from looking at the data booklet I would say the answer is C, thymol blue (base) since the pH range for thymol blue (base) is 8.0-9.6 and 9 is in that range (none of the other options have a suitable range). Hope that helps, @dinaa
can someone help me with percentage uncertainty
a question for the set 2 papers of the new spec is confusing me
because usually %uncertainty=error?reading

the eroor is 5mg for would it just be 0.005g/ the reading x 100

but it shows eroor to be 0.001
Original post by dpoojaraa
I would disagree since we aren't looking at the end point but in fact the equivalence point. The equivalence point for the titration between a weak acid and strong base is 9 so from looking at the data booklet I would say the answer is C, thymol blue (base) since the pH range for thymol blue (base) is 8.0-9.6 and 9 is in that range (none of the other options have a suitable range). Hope that helps, @dinaa


Yep, I agree with you, i was looking at end point. Woopsies.


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Reply 574
Avatar for Dinaa
Dinaa
17
Original post by Bloom77
Is the answer d) alizRin yellow R
Because you are adding NaOH into ethanoic acid and NaOH is a strong base so the pH would go very high and by looking at the data booklet, the highest pH indication is D) alizarin yellow R


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Oops I meant q18 and naw, the answer to 8 is C:3
Original post by Dinaa
Oops I meant q18 and naw, the answer to 8 is C:3


What have you tried so far? Is it the whole of 18 or just some parts
Reply 576
Avatar for Dinaa
Dinaa
17
Original post by samb1234
What have you tried so far? Is it the whole of 18 or just some parts


I found the order wrt to [H+]

It is [Br-] that is confusing me.

Using experiment 1 and 3, the [H+] stays constant, [BrO3-] triples and [Br-] doubles. The rate is x6.
I don't know how to link all this information together to get an order wrt [Br-].
Original post by Thisshouldbeapun
Does anyone know what Kc actually is?
I'm so confused because i thought it showed where the position of equilibrium was for a reaction but if that is the case then surely it would change with pressure/concentration because these move the position of equilibrium?
Thanks :smile:




Kc actually stands for concentration constant ........questions related to Kc and Kp are usually calculations i mean i think i never solved a paper where they ask for the definition or something and in multiple choice questions its usually like they ask which factor effect the Kc or Kp and the answer always stays the same that is temperature ....anyways generally Kc and Kp are called equilibrium constant ...value of Kc is calculated using the equilibrium concentrations of the products and reactant.
Original post by Dinaa
I found the order wrt to [H+]

It is [Br-] that is confusing me.

Using experiment 1 and 3, the [H+] stays constant, [BrO3-] triples and [Br-] doubles. The rate is x6.
I don't know how to link all this information together to get an order wrt [Br-].


Well we are told in the question that the reaction is first order wrt [BrO3-], so if we just tripled BrO3 what would we expect the rate to be? Therefore what factor is the bromine responsible for
[QUOTE="Nightinwind;64956675"]
Original post by alyoan tariq


so as it reduces chromium from +6 to +3 (orange to green) then the substance must be oxidised so it cannot be a ketone ( not c ) it reacts with HCN and KCN so must be a carbonyl but not ketone so has to be an aldehyde and as one mole of it reacts with two moles of HCN then there must be 2 aldehyde groups so answer is D ?

thank u so much i got it .....yes the answer is D ......yea I already knew that it would be an aldehyde but I was confused on how it reacted in terms of the structural formula ....thank u once again ...:smile:

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