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Edexcel A2 Chemistry 6ch04/05 JUNE 2015

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For question 2 here, as the Kc equation is being flipped, would the value be 1/40?


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Original post by bluegreenjade
Because the stoichiometry of NaOH to the acid and the sodium salt is the same. So depending on the no. of moles of NaOH used in the question find the how much volume would give you the conc. of salt (0.075 moles per diameter cube) and find the volume (of NaOH) in the x axis of the graph and find the corresponding y value (pH).I could be wrong though :/


Thank you, I've been reading it as 0.75mol all alone so I had no clue where the extra salt came from thanks!


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Original post by rowanbarnesss
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For question 2 here, as the Kc equation is being flipped, would the value be 1/40?


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i think so because lets forexample tae conc of pcl5 as 2 and conce of pcl3 as 3 and conc of cl2 as 2
then first kc wheb pcl5 is in the product side will be 2/6 which is 1/3
and if pcl5 was in the reactant side then kc=6/2 which equals 3
correct me if im wrong
uhmm.. quick help , can you please tell me how this works ?
Aluminium carbide, Al 4C3, reacts readily with aqueous sodium hydroxide. The two products of thereaction are NaAlO2 and a hydrocarbon. Water molecules are also involved as reactants.What is the formula of the hydrocarbon?
A CH4 B C2H6 C C3H8 D C6H12
[QUOTE="AnishaJayne;56736451"]Is anyone able to explain this please?iv

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The conc of acid is 0.15. So when half of it is neutralised 0.075 mol dm^-3 of products will form. The acid and alkali are presnt in equal amounts halfway up the vertical section of the graph

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(edited 8 years ago)
Could anyone briefly explain this question? I think the answer is that in both cases there would be no effect but it's seems odd for that to be correct

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Original post by NotYourType
uhmm.. quick help , can you please tell me how this works ?
Aluminium carbide, Al 4C3, reacts readily with aqueous sodium hydroxide. The two products of thereaction are NaAlO2 and a hydrocarbon. Water molecules are also involved as reactants.What is the formula of the hydrocarbon?
A CH4 B C2H6 C C3H8 D C6H12


Is the ans C ?
Original post by Undisclosed 15
Could anyone briefly explain this question? I think the answer is that in both cases there would be no effect but it's seems odd for that to be correct

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i think ur correct no effect in both because there is no argon in either the reactants or product and it is inert so wont react with anything so wont affect the conc of reactants or products
Original post by Raghadabuali
Is the ans C ?


No , it's A for some odd reason.
Original post by NotYourType
No , it's A for some odd reason.

Woww i realldy dont get it
Original post by Raghadabuali
Woww i realldy dont get it


Btw where did u get this question from ?
well.. this is a CIE paper.
Original post by Raghadabuali
Btw where did u get this question from ?
Original post by NotYourType
well.. this is a CIE paper.

Well maybe they have it in their syllabus
but we dont
Original post by Undisclosed 15
Could anyone briefly explain this question? I think the answer is that in both cases there would be no effect but it's seems odd for that to be correct

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in part a I think the equilibrium yield of ammonia would increase, as adding another gas into the mix, while maintaining the same volume, should increase the pressure of the system, as more gas molecules are occupying the same amount of space. Increasing pressure of a system favours the side of fewer moles of gases, which is the right hand side where ammonia is, therefore the yield increases. For part b I think there's no effect.


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(edited 8 years ago)
Original post by Raghadabuali
Well maybe they have it in their syllabus
but we dont


Oh , probabbly..Sorry.
This is really simple but I can never get my head around it, why is it at 12.5cm?

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pl.png Hey guys, in this question I get that you'd use 50cm^3 x 4.2 x 22.5 since theyve given a formula to use but dont you usually in these types of questions, times the mass of water in kg by the temp change and 4.2 (mass of water x 4.2 x temp change)?
(edited 8 years ago)
Original post by AnishaJayne
This is really simple but I can never get my head around it, why is it at 12.5cm?

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Hiya!
Take the vol of vinegar to be the vol of ethanoic acid

moles of ethanoic acid= 25x0.125/1000= 3.125x10^-3

vol of sodium hydroxide=
1:1 ratio so...
3.125x10^-3 x1000/0.250=12.5
Does anyone know where to find answers for end of chapter questions for the Phillip Allan George facer tb? I really need to do calculation practise but there are no answers :/
Original post by MeeraP07
Hiya!
Take the vol of vinegar to be the vol of ethanoic acid

moles of ethanoic acid= 25x0.125/1000= 3.125x10^-3

vol of sodium hydroxide=
1:1 ratio so...
3.125x10^-3 x1000/0.250=12.5


Ah thanks massive help


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