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    Hi can someone please show me how to do rate-determining steps, and maybe give two different examples?
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    (Original post by bluedate)
    Hi can someone please show me how to do rate-determining steps, and maybe give two different examples?
    I watched videos on youtube
    I also remember to start by writing out the final equation and writing the starting of the first line with the rate equation components, then play around with stuff until I get the same on each side totally to the final equation and other stuff cancelling out.
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    (Original post by otrivine)
    Exactly!

    you know the legacy papers right , are they the ones on the OCR website?
    Yeah as far as I know, that's where I downloaded mine from too
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    Does anybody know how to do spoilers on here? I've answered Gulzar's question but the answer is quite large - I don't want it to take up the thread!
    Thank you
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    (Original post by Funtry)
    I use Chemguide, Knockhardy and These questions primarily, as well as my own notes.
    Yeah I've got those sources down too

    Cool, thank you
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    Hey,

    Regarding the topic of how non-standard conditions affect electrode potenital values, I think I understand the concept, but does anyone know what the best way is to show understanding in an exam as in the Heinemann I feel the wording on this is slightly weird.

    This is on page 189 - the example on concentration.

    Thanks in advance
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    Ughhh retaking this exam after getting an E in January. Started looking at it yesterday and have scheduled around 5-6 days of full on revision/re-learning before the exam. I'm just totally ****ed. I hate F325 with a passion lol, F321 was very basic, I like organic chemistry but I'm really struggling with F325 o.O

    I can remember the simple things like [H+]=10^pH, pH= -log10 [H+], Kc = Products/Reactants, Ka = [H+] [A-] / [HA] etc etc but when it comes to the question in the exams I just have no clue and can't apply what I know. I think one of the main problems I have is that because I don't take maths, when rearranging the formula eg. if Ka = [H+] [A-] / [HA] then [H+] = square root of (Ka * [HA] ) I just have to try and remember both instead of understanding why it can be rearranged like that.

    No real reason for posting this but oh well, just felt like ranting after having a stressful few hours trying to get to grips with it lol.
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    (Original post by zef1995)
    Yeah as far as I know, that's where I downloaded mine from too
    Spoiler:
    Show
    You can do a spoiler by putting [SPOILER.] your text and then [/SPOILER.] removing the full stops.


    quoted wrong post, sorry
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    (Original post by Funtry)
    Spoiler:
    Show
    You can do a spoiler by putting [SPOILER.] your text and then [/SPOILER.] removing the full stops.


    quoted wrong post, sorry
    Ah thank you! I will be sure to remove the full stops :P
    Thanks again
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    (Original post by Gulzar)
    Sure
    I've got it down as 5V2+ + 3MnO4- + 3H2O ------> 5VO3- + 3Mn2+ + 6H+
    Spoiler:
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    Thank you very much

    I've had another go at the question, and I've got the answer you posted (Yay!).

    Now we know that n=2, we can construct the equation for the reaction:

    MnO4- +8H+ +5e- -> Mn2+ +4H2O

    As n=2, vanadium is oxidised from V2+ to VO3-, and we need to construct a half equation for this reaction.

    First we can put down what we know: V2+ -> VO3- + 3e- (I've put 3e- because vanadium increases its oxidation state by 3)
    As you can see, VO3- contains oxygen - but where did these come from? As there are no oxygens in V2+, we can add oxygen by adding water on the left side of the equation:

    V2+ + H2O -> VO3- +3e-

    However, although we have oxygen on both sides, we have no hydrogens! We can sort this by adding H+ to the right hand side:

    V2+ + H2O -> VO3- +3e- + H+

    We now have all the components of the half equation, we just need to balance them:

    We know there are 3 oxygens on the right, so we have 3 H2O. As we have 3 H2O, there are 6 hydrogens (so 6H+).

    Final half equation: V2+ +3H2O -> VO3- + 6H+ +3e-

    Now we have both half equations, we can work out an equation for the whole reaction:

    V2+ +3H2O -> VO3- + 6H+ + 3e-
    MnO4- + 5e- + 8H+ -> Mn2+ + 4H2O

    3e- for Vanadium equation, and 5e- for the Manganese one. To balance out the electrons, multiply Vanadium equation by 5 and Manganese's by 3 to give 15e- in both equations.

    5V2+ + 15H2O -> 5VO3- + 30H+ + 15e-
    3MnO4- + 15e- + 24H+ -> 3Mn2+ + 12H2O

    Combined to give: 5V2+ + 15H2O + 3MnO4- +15e- +24H+ -> 3Mn2+ + 12H2O + 15e- + 30H+ + 5VO3-

    As you can see, we have 15e- on each side (these cancel), and water and protons on both sides. The equation can therefor be simplified to:

    5V2+ + 3MnO4- +3H2O -> 3Mn2+ + 6H+ + 5VO3-

    I know that was very long winded, I just wanted to make sure I covered everything....I hope that's useful!

    If you have any questions, please feel free to ask

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    do we need to know colours of the transition metal ions?
    The book says "you dont need to know why they are coloured but be aware of the colours" in the examiners' tip blue box but there is nothing in the spec... :/
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    (Original post by ofudge)
    do we need to know colours of the transition metal ions?
    The book says "you dont need to know why they are coloured but be aware of the colours" in the examiners' tip blue box but there is nothing in the spec... :/
    Not all, but I've learnt Iron, Cobalt, Copper and Manganese's. Also going to learn Chromium's and Vanadium's as they tend to appear quite a bit then might dedicate some time to learning the last two
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    How many past papers has everyone done, and what are your average scores??

    And same for Chem 4 please?
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    (Original post by Better)
    How many past papers has everyone done, and what are your average scores??

    And same for Chem 4 please?
    no point comparing yourself to others...as we all work at our own pace and abilities. I've done none since I want to do them well my knowledge is solid as there is only 5 for f325 but I've done the mock (jan paper) and my friends done 2 of them and doing the rest these hols.
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    (Original post by MedMed12)
    no point comparing yourself to others...as we all work at our own pace and abilities. I've done none since I want to do them well my knowledge is solid as there is only 5 for f325 but I've done the mock (jan paper) and my friends done 2 of them and doing the rest these hols.

    I am sure once you did the packs of question you will get a good mark

    They are really excellent
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    (Original post by otrivine)
    I am sure once you did the packs of question you will get a good mark

    They are really excellent
    need to start those :P but thank you!
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    (Original post by MedMed12)
    no point comparing yourself to others...as we all work at our own pace and abilities. I've done none since I want to do them well my knowledge is solid as there is only 5 for f325 but I've done the mock (jan paper) and my friends done 2 of them and doing the rest these hols.
    Examinations are competitive events which can be made comparable to any Sporting Eventing where there is an Objective Standard. This is a the very foundation of our modern Western Society, hence it is important to compare yourself to others to ensure you are sufficiently prepared to get the result you desire or rather need to get into University.

    It really is not rocket science, anyone who plays sports understands you need to collaborate with others to improve; it is imperative for success. If you don't do this you are truly selling yourself short and will increase your propensity to get a lower grade.

    Edit: As a side note, the people who I know at Oxbridge and other Top 5 Universities, all went to schools where they were told to finish all the past papers at least 1-2 weeks before at an A* level. There is a clear reason Private/Grammar Schools do significantly better - they understand how the system works, sorry but I don't agree with you at all.

    Has anyone else done all the Past Papers? And what type of scores are you getting. Thanks in advance
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    Is anyone able to explain to me how the equilibrium will shift with the change of pressure and concentration? I'm never really sure how to answer the shift changes part of the question but can easily do the equation and any calculating things too as I'm better in the Maths part than the understanding of the actual Chemistry :-/


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    (Original post by MathsNerd1)
    Is anyone able to explain to me how the equilibrium will shift with the change of pressure and concentration? I'm never really sure how to answer the shift changes part of the question but can easily do the equation and any calculating things too as I'm better in the Maths part than the understanding of the actual Chemistry :-/


    Posted from TSR Mobile
    Sure, it would me my pleasure to explain to you


    Pressure

    In equilibrium when in the question they ask for decrease/increase pressure, you always have to look at the number of moles.

    For example

    N2 + 3H2 equilibrium sign 2NH3

    the harber process for instance,

    If you want to increase the pressure , the equilibrium will shift to the fewer side of moles as you are applying Le Chateliers principle, the system always opposes the change.

    If you want to decrease the pressure the equilibrium will shift to the more side of moles .

    In the harber process you want to decrease the pressure as equilibrium will shift to the side which has less number of moles in this case being the NH3 and so product side is favoured and equilibrium yield is more.

    do you get the pressure now?
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    (Original post by otrivine)
    Sure, it would me my pleasure to explain to you


    Pressure

    In equilibrium when in the question they ask for decrease/increase pressure, you always have to look at the number of moles.

    For example

    N2 + 3H2 equilibrium sign 2NH3

    the harber process for instance,

    If you want to increase the pressure , the equilibrium will shift to the fewer side of moles as you are applying Le Chateliers principle, the system always opposes the change.

    If you want to decrease the pressure the equilibrium will shift to the more side of moles .

    In the harber process you want to decrease the pressure as equilibrium will shift to the side which has less number of moles in this case being the NH3 and so product side is favoured and equilibrium yield is more.

    do you get the pressure now?
    Okay thanks, I think my lack of understanding was because I never really took the time last year to understand Le Chatelier's principle, I guess I'll just have to read the question carefully and see what it's asking for before being able to respond with it. How would concentration effect the equilibrium though? Thanks for the help by the way
 
 
 
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