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OCR Chemistry F322~ 4th June 2013~ AS Chemistry Watch

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    Hard A= 72
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    Easy A= 79
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    Very Easy A=82
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    (Original post by Simone739)
    Can someone explain boltzmann distribution?
    The Boltzmann distribution is a graph showing the number of molecules that have a certain energy the area under the curve is the number of molecules the graph always starts at the origin as no molecules have 0 energy and the graph never touches the x axis okay the basics out of the way now if you increase the temperature the graph shifts to the right as more molecules now have more energy and the peak drops a little also if you increase the temperature more molecules exceed the activation energy so there are more collisions and the rate is quicker remember that the activation energy doesn't change with temperature but more molecules exceed it now if you add a catalyst the activation energy now decreases (Ec) so more molecules exceed the activation energy more collisions and yeh so on thats basically it unless i have missed something out but good luck and hope it goes well


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    (Original post by Jimmy20002012)
    If you were meant to draw a molecule such as 1.4-dicholobutane, would they accept the vertical to be placed on the chlorines?


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    I don't understand your question? What do you mean by the vertical?
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    When drawing skeletal form does it matter where I start drawing the zig zag from? For example if I wanted to draw pentane, would it matter if it was a 'W' shape or an 'M' shape?
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    (Original post by Kits)
    When drawing skeletal form does it matter where I start drawing the zig zag from? For example if I wanted to draw pentane, would it matter if it was a 'W' shape or an 'M' shape?
    Not at all
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    (Original post by needtosucceed=))
    They're not both at the start of the chain as such, it's just that's how you write structual formula when the two methyl groups are on one carbon. I've attached a drawing

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    Hey thanks
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    (Original post by needtosucceed=))
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    I don't understand your question? What do you mean by the vertical?
    Sorry, didn't probably explain the question well Just look at the attachment posted, would they accept 1,4-dichlorobutane in both ways?


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    (Original post by Jimmy20002012)
    Sorry, didn't probably explain the question well Just look at the attachment posted, would they accept 1,4-dichlorobutane in both ways?


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    Yes, I answered this question already yesterday. The atoms rotate around carbon atom so it doesn't matter
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    (Original post by .raiden.)
    Yes, I answered this question already yesterday. The atoms rotate around carbon atom so it doesn't matter
    Cool, thanks How's revision going?


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    (Original post by That_Clever_Guy)
    The Boltzmann distribution is a graph showing the number of molecules that have a certain energy the area under the curve is the number of molecules the graph always starts at the origin as no molecules have 0 energy and the graph never touches the x axis okay the basics out of the way now if you increase the temperature the graph shifts to the right as more molecules now have more energy and the peak drops a little also if you increase the temperature more molecules exceed the activation energy so there are more collisions and the rate is quicker remember that the activation energy doesn't change with temperature but more molecules exceed it now if you add a catalyst the activation energy now decreases (Ec) so more molecules exceed the activation energy more collisions and yeh so on thats basically it unless i have missed something out but good luck and hope it goes well


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    Thank you so much!
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    (Original post by supernova99)
    Just thought i'd say that i would advise people to draw the hess's law cycle rather than remembering the equations of products-reactants/reactants-products just because it is SO easy to make a mistake when you try to remember it. I never bothered to learn it properly the first time i did it last year and it was always hit or miss as to whether i got the values right. Now I can do the cycle it's much easier to work out enthalpy changes and even if you get it wrong it's much easier for examiners to award marks from a cycle rather than just a long list of calculations that aren't clear. Just a tip !

    ok so i basically know how to do the hess cycle but the only problem is i dont understand where i put the unknown value symbol , i.e on the product or reactant side of the equation. i mean where i put it determines whether i get a positive or negative value and its confusing me . i don't know if that made sense but heres an example. -1576 + -1430 + x = -2877
    should the x be there or on the other side when working out the calculation? this is for the standard enthalpy change of formation of butane. someone please help!!! got exam in four days and freaking out :eek:
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    In the jan 13 paper, q7 involved both electrophilic addition and nucleophilic substitution, so do i do the mechanisms separate or together? because the mark scheme was too confusing to follow
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    (Original post by Sara_A)
    In the jan 13 paper, q7 involved both electrophilic addition and nucleophilic substitution, so do i do the mechanisms separate or together? because the mark scheme was too confusing to follow
    Do them separately, just to organise and structure your answer


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    Also, in a few papers, they asked to give stereoisomers for A and B then a structural isomer for C. I get how to do the stereoisomers but im not sure on what to draw when they asked for the structural isomer
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    (Original post by Jimmy20002012)
    Do them separately, just to organise and structure your answer


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    Ok thank you
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    Hey guys!

    I was wondering if anyone could help with a little prediction of the UMS conversion?
    I just completed a mock and scored 97/100.
    Do you think this is full UMS!?

    You just never know when OCR will f**k us students over!
    If you could let me know, I would be grateful.



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    (Original post by Ché.)
    Hey guys!

    I was wondering if anyone could help with a little prediction of the UMS conversion?
    I just completed a mock and scored 97/100.
    Do you think this is full UMS!?

    You just never know when OCR will f**k us students over!
    If you could let me know, I would be grateful.



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    Yupp that will deffo be full ums
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    How are you guys revising for this? I can't get anything to stick im my head LOL
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    (Original post by Sara_A)
    Also, in a few papers, they asked to give stereoisomers for A and B then a structural isomer for C. I get how to do the stereoisomers but im not sure on what to draw when they asked for the structural isomer
    The double bond will be in a different position in the molecule.

    So A and B might be the E and Z isomers of but-2-ene and therefore C would be but-1-ene.
    Remember structural isomerism = same molecular formula but different structural formula. So A and B are both CH3CH=CHCH3 whereas C is CH3CH2CH=CH2.
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    (Original post by Smko)
    Yupp that will deffo be full ums
    Thank you ever so much!
    I just need to smash Tuesday JUST how I done it now!

    Thanks, again.
    Additionally, how is revision going for everyone on TSR!?



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    (Original post by Smko)
    Yupp that will deffo be full ums
    What mark is full ums?


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