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    (Original post by buzzing22)
    anybody know how to answer 3.a.i)
    http://www.ocr.org.uk/images/61538-q...y-for-life.pdf
    From the table of the bond lengths, you can see a pattern that as you go down the group, the bond length increases so you just predict a bond length for the Pb-C bond to being a little higher for example 0.245.
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    (Original post by buzzing22)
    anybody know how to answer 3.a.i)
    http://www.ocr.org.uk/images/61538-q...y-for-life.pdf
    difference between the first two is 0.040, then 0.005 , then 0.018
    i would either to do it like this: 0.040+0.005+0.018/3 and then add your answer to the bond length of 0.217... OR...
    i would add the number thats not the biggest nor the smallest from the difference we worked out.. so if you have 0.040 (biggest) 0.005 (smallest) and 0.018 (middle) that way you wont be over or under the ranges theyve set ...if you get me
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    (Original post by Ali_Ludley)
    First question:
    I think I have a rough idea. You can see the question already gives you the ratio which is 1:12.5 and now you need to figure out the moles of iso-octane with moles=mass/fm and from there you can figure out the dm^3 of oxygen required using Avagadro's law and then that answer equals 21% (Only oxygen volume and you need air) so find what 100% is. That should be your answer of air in dm^3. Try following those steps, I 'think' its right.

    Second question:
    Ga + As -> GaAs
    This gets you like 1 mark but its the state symbols that matter. It gives you the melting point of each element which are all above 298 kelvin which is 25 degrees Celsius (Standard conditions) so the elements will all be solid under standard conditions which enthalpy change of formation is under!
    i still dont get the % question
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    (Original post by Izar)
    i still dont get the % question
    Just messed up this question for the third time - I think it's really confusing in what it wants from you

    anyway,

    Ratio of Isooctane to oxygen is 1:12.5

    So 60cm^3 isooctane gives 12.5 times this of oxygen

    60 x 12.5 = 750cm^3 (this is the 21%)

    so just divide by 21 to get 1% - and times by 100 to get 100% (air)

    750 x 100/21 = 3571.42...cm^3

    answer is wanted in dcm^3 so divide by 1000 to get 3.57dm^3 (2dp)

    ---

    now can anyone help me in working out whether an element will be simple molecular or giant covalent network??

    like what would P and As be?
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    (Original post by Izar)
    i still dont get the % question
    annoyingly it doesn't want anything to do with moles (although you maybe could work it out like this) Although it states that it is not standard temp. and pressure so you can't assume 1 mole of isooctane is 24dm^3

    which I think I did in the actual exam -.-
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    June 2012 mark scheme?
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    Btw, if anyone needs help on knowing which state symbol to apply, see
    http://www.thestudentroom.co.uk/show...2#post40947132
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    The exam is over. How did you guys find it?
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    (Original post by EstebanK0)
    The exam is over. How did you guys find it?
    Smashed it iI think, easier than in the summer. How about you?

    I think grade boudries will be 49/50 for an A this exam
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    (Original post by EstebanK0)
    The exam is over. How did you guys find it?
    Went well for me! What did you get for the empirical formula or the energy density question?
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    I got CeO2 for empirical (not sure if its right, most people seemed to have CeO that i asked) and energy density was it -4.7x104, apart from calculations i thought the paper was reasonable
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    I got Ce2O4 for empirical...

    Easier paper than June!
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    I got something like -534 for the energy density and CE02 for empirical. What did every1 get for the bond angle? 120?
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    (Original post by Jozzhell)
    I got Ce2O4 for empirical...

    Easier paper than June!
    Should'nt the simplest ratio then be CeO2 (either way the wording on that question wasnt great), what did you get for bond angles a and b if you dont mind me asking?
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    (Original post by exams567)
    I got something like -534 for the energy density and CE02 for empirical. What did every1 get for the bond angle? 120?
    yeah i got the same empirical, first bond angle i had 120 2nd was 109
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    (Original post by ThePanoramicGame)
    Should'nt the simplest ratio then be CeO2 (either way the wording on that question wasnt great), what did you get for bond angles a and b if you dont mind me asking?
    Yeah, some parts of the paper were very vague

    I got 120 degrees for both, due to three sets of electrons.

    Yourself?
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    (Original post by exams567)
    I got something like -534 for the energy density and CE02 for empirical. What did every1 get for the bond angle? 120?
    For the energy density I think you divide 1000 by mr and then times by the enthalpy value. Bond angles was 120 then 109
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    first i had 120 but for the 2nd i wrote 109 (wasnt too sure about it at the time)
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    (Original post by Isaacymoose)
    For the energy density I think you divide 1000 by mr and then times by the enthalpy value. Bond angles was 120 then 109
    yeah i did that but i divided by enthalpy value rather than multiply
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    (Original post by ThePanoramicGame)
    yeah i got the same empirical, first bond angle i had 120 2nd was 109
    Ah, dang it was 109! Totally forgot about the hydrogen!

    What did you all get as the extra piece of information he needed to calculate enthalpy change?

    I remember I got 47000 for something too haha
 
 
 
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