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    (Original post by Davelittle)
    Well the longest carbon chain was 8, the double bond was between the third and the 4th carbon, there were methyl groups on the 3rd and 7th Carbon.

    Therefore 3,7-dimethyloct-3-ene
    I believe if you draw out 2,6- dimethyloct-5-ene it's exactly the same- are there any specific naming rules for this, or are they both allowed? How many marks was it, in case they aren't
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    (Original post by ambbs)
    Anyone else hate that last 6 mark question on the advance notice? Thought the paper was going alright up until I got to that question, I didn't have a clue!


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    I didn't like it either!! I didn't see the N2O equations on the 2nd page.... D:
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    (Original post by King Hotpie)
    I believe if you draw out 2,6- dimethyloct-5-ene it's exactly the same- are there any specific naming rules for this, or are they both allowed? How many marks was it, in case they aren't
    I think you had to use the lowest numbers. (They may accept both though)


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    (Original post by King Hotpie)
    I believe if you draw out 2,6- dimethyloct-5-ene it's exactly the same- are there any specific naming rules for this, or are they both allowed? How many marks was it, in case they aren't
    It was 2 marks
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    (Original post by Davelittle)
    Did you get 41025 or around that?

    You didn't need to multiply the Mr of NaCl by 2!
    I got something like that, I remember putting 4.10 x 10 to the something, to 3 sig fig
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    (Original post by Jooost)
    what was the answer for the ppm to % questionhoiw many times larger was it???I got 7.8i think its wrong
    I got 7.8x10^8
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    (Original post by Sumi Prakash)
    I got 7.8x10^8
    I got that as well
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    (Original post by abzy1234)
    I thought it was a nice 6-marker. The answers are mainly on the bottom right paragraph of page 3, with some of the equations already there. I talked about how N2O can be broken down in the atmosphere by high frequency/energy UV radiation, giving the N2O -> N2 + O equation. Plus N2O itself can react with existing oxygen radicals in the atmosphere, so that may break it down also...

    I'm sure you would've picked up marks
    Thanks I waffled too much - I think half of what I wrote was crap but the other half sounds similar to what you put so that's good I think I just don't like the idea of the advance notice in general, I either feel like I'm using it too much in a question (and copying it word for word like I mostly did for the last question) or not enough!


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    (Original post by super121)
    I think you had to use the lowest numbers. (They may accept both though)


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    Well 2 is lower than 3! Hopefully they'll except both, it would be incredibly harsh not to. Does anybody remember any similar scenarios from a past paper? (Doesn't have to be F332, and Salters paper I expect)
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    2methyl7methyl oct3ene will they allow it?
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    Name:  20130604_162301.jpg
Views: 202
Size:  375.4 KBI got 2ethyl6methylhept-2-ene too
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    (Original post by Waki)
    2methyl7methyl oct3ene will they allow it?
    You should get a mark for the oct-3-ene bit
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    (Original post by ambbs)
    Thanks I waffled too much - I think half of what I wrote was crap but the other half sounds similar to what you put so that's good I think I just don't like the idea of the advance notice in general, I either feel like I'm using it too much in a question (and copying it word for word like I mostly did for the last question) or not enough!


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    That's alright, you would've still picked up majority of the marks then And ikr. Salters has to be extra At least it's over now!
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    Some questions;

    1. Was the alcohol tertiary?

    2. Will 0.78 x10^9 be ok...

    3. For the curly arrows... what did you have to do?
    I drew an arrow from the lone pair of F to the C
    then 4 arrows from the covalent bonds to the Chlorine atoms...
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    (Original post by Waki)
    2methyl7methyl oct3ene will they allow it?
    Yea, they should if they accept it that way around.


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    (Original post by super121)
    780 000 000 or something along those lines


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    how did you work it out?
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    Urgh, now time to prepare for F335 and F334
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    (Original post by Jooost)
    how did you work it out?
    You either had to convert them both to ppm or percentage then do nitrogen over the other one.


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    What was the inorganic product formed? if anyone remembers this question..
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    (Original post by steviep14)
    Some questions;

    1. Was the alcohol tertiary?

    2. Will 0.78 x10^9 be ok...

    3. For the curly arrows... what did you have to do?
    I drew an arrow from the lone pair of F to the C
    then 4 arrows from the covalent bonds to the Chlorine atoms...
    I got tertiary
    It should be cos thats the same as 780,000,000
    I did the same for the first arrow but then I drew only one arrow from one covelant bond to the chlorine :/
 
 
 
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