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    Aww man i was quite confident about that test aswell when i walked out of it..now i read the questions and realised ive made some stuuuuuuuuuuuuuupid pretty bad mistakes....well there goes my A down the drain
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    (Original post by Mr_Muffin_Man)
    Aww man i was quite confident about that test aswell when i walked out of it..now i read the questions and realised ive made some stuuuuuuuuuuuuuupid pretty bad mistakes....well there goes my A down the drain
    That's the drawback of looking at papers and tallying answer
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    (Original post by Monkeyy)
    But when you remove an electron from any subshell, the Ionisation energy becomes higher, right?
    Why is it, when you remove a second electron, Lithium suddenly has the highest I.E??? But.... I don't know... Maybe you're right, maybe I am...
    No I'm right!
    Look, I'll explain:
    When you remove an electron from each of these atoms, you're left with this electron configuration
    He --> 1s1
    Li --> 1s2
    Be --> 1s2 2s1
    B --> 1s2 2s2

    Now, obviously, lithium would have the highest ionisation energy because you need A LOT of energy to remove an electron from a complete subshell.

    Does that make sense?
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    :ahee:
    (Original post by Narik)
    No I'm right!
    Look, I'll explain:
    When you remove an electron from each of these atoms, you're left with this electron configuration
    He --> 1s1
    Li --> 1s2
    Be --> 1s2 2s1
    B --> 1s2 2s2

    Now, obviously, lithium would have the highest ionisation energy because you need A LOT of energy to remove an electron from a complete subshell.

    Does that make sense?
    :ahee: AHAAAAA!!! okay... I think i'm getting it. Yep, I guess you're right. Lithium will have the configiration of 1s2 which is the same as Helium, which is the Highest. Right, right.. Thanks..
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    (Original post by Monkeyy)
    :ahee:

    :ahee: AHAAAAA!!! okay... I think i'm getting it. Yep, I guess you're right. Lithium will have the configiration of 1s2 which is the same as Helium, which is the Highest. Right, right.. Thanks..
    ahah no problem
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    ok im quite annoyed about question 5
    and its bugging me loads :P

    Which equation represents the reaction for which the enthalpy change, ΔH, is the mean
    bond enthalpy of the C–H bond?
    A ¼CH4(g) → ¼C(g) + H(g)
    B CH4(g) → C(s) + 2H2(g)
    C CH4(g) → C(g) + 4H(g)
    D CH4(g) → C(g) + 2H2(g)


    im gonna try using my explanation...
    the question is asking us which reaction will have an enthalpy change that has a value for the mean bond enthalpy of the C-H bond

    we can scrap B and D...
    leaving us with A and C

    From the Edexcel data book... the mean bond enthalpy of the C-H bond is 412 kJ/mol

    The enthalpy change of reaction for option C is 1648 kJ/mol ... because u need to break 4 C-H bonds...

    However the question asks for a value that should be 412 kJ/mol

    the only way that you can achieve that is if you break 1/4 of a methane molecule.... namely ONE C-H bond....

    does no one else see my logic?
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    (Original post by killfestab)
    ok im quite annoyed about question 5
    and its bugging me loads :P

    Which equation represents the reaction for which the enthalpy change, ΔH, is the mean
    bond enthalpy of the C–H bond?
    A ¼CH4(g) → ¼C(g) + H(g)
    B CH4(g) → C(s) + 2H2(g)
    C CH4(g) → C(g) + 4H(g)
    D CH4(g) → C(g) + 2H2(g)


    im gonna try using my explanation...
    the question is asking us which reaction will have an enthalpy change that has a value for the mean bond enthalpy of the C-H bond

    we can scrap B and D...
    leaving us with A and C

    From the Edexcel data book... the mean bond enthalpy of the C-H bond is 412 kJ/mol

    The enthalpy change of reaction for option C is 1648 kJ/mol ... because u need to break 4 C-H bonds...

    However the question asks for a value that should be 412 kJ/mol

    the only way that you can achieve that is if you break 1/4 of a methane molecule.... namely ONE C-H bond....

    does no one else see my logic?
    I don't understand what's so hard about it. You're trying to find the reaction that represents the MEAN bond enthalpy. Using reaction C, you will get four values of the C-H bond, which is what you want, so you can divide your answer by 4, and work out the AVERAGE OR MEAN bond enthalpy for a C-H bond.
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    (Original post by Narik)
    I don't understand what's so hard about it. You're trying to find the reaction that represents the MEAN bond enthalpy. Using reaction C, you will get four values of the C-H bond, which is what you want, so you can divide your answer by 4, and work out the AVERAGE OR MEAN bond enthalpy for a C-H bond.
    Agreeed.
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    can someone tell me what grade will be 43 or 44/80??
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    (Original post by Narik)
    I don't understand what's so hard about it. You're trying to find the reaction that represents the MEAN bond enthalpy. Using reaction C, you will get four values of the C-H bond, which is what you want, so you can divide your answer by 4, and work out the AVERAGE OR MEAN bond enthalpy for a C-H bond.

    C: CH4(g) → C(g) + 4H(g)

    Dividing it by 4

    CH4(g) / 4 → C(g)/4 + H(g)

    or

    1/4 CH(4) → 1/4 C (g) + H (g)
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    (Original post by killfestab)
    C: CH4(g) → C(g) + 4H(g)

    Dividing it by 4

    CH4(g) / 4 → C(g)/4 + H(g)

    or

    1/4 CH(4) → 1/4 C (g) + H (g)
    This represents the bond enthalpy NOT the MEAN bond enthalpy
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    the bond enthalpy would be

    CH4(g) → CH3(g) + H (g)
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    The question asked you which equation will give an enthalpy change of reaction that is equal to the mean bond enthalpy of a C-H bond. The enthalpy of reaction for answer C will clearly give you the enthalpy change for 4 bonds of C-H. You cannot decide to divide answer C by 4. This has been done on answer A, which is the correct answer.
    As for question 7. the secondionisation energy of Lithium is 75.614 and that for helium is 54.403, therefore the answer is B.
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    can someone tell me what grade will be 43 or 44/80??
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    (Original post by moh2009)
    Unit 2 with A2 Units? Wow man. I could never do that unless I had no choice, any particular reason you're doing it in summer instead of January? Mines next Thursday.
    LOL... I dont have a choice...im doing 1yr intensive a-levels but because it's been a while since im doing chemistry again, my understanding of unit 2 topics was sooo bad :-s and i became a wimp and decided to do it in june instead lol, so I have enough time to practise etc.

    Few ppl on the intensive course are doing it next thursday too, but in unit 2 mock exams they've been getting decent marks :-s... gooood luck for it!
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    (Original post by pingu09)
    LOL... I dont have a choice...im doing 1yr intensive a-levels but because it's been a while since im doing chemistry again, my understanding of unit 2 topics was sooo bad :-s and i became a wimp and decided to do it in june instead lol, so I have enough time to practise etc.

    Few ppl on the intensive course are doing it next thursday too, but in unit 2 mock exams they've been getting decent marks :-s... gooood luck for it!

    Oh I seeeee. Fair dues. And all the best to you as well!
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    i need chem unit 2 question papers..urgent!!
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    (Original post by killfestab)
    C: CH4(g) → C(g) + 4H(g)

    Dividing it by 4

    CH4(g) / 4 → C(g)/4 + H(g)

    or

    1/4 CH(4) → 1/4 C (g) + H (g)
    Yeah but that isn't the reaction to work out the mean bond enthalpy. You don't divide the reaction by 4, you divide you're value from the reaction by 4.
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    (Original post by Marsha2112)
    Word for word SNAP!

    High Five! :five:
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    (Original post by RA92)
    can someone tell me what grade will be 43 or 44/80??
    I would guess B. It was a hard paper, and the lowest chemistry has been in the past is about 49 for an A, so i would assume high 40s is A
 
 
 
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