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    Hiii couldn't find a thread for this exam so I created this one.
    Does anybody have the question paper and mark scheme for the Jan 2013 unit 1 exam ??
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    Yeah I do


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    (Original post by Ifa_94)
    Hiii couldn't find a thread for this exam so I created this one.
    Does anybody have the question paper and mark scheme for the Jan 2013 unit 1 exam ??
    As per usual, Free Exam Papers does. I've attached it for convenience
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  1. File Type: pdf 6CH01_01_que_20130110.pdf (475.5 KB, 2822 views)
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    (Original post by adi19956)
    As per usual, Free Exam Papers does. I've attached it for convenience
    Thank You so much!!
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    Is this everyone's first time or are they resitting it? I'm resitting it as I got a C in January...hopefully I'll get an A this time round!
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    Hi guys. I sat this paper in January and failed It was awful compared to past papers. There were some stupid mc questions and I was wondering if anyone could help explain this one to me. Apparently the answer is A (z+4 neutrons) but how if neutrons = mass number - atomic number then surely 2z+4 - z is not z + 4.

    Im very bad at maths so it might just be me

    An isotope of an element, atomic number z, has mass number 2z + 4. How many
    neutrons are in the nucleus of the element?
    A z + 4
    B z + 2
    C z
    D 4
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    (Original post by felicity95)
    Hi guys. I sat this paper in January and failed It was awful compared to past papers. There were some stupid mc questions and I was wondering if anyone could help explain this one to me. Apparently the answer is A (z+4 neutrons) but how if neutrons = mass number - atomic number then surely 2z+4 - z is not z + 4.

    Im very bad at maths so it might just be me

    An isotope of an element, atomic number z, has mass number 2z + 4. How many
    neutrons are in the nucleus of the element?
    A z + 4
    B z + 2
    C z
    D 4

    Well firstly an isotope has the same number of protons but different number of neutrons. Replace z for any number i.e 6 and since z is the atomic number this element would be Carbon. then it says it has mass number 2z + 4, so when you replace z with 6 you get 16. and 16 is 4 more than 12 (which is Carbon's mass number) so it must be z + 4.

    Hope this helped
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    (Original post by kevsamuel)
    Well firstly an isotope has the same number of protons but different number of neutrons. Replace z for any number i.e 6 and since z is the atomic number this element would be Carbon. then it says it has mass number 2z + 4, so when you replace z with 6 you get 16. and 16 is 4 more than 12 (which is Carbon's mass number) so it must be z + 4.

    Hope this helped
    Thank you! much better x
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    (Original post by kevsamuel)
    Well firstly an isotope has the same number of protons but different number of neutrons. Replace z for any number i.e 6 and since z is the atomic number this element would be Carbon. then it says it has mass number 2z + 4, so when you replace z with 6 you get 16. and 16 is 4 more than 12 (which is Carbon's mass number) so it must be z + 4.

    Hope this helped
    Very well explained! I was having the same problem. Thanks


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    Hi guys
    I was just wondering if someone could tell me how to answer the multiple choice questions when it asks you to work out the number of moles of a substance and gives you the Avogadro constant?
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    (Original post by jemma01)
    Hi guys
    I was just wondering if someone could tell me how to answer the multiple choice questions when it asks you to work out the number of moles of a substance and gives you the Avogadro constant?
    You work it out using:
    Moles= no. of particles /Avogadro constant

    then tick the box with the answer that matches yours (sorry, couldn't resist) .
    Be on the lookout for whether they have given you atoms or molecules to work with.
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    If you want the January 2013 question paper and mark sheet, here they are:
    Attached Images
  2. File Type: pdf Jan 2013 Chem Unit 1 QP.pdf (466.7 KB, 368 views)
  3. File Type: pdf Jan 2013 Chem Unit 1 MS.pdf (245.5 KB, 1300 views)
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    (Original post by kevsamuel)
    Is this everyone's first time or are they resitting it? I'm resitting it as I got a C in January...hopefully I'll get an A this time round!
    I'm a resitter, simply to boost up my grade for A2.
    I'm doing Unit 5 in a month too, so Unit 1 is a good like memory jogger.
    I really need full UMS in this if I want an A, so it's a tight margin.

    Any questions, you can ask me and I'll do my best to try and answer them
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    I'm resitting too

    Will be starting revision tomorrow for this... I need as many UMS as I can get As at the moment I'm on a D & need a B !

    But still it's good practice anyway I guess I hope this thread become more active!

    Good luck with revision everyone !
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    got 77 marks in jan dont know why im resitting lol
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    Are u serious


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    (Original post by Pirateprincess)
    You work it out using:
    Moles= no. of particles /Avogadro constant

    then tick the box with the answer that matches yours (sorry, couldn't resist) .
    Be on the lookout for whether they have given you atoms or molecules to work with.
    Thanks!
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    Hi

    In January there was this question on bond enthalpies and can someone please explain how you calculate them. I always thought you multiply each value by the amount of that type of bond e.g. for C2H6 there would be 6 x C-H bond. Ive attached the question and answer so id appreciate any help. thanks
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    (Original post by felicity95)
    Hi

    In January there was this question on bond enthalpies and can someone please explain how you calculate them. I always thought you multiply each value by the amount of that type of bond e.g. for C2H6 there would be 6 x C-H bond. Ive attached the question and answer so id appreciate any help. thanks
    You were on the right track, so you just need to follow up on it.
    On the left hand side (LHS), you have 6 C-H bonds and 1 Cl-Cl bond which equates to-> 6(413)+243
    On the RHS, you have 5 C-H bonds, 1 C-Cl bond, and 1 H-Cl bond which equals to-> 5(413)+346+432
    LHS=2721
    RHS=2843
    Enthalpy Change=LHS-RHS
    2721-2843=-122 kJ mol^-1
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    (Original post by felicity95)
    Hi

    In January there was this question on bond enthalpies and can someone please explain how you calculate them. I always thought you multiply each value by the amount of that type of bond e.g. for C2H6 there would be 6 x C-H bond. Ive attached the question and answer so id appreciate any help. thanks
    Bonds broken subtract bonds made.
    It's an exothermic energy change, so

    You're right for the 6 x C-H bond which equals 2478 KJmol-1
    Then 243 for the Cl-Cl bond KJmol-1

    The sum of these is 2721.

    On the products side it's 5 x 413 = 2065 KJmol-1
    +346 + 432 = 2843 KJmol-1

    Then, we do 2843 - 2721 = 122 KJmol-1

    It's an exothermic energy change because more more energy is used to break bonds, than make the bonds, so we change the sign to -, so it's -122 KJmol-1
 
 
 
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