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    (Original post by Davelittle)
    I got 29/38 when I did that question last night

    I have no idea how to do Q2, would you mind walking me through it? :/
    Q2)
    Parts (a) to (d) are basic AS chemistry, so I assume you’ve done them
    e) i)Count the number of atoms within the cube: 4 atoms inside the cube, 6 atoms each in one face (half of each atom is inside the cube so we count them as 6 half-atoms). We also have 8 atoms each at a corner (only one eighth of each atom is inside the cube- this is 8 1/8-atoms). In total, we have 4+6*1/2+8*1/8=8 atoms in one cube.
    ii) Work out the volume of the cube in m3: you’re given its length in pm. The number of cubes in the sphere is its volume divided by the volume of one cube. The total number of atoms is the total number of cubes times the number of atoms in each cube (n).
    iii) Avogadro constant is the number of atoms per mole. The number of moles is mass/molar mass (m/Ar), so Avogadro constant it the number of atoms (the expression in part ii) divided by the number of moles (m/Ar).
    f) Bond length is the distance from the centre of a small cube to one of its corners. This is half the distance between two opposite corners. This can be calculated using Pythagoras twice.
    g) A basic AS question- though very long!
    h) Using the expression from e)iii and by substituting thevalues given, you would get the answer.
    Hope this helps- tell me if anything needs to be clearer
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    (Original post by Naalm)
    Q2)
    Parts (a) to (d) are basic AS chemistry, so I assume you’ve done them
    e) i)Count the number of atoms within the cube: 4 atoms inside the cube, 6 atoms each in one face (half of each atom is inside the cube so we count them as 6 half-atoms). We also have 8 atoms each at a corner (only one eighth of each atom is inside the cube- this is 8 1/8-atoms). In total, we have 4+6*1/2+8*1/8=8 atoms in one cube.
    ii) Work out the volume of the cube in m3: you’re given its length in pm. The number of cubes in the sphere is its volume divided by the volume of one cube. The total number of atoms is the total number of cubes times the number of atoms in each cube (n).
    iii) Avogadro constant is the number of atoms per mole. The number of moles is mass/molar mass (m/Ar), so Avogadro constant it the number of atoms (the expression in part ii) divided by the number of moles (m/Ar).
    f) Bond length is the distance from the centre of a small cube to one of its corners. This is half the distance between two opposite corners. This can be calculated using Pythagoras twice.
    g) A basic AS question- though very long!
    h) Using the expression from e)iii and by substituting thevalues given, you would get the answer.
    Hope this helps- tell me if anything needs to be clearer
    You're a legend! Thankyou very much
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    Hey there For Q.1 c(iii) on the 2012 paper, How do you work out the Volume required for 1 atom? Could someone work me through it Please?
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    Does anyone know where I can find the past papers, mark schemes and the pass marks?

    It's alright I found it.
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    (Original post by Me123456789)
    Does anyone know where I can find the past papers, mark schemes and the pass marks?

    It's alright I found it.
    Past papers etc. are on www.C3L6.org

    I'm not sure if it was you who quoted or asked though!
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    (Original post by Davelittle)
    Past papers etc. are on www.C3L6.org

    I'm not sure if it was you who quoted or asked though!
    It was me, thanks.
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    (Original post by Naalm)
    Q2)
    Parts (a) to (d) are basic AS chemistry, so I assume you’ve done them
    e) i)Count the number of atoms within the cube: 4 atoms inside the cube, 6 atoms each in one face (half of each atom is inside the cube so we count them as 6 half-atoms). We also have 8 atoms each at a corner (only one eighth of each atom is inside the cube- this is 8 1/8-atoms). In total, we have 4+6*1/2+8*1/8=8 atoms in one cube.
    ii) Work out the volume of the cube in m3: you’re given its length in pm. The number of cubes in the sphere is its volume divided by the volume of one cube. The total number of atoms is the total number of cubes times the number of atoms in each cube (n).
    iii) Avogadro constant is the number of atoms per mole. The number of moles is mass/molar mass (m/Ar), so Avogadro constant it the number of atoms (the expression in part ii) divided by the number of moles (m/Ar).
    f) Bond length is the distance from the centre of a small cube to one of its corners. This is half the distance between two opposite corners. This can be calculated using Pythagoras twice.
    g) A basic AS question- though very long!
    h) Using the expression from e)iii and by substituting thevalues given, you would get the answer.
    Hope this helps- tell me if anything needs to be clearer
    Small thing, Q2. c) how did you know what the two products formed were when the CaH2 reacted with the SiH4?
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    (Original post by Davelittle)
    Small thing, Q2. c) how did you know what the two products formed were when the CaH2 reacted with the SiH4?
    When SiF4 reacts with CaH2, there are only two possible products (by swapping the Hs & the Fs). This forms 2CaF2 + SiH4
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    (Original post by Naalm)
    When SiF4 reacts with CaH2, there are only two possible products (by swapping the Hs & the Fs). This forms 2CaF2 + SiH4
    Oh yeah, thanks!

    People from my school are saying it is extremely hard
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    (Original post by livium)
    Anyone done this paper yet? I did it today

    How did you find it, any tricky bits?
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    (Original post by geor)
    Same, I'm so scared for it! If I drop down from a gold on the harder Olympiad i'll feel like the biggest moron.
    You got a gold on the Olympiad? Well done!

    What school year did you do the Olympiad in?
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    I did it today, bloody difficult haha! Barely answered the questions, and the ones I did answer I'm not even sure are right...
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    (Original post by geor)
    Oh dear! Have you looked at/done the 2011/2012 papers; Would you say it is harder than them or about the same?
    Yeah, I had a little go at some of the questions on each and they were way more accessible. This years paper was very challenging.
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    (Original post by geor)
    Ah how annoying, at least it means the boundaries will be very very low!
    Yeah, I think they will be quite low for a copper! Heres hoping xD
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    No sorry, please do not discuss any questions until the weekend, the last date for doing it is tmmrw,
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    Well I failed that, did the first few qurstions fine and then none of our class could do the rest
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    Hey everybody!!

    Okay, so does anyone have any idea as to what is the prize and stuff for C3L6?? I read the page on scoring and prizes (http://www.c3l6.com/honours/scoring/) but it mentions 2012 and not 2013 so are the rules still the same or have they been altered?? And would you still get a prize if you are among the 10 in the country under all age groups category and in the top 5 in the under 18 category?? :confused:
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    Wow can't believe it's been a year since I did this! Good luck to everyone waiting for results!


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    (Original post by samsimmons)
    Thanks Can you remember how long it took to get the results?
    Think it took about 2 weeks or so .


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    Hey how everyone find question 2?
 
 
 
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