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    does anybody know what the grade boundaries might be like?

    In june the grade boundaries do get quite high because a D in June 2012 was apparently an A in Jan 13....

    Jan 13 grade boundaries were so low..I hope they stay the same
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    Thank you Jayqwe an Mollymod, that really helps. I was so confused haha
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    (Original post by Ifa_94)
    does anybody know what the grade boundaries might be like?

    In june the grade boundaries do get quite high because a D in June 2012 was apparently an A in Jan 13....

    Jan 13 grade boundaries were so low..I hope they stay the same
    Theres no way to predict the grade boundaries. It depends on the difficulty of the paper, and then the marks are UMS'd. for example, I did this exam in june 2012 and got 53 which was a C borderline D and around 70 UMS. In January I found the paper awful and got 39 (bad I know) and still got 70 UMS which is a D. theres no way of predicting. Just aim to do as well as you can
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    Is this right:

    DeltaHF= DeltaHC(reactants) - DeltaHC(products)
    DeltaHC= DeltaHF(products) - DeltaHF(reactants)
    DeltaHR= DeltaHF(products) - DeltaHF(reactants)
    DeltaHF= DeltaH(bonds brocken) - DeltaH(bonds made)

    It worked for me when i practised questions.
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    (Original post by felicity95)
    Theres no way to predict the grade boundaries. It depends on the difficulty of the paper, and then the marks are UMS'd. for example, I did this exam in june 2012 and got 53 which was a C borderline D and around 70 UMS. In January I found the paper awful and got 39 (bad I know) and still got 70 UMS which is a D. theres no way of predicting. Just aim to do as well as you can
    yeh that's true I did this exam last june as well and got 78 ums... really need to do well this time.

    good luck
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    (Original post by Ifa_94)
    yeh that's true I did this exam last june as well and got 78 ums... really need to do well this time.

    good luck
    Thank you! good luck to you too!
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    explain why the first ionisation energy of aluminium is lower than that of magnesium? someone please help me
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    (Original post by newyork newyork)
    explain why the first ionisation energy of aluminium is lower than that of magnesium? someone please help me
    The outer electron of Al occupies the 3P sub shell which is a higher energy level compared to Magnesium in which the outer electron occupies the 3S sub shell. The outer electron in Al is further away from the attraction of the nucleus due to greater distance and also greater inner shielding from electrons which again reduces the attractive force of the nucleus on the outer electron therefore the first ionisation energy for Al is lower than Mg.

    Hope this helps


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    Any predictions on what might come up in the paper on Thursday?
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    Hey guys

    I'm stuck with two things that have been stated in the George Facer book in the the "Bonding" chapter.

    Firstly... "There are slight spaces between the metal ions, which are occupied by the delocalised electrons. Therefore, the metallic radius is slightly larger than the ionic radius"

    I have no idea why that is, could someone explain

    "d-clock metals use their (n-1)d electrons as well as their ns^2 electrons in bonding" ... does this mean what I think it means that both d & s electrons are used in bonding? Therefore making d-block metals super hard Was just confused with the term (n-1)d electrons, what do they mean by that?

    And does this mean d-block metals can have like 10 electrons localized... therefore with ions with up to like 10+ charge. It seems unlikely, I think I'm definitely getting something wrong here help !

    Thanks in advance to any responses
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    (Original post by posthumus)
    Hey guys

    I'm stuck with two things that have been stated in the George Facer book in the the "Bonding" chapter.

    Firstly... "There are slight spaces between the metal ions, which are occupied by the delocalised electrons. Therefore, the metallic radius is slightly larger than the ionic radius"

    I have no idea why that is, could someone explain

    "d-clock metals use their (n-1)d electrons as well as their ns^2 electrons in bonding" ... does this mean what I think it means that both d & s electrons are used in bonding? Therefore making d-block metals super hard Was just confused with the term (n-1)d electrons, what do they mean by that?

    And does this mean d-block metals can have like 10 electrons localized... therefore with ions with up to like 10+ charge. It seems unlikely, I think I'm definitely getting something wrong here help !

    Thanks in advance to any responses
    George Facer books sucks, study edexcel book
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    Are there any video tutorials available for AS chemistry..especially for the past papers?
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    (Original post by felicity95)
    Theres no way to predict the grade boundaries. It depends on the difficulty of the paper, and then the marks are UMS'd. for example, I did this exam in june 2012 and got 53 which was a C borderline D and around 70 UMS. In January I found the paper awful and got 39 (bad I know) and still got 70 UMS which is a D. theres no way of predicting. Just aim to do as well as you can
    Exactly the same as me, but didn't do the January 2013 thankfully.
    This is my only retake of unit 1 and I dont need to retake it to get a B, but well I dont want to settle. I'll be happy if I get a B but while it's still possible i still want to aim for the A
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    Do you guys know any of the expected questions for thursday's exam?
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    How do you calculate percentage errors?!
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    Hi!

    Just another bond enthalpy question. why is the answer to this question A and not C. Surely the mean bond enthalpy is the energy required to break one mole of bonds in the gas phase averaged over many compounds. so then isn't CH4 1 mole, and 1/4CH4 is just a 1/4 mole?

    Also if the question had said bond enthalpy as opposed to mean bond enthalpy would it then have been C?

    Sorry to ramble on, last minute panic.
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    Hey

    guys quick questions... can sumone plzzzz explain me question 6,9 and 13.... in the january 2013 unit 1 chemistry paper
    these are the mcq questions (clearly)
    thanks alot....
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    (Original post by felicity95)
    Hi!

    Just another bond enthalpy question. why is the answer to this question A and not C. Surely the mean bond enthalpy is the energy required to break one mole of bonds in the gas phase averaged over many compounds. so then isn't CH4 1 mole, and 1/4CH4 is just a 1/4 mole?

    Also if the question had said bond enthalpy as opposed to mean bond enthalpy would it then have been C?

    Sorry to ramble on, last minute panic.
    I'm not sure, but it seems C implies that 4 C-H bonds have been broken at once?
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    (Original post by amber206)
    How do you calculate percentage errors?!
    % error = Uncertainty in equipment / the reading x 100

    the uncertainty is usually given in the question e.g. 0.01, and the for the reading you have to look for what you measured so if you measured the temperature using a thermometer that had a uncertainty of 0.01 and the temperature you measured was say 37.8 then your calculation would be 0.01/37.8 x 100 = + or - 0.026%

    If you did more than one measurement i.e. you took two readings etc. you must do 2 x the whole thing.

    Hope that made sense
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    (Original post by felicity95)
    % error = Uncertainty in equipment / the reading x 100

    the uncertainty is usually given in the question e.g. 0.01, and the for the reading you have to look for what you measured so if you measured the temperature using a thermometer that had a uncertainty of 0.01 and the temperature you measured was say 37.8 then your calculation would be 0.01/37.8 x 100 = + or - 0.026%

    If you did more than one measurement i.e. you took two readings etc. you must do 2 x the whole thing.

    Hope that made sense
    Thank you
 
 
 
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