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OCR Chemistry A Exam Thread (Breadth - May 27 2016 and Depth - June 10 2016) Watch

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    (Original post by 11...)
    safe to revise everything basically
    Yeah definitely.. There will definitely be miles and volumes (which came up on the first paper but can come again) there might be titrations,
    Just hoping for a 6 marker on mass and IR spectroscopy in one question
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    Anyone know how to do this, I know its pv=nrt but can't get the answer when I work it out

    2.15 dm3 of an unknown gas had a mass of 5.2 g at 18°C and a pressure of 101 kPa.Calculate the relative molecular mass of the gas and suggest what gas it could be.
    Spoiler:
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    the answer is 58 btw
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    (Original post by AKK6199)
    Anyone know how to do this, I know its pv=nrt but can't get the answer

    2.15 dm3 of an unknown gas had a mass of 5.2 g at 18°C and a pressure of 101 kPa.Calculate the relative molecular mass of the gas and suggest what gas it could be.
    Re-arrange pV = nRT to find n (the number of moles of the gas). Then, use mass = mr * moles to find the Mr of the gas.
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    (Original post by AKK6199)
    Anyone know how to do this, I know its pv=nrt but can't get the answer

    2.15 dm3 of an unknown gas had a mass of 5.2 g at 18°C and a pressure of 101 kPa.Calculate the relative molecular mass of the gas and suggest what gas it could be.
    do you know what the answer was???
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    (Original post by 11...)
    do you know what the answer was???
    57.9357 so the Mr is 58
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    (Original post by marioman)
    Re-arrange pV = nRT to find n (the number of moles of the gas). Then, use mass = mr * moles to find the Mr of the gas.
    Yeah thats what I tried, probably got the conversions wrong, thanks anyway
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    (Original post by AKK6199)
    57.9357 so the Mr is 58
    that is exactly what I got.

    = 101000 x (2.16 x 10^-3)
    ---------------------------------------------
    8.314 x 291

    = 0.089755 moles

    5.2
    ---------------
    0.089755
    = 57.935

    but on the periodic table that is not a gas.
    do u think it might be a diatomic gas or co2 etc

    is there any other info given in that question??
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    (Original post by 11...)
    that is exactly what I got.

    = 101000 x (2.16 x 10^-3)
    ---------------------------------------------
    8.314 x 291

    = 0.089755 moles

    5.2
    ---------------
    0.089755
    = 57.935

    but on the periodic table that is not a gas.
    do u think it might be a diatomic gas or co2 etc
    OHHH that was actually so simple lol woops, and its continued off this long question so its butane gas (its from the ocr b specimen depth paper if you're curious)
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    (Original post by AKK6199)
    Anyone know how to do this, I know its pv=nrt but can't get the answer

    2.15 dm3 of an unknown gas had a mass of 5.2 g at 18°C and a pressure of 101 kPa.Calculate the relative molecular mass of the gas and suggest what gas it could be.
    First convert all the units to what they need to be:
    2.15dm^3=0.00215m^3 (divided by 1000 to get from dm^3-m^3)
    18°C=273+18K=291K
    101kPa=101,000Pa

    Now rearrange the equation to make n the subject:
    n = pv/rt = (101000*0.00215)/(291*8.314) = 0.08975...
    now use the formula n=m/Mr
    so Mr = m/n = 5.2/0.08975 = 57.9
    so the gas could be butane C4H10 as it has Mr of 58
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    (Original post by AKK6199)
    OHHH that was actually so simple lol woops, and its continued off this long question so its butane gas (its from the ocr b specimen depth paper if you're curious)
    oh right. yeah cos i was wondering how am I supposed to find out the name of the gas with no other information, anyway thanks

    do u know at what point the alkane is a gas, liquid etc. thanks
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    (Original post by 4nonymous)
    First convert all the units to what they need to be:
    2.15dm^3=0.00215m^3 (divided by 1000 to get from dm^3-m^3)
    18°C=273+18K=291K
    101kPa=101,000Pa

    Now rearrange the equation to make n the subject:
    n = pv/rt = (101000*0.00215)/(291*8.314) = 0.08975...
    now use the formula n=m/Mr
    so Mr = m/n = 5.2/0.08975 = 57.9
    so the gas could be butane C4H10 as it has Mr of 58
    Thank you!!
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    (Original post by 4nonymous)
    First convert all the units to what they need to be:
    2.15dm^3=0.00215m^3 (divided by 1000 to get from dm^3-m^3)
    18°C=273+18K=291K
    101kPa=101,000Pa

    Now rearrange the equation to make n the subject:
    n = pv/rt = (101000*0.00215)/(291*8.314) = 0.08975...
    now use the formula n=m/Mr
    so Mr = m/n = 5.2/0.08975 = 57.9
    so the gas could be butane C4H10 as it has Mr of 58
    how well do u think u did in the chemistry breadth paper? and also do u do biology?
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    (Original post by 11...)
    Nobody has got a copy of the breadth, since it has to be used for mock exams next year. But i have no idea how there are proper unofficial mark schemes for maths. but arsey no longer writes them like the proper ones with the questions and answers
    You can actually receive a paper 24 hours after an exam has finished


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    (Original post by 11...)
    how well do u think u did in the chemistry breadth paper? and also do u do biology?
    I think I did ok. messed up a bit on the MCQ's cos I dont like them. And yes I do biology. I thought chem breadth was easier than bio.
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    Good luck for tomorrow everyone
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    when explaining the boiling point trend in alcohols, should you talk about induced dipole-dipole interactions (London Forces?) as well as the hydrogen bonding?
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    (Original post by Book-Girl)
    when explaining the boiling point trend in alcohols, should you talk about induced dipole-dipole interactions (London Forces?) as well as the hydrogen bonding?
    Yes, mention both to reinforce the fact that energy is required not only to overcome and break just london forces but also the hydrogen bond formed by the hydroxyl groups of the alcohols. They could ask you to compare the bp trend in alcohols related to alkanes and in this case you state the bonds needed to be overcome in both.
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    With electronegativity questions, would it be better I wrote about the specific dipole-dipole interactions or just say van der Waals' or London forces?

    Also, just to clarify, are van der Waals' referring to induced and permanent and London only induced?
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    (Original post by Internets)
    With electronegativity questions, would it be better I wrote about the specific dipole-dipole interactions or just say van der Waals' or London forces?

    Also, just to clarify, are van der Waals' referring to induced and permanent and London only induced?
    The way we have been taught is that to state the interactions between the molecules depending on the difference in electronegativity, I don't use van der waals. I either use london forces (induced), permanent dipole-dipole interactions or hydrogen bonding
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    (Original post by ashbawa)
    Yes, mention both to reinforce the fact that energy is required not only to overcome and break just london forces but also the hydrogen bond formed by the hydroxyl groups of the alcohols. They could ask you to compare the bp trend in alcohols related to alkanes and in this case you state the bonds needed to be overcome in both.
    ahh okay, thank you!
 
 
 
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