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    (Original post by TeachChemistry)
    No.

    How many moles of O2 do you need for 0.1 mol of cyclohexane?
    1:9 ratio

    0.9 moles of oxygen

    multiply by Mr

    28.8?
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    (Original post by Jassy16)
    1:9 ratio

    0.9 moles of oxygen Yes

    multiply by Mr ????

    28.8?
    How much volume does 1 mol of a gas occupy at room temperature and pressure?
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    (Original post by TeachChemistry)
    How much volume does 1 mol of a gas occupy at room temperature and pressure?
    22.4dm^3

    btw we haven't started this topic so im unfamiliar with these constants

    so is it 18.144dm^3
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    (Original post by Jassy16)
    22.4dm^3

    btw we haven't started this topic so im unfamiliar with these constants
    It's 22.4 at standard temperature and pressure but 24.0 at room temperature and pressure. Are you in the UK because you should have done this by now? So go with the 24.0. What's the answer?
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    (Original post by TeachChemistry)
    It's 22.4 at standard temperature and pressure but 24.0 at room temperature and pressure. Are you in the UK because you should have done this by now? So go with the 24.0. What's the answer?
    21.6dm^3 ?
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    (Original post by Jassy16)
    21.6dm^3 ?
    Correct. Understand it?
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    (Original post by TeachChemistry)
    Correct. Understand it?
    yeah but kinda confused why my teacher hasn't covered it by now

    out of interest do you know where this information is within the aqa nelson thornes book

    and do you have any recomendations for AS/A2 chemistry text books to buy for aqa
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    (Original post by Jassy16)
    yeah but kinda confused why my teacher hasn't covered it by now

    out of interest do you know where this information is within the aqa nelson thornes book

    and do you have any recomendations for AS/A2 chemistry text books to buy for aqa
    I would have thought you'd have done this in autumn term. Look under molar volume of a gas in the index.

    I teach OCR course so can't comment on AQA textbooks with any good advice.
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    (Original post by RMNDK)
    I wouldn't say purple, the end-point is the first hint of a pink tinge colour.

    The reason why it's colourless to pink is because exactly what you said.
    If you add Mn7+ to Fe2+, they will be reduced to Mn2+ which is colourless.

    As I keep adding and adding the Mn7+, all the Fe2+ will eventually become oxidised to Fe3+.

    There is a single moment when I have fully oxidised all the Fe2+ in the conical flask, and so at that moment, there is no more Fe2+.

    Thus, the instant I add more Mn7+, it's going to remain as Mn7+ because there's nothing to oxidise. Thus we get this pink tinge colour. Add some more and we get that intense purple colour.

    Unless you wanted to use some sort of indicator which changes colour around the neutralisation point, you would look for this colour change. And we tend not to because it is a prominent change.
    ahh that makes sense now!! thank you!
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    Trying to find some people who can help me out with an F335 OCR Chemistry question - don't know if you're able to help! (No worries if not!)
    Name:  Screen Shot 2016-04-18 at 20.26.17.png
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    I have managed to get the first two questions: pH = 3.33 & volume = 4.5
    However I cannot for the life of me manage the last question! Thank you in advance
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    (Original post by NivvyLeaver)
    Trying to find some people who can help me out with an F335 OCR Chemistry question - don't know if you're able to help! (No worries if not!)
    Name:  Screen Shot 2016-04-18 at 20.26.17.png
Views: 173
Size:  32.0 KB
    I have managed to get the first two questions: pH = 3.33 & volume = 4.5
    However I cannot for the life of me manage the last question! Thank you in advance
    There's a REALLY LONG way, and there's a much shorter way.

    Here's the short way, because **** the long way.
    Lol I say short, but my post is long. That's only because I just really really want you to understand it sorry. The working out when you come to write it is short.

    You agree that you have some moles of HA.
    The question is saying that the student adds some NaOH. This amount is equal to a third of the amount needed to neutralise it.

    In other words, a third of the HA has reacted with the NaOH. If it said, half the amount was added, then half of the HA would have been neutralised, and so on.

    When we react the HA and NaOH, this forms our A-.
    Spoiler:
    Show
    Cos remember, NaOH + HA ---> NaA + H2O
    and the NaA ---> Na+ + A-
    So as a result, the amount of A- we formed is equal to one third of the HA we started with.

    In other words...

     \mathrm{A}^- = \dfrac{1}{3}\times \mathrm{HA}.

    If you want to represent this as concentration, by all means divide it by the volume. But the volume will cancel out, which allows me to just use the moles.

    In fact I don't need to do that because...

     \mathrm{H}^+ = \dfrac{\mathrm{K}_c \times \mathrm{HA}}{\mathrm{A}^-} \ \ \ \ \ becomes  \ \ \ \mathrm{H}^+ = \dfrac{\mathrm{K}_c \times \mathrm{HA}}{\dfrac{1}{3} \times \mathrm{HA}}

    And now the HAs cancel out, which just leaves...

     \mathrm{H}^+ = \mathrm{K}_c \times 3

    This is now solvable for pH
    It's just like that first question of where the HA is half the number of A
    Spoiler:
    Show
    The long **** way would be to calculate the moles of each reagent, and then the concentration, then dividing one by the other, and you get funny numbers.
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    Hey.
    Does anyone here do the OCR B salters syllabus?
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    (Original post by SamuelN98)
    Hey.
    Does anyone here do the OCR B salters syllabus?
    I do
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    (Original post by SamuelN98)
    Hey.
    Does anyone here do the OCR B salters syllabus?
    Me!

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    Can someone please explain the meaning of the Latin numbers next to ions (e.g Sulphate(IV)) and what effect they have on the number of oxygen atoms?
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    (Original post by TheGreatPumpkin)
    Can someone please explain the meaning of the Latin numbers next to ions (e.g Sulphate(IV)) and what effect they have on the number of oxygen atoms?
    The latin numerals show the oxidation states. For example, sulphate (iv) means that sulphur has a oxidation state of +4. But we see ''ate'' which shows there is an oxygen but we don't know how many.

    So if sulphur is +4 then there must be two oxygen. Because each oxygen is -2 and -2*2 is -4. So if we have sulphur 4+ and 2 oxygen being 4- (remember there is two oxygens each being -2 and -2*2 is -4) the charges will balance to give zero.
    Therefore the formula for this compound is
    SO2
    Hope I helped you!
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    (Original post by KatieAlicexxx)
    Me!

    Posted from TSR Mobile
    (Original post by NivvyLeaver)
    I do
    Do either of you have the answers to the questions from the practical book?
    i cant find them anywhere.
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    hi
    is anyone retaking chem3x empa (aqa)??
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    Hi,

    I was wondering whether any of you may have the 2016 New AS chemistry specimen paper? If you do could someone please attach it please.
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    (Original post by farihak97)
    Hi,

    I was wondering whether any of you may have the 2016 New AS chemistry specimen paper? If you do could someone please attach it please.
    For OCR?

    http://www.ocr.org.uk/qualifications...432-from-2015/
 
 
 
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