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    (Original post by h. radiation)
    Oi fam

    Why you cheating for

    Shall i report you to the exam board so they can disqualify you

    Do you really want that to happen :hmmm:

    Posted from TSR Mobile
    How on earth is that cheating ? i will not provide anyone with answers that would be breaking my profession . help doesn't mean cheat it means i can tell them what topics to revise on .So go ahead with no proof be my guest.
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    Does anyone know how to ace isa's I'm failing so bad on them!
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    (Original post by samb1234)
    Its easier than drawing the 30+ isomers this most likely has

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    Apparently 20 :dontknow:

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    Right doing task 2 of the EMPA tomorrow. I have a few questions.

    Is the table essentially just |Test| Observations|

    And you number test and write down what you say.

    If the solution is white and cloudy does that mean there is a precipitate formed or just that is is cloudy.

    If it just say yellow with no visible precipitate is it just a yellow solution

    I can't think of any other instances but do you have to mention anything to do with the temperature and that or not?

    Thanks
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    (Original post by Super199)
    If the solution is white and cloudy does that mean there is a precipitate formed or just that is is cloudy.

    If it just say yellow with no visible precipitate is it just a yellow solution


    Thanks
    I don't do AQA so I can't tell you specifically but

    If white and cloudy it's white precipitate
    If yellow and clear it is just a yellow solution like bromine water.


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    (Original post by samb1234)
    I feel even more sorry for the people who have to mark mine. .my handwriting is pretty bad when I write fast and I write so much that I write 2 lines per line they give me.

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    My writing changes on every page.

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    (Original post by C0balt)
    I don't do AQA so I can't tell you specifically but

    If white and cloudy it's white precipitate
    If yellow and clear it is just a yellow solution like bromine water.


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    So whenever something is cloudy there is a precipitate?
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    (Original post by VetVikki)
    It's more of a general thing that I don't understand when to add H+ and H2O and is there an OH- too!?
    However for example:
    Write the full redox equation for the reaction between sulphuric acid and sodium iodide?
    What are the steps?
    I have this problem too!

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    (Original post by Super199)
    So whenever something is cloudy there is a precipitate?
    Yes

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    does anyone do edexcel chemistry?
    the crystallisation practical?
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    Hey guys, i'm really stuck on this question.

    MgSO4*7H2O has a mass of 4.2g

    MgSO4 alone is 2g

    Work out the percentage by mass of water? Does anyone know how to do this? Could list some steps into how I can get the answer?

    Thanks
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    (Original post by JackP32)
    Hey guys, i'm really stuck on this question.

    MgSO4*7H2O has a mass of 4.2g

    MgSO4 alone is 2g

    Work out the percentage by mass of water? Does anyone know how to do this? Could list some steps into how I can get the answer?

    Thanks
    Isn't it just subtracting 2 from 4.2 to get mass of water and dividing it by 4.2 then times 100

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    Is it as easy as that? I thought I had to use molar masses of water? :confused:
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    (Original post by JackP32)
    Is it as easy as that? I thought I had to use molar masses of water? :confused:
    I don't know but where else could that extra mass come from

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    (Original post by JackP32)
    Is it as easy as that? I thought I had to use molar masses of water? :confused:

    (Original post by C0balt)
    Isn't it just subtracting 2 from 4.2 to get mass of water and dividing it by 4.2 then times 100

    Posted from TSR Mobile
    It's more accurate to use the Mr

    Mr of the waters = (7x18)
    Mr of the whole hydrated compound = 24.3+32.1+64+(7x18)=246.4

    (7x18)/246.4 = 0.511... x100 = 51.1%

    Notice how this is different to the experimental value of 2.2/4.2 x 100 = 52.3%
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    I have my coursework tomorrow, which method do you suggest I use? or does it matter?
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    (Original post by Dylann)
    It's more accurate to use the Mr

    Mr of the waters = (7x18)
    Mr of the whole hydrated compound = 24.3+32.1+64+(7x18)=246.4

    (7x18)/246.4 = 0.511... x100 = 51.1%

    Notice how this is different to the experimental value of 2.2/4.2 x 100 = 52.3%
    I just noticed that
    I also did working amount of mole of 2g MgSO4 out and multiplying with 7 then dividing which gives 50.0% lol it's interesting

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    (Original post by JackP32)
    MgSO4 alone is 2g

    Work out the percentage by mass of water? Does anyone know how to do this? Could list some steps into how I can get the answer?

    Thanks
    (Original post by C0balt)
    I just noticed that
    I also did working amount of mole of 2g MgSO4 out and multiplying with 7 then dividing which gives 50.0% lol it's interesting

    Posted from TSR Mobile
    It's probably due to the assumption that every hydrated MgSO4 contains 7 water molecules...when actually it's likely some are 7 and some are 8.

    If you like a bit of simple maths, you can work out what the average number of water molecules are given the experimental data...I've put the calculations in a spoiler if you wanna have a go before you look at the answer

    Spoiler:
    Show
    If the mass of MgSO4 is 2 grams, then the moles of MgSO4 is

    mol = mass/mr (note how I got the correct equation this time lol)
    mol = 2/120.4
    mol = 5/301 moles of MgSO4

    If the mass of H2O is 2.2 grams, then the moles of H2O is

    mol = 2.2/18
    mol = 0.12222....

    since 5/301 is smaller than 2.2/18, we can divide them both by 5/301 to find a molar ratio

    2.2/18 divided by 5/301 is 7.35777....

    So for every 1 mole of MgSO4, there are 7.3577 moles of H2O. That means that most of compounds contain 7 waters, but a considerable number contain 8.

    We can prove this further by repeating the Mr calculation, using our new number of waters:

    (7.3577 x 18)/120.4 + (7.3577 x 18) = 0.05238....x100 = 52.38% which was the same as the experimental calculation (2.2/4.2 x 100)


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    (Original post by Dylann)
    If the mass of MgSO4 is 2 grams, then the moles of MgSO4 is

    mol = mass/mr (note how I got the correct equation this time lol)
    mol = 2/120.4
    mol = 5/301 moles of MgSO4

    If the mass of H2O is 2.2 grams, then the moles of H2O is

    mol = 2.2/18
    mol = 0.12222....

    since 5/301 is smaller than 2.2/18, we can divide them both by 5/301 to find a molar ratio

    2.2/18 divided by 5/301 is 7.35777....

    So for every 1 mole of MgSO4, there are 7.3577 moles of H2O. That means that most of compounds contain 7 waters, but a considerable number contain 8.

    We can prove this further by repeating the Mr calculation, using our new number of waters:

    (7.3577 x 18)/120.4 + (7.3577 x 18) = 0.05238....x100 = 52.38% which was the same as the experimental calculation (2.2/4.2 x 100)


    [/SPOILER]
    Yup, looks like you are awake this time lol

    I didn't know it could contain 8, I just assumed it to be some experimental error or impurity so thanks for that :lol:

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    (Original post by C0balt)
    Yup, looks like you are awake this time lol

    I didn't know it could contain 8, I just assumed it to be some experimental error or impurity so thanks for that :lol:

    Posted from TSR Mobile


    And yeah, impurity or weighing error is probably quite likely too. BTW which exam board are you on?
 
 
 
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