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    How would you conduct an experiment to measure enthalpy of solution (4 marks)
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    (Original post by lai812matthew)
    my last attempt on A* after i flopped f324 if this doesn't go well i'll have no uni to go to.......
    Same.... F324 was too wordy for me, I wanted more synthesis and a long NMR. I made many incredibly silly errors - I completely ignored the carbon-13 spectrum until I left the exam hall and realised why it was there etc.... F325 tends to go better for me. More maths = more marks haha. Good luck!
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    (Original post by mechanism)
    To do with acid and bases where HA=A-.
    how does that work? Why would it be only a jhalf neutralisation
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    (Original post by ToLiveInADream)
    How would you conduct an experiment to measure enthalpy of solution (4 marks)
    you can't measure the lattice enthalpy directly so how would that work?
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    (Original post by tcameron)
    how does that work? Why would it be only a jhalf neutralisation
    This is what I'm confused about, however I think I get it - when half of HA reacts, that amount of A- has been formed. so the [HA] left = [A-] formed
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    (Original post by M.Branson98)
    How do you know if an indicator is suitable for a certain acid-base titration?
    find the pH range of the indicator - say for example it's given as 4.0 - 8.0. Then look at the curve for the acid base titration and spot the end point - so like the centre of the sharp upward line where neutralisation occurs. If the end point lies about the middle of the pH's range (so around 6.0 in this case) then it's suitable
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    (Original post by tcameron)
    anyone have a proper definition of Kstab? The one in the book makes no sense to me
    It's basically Kc without water, the higher the kstab the more stable a complex is.
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    Can somebody run through those standard 3 markers on energy comparisons of e.g. RbF vs RbCl or MgCl2 vs NaCl Usually accompanied by a table with some enthalpy values of hydration/solution/lattice enthalpy.
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    (Original post by Arima)
    find the pH range of the indicator - say for example it's given as 4.0 - 8.0. Then look at the curve for the acid base titration and spot the end point - so like the centre of the sharp upward line where neutralisation occurs. If the end point lies about the middle of the pH's range (so around 6.0 in this case) then it's suitable
    Is the end point the stationary point?
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    (Original post by mechanism)
    This is what I'm confused about, however I think I get it - when half of HA reacts, that amount of A- has been formed. so the [HA] left = [A-] formed
    but if the equilibrium lies so far to the acid side why would so much [A-] be formed?
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    (Original post by Dinasaurus)
    Is the end point the stationary point?
    its bang in the middle of the sharp rise in pH- not sure if it's classified as stationary
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    (Original post by TOMWIGHT)
    Hi has anyone got the 2015 F325 mark scheme??
    K is group 1, so NO3 is -1 charge.

    Isn't the only F(VI), FeO4?
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    (Original post by Rust Cohle)
    Can somebody run through those standard 3 markers on energy comparisons of e.g. RbF vs RbCl or MgCl2 vs NaCl Usually accompanied by a table with some enthalpy values of hydration/solution/lattice enthalpy.
    say which one is the smaller ion/has the smaller ionic radius
    which one has the higher ionic charge if they're in different groups
    say about being more strongly attracted to the oppositely charged ion or to water molecules
    say more/less energy energy is required to break the attraction
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    (Original post by tcameron)
    but if the equilibrium lies so far to the acid side why would so much [A-] be formed?
    It's only a half neutralisation when base is added. Remember, it's not just the acid dissociating - it's reacting with the base, so it's no longer true that [H+]=[A-]
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    (Original post by tcameron)
    say which one is the smaller ion/has the smaller ionic radius
    which one has the higher ionic charge if they're in different groups
    say about being more strongly attracted to the oppositely charged ion or to water molecules
    say more/less energy energy is required to break the attraction
    Thanks. If you need any last min assistance quote or pm
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    (Original post by mechanism)
    This is what I'm confused about, however I think I get it - when half of HA reacts, that amount of A- has been formed. so the [HA] left = [A-] formed
    That is pretty convoluted but when you phrase it like that it does make sense.
    In answer to your original question I am 99% that is not in the spec, I remember my chem teacher really briefly mentioning it but I'm fairly certain he didn't even go into detail as it was not in the course.
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    (Original post by ToLiveInADream)
    How would you conduct an experiment to measure enthalpy of solution (4 marks)
    Is that an actual exam question? If so, from which year?
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    (Original post by VMD100)
    That is pretty convoluted but when you phrase it like that it does make sense.
    In answer to your original question I am 99% that is not in the spec, I remember my chem teacher really briefly mentioning it but I'm fairly certain he didn't even go into detail as it was not in the course.
    It's came up before (on several legacy papers), it's on the spec as an application question. Could come up easily. Don't leave anything to chance!
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    (Original post by tcameron)
    you can't measure the lattice enthalpy directly so how would that work?
    You basically measure the temperature change when 1 mole of that substance is dissolved in water . With a thermometer and the conditions , calculate the enthalpy of solution by temperature change times c times , m
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    (Original post by luckycactus1337)
    You basically measure the temperature change when 1 mole of that substance is dissolved in water . With a thermometer and the conditions , calculate the enthalpy of solution by temperature change times c times , m
    so you'd measure the temperature increase or decrease?
 
 
 
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