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OCR Salters Chemistry F335 12th June 2013 Exam revision thread Watch

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    (Original post by Harantony)
    Nope, they've been removed from the specification.
    Hmmm. There's two thirds of a page about this on page 78 in the CGP revision guide. However, it doesn't say anything about partial pressures within the F335 spec.
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    (Original post by AmirHabeeb)
    I couldn't find anything but just remember that concentration is proportional to number of moles and if they don't give you any volumes you can't be expected to work them out.



    You're right in saying that it provides H+ but that protonates basic groups after the amide group has broken so it would form an -NH3+ group ( as amine groups are basic ) and amide hydrolysis always leads to an amine + carboxylic acid. If it were alkali conditions, the carboxyl group will lose an H to form COO-.
    ahh right so under alkali conditions what would happen ? would we get a carboxylate salt?

    How do you know which groups are basic? (I'm really bad at this acid/base/alkali stuff sorry :P)

    thanks very much by the way that was bugging me alot
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    (Original post by Tikara)
    ahh right so under alkali conditions what would happen ? would we get a carboxylate salt?

    How do you know which groups are basic? (I'm really bad at this acid/base/alkali stuff sorry :P)

    thanks very much by the way that was bugging me alot
    Under alkali conditions your acidic group would be affected so the carboxyl group would deprotonate to form COO- which could form a salt yes. Basic groups are ones which have lone pairs of electrons so they are willing to accept a proton by forming a dative covalent bond.
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    guys, please help, i dont know how to begin revising!
    should i make notes, and do q's from revision guide, or straight at it with past papers?
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    Hey guys, on the Jan 2010 (Legacy) paper, could anyone look at question 2 (v), and explain how they would've worked out the Mr for Alizarin? It has quite a complicated looking structure.
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    In the example on page 69 of the official revision guide, could someone please explain why they've halved the concentrations. Thanks
    Also, In the example on page 79, it says the molecular formula is C3H6O, so why are there 8 hydrogens?
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    (Original post by super121)
    In the example on page 69 of the official revision guide, could someone please explain why they've halved the concentrations. Thanks


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    yeah that is a bit confusing and it doesn't go into any detail but the way my teacher explained it was:

    C is always halved if you double the volume it's in

    C = n/V

    as volume is the number on the bottom - as volume increases, the concentration decreases.

    As they are in equal volumes - each concentration will be n / both their volumes added together (C = n/2V) which is the same essentially as halving the concentration.

    The best way to illustrate this is to just pick any number for the volume and then find the moles and then double the volume and you'll find the concentrations halved. like picking randomly 1050 for the volume 0.1 = n / 1050 so n = 105. if the same conc is mixed in 1050 of something else it'll double the volume so 0.1 = n/2100 = 105 /2100 = 0.05

    there is probably some mathematical and/or easier way to explain that better but thats how I understood it anyways and you can try with any number and the C is always halved if you double the volume

    hope that helped !!

    also: looking at page 79 it seems like thats a mistake lmao and meant to say C3H8O
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    (Original post by Tikara)
    yeah that is a bit confusing and it doesn't go into any detail but the way my teacher explained it was:

    C is always halved if you double the volume it's in

    C = n/V

    as volume is the number on the bottom - as volume increases, the concentration decreases.

    As they are in equal volumes - each concentration will be n / both their volumes added together (C = n/2V) which is the same essentially as halving the concentration.

    The best way is to just pick any number for the volume and then find the moles and then double the volume and you'll find the concentrations halved. like picking randomly 1050 for the volume 0.1 = n / 1050 so n = 105. if the same conc is mixed in 1050 of something else it'll double the volume so 0.1 = n/2100 = 105 /2100 = 0.05

    there is probably some mathematical and/or easier way to explain that better but thats how I understood it anyways and you can try with any number and the C is always halved if you double the volume

    hope that helped !!
    Sorry, but what's n and how do you know that the volume has been doubled? We're not given any information on volume
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    (Original post by Tikara)
    yeah that is a bit confusing and it doesn't go into any detail but the way my teacher explained it was:

    C is always halved if you double the volume it's in

    C = n/V

    as volume is the number on the bottom - as volume increases, the concentration decreases.

    As they are in equal volumes - each concentration will be n / both their volumes added together (C = n/2V) which is the same essentially as halving the concentration.

    The best way is to just pick any number for the volume and then find the moles and then double the volume and you'll find the concentrations halved. like picking randomly 1050 for the volume 0.1 = n / 1050 so n = 105. if the same conc is mixed in 1050 of something else it'll double the volume so 0.1 = n/2100 = 105 /2100 = 0.05

    there is probably some mathematical and/or easier way to explain that better but thats how I understood it anyways and you can try with any number and the C is always halved if you double the volume

    hope that helped !!

    also: looking at page 79 it seems like thats a mistake lmao and meant to say C3H8O
    I thought so, but they've done it twice so I thought I must just be going mad!
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    (Original post by super121)
    Sorry, but what's n and how do you know that the volume has been doubled? We're not given any information on volume
    n is number of moles you should be familiar with n = c x v

    You are given that there are equal volumes of two solutions. So then when mixed together the volume of both together is double either one on it's own (sorry I threw in unnecessary numbers just to try and illustrate through the formula)

    as each solution is diluting the other by a factor of 2, you therefore need to halve the concentration.

    (if you do maths I've just figured out a way to show it mathematically lmao)
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    (Original post by super121)
    I thought so, but they've done it twice so I thought I must just be going mad!
    yeah I know haha I had to check a few times- I think who ever was typing it must've been very tired and accidentally started putting the formula for the one on the previous page :P
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    (Original post by AmirHabeeb)
    Under alkali conditions your acidic group would be affected so the carboxyl group would deprotonate to form COO- which could form a salt yes. Basic groups are ones which have lone pairs of electrons so they are willing to accept a proton by forming a dative covalent bond.
    thanks very much for the help - you prompted me to thoroughly review hydrolysis I'm not completely confident but getting there, I'm just confused as to how the mechanism works it doesn't have it in the book
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    (Original post by Tikara)
    n is number of moles you should be familiar with n = c x v

    You are given that there are equal volumes of two solutions. So then when mixed together the volume of both together is double either one on it's own (sorry I threw in unnecessary numbers just to try and illustrate through the formula)

    as each solution is diluting the other by a factor of 2, you therefore need to halve the concentration.

    (if you do maths I've just figured out a way to show it mathematically lmao)
    Oh right, I usually just use m
    No, the formula makes sense, but I just don't understand how you know to half the volumes :/
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    (Original post by super121)
    Oh right, I usually just use m
    No, the formula makes sense, but I just don't understand how you know to half the volumes :/
    well it makes sense using the formula and to remember if you times the volume by 2 you divide the concentrations by two
    just like if you had 4 different solutions of equal volumes mixed, you'd divide each concentration by four and if you had 5, divide by 5 etc.

    Spoiler:
    Show

    using m

    C before mixing = m/v
    C after mixing = m/2v
    = 1/2(m/v)
    = (1/2)(C before mixing)
    therefore C after mixing = C before mixing / 2
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    (Original post by Tikara)
    well it makes sense using the formula and to remember if you times the volume by 2 you divide the concentrations by two
    just like if you had 4 different solutions of equal volumes mixed, you'd divide each concentration by four and if you had 5, divide by 5 etc.

    Spoiler:
    Show

    using m

    C before mixing = m/v
    C after mixing = m/2v
    = 1/2(m/v)
    = (1/2)(C before mixing)
    therefore C after mixing = C before mixing / 2
    This is where i'm confused, where does it tell us that the volume was multiplied by 2?
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    can someone PLEASE help me with this question 5(f)(ii)

    http://www.ocr.org.uk/Images/79341-q...-by-design.pdf

    thanks in advance!


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    (Original post by super121)
    This is where i'm confused, where does it tell us that the volume was multiplied by 2?
    sorry I'm not very good at explaining xD

    if two volumes are equal and put together then it is the same as timesing one of the volumes by 2

    the same way 10 + 10 = 10 x 2 :P

    so it's the "two volumes that are equal" part which lets us know that the volume of each concentration is doubled when they're thrown together like if you mixed two seperate jugs of water of 4L each you'd have a final volume of 8L. So the volume of the mixture (the buffer solution) is therefore each volume added together, and if they are equal then it is the same as double the volume of one. each solution is basically diluting the other - does any of that help?
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    (Original post by Tikara)
    sorry I'm not very good at explaining xD

    if two volumes are equal and put together then it is the same as timesing one of the volumes by 2

    the same way 10 + 10 = 10 x 2 :P

    so it's the "two volumes that are equal" part which lets us know that the volume of each concentration is doubled when they're thrown together like if you mixed two seperate jugs of water of 4L each you'd have a final volume of 8L. So the volume of the mixture (the buffer solution) is therefore each volume added together, and if they are equal then it is the same as double the volume of one. each solution is basically diluting the other - does any of that help?
    This bit has made me understand it. So if you have 10L of 6moldm orange juice and add 10L of water, you end up with 20L of 3moldm of orange juice, correct?
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    (Original post by super121)
    This bit has made me understand it. So if you have 10L of 6moldm orange juice and add 10L of water, you end up with 20L of 3moldm of orange juice, correct?
    yeah man thats the idea - and if the water was instead 10L of 4moldm apple juice you'd have 3moldm of orange and 2moldm of apple in the 20L - I'm so sorry for confusing you with the rest of the stuff- I think you should look earlier in this thread cause someone explained it to me but I didn't understand in terms of diluting
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    (Original post by Tikara)
    yeah man thats the idea - and if the water was instead 10L of 4moldm apple juice you'd have 3moldm of orange and 2moldm of apple in the 20L - I'm so sorry for confusing you with the rest of the stuff- I think you should look earlier in this thread cause someone explained it to me but I didn't understand in terms of diluting
    Thank you!
    Yea, i'll have a look
 
 
 
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