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    (Original post by kiiten)
    Thanks - I wasn't sure if it was ethyl or methyl
    The longest carbon chain has 4 carbon atoms and the side chain is C2H5 so it's ethyl. If the side chain is CH3 instead of C2H5, then it would be methyl.
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    (Original post by jyyl)
    The longest carbon chain has 4 carbon atoms and the side chain is C2H5 so it's ethyl. If the side chain is CH3 instead of C2H5, then it would be methyl.

    Yeah, I wasn't sure if I was counting the longest chain or not. I have another question:

    Is the structural formula for pentanoic acid
    CH3CH2CH2C=OCH2(OH) - I'm not sure if there are meant to be brackets on the O or if you show the = bond
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    (Original post by kiiten)
    Yeah, I wasn't sure if I was counting the longest chain or not. I have another question:

    Is the structural formula for pentanoic acid
    CH3CH2CH2C=OCH2(OH) - I'm not sure if there are meant to be brackets on the O or if you show the = bond
    For pentanoic acid, write like this will do CH3CH2CH2CH2COOH
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    (Original post by jyyl)
    For pentanoic acid, write like this will do CH3CH2CH2CH2COOH
    Right and if you oxidise then immediately distil butan-1-ol would the product be butanal or butan-1-al?
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    (Original post by kiiten)
    Right and if you oxidise then immediately distil butan-1-ol would the product be butanal or butan-1-al?
    Butanal and butan-1-al are the same thing but when naming aldehydes you don't need write the number to show which carbon the functional group is on because the aldehyde group is always on carbon one, so you can just write Butanal

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    (Original post by LThomas694)
    Butanal and butan-1-al are the same thing but when naming aldehydes you don't need write the number to show which carbon the functional group is on because the aldehyde group is always on carbon one, so you can just write Butanal

    Posted from TSR Mobile
    Oh yeah, of course its alcohols that can be on different carbons not aldehydes. Thanks
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    (Original post by FemaleBo55)
    Lol I really like your mind maps, what other ones have you made for chemistry?
    All that I have! (so far anyway)
    There are some more things I need to add one though, like in kinetics when a question says 'equimolar' it's a hint to say pH=pKa .
    And obviously the mechanisms one isnt finished becuase AS mechanisms can also come up.

    so yh, quite a bit left to do!

    do you have anything that i could use??
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    (Original post by P____P)
    All that I have! (so far anyway)
    There are some more things I need to add one though, like in kinetics when a question says 'equimolar' it's a hint to say pH=pKa .
    And obviously the mechanisms one isnt finished becuase AS mechanisms can also come up.

    so yh, quite a bit left to do!

    do you have anything that i could use??
    They look a bit like AS notes? Btw TSR resources are free. Have you checked them out?
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    Please could someone explain why the answer is 1.5 and not 3?

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    Please could someone explain this too thanks

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    (Original post by kiiten)
    Please could someone explain why the answer is 1.5 and not 3?

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    In the question it says 50%
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    Hi could someone please help me with this question...

    A 1.575g sample of ethanedioic acid crystals, H2C2O4.nH2O, was dissolved in water and made up to 250cm3. One mole of the acid reacts with two moles of NaOH. In a titration, 25.0cm3 of this solution of acid reacted with exactly 15.6cm3 of 0.160 mol dm-3 NaOH. Calculate the value of n.

    Any help would be appreciated
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    Also could someone please help with this question...

    3.88g of a monoprotic acid was dissolved in water and the solution was made up to 250cm3. 25.0cm3 of this solution was titrated with 0.095 mol dm-3 NaOH solution, requiring 46.5 cm3. Calculate the relative molecular mass of the acid.

    The answer is 87.8 but I got 878...?
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    (Original post by P____P)
    All that I have! (so far anyway)
    There are some more things I need to add one though, like in kinetics when a question says 'equimolar' it's a hint to say pH=pKa .
    And obviously the mechanisms one isnt finished becuase AS mechanisms can also come up.

    so yh, quite a bit left to do!

    do you have anything that i could use??
    do you have any AS revision notes, similar to the mind maps??
    If so , could I possibly use them?, they seem amazing sources of revision
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    (Original post by Samistrawberry)
    They look a bit like AS notes? Btw TSR resources are free. Have you checked them out?
    The spec. changed this year, and I found out they do some A2 kinetics stuff in AS like Kc, Kw , and some other stuff. But these mindmaps are for A2.
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    (Original post by Science_help)
    do you have any AS revision notes, similar to the mind maps??
    If so , could I possibly use them?, they seem amazing sources of revision
    Sorry, I don't have any of my notes from AS.
    I threw them all away. I didn't make any mindmaps last year.
    Spoiler:
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    It's easy to make your own though! Just have the Spec with you while you make it on A3 paper.
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    (Original post by kiiten)
    Please could someone explain this too thanks

    Posted from TSR Mobile
    1- (black ink in the image) you start with 3.0 mol and a 1.0 mol of ethanoic acid and the alcohol.you end with 0.9 mol of water
    2- (red ink in the image) you must have started with 0 mol on the right for all (because nothing has been formed)
    3-(purple ink in the image) looking at the ratios of the products, you see that water and the ester are of the ratio 1:1, so if 0.9 mol of water has formed at equilibrium , then 0.9 mol of the ester must have formed too.
    4- (mol of ester/mol of ethanoic acid) x 100= your answer


    I find it a lot easier if you draw up a table. It's a lot clearer to see what you're dealing with.
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    i.e. (0.9/3.0) x 100 = 30%
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    (Original post by ilikechemistry)
    Hi could someone please help me with this question...

    A 1.575g sample of ethanedioic acid crystals, H2C2O4.nH2O, was dissolved in water and made up to 250cm3. One mole of the acid reacts with two moles of NaOH. In a titration, 25.0cm3 of this solution of acid reacted with exactly 15.6cm3 of 0.160 mol dm-3 NaOH. Calculate the value of n.

    Any help would be appreciated
    Hi- I've tried to give it a shot, forgive me if there are any mistakes...

    Okay so first I calculated the moles of NaOH: which is conc x (vol/1000) to get me 2.496 x 10-3 moles
    We know its a 1:2 ratio, for every one mole of acid, two moles of NaOH is reacted, so we divide the moles of NaOH we have calculated by 2, to get 1.248 x 10-3 moles of the acid (in 25cm3)
    Now we have the moles of the acid in 25cm3 we need to calculate moles in 250cm3 so we multiply the moles by 10 to get 0.01248 moles of the acid
    Now we have the moles of the acid in 250cm3. We can now divide the mass of the acid crystals by the moles to give us the Mr which is 1.575/0.01248= 126

    Now we have the Mr of H2C2O4.nH2O we then first find out the Mr of the first part: H2C2O4= which is 90. 126 - 90 is 36 so H20 is 36. The Mr of H20 is 18, so we divide 36/18 to get us n which is 2.
    So n is 2. I think.
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    (Original post by ilikechemistry)
    Also could someone please help with this question...

    3.88g of a monoprotic acid was dissolved in water and the solution was made up to 250cm3. 25.0cm3 of this solution was titrated with 0.095 mol dm-3 NaOH solution, requiring 46.5 cm3. Calculate the relative molecular mass of the acid.

    The answer is 87.8 but I got 878...?
    For this question, we first find out the moles of NaOH which is: (46.5/1000) x 0.095= to give us 4.4175 x 10^-3. REMEMBER THIS IS IN 25 CM3. You HAVE to multiply the moles by 10 (which i think you forgot to do) to get 0.044175 moles of acid.
    I assumed it's a 1:1 ratio so I then, divided the mass of the acid by the moles, so 3.88/0.044175= 87.8

    Just remember to not forget the volumes of the solutions- we may have to convert the moles in 25cm3 to 250cm3 which is why we had to multiply by 10
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    So I've been stuck on this AS past paper chem question for too long and I would be very, very grateful if anyone could please help me with it, thanks. Here it is:
    Anhydrous calcium chloride, CaCl2, can be used to dry some organic liquids. During this process, hydrated calcium chloride, CaCl2.2H20 is formed.
    CaCl2 + 2H20 --> CaCl2.2H20
    Mr 111
    In a drying process, 5.55g of anhydrous calcium chloride, CaCl2, is used. Calculate the amount of water that can be removed from the organic liquid. (2 marks)

    Thank you!!
 
 
 
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