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    (Original post by pineneedles)
    Cu+ : 1s2 2s2 2p6 3s2 3p6 3d10 4s0
    Cu2+ : 1s2 2s2 2p6 3s2 3p6 3d9 4s0
    Cu2+ has an incomplete d subshell whereas Cu+ doesn't. The incomplete d subshell is allows transition metal ions to have colour. When a photon of light is absorbed by Cu2+, electrons in the lower energy levels of the 3d subshell jump up to the higher energy levels, and gain energy of a specific wavelength which corresponds to a colour. This colour is absorbed and therefore not seen. The colours that aren't absorbed are seen. The electrons of Cu+ can't jump up that energy level because the 3d subshell is full, there's no space to move up to. Therefore no energy is absorbed, all the colours are reflected and seen as white light.
    You don't need to know that, however, in the question they probably just want you to link the incomplete d subshell to the ability to form coloured complexes.

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    Thanks for the reply! I figured it wasn't in spec, thanks for detailed response!
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    (Original post by ToLiveInADream)
    Legacy paper right? Probably want the answer in terms of of wavelengths and stuff that isn't on the spec anymore.
    Yes budd, thanks for the reply!

    (Yes Ranz, it was a legacy paper)
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    Could someone also help with this sketching titration question. Not sure how to get the marks. Where would the equivalence point? Where would the end Ph be?
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    Could someone explain to me why benzene will only undergo substitution reactions but not addition reactions?
    I've read the explanation in the book and it didn't help
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    (Original post by TrentAlistar)
    Could someone explain to me why benzene will only undergo substitution reactions but not addition reactions?
    I've read the explanation in the book and it didn't help
    because benzene is more stable structure than if something were to be added, cause addition reactions it disrupts the delocalised pi electron, idk how correct this is though to be honest


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    (Original post by zirak46)
    Yes budd, thanks for the reply!

    (Yes Ranz, it was a legacy paper)
    yh we dont need to know that then! its not in our spec


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    (Original post by zirak46)
    Could someone also help with this sketching titration question. Not sure how to get the marks. Where would the equivalence point? Where would the end Ph be?
    Hey, the equivalence point is the volume (of the solution being added) at which the number of moles of both solutions are the same.

    To work out the end pH: it says that 0.1 moldm-3 of NaOH is added (obviously in excess) so the end pH will be the pH of the NaOH solution.

    Hint pH = 14 - pOH

    Strong bases, i.e NaOH, dissociate fully in water
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    (Original post by Serine Soul)
    Hey, the equivalence point is the volume (of the solution being added) at which the number of moles of both solutions are the same.

    To work out the end pH: it says that 0.1 moldm-3 of NaOH is added (obviously in excess) so the end pH will be the pH of the NaOH solution.

    Hint pH = 14 - pOH

    Strong bases, i.e NaOH, dissociate fully in water
    Ahhh thank you!

    I completely forgot that NaOH was a strong base. Derp.

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    Could someone please help with part ii), on what the ph of a diluted acid would be if conc of acid is given?

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    (Original post by L'Evil Wolf)
    is the nitration of phenol the same as the nitration of benzene?
    No but we do not need to know bout it. There's an application question in a past paper which gives the different conditions for the nitration of phenol and then you have to compare them and say why phenol does not need concentrated acids or a h2so4 catalyst and can occur at lower temperatures but we do not need to know this but somehow I do
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    (Original post by zirak46)
    Ahhh thank you!

    I completely forgot that NaOH was a strong base. Derp.

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    No problem!

    (Original post by zirak46)
    Could someone please help with part ii), on what the ph of a diluted acid would be if conc of acid is given?

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    Okay so we have added an equal volume of water, so let's say, with 1 dm3 of the acid, we added 1 dm3 of water. The total volume of the solution has doubled but the number of moles of acid remains the same

    So the new concentration is 0.015/2

    HNO3 is still a strong acid so the pH will be....
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    Name:  ImageUploadedByStudent Room1465601417.708364.jpg
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    in qs like thse i neevr know if to for example: make all NH2 into NH3+ or just the ones who were involved in the bonds? if that makes sense
    same goes for zwitter ions do i make all cooh coo- and all nh2 nh3+, or with alkali or acid hydrolysis i just get confused. help!? i hope i made sense.


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    (Original post by ranz)
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    in qs like thse i neevr know if to for example: make all NH2 into NH3+ or just the ones who were involved in the bonds? if that makes sense
    same goes for zwitter ions do i make all cooh coo- and all nh2 nh3+, or with alkali or acid hydrolysis i just get confused. help!? i hope i made sense.


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    All amines (if they can, i.e primary or secondary amines) accept a H+ ion in acid hydrolysis (lone pair of e- on N atom) as there is an abundance of H+ ions, supplied by the acid

    Similarly, all carboxylic acid groups become carboxylate ions and form a salt in alkali hydrolysis.
    So all COOHs turn into COO-Na+ with NaOH
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    (Original post by Serine Soul)
    Hey, the equivalence point is the volume (of the solution being added) at which the number of moles of both solutions are the same.

    To work out the end pH: it says that 0.1 moldm-3 of NaOH is added (obviously in excess) so the end pH will be the pH of the NaOH solution.

    Hint pH = 14 - pOH

    Strong bases, i.e NaOH, dissociate fully in water
    I learnt that the end pH would be different as the alkali will be diluted as it has been added to the acid. Don't really know though.
    To work it out I was told to make the volume equal to the acid plus the alkali added. And use the normal nber of moles.
    Sub this into moles=CV/1000 and solve for concentration which you then put into -log(ans) and then do 14 minus whatever you got.
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    (Original post by zirak46)
    Could someone also help with this sketching titration question. Not sure how to get the marks. Where would the equivalence point? Where would the end Ph be?
    Hi
    What paper is this from if you don't mind me asking? Thanks
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    (Original post by pineneedles)
    If we consider an ester, hydrogen bonds can't form between molecules of the ester because although there is lone pairs on the oxygen atom there is no partially positive hydrogen to be attracted to them. Hydrogen bonds can form between an ester and water, however. Just remember that to have a hydrogen bond, there has to be nitrogen, oxygen, or fluorine in the molecule and a partially positive hydrogen atom. As for the second question, that's right

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    Thanks! so Hydrogen bonds may form between the C=O bond and water, and OH with water?
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    (Original post by ImNervous)
    I learnt that the end pH would be different as the alkali will be diluted as it has been added to the acid. Don't really know though.
    To work it out I was told to make the volume equal to the acid plus the alkali added. And use the normal nber of moles.
    Sub this into moles=CV/1000 and solve for concentration which you then put into -log(ans) and then do 14 minus whatever you got.
    Oh yeah, you're right! Sorry about that it was a little too late for me I think
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    Hey guys on the f324 papers I've got an A* on every one apart from the 2015 one in which I got a C :/, do you think it'll be as hard as 2015 and if so what can I do now to get even better? I've done the past papers nearly twice
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    With zwitterions, if there is a COOH/NH2 group on the R group, you don't protonate/deprotonate it?

    my teachers been teaching us that you do but on a ms, I saw only one NH2 being protonated? Can anyone explain this pls???
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    (Original post by HarveySpecter97)
    Hey guys on the f324 papers I've got an A* on every one apart from the 2015 one in which I got a C :/, do you think it'll be as hard as 2015 and if so what can I do now to get even better? I've done the past papers nearly twice
    The 2015 paper wasn't that hard for me, I just thought there were too many questions to do within the time limit.

    Have you tried the legacy papers?
 
 
 
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