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    And if you do OCR A, this will really help you out alot!!!!!!!!!!!!!!!


    https://www.tes.co.uk/teaching-resou...ksheet-6423088
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    (Original post by leaveyusufalone)
    Hey guys,
    I do OCR A.
    I was just wondering why my teacher stressed on the importance of the 5 ions (Carbonate, Nitrate, Sulphate, Hydroxide and Ammonium)?
    Why do we NEED to learn their formula and names?
    You should know their formula and charge because you're gonna need it when it comes to oxidation states.


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    hey guys just a quick one here thanks,
    for the AQA as june 2012 unit 2 paper question 9b(i) you have to get an ionic equation - i got the right answer
    however now i don't think it makes much sense
    the full equation i believe is KCl (s) + H2SO4(l)===> KHSO4(s) + HCl(g) however to get an ionic equation from this I'm not sure as no reactants or products are aqueous
    my possible explanation is that adding the acid dissolves the solid KCL?? making it aqueous. However this is not very convincing
    Anyone know?? any help would be appreciated
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    (Original post by Super199)
    Need some help with an enthalpy change question

    Calculate the enthalpy changes of reaction for each of the following reaction

    a). C2H4(g) + H2(g) ----> C2H6(g)

    Right so it gives you a ton of values. So since they are in the gas state it is Sum of bonds broken - Sum of bonds made.

    There is a C=C double bond which is 612Kjmol^-1
    But it says there is one H-H bond that gives the 436Kjmol^-1
    I don't really understand that bit isn't there 4 C-H bonds? Can someone explain this. I get the rest of the question apart from this bit.

    Thanks

    Hi , what helps is to draw the displayed formula e.g. Ethane has 6 C-H and one C-C ( single bond ) in that you can see the bonds and so forth. Hope that helps ;P
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    What functional group is C=O in?
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    (Original post by Charlblackburn)
    What functional group is C=O in?
    The C=O (carbonyl group) is a functional group
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    (Original post by ethanoylchloride)
    The C=O (carbonyl group) is a functional group
    Thanks! The notes were a little messed up but I get it now
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    (Original post by Charlblackburn)
    Thanks! The notes were a little messed up but I get it now
    No problem, glad i could be of help.
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    (Original post by Super199)
    Anyone care to explain the propagation step in free radical substitution. With an example, other than the methane and chlorine one? I understand the initiation and termination step but this one is a bit hit and miss...
    How about ethane and bromine?

    If you understand the initation and termination in these steps, so I can omit them in my explanation.

    In propagation, the former bromine molecule became radicals, you know? so ions with the same charge. One of this radical is binding to an H-Atom, so the bromo radical become a hydrogen bromide molecule, while the former ehtane gets a charge, just because it has lost an H-Atom to the bromo radical. That is why the former ethane becomes a radical, an ethyl radical to be exact. This ethyl radical may react with a bromine molecule or with an ethane one. In the case of bromine molecule, a Br-Atom is binding to the ehtyl radical, so it comes to a Bromo ethane molecule, while the former bromine molecule gets a charge, as a Br-Atom got lost, so a new bromo radical comes into being. In the case of ethane molecule, an H-Atom is binding to the ethyl radical, it comes to an ethane molecule, while an ethyl radical remains again (loss of H-Atom, you know?). And so the propagation repeats itself and goes on, till termination.
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    (Original post by Kallisto)
    How about ethane and bromine?

    If you understand the initation and termination in these steps, so I can omit them in my explanation.

    In propagation, the former bromine molecule became radicals, you know? so ions with the same charge. One of this radical is binding to an H-Atom, so the bromo radical become a hydrogen bromide molecule, while the former ehtane gets a charge, just because it has lost an H-Atom to the bromo radical. That is why the former ethane becomes a radical, an ethyl radical to be exact. This ethyl radical may react with a bromine molecule or with an ethane one. In the case of bromine molecule, a Br-Atom is binding to the ehtyl radical, so it comes to a Bromo ethane molecule, while the former bromine molecule gets a charge, as a Br-Atom got lost, so a new bromo radical comes into being. In the case of ethane molecule, an H-Atom is binding to the ethyl radical, it comes to an ethane molecule, while an ethyl radical remains again (loss of H-Atom, you know?). And so the propagation repeats itself and goes on, till termination.
    Is the propagation step this then:

    CH4 + .Br ---> .CH3+HBR
    .CH3+ Br2 --> ? + .Br

    I don't understand what happens in the question mark bit. With the first line I get that the Br radical knocks of a hydrogen forms a methyl radical and HBR. It's the second bit which is confusing me. When explaining it do you mind spacing out your explanation, find it hard to acknowledge what you are saying when it is all in one paragraph.
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    AQA and ISA





















































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    Lovely helpers!

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    Hello all! Just a quick reminder and message from our lovely helpers:

    When asking a question, please try to include all relevant information, which should mean you are given assistance quicker. And also, try to include what you have already attempted/tried to do(though a lot of people do this already).

    Thank you very much! :hugs:



    :elefant::elefant::elefant::elefant::elefant::elefant::elefant::elefant::elefant::elefant:[/QUOTE]
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    (Original post by Super199)
    Is the propagation step this then:

    CH4 + .Br ---> .CH3+HBR
    .CH3+ Br2 --> ? + .Br

    I don't understand what happens in the question mark bit. With the first line I get that the Br radical knocks of a hydrogen forms a methyl radical and HBR. It's the second bit which is confusing me. When explaining it do you mind spacing out your explanation, find it hard to acknowledge what you are saying when it is all in one paragraph.
    Br2-----> 2Br•
    Br• + CH4 ----> CH3• + HBr
    CH3• + Br2----> CH3Br + Br•

    The bromine free radical has been regenerated.


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    (Original post by Super199)
    Is the propagation step this then:

    CH4 + .Br ---> .CH3+HBR
    .CH3+ Br2 --> ? + .Br

    I don't understand what happens in the question mark bit. With the first line I get that the Br radical knocks of a hydrogen forms a methyl radical and HBR. It's the second bit which is confusing me. When explaining it do you mind spacing out your explanation, find it hard to acknowledge what you are saying when it is all in one paragraph.
    What you wrote is methane, not ethane. Ethane has the formula C2H6...

    Sorry, that I was not understandable enough for you. That is why, I will explain the propagation for you again in segregated lines - and reaction equations!

    First you got two bromo radicals ( brom ions with the same charge).

    Br* + Br* (the * stands for the radicals).

    One of these radicals reacts with ethane by knocking of an H-Atom, as you said.

    Br* + C2H6 -> HBr + C2H5*

    An ethyl radical comes into being. So there are two possible reactions now:

    1.) C2H5* + Br2 -> BrC2H5 + Br* (knocking of an Br-Atom) or
    2.) C2H5* + C2H6 -> C2H6 + C2H5* (knocking of an H-Atom)

    in 1.) a bromo radical comes into being, in 2.) its an ethyl one. Is that better?
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    (Original post by Kallisto)
    What you wrote is methane, not ethane. Ethane has the formula C2H6...

    Sorry, that I was not understandable enough for you. That is why, I will explain the propagation for you again in segregated lines - and reaction equations!

    First you got two bromo radicals ( brom ions with the same charge).

    Br* + Br* (the * stands for the radicals).

    One of these radicals reacts with ethane by knocking of an H-Atom, as you said.

    Br* + C2H6 -> HBr + C2H5*

    An ethyl radical comes into being. So there are two possible reactions now:

    1.) C2H5* + Br2 -> BrC2H5 + Br* (knocking of an Br-Atom) or
    2.) C2H5* + C2H6 -> C2H6 + C2H5* (knocking of an H-Atom)

    in 1.) a bromo radical comes into being, in 2.) its an ethyl one. Is that better?
    Ah I've got it. Thank you
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    hey guys just a quick one here thanks,
    for the AQA as june 2012 unit 2 paper question 9b(i) you have to get an ionic equation - i got the right answer
    however now i don't think it makes much sense
    the full equation i believe is KCl (s) + H2SO4(l)===> KHSO4(s) + HCl(g) however to get an ionic equation from this I'm not sure as no reactants or products are aqueous
    my possible explanation is that adding the acid dissolves the solid KCL?? making it aqueous. However this is not very convincing
    Anyone know?? any help would be appreciated
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    (Original post by Anony1234)
    hey guys just a quick one here thanks,
    for the AQA as june 2012 unit 2 paper question 9b(i) you have to get an ionic equation - i got the right answer
    however now i don't think it makes much sense
    the full equation i believe is KCl (s) + H2SO4(l)===> KHSO4(s) + HCl(g) however to get an ionic equation from this I'm not sure as no reactants or products are aqueous
    my possible explanation is that adding the acid dissolves the solid KCL?? making it aqueous. However this is not very convincing
    Anyone know?? any help would be appreciated
    The ionic equation would be H2SO4 + Cl- ===> HSO4- + HCl
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    If anyone here doing AQA Unit 1 Chem if they can bother sharing their notes, it'd be really appreciated.
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    (Original post by ethanoylchloride)
    The ionic equation would be H2SO4 + Cl- ===> HSO4- + HCl
    could you explain why please thanks
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    (Original post by mcmuzzi98)
    If anyone here doing AQA Unit 1 Chem if they can bother sharing their notes, it'd be really appreciated.
    chemrevise revision guides are really good
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    (Original post by Super199)
    Ah I've got it. Thank you
    You are welcome, Super199. I do what I can to enable a better understanding in chemistry.
 
 
 
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