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Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread watch

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    (Original post by genevievelaw)
    include water if all the reactants are liquids
    don't include water if water is the solvent (i.e. if the reactants are (aq))
    hope I helped
    OMG!!! Thanks so much!!!

    Could you also explain 2)c)i) of this paper?

    https://ks5chemistry.files.wordpress...-acid-lots.pdf
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    (Original post by Sandy_Vega30)
    Hey! COuld someone tell me when to and when not to include H2O in the equilibrium constant expression? Because in the CHem U6 exam, we were supposed to include water in the expression, I thought the Ionization of water is a constant. Isn't that so?
    You need to include water in the Kc expression when it is a product of the reaction, for example in an esterification reaction. The reason water is mostly not included is that it's normally the solvent, meaning that it's in such an excess that the concentration doesn't change at all during the reaction.
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    (Original post by Spoderman14)
    You need to include water in the Kc expression when it is a product of the reaction, for example in an esterification reaction. The reason water is mostly not included is that it's normally the solvent, meaning that it's in such an excess that the concentration doesn't change at all during the reaction.
    When you say 'Solvent' do you mean we don't include water when it is a reactant?
    How about when it is a gas and is a reactant?
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    (Original post by Supermanxxxxxx)
    Attachment 549197Attachment 549197549199
    any hep with this one
    What paper is this from? Or do you have the answers?
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    (Original post by Sandy_Vega30)
    OMG!!! Thanks so much!!!

    Could you also explain 2)c)i) of this paper?

    https://ks5chemistry.files.wordpress...-acid-lots.pdf
    1. Calculate decrease in moles in HCOOH
    2. Subtract this figure from the original mole value for C2H5OH
    3. Add the figure from step 1 to the 0 for the ester
    4. Do the same for H2O

    You do this because there is 1 mol of each reactant and of each product
    If there were to be 1 mole of a reactant and 2 moles of a product, you would calculate the increase in product moles by adding double the value of the decrease in the moles of reactants

    Sorry if this doesn't make much sense
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    (Original post by genevievelaw)
    OMG I did this question like 5 minutes ago! Let me just take a photo of what I wrote
    Lol really? I'd appreciate that thanks. I'm marking that exam atm and I didn't like it at all but I guess there are worse exams out there. I just can't get my head around NMR questions

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    (Original post by genevievelaw)
    1. Calculate decrease in moles in HCOOH
    2. Subtract this figure from the original mole value for C2H5OH
    3. Add the figure from step 1 to the 0 for the ester
    4. Do the same for H2O

    You do this because there is 1 mol of each reactant and of each product
    If there were to be 1 mole of a reactant and 2 moles of a product, you would calculate the increase in product moles by adding double the value of the decrease in the moles of reactants

    Sorry if this doesn't make much sense
    I get it now. Thanks a lot!
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    (Original post by TeaAndTextbooks)
    Help, I have no idea how you do this.

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    (Original post by genevievelaw)
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    Thanks. You condense your answer down so well. When I answer these questions I end up writing a full paragraph on one peak :dontknow:. This is super helpful especially for structuring the answer correctly. Good luck tomorrow.

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    International papers seem a lot harder
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    Any help on this one please? Lool I've seen so many water questions I am now starting to confuse myself!
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    (Original post by TeaAndTextbooks)
    Thanks. You condense your answer down so well. When I answer these questions I end up writing a full paragraph on one peak :dontknow:. This is super helpful especially for structuring the answer correctly. Good luck tomorrow.

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    Np
    Lol normally I waffle too much You too!
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    (Original post by Funky_Giraffe)
    Hey guys, can someone please explain what the heck this question is on about? How are we supposed to know to compare experimental and theoretical lattice energies?!

    Hi, do you know which past paper this is from?
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    (Original post by SRR3456)
    Hi, do you know which past paper this is from?
    June 2012 I think
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    (Original post by Funky_Giraffe)
    June 2012 I think
    Thanks - I haven't revised any AS stuff so I hope the paper isn't really synoptic!
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    (Original post by iStudent96)
    Any help on this one please? Lool I've seen so many water questions I am now starting to confuse myself!

    The pH is temperature dependent. If we increase the temperature, the value for Kw will increase(Kw is a type of equilibrium constant) as this is an endothermic reaction so increasing the temperature would favour the forward direction hence increasing the concentration of hydrogen ions and thus lowering the pH. However, it is not acidic because concentration of hydrogen and hydroxide ions are equal.
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    do we need to know the mechanism for the iodination of propanone? or just be able to guess one?
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    (Original post by Wunderbarr)
    Question 2: When talking about thermodynamic feasibility, this is ALWAYS talking about entropy. You chose to talk about activation energy, which is to do with kinetics.

    So something that is kinetically stable will have a high activation energy.

    Something thermodynamically stable will have a more negative ΔS(total).

    Question 5: count the moles of gas on each side of the equilibrium equation, and you should be able to figure that one out. Why it isn't A is because it isn't B.

    Lol jk but if a gas is more pressured then surely it helps it go through the reactor anyway? Much like pushing harder to get onto a packed train will make people want to move out of the way so they aren't so squashed.
    Oh yeah thanks, I'm an idiot for not getting 5.
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    (Original post by sabahshahed294)
    The pH is temperature dependent. If we increase the temperature, the value for Kw will increase(Kw is a type of equilibrium constant) as this is an endothermic reaction so increasing the temperature would favour the forward direction hence increasing the concentration of hydrogen ions and thus lowering the pH. However, it is not acidic because concentration of hydrogen and hydroxide ions are equal.
    You see I am lost when you say 'so increasing the temperature would favour the forward direction hence increasing the concentration of hydrogen ions and thus lowering the pH', but the forward direction produces both H+ and OH-. Why would it only increase conc of H+ and not OH-?
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    (Original post by Sandy_Vega30)
    When you say 'Solvent' do you mean we don't include water when it is a reactant?
    How about when it is a gas and is a reactant?
    No, when water is the solvent (the reactants are aq), then it's concentration hardly changes during the reaction, so it won't be present in the kc expression. If water is a reactant or product, then it will be in the kc equation if it isnt in excess
 
 
 

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