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    Can anyone show how you would work out the enthalpy change of neutralisation for a reaction between 50cm3 of 0.1M H2SO4 and 50cm3 of of 0.1M NaOH with specific heat capacity of 4.18 and temp change from 20 to 30?

    I know the equation would be H2S04 + 2NaOH -> NA2SO4+ 2H20 and Q = -4180J
    What would you do next?
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    (Original post by Timon512)
    nN=m/Mr = 63.64/14 (Take % as being the mass because it is a ratio.) = 4.55mol nO=m/Mr = 100-63.64/16 = 36.36/16 = 2.27 mol

    Hence, nN:nO = 2:1

    Hence N2O.
    You have to admit, that was some sweaty question...
    to work out its molar mass took a lot of work
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    (Original post by aegonsconquest)
    why haven't they given an indication of the +- value on 2.71? How do we know if its -2.71 or + 2.71?

    I mean, obvs you assume its +2.71 bc they wouldve stated otherwise no?
    if there is no negative sign in front of it then it must be +2.71V
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    (Original post by lai812matthew)
    when you hydrolyse peptides with acid you get NH3+
    I cant remember the Q, but what she is saying is correct (if it were on the exam). but From what i remember i deffo put down NH3+.
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    (Original post by cr7alwayz)
    Can anyone show how you would work out the enthalpy change of neutralisation for a reaction between 50cm3 of 0.1M H2SO4 and 50cm3 of of 0.1M NaOH with specific heat capacity of 4.18 and temp change from 20 to 30?

    I know the equation would be H2S04 + 2NaOH -> NA2SO4+ 2H20 and Q = -4180J
    What would you do next?
    isn't Q 4.18?

    Also by 0.1M you mean concentration right?

    Actually, Do you know the QP this Q came from? Can you upload a pic of the Q?
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    (Original post by itsConnor_)
    Is my two step mech wrong? If so why? Diff to ms

    Posted from TSR Mobile
    did you get a response to this? I think your answer is correct since H2O2 cancels out giving you the correct overall equation. In fact it is correct, it doesnt matter if your two step reactions are false, as long as it makes sense and cancels accordingly you should get the mark.
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    With titration ph graphs how do you know which ph it starts and ends with?
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    Anyone know how to figure out what's happening with equilibrium graphs?


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    (Original post by aegonsconquest)
    isn't Q 4.18?

    Also by 0.1M you mean concentration right?

    Actually, Do you know the QP this Q came from? Can you upload a pic of the Q?
    I just made the question up myself but I think I've seen something similar before.
    Q is the heat that is released or absorbed by the system.
    The 4.18 has the symbol C which is the specific heat capacity.
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    (Original post by cr7alwayz)
    I just made the question up myself but I think I've seen something similar before.
    Q is the heat that is released or absorbed by the system.
    The 4.18 has the symbol C which is the specific heat capacity.
    I think you do q divided by moles of water formed for deltaH. Not sure though

    Posted from TSR Mobile
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    Does anyone know where you can get some difficult practice questions for this unit? I've done all the past papers and am trying to find some more questions to do
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    What topic do you guys find the hardest?

    I personally dont like electrode potential- the half equations part
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    (Original post by maisym00)
    Does anyone know where you can get some difficult practice questions for this unit? I've done all the past papers and am trying to find some more questions to do
    The website Alevelchemistry has questions sorted into individual topics. I am not sure about difficulty but they are good practice. They might be from previous past papers tho.
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    Why are [Cr(H2O)6]3+'s ligands ALL substituted by NH3 rather than only four like in Cu 2+?
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    (Original post by Ali1998HRM)
    Enthalpy change of solution= (enthalpy change of hydration of cation + enthalpy change of hydration of anion) - lattice enthalpy

    Enthalpy change of hydration of cation K+ will be more exothermic than the value of the cation Rb+ (as potassium ions have a higher charge density so are better able to attract and form bonds with water molecules). Enthalpy change of hydration of anion I- will be the same for both ionic compounds. Lattice enthalpy will be more negative for KI due to stronger electrostatic attraction between oppositely charged ions. Therefore in the equation you will be obtain a more positive number for enthalpy change of solution for KI.

    Hopefully this makes sense and coincides with the mark scheme. If there is a mistake some where in there do correct me 😊
    ah. Okay. I kinda get it now. I didn't know enthalpy solution= hydration enthalpy - lattice enthalpy. But it makes sense using a born haber cycle.

    You get 3 marks for talking about lattice enthalpy and hydration enthalpy and how it's both more -ev for KF and the normal explanation thing we're used to. But last mark is for saying that lattice enthalpy has a larger effect than hydration enthalpy. Does anyone know why? Is it because it's just a larger negative number than hydration enthalpy?
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    (Original post by cr7alwayz)
    Can anyone show how you would work out the enthalpy change of neutralisation for a reaction between 50cm3 of 0.1M H2SO4 and 50cm3 of of 0.1M NaOH with specific heat capacity of 4.18 and temp change from 20 to 30?

    I know the equation would be H2S04 + 2NaOH -> NA2SO4+ 2H20 and Q = -4180J
    What would you do next?
    It's not -4180, it's +4180J. Also you convert this into KJ by dividing by 1000 so it's 4.180KJ then divide this value by the number of moles of one of the reactants. Should've worked this out in a previous part of the question I assume
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    (Original post by charnot1997)
    With titration ph graphs how do you know which ph it starts and ends with?
    If they give you values then you should be able to work out the starting pH using -log[h+] but otherwise guesstimate. If it's a strong acid weak base, start at 2.5 end at 9 etc. There are examples in the textbook
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    fuel cells are the bane of my existence
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    (Original post by maisym00)
    Does anyone know where you can get some difficult practice questions for this unit? I've done all the past papers and am trying to find some more questions to do
    do past papers for old spec, do past papers for other exam board, do them again,
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    (Original post by HFancy1997)
    Kw=[OH-][H+]
    [OH-]=100[H+]
    10^-14=100[H+]^2
    [H+]=10^-8
    PH=-Log(10^-8)
    PH=8

    Hope that helps
    I still don't quite get it. Why are we multiplying [H+] by a 100 when the conc of [OH-] is a 100 times greater?
 
 
 
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