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    (Original post by thesuperdark)
    Why is it difficult to use Born-Haber cycles to find the lattice enthalpy of sodium carbonate? I know it's something to do with the formation of the carbonate ions right?
    Cos carbonate contain 2 different atoms, you would have to work out the enthalpy change of formation of the carbonate ions first.
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    Can anyone explain how you work out the element from the 'average mass' like in this question:
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    (Original post by CalistaJupiter)
    Can anyone explain how you work out the element from the 'average mass' like in this question:
    Just times that value by the Avogadro constant to get the Mr of the element
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    Jan 13 Q8(c)(ii) explained and worked through

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    (Original post by AqsaMx)
    2SO2 + O2 > 2SO3

    Explain what would happen to the pressure as the system was allowed to reach equilibrium

    I wrote the pressure would decrease as wquilibrium would shift to the left, but the MS says pressure decreases and fewer moles? Could anyone explain what Is meant by this?
    you start off with no SO3 so as it reaches equilibrium, more SO3 forms. There's fewer moles of gas on this side which also means pressure decreases (remember how if you increase the pressure, the position of equilibrium shits to the side with fewest moles of gas to decrease the pressure). hope that makes sense
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    (Original post by AqsaMx)
    2SO2 + O2 > 2SO3

    Explain what would happen to the pressure as the system was allowed to reach equilibrium

    I wrote the pressure would decrease as wquilibrium would shift to the left, but the MS says pressure decreases and fewer moles? Could anyone explain what Is meant by this?
    Yes, pressure would decrease because position of equilibrium would shift to the right, which has fewer moles of gas.
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    Does anyone have any tips on how to put together ionic equations? It's pretty much the one thing I have no idea about, I never know which ions to use and which ones get left out

    As an example Q;

    http://www.ocr.org.uk/Images/144762-...d-elements.pdf

    Question 7bii

    Any tips would be greatly appreciated
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    (Original post by megafidget)
    Gonna join you on that boat after the ocr core 3 disaster... Literally our entire uni offer depends on tomorrows paper wow
    Lmao I don't even have an offer. I'm taking a gap year but still, I need the grades to maximise my chances. And it just feels like after all this hard work, I'm still not gonna get the grades I want. If only I was born 4 years earlier...the 2012 papers for all my subjects were so easy, I would've had 3A*. But Life's a b****. Good luck to you and everyone else for tomorrow anyway.
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    (Original post by AqsaMx)
    2SO2 + O2 > 2SO3

    Explain what would happen to the pressure as the system was allowed to reach equilibrium

    I wrote the pressure would decrease as wquilibrium would shift to the left, but the MS says pressure decreases and fewer moles? Could anyone explain what Is meant by this?

    The pressure would decrease because at equilibrium there are only 2 moles of gas on the right hand side. If you can imagine having 2SO2 + O2 before any reaction there are 3 moles, however some of that 3 moles is converted into 2SO3 (Le Chatelier's principle) so overall there is a decrease in pressure.
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    anyone understand 7b from june 2012??????
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    (Original post by KB_97)
    Lmao I don't even have an offer. I'm taking a gap year but still, I need the grades to maximise my chances. And it just feels like after all this hard work, I'm still not gonna get the grades I want. If only I was born 4 years earlier...the 2012 papers for all my subjects were so easy, I would've had 3A*. But Life's a b****. Good luck to you and everyone else for tomorrow anyway.
    Haha ah fair enough, yeh I completely agree it feels like all the past papers/revision you do were for nothing when an exam goes like that arghhh. Oh well we go again tomoz, cheers and good luck to you too.
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    I feel strangely unprepared for this exam... anyone else?
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    (Original post by Sab_007)
    anyone understand 7b from june 2012??????
    CO2 reacts with water and forms carbonic acid and H+ ions (I only know this because of biology)
    So basically CO2 is acidic in solution and the H+ would react with the OH- ions in the equation so equilibrium shifts to the right to restore the OH- ions
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    Could someone help me understand how to do this question please? Name:  image.jpg
Views: 146
Size:  508.5 KB
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    (Original post by tcameron)
    is the cathode the positive electrode?
    CPR
    Cathode = Positive = Reduction
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    (Original post by rory58824)
    I feel strangely unprepared for this exam... anyone else?
    (Original post by Rust Cohle)
    Jan 13 Q8(c)(ii) explained and worked through

    I still don't understand how you got the 5:3 ratio
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    Could someone pls explain biii? https://gyazo.com/9b6f88ef2b4b5858ad75ea9f3f3bfd36

    Also for enthalpy change of neut do you DOUBLE volumes for m (in Q=mcdeltaT) and only do one volume for moles? ty vm
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    for ci https://gyazo.com/2d3b610ec0ec6ae199ede503f4da0794

    how do you get first equation here:?

    https://gyazo.com/053605e63dbfd229d81453c968d2033e

    I thought I- would react with Fe3+ with accordance to electrode potentials?
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    https://gyazo.com/49069f767bf5ea95d7e870c6a979a5e2 How do we know that 2 mol X reacts with 2 mol Cr? 1.456/1.021 = 1.43, do we just round 1.43 up to 1.5? seems a bit off :/
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    (Original post by itsConnor_)
    Could someone pls explain biii? https://gyazo.com/9b6f88ef2b4b5858ad75ea9f3f3bfd36

    Also for enthalpy change of neut do you DOUBLE volumes for m (in Q=mcdeltaT) and only do one volume for moles? ty vm
    Use your Q from previous question in this rearranged Q = mCdT

    For iii:
    dT = Q/(mc). Then x1000 if you used kJ (as opposed to J) for Q.
    Same number of moles as in (ii) except occupying a larger volume.

    For ii:
    m = sum of reactants = 35+35 = 70.
    For the n, yes, you use only the moles of the limiting reagent (whichever has lower moles). Here they're both equal moles (which is common for acid+base) so irrelevant.

    -Q is always around -54.6 kJ for future reference

    -Q/n = -4.827 kJ (exothermic)
 
 
 
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