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    (Original post by cheesypuff)
    this chemy paper better be easy.

    someone told me it's the trickiest
    I've always found chem 5 much easier than 1, 2 and 4
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    (Original post by Beth_L_G)
    I've always found chem 5 much easier than 1, 2 and 4
    I find it harder than those!
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    (Original post by mackie0088)
    NiO(OH)(s) + H2O(I) + e–→ Ni(OH)2(s) + OH–(aq) (+0.52) Higher E so would normally be reduced (more positive so more likely to gain electrons) BUT this is RECHARGING and so it be oxidised, the equation will go the opposite way to regenerate electrode.
    Ni(OH)2(s) + OH–(aq) →NiO(OH)(s) + H2O(I) + e–

    Cd(OH)2(s) + 2e–→ Cd(s) + 2OH–(aq) (–0.88) Lower E (more negative so more likely to lose electrons) so would normally be oxidized like so:

    Cd(s) + 2OH–(aq)→ Cd(OH)2(s) + 2e–

    BUT this is RECHARGING so will go the opposite way to regenerate electrode.Cd(OH)2(s) + 2e–→ Cd(s) + 2OH–(aq)



    Next balance for electrons

    Ni(OH)2(s) + OH–(aq) →NiO(OH)(s) + H2O(I) + e– ( x2)

    Cd(OH)2(s) + 2e–→ Cd(s) + 2OH–(aq)

    So
    2Ni(OH)2(s) + 2OH–(aq) →2NiO(OH)(s) + 2H2O(I) + 2e–

    Cd(OH)2(s) + 2e–→ Cd(s) + 2OH–(aq)

    Overall

    2Ni(OH)2(s) + 2OH–(aq) + Cd(OH)2(s) + 2e→2NiO(OH)(s) + 2H2O(I) + 2e–+Cd(s) + 2OH–(aq)

    Cancel things on both sides (OH- and e-)
    2Ni(OH)2(s) +Cd(OH)2(s) →2NiO(OH)(s) + 2H2O(I) +Cd(s)

    Hope this helps, not great at explaining things sorry
    aah i see, thanks so much that makes sense now.. yeh because my equation was the wrong way round sort of because I never knew they swapped if its recharging.. thanks again!
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    can someone please explain thermodynamics to me, i have attempted calculations and my answers keep coming out wrong. I understand entropy, im just confused with the born harber cycle, and enthalpy of hydration and solution?
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    Hate enthalpy and fuel cells!
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    (Original post by nadiaalq)
    can someone please explain thermodynamics to me, i have attempted calculations and my answers keep coming out wrong. I understand entropy, im just confused with the born harber cycle, and enthalpy of hydration and solution?
    Same!! I thought I was fine with all the thermodynamic stuff, but then we did the jan 2013 paper as a mock in school today and I couldn't really answer any of the thermodynamics questions.
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    (Original post by brittanna)
    Same!! I thought I was fine with all the thermodynamic stuff, but then we did the jan 2013 paper as a mock in school today and I couldn't really answer any of the thermodynamics questions.
    The Jan 13 was a really strange paper :s much harder to begin with and easier towards the end..but those thermodynamics questions were awful. I think thats as hard as they can come though..at least we now know all about bond enthalpies!
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    (Original post by erniiee)
    The Jan 13 was a really strange paper :s much harder to begin with and easier towards the end..but those thermodynamics questions were awful. I think thats as hard as they can come though..at least we now know all about bond enthalpies!
    I still don't know how you were supposed to most of the calculation ones though. I just tried randomly putting the numbers together until I got somewhere reasonably close to the answer they give if you can't do the question . How are you actually supposed to do those enthalpy of hydration ones though? Is it using a born haber cycle? Even if it is, I don't know how I would put it together for those enthalpies ;
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    (Original post by brittanna)
    I still don't know how you were supposed to most of the calculation ones though. I just tried randomly putting the numbers together until I got somewhere reasonably close to the answer they give if you can't do the question . How are you actually supposed to do those enthalpy of hydration ones though? Is it using a born haber cycle? Even if it is, I don't know how I would put it together for those enthalpies ;
    I think they just wanted us to read the question really carefully..it tests the first part of 12.4 which seems so niche to have a whole question on argh

    Yea you do have to use a cycle, althoughit doesn't have all of the steps that a born harber one does. You have to use the lattice enthalpy and hydration of each ion e.g.
    exothermic hydration
    endothermic hydration!

    e.g. sodium chloride - an endothermic hydration.
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    (Original post by mathsguy)
    Hate enthalpy and fuel cells!
    so the whole of chem5 then ...
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    Good luck everyone with revision!
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    Hi, would anybody be able to help me with this question from Jan 2010:

    8 (c) A chemical company has a waste tank of volume 25 000dm3. The tank is full of
    phosphoric acid (H3PO4) solution formed by adding some unwanted phosphorus(V)
    oxide to water in the tank.
    A 25.0cm3 sample of this solution required 21.2cm3 of 0.500 mol dm–3 sodium
    hydroxide solution for complete reaction.
    Calculate the mass, in kg, of phosphorus(V) oxide that must have been added to the
    water in the waste tank.
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    how come the manganate ions cant oxidise sulfate ions?
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    I have a few q:

    when the potential difference is measured, is there a current aswell?
    does the voltmeter allow a current to flow?
    if there was no voltmeter and just a metal wire and saltbridge connecting both half cells, would the current flow?
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    has everyone started doing past papers ? what kind of marks are everyone getting at this stage, if you guys don't mind me asking? Just wanted to get an idea!
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    (Original post by laurawoods)
    has everyone started doing past papers ? what kind of marks are everyone getting at this stage, if you guys don't mind me asking? Just wanted to get an idea!
    I've done 3 and got high 70s/100. Need more than that for the grade I want. Havent checked the UMS conversions because I darent :L

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    (Original post by Sherlockedd)
    I've done 3 and got high 70s/100. Need more than that for the grade I want. Havent checked the UMS conversions because I darent :L

    Posted from TSR Mobile
    Ok cool...so are u aiming for an A star grade in the summer?
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    (Original post by mackie0088)
    NiO(OH)(s) + H2O(I) + e–→ Ni(OH)2(s) + OH–(aq) (+0.52) Higher E so would normally be reduced (more positive so more likely to gain electrons) BUT this is RECHARGING and so it be oxidised, the equation will go the opposite way to regenerate electrode.
    Ni(OH)2(s) + OH–(aq) →NiO(OH)(s) + H2O(I) + e–

    Cd(OH)2(s) + 2e–→ Cd(s) + 2OH–(aq) (–0.88) Lower E (more negative so more likely to lose electrons) so would normally be oxidized like so:

    Cd(s) + 2OH–(aq)→ Cd(OH)2(s) + 2e–

    BUT this is RECHARGING so will go the opposite way to regenerate electrode.Cd(OH)2(s) + 2e–→ Cd(s) + 2OH–(aq)



    Next balance for electrons

    Ni(OH)2(s) + OH–(aq) →NiO(OH)(s) + H2O(I) + e– ( x2)

    Cd(OH)2(s) + 2e–→ Cd(s) + 2OH–(aq)

    So
    2Ni(OH)2(s) + 2OH–(aq) →2NiO(OH)(s) + 2H2O(I) + 2e–

    Cd(OH)2(s) + 2e–→ Cd(s) + 2OH–(aq)

    Overall

    2Ni(OH)2(s) + 2OH–(aq) + Cd(OH)2(s) + 2e→2NiO(OH)(s) + 2H2O(I) + 2e–+Cd(s) + 2OH–(aq)

    Cancel things on both sides (OH- and e-)
    2Ni(OH)2(s) +Cd(OH)2(s) →2NiO(OH)(s) + 2H2O(I) +Cd(s)

    Hope this helps, not great at explaining things sorry
    Hello, I had a couple of chem5 questions to ask you...pls can I fire away...
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    (Original post by mackie0088)
    NiO(OH)(s) + H2O(I) + e–→ Ni(OH)2(s) + OH–(aq) (+0.52) Higher E so would normally be reduced (more positive so more likely to gain electrons) BUT this is RECHARGING and so it be oxidised, the equation will go the opposite way to regenerate electrode.
    Ni(OH)2(s) + OH–(aq) →NiO(OH)(s) + H2O(I) + e–

    Cd(OH)2(s) + 2e–→ Cd(s) + 2OH–(aq) (–0.88) Lower E (more negative so more likely to lose electrons) so would normally be oxidized like so:

    Cd(s) + 2OH–(aq)→ Cd(OH)2(s) + 2e–

    BUT this is RECHARGING so will go the opposite way to regenerate electrode.Cd(OH)2(s) + 2e–→ Cd(s) + 2OH–(aq)



    Next balance for electrons

    Ni(OH)2(s) + OH–(aq) →NiO(OH)(s) + H2O(I) + e– ( x2)

    Cd(OH)2(s) + 2e–→ Cd(s) + 2OH–(aq)

    So
    2Ni(OH)2(s) + 2OH–(aq) →2NiO(OH)(s) + 2H2O(I) + 2e–

    Cd(OH)2(s) + 2e–→ Cd(s) + 2OH–(aq)

    Overall

    2Ni(OH)2(s) + 2OH–(aq) + Cd(OH)2(s) + 2e→2NiO(OH)(s) + 2H2O(I) + 2e–+Cd(s) + 2OH–(aq)

    Cancel things on both sides (OH- and e-)
    2Ni(OH)2(s) +Cd(OH)2(s) →2NiO(OH)(s) + 2H2O(I) +Cd(s)

    Hope this helps, not great at explaining things sorry

    http://filestore.aqa.org.uk/subjects...W-QP-JAN10.PDF

    1) In 1ci ) , for the equation 2 that they are asking , would they accept,
    V2O4 + 1/2 O2 --> V2O5

    because in the MS they have actually doubled it all up?

    2) In question number 3c ) is my answer , given below, right?

    Pt (s) /H2(g) /H2O (l) // O2 (g) / OH- (aq) / Pt (s)

    The MS is confusing me...

    3) For 3h) would they accept my answer: Hydrogen could be produced by the electrolysis of water, and this requires electricity which is generated by burning fossil fuels. This burning of fossil fuels will release CO2 into the envt.

    4) For 6a) would you get the second mark , with a wrong product?

    5) Pls can u explain the calculation question in 8c ?

    6) For 9a) would this get 3 marks for the colours explanation?

    Ti (III) ions are coloured because it has got partially filled 3d orbitals. Therefore, d-d transitions can take place which means electrons are excited from ground state to higher energy level. Light is absorbed from the visible range. Ti (IV) has empty 3d orbitals. So no d-d electron transitions taking place. No light energy from visible range absorbed.

    Would this get the marks, because the MS seems so different?

    7) For that very last question , how does the graph actually look? I have drawn a graph , but i don;t know if that would pick up the marks?

    phew and thanks!
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    (Original post by dizzy17)
    Hi, would anybody be able to help me with this question from Jan 2010:

    8 (c) A chemical company has a waste tank of volume 25 000dm3. The tank is full of
    phosphoric acid (H3PO4) solution formed by adding some unwanted phosphorus(V)
    oxide to water in the tank.
    A 25.0cm3 sample of this solution required 21.2cm3 of 0.500 mol dm–3 sodium
    hydroxide solution for complete reaction.
    Calculate the mass, in kg, of phosphorus(V) oxide that must have been added to the
    water in the waste tank.

    equation for the reaction of sodium hydroxide with phosphoric acid:
    H3PO4 + 3NaOH = Na3PO4 + 3H2O

    moles of naoh = 0.5 x ( 21.2/1000) = 0.0106
    moles of h3po4 used in titre = 0.0106/3 =3.53x10-3

    however this is the moles in 25cm3 the tank has 25000 dm3
    25cm3 = 25/1000 =0.0025dm3

    ratio of volume in tank to volum titrated = 25000/0.025 =1x106

    so the moles on h3po4 in tank will be 1x106 times that we used for the tiration
    therefor moles of h3po4 in tank = 3.53x10-3 x 1x106= 3530 moles

    now we need to find the moles of phosphorus oxide
    equation for phosphorus oxide reacting with water to make phosphoric acid is:

    P4O10 + 6H20 = 4H3PO4

    we know the moles of h3po4 is 3530
    moles of p4o10 = 3530/4 = 882.5 moles

    find the mass of p4o10
    mass = moles x mr
    = 882.5 x 248 = 250630g
    = 250630 / 1000 = 250.63kg
 
 
 
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