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    (Original post by jlcf)
    Someone posted this in this thread, a few pages back,

    http://www.youtube.com/watch?v=AgJDp...feature=relmfu
    Haha my teacher Brady, he's basically one of us.
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    (Original post by haydyb123)
    Haha my teacher Brady, he's basically one of us.
    You are lucky to have him, he's brillant
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    (Original post by Flux_Pav)
    in your example, wasn't the NaOH in excess ?

    H2S04 + 2NAOH = NA2S04 2H20

    so if u have 0.5 moles of acid and 0.5mol for the alkali, you know that the naoh is in excess so you divide 0.5 by 2 as it will form 1 mol of water

    is it possible in neutralization for the no. of moles of acid and alkali to be different?
    what would happen then?

    LOL this part is doing my head in.
    No because H2SO4 dissociates twice, so the H2SO4 was in excess by 0.5, and the moles were different
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    (Original post by Flux_Pav)
    in your example, wasn't the NaOH in excess ?

    H2S04 + 2NAOH = NA2S04 2H20

    so if u have 0.5 moles of acid and 0.5mol for the alkali, you know that the naoh is in excess so you divide 0.5 by 2 as it will form 1 mol of water

    is it possible in neutralization for the no. of moles of acid and alkali to be different?
    what would happen then?

    LOL this part is doing my head in.
    say you've got 50cm3 of NaOH reacting with 50cm3 of H2SO4, both are 1 mol dm3
    the moles of them both is gonna be 0.05
    the stoichiometrical equation is 2NaOH + H2SO4 -> Na2SO4 + 2H2O
    say the change in temp is 13C
    so Q=100x4.18x13 = -5.343 kJ
    but you need to get this into kJ mol-1 so you divide by the amount of moles used, which is 0.025 (because 2:1 for NaOH2SO4) so for every 2 moles of NaOH, 1 mol of H2SO4 is reacted. so even if you have 0.05 moles of both, only 0.025 moles of the sulphuric acid has reacted, so you divide by that.
    so its -217.36 kJ mol -1

    i don't even know if im right anymore, you've confused me why can't you just stick with HCl!!!! haha
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    (Original post by jlcf)
    Your are lucky to have him, he's brillant
    He mostly teaches AS, as a kind-of get them to love Chemistry role.
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    Can someone explain this to me:

    If you have this reaction (I put the charge in the bracket):

    MnO4(-) + 8H(+) + 5Fe(2+) -------------> Mn(2+) + 4H2O + 5Fe(3+)

    What is the colour change?

    Fe2+ is pale green and Fe3+ is yellow
    but
    MnO4- is purple and [Mn(H20)6](2+) is colourless

    So whats the colour change?
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    I think everyone should watch this video:

    http://www.youtube.com/watch?v=AgeCz...feature=relmfu
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    Im gonna seriously start revision now!!!
    up till now I was demotivated by F215 exam from yesterday...
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    (Original post by haydyb123)
    I don't think you would ever get it as 3 moles of water forming as by definition, enthalpy of neutralisation is the formation of one mole of water.
    thats what i wrote
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    (Original post by M_I)
    Can someone explain this to me:

    If you have this reaction (I put the charge in the bracket):

    MnO4(-) + 8H(+) + 5Fe(2+) -------------> Mn(2+) + 4H2O + 5Fe(3+)

    What is the colour change?

    Fe2+ is pale green and Fe3+ is yellow
    but
    MnO4- is purple and [Mn(H20)6](2+) is colourless

    So whats the colour change?
    purple to yellow (I could be wrong as i didn't revise this module yet)
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    (Original post by monkeyDace)
    purple to yellow (I could be wrong as i didn't revise this module yet)
    purple to very pale pink.
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    (Original post by Limesasquatch)
    thats what i wrote
    Sorry I was meant to quite flux pav!
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    (Original post by wilsea05)
    say you've got 50cm3 of NaOH reacting with 50cm3 of H2SO4, both are 1 mol dm3
    the moles of them both is gonna be 0.05
    the stoichiometrical equation is 2NaOH + H2SO4 -> Na2SO4 + 2H2O
    say the change in temp is 13C
    so Q=100x4.18x13 = -5.343 kJ
    but you need to get this into kJ mol-1 so you divide by the amount of moles used, which is 0.025 (because 2:1 for NaOH2SO4) so for every 2 moles of NaOH, 1 mol of H2SO4 is reacted. so even if you have 0.05 moles of both, only 0.025 moles of the sulphuric acid has reacted, so you divide by that.
    so its -217.36 kJ mol -1

    i don't even know if im right anymore, you've confused me why can't you just stick with HCl!!!! haha
    LOL haha - lets just hope this doesn't come up
    anyways if it does - i hope its not as complex as i'm expecting it to be!
    thanks for the explanation though - its ever so slowly increasing my understanding lol

    TIME TO REVISE WHOLE OF MODULE 3 IN THE NEXT FEW HOURS.
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    (Original post by haydyb123)
    Sorry I was meant to quite flux pav!
    Thanks for the correct answer
    Now at least I wont make that
    mistake in the exam
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    Definitions we have to learn/ useful to learn
    Code:
    Rate of Reaction
    Rate constant
    Order
    Half Life
    RD Step
    Equilibrium Law
    Dynamic Equilibrium
    Bronsted Lowry Acid and Base
    Conjugate acid-base pairs
    pH
    [H+]
    Strong & weak acid
    Acid dissociation constant 
    Ionic product of water
    Buffers
    Equivalence point 
    End point
    Standard enthalpy change of neutralisation
    Standard enthalpy change of hydration
    Standard enthalpy change of formation
    Standard enthalpy change of atomisation
    Standard enthalpy change of solution
    Lattice Enthalpy
    Ionisation energy – First and Second
    Electron affinity – First and Second
    Entropy
    Enthalpy
    Standard entropy change of reaction
    Free energy change
    Oxidation & reduction & their agents
    Standard electrode potential
    Fuel cell
    Half cell
    Transition element
    Precipitation reactions
    Complex ion
    Coordinate number
    Ligand
    Stereoisomers
    cis-platin
    Mono/BI/Hexa – dentate ligands
    Ligand substitution 
    Stability constant
    Anymore?
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    (Original post by Rogercbinboy)
    Anyone know if we're meant to remember the names/pH ranges of indicators? Can't find a spec' sheet anywhere to find out.

    Cheers
    Re-quoting that, anybody know?

    Good luck people anyway, prep your tits off and you'll do fine.
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    (Original post by haydyb123)
    I think the rule is, everything other than Pt. complexes (with a co-ordination no. of 4) is tetrahedral, where as the Pt. complexes are square planar
    definitly don't need to know this, only that they could be either
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    (Original post by wilsea05)
    say you've got 50cm3 of NaOH reacting with 50cm3 of H2SO4, both are 1 mol dm3
    the moles of them both is gonna be 0.05
    the stoichiometrical equation is 2NaOH + H2SO4 -> Na2SO4 + 2H2O
    say the change in temp is 13C
    so Q=100x4.18x13 = -5.343 kJ
    but you need to get this into kJ mol-1 so you divide by the amount of moles used, which is 0.025 (because 2:1 for NaOH2SO4) so for every 2 moles of NaOH, 1 mol of H2SO4 is reacted. so even if you have 0.05 moles of both, only 0.025 moles of the sulphuric acid has reacted, so you divide by that.
    so its -217.36 kJ mol -1

    i don't even know if im right anymore, you've confused me why can't you just stick with HCl!!!! haha
    I am not sure now either
    Isn't what you have done the enthalpy change of neutralisation per mole of acid?
    Since one mole of H2SO4 produces 2H2O I think you may need to divide by 2 to get the enthalpy change of neutralisation per mole of water.
    Surely they will just stick to monobasic acids in the exam
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    (Original post by jlcf)
    Definitions we have to learn/ useful to learn
    Code:
    Rate of Reaction
    Rate constant
    Order
    Half Life
    RD Step
    Equilibrium Law
    Dynamic Equilibrium
    Bronsted Lowry Acid and Base
    Conjugate acid-base pairs
    pH
    [H+]
    Strong & weak acid
    Acid dissociation constant 
    Ionic product of water
    Buffers
    Equivalence point 
    End point
    Standard enthalpy change of neutralisation
    Standard enthalpy change of hydration
    Standard enthalpy change of formation
    Standard enthalpy change of atomisation
    Standard enthalpy change of solution
    Lattice Enthalpy
    Ionisation energy – First and Second
    Electron affinity – First and Second
    Entropy
    Enthalpy
    Standard entropy change of reaction
    Free energy change
    Oxidation & reduction & their agents
    Standard electrode potential
    Fuel cell
    Half cell
    Transition element
    Precipitation reactions
    Complex ion
    Coordinate number
    Ligand
    Stereoisomers
    cis-platin
    Mono/BI/Hexa – dentate ligands
    Ligand substitution 
    Stability constant
    Anymore?
    you mentioned half life twice! lol! could you please tell me the difference between equivalence point and end point? Equivalence point is the point where both the acid and base are found in same quantities, whereas end point is when you reach full neutralization? is this right?
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    (Original post by jlcf)
    Anymore?
    Le Chatelier's Principle? and are you making revision cards out of this?
 
 
 
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