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    (Original post by Bloom77)
    And how would you do that
    I normally combine them to find the ratio like you know for those maths chem questions at the end


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    Gibbs free energy matches to it. G is y, -DeltaS is gradient, T is x and DeltaH is the y intercept,
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    is a reaction spontaneous if delta G is below or equal to zero or just below 0???
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    (Original post by GO97)
    hm im pretty sure bonds forming release energy and u have put in energy to break bonds so bonds breaking would take up energy
    I meant it uses up energy. It's still endothermic
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    (Original post by thomaarrss)
    is a reaction spontaneous if delta G is below or equal to zero or just below 0???
    Equal and below.
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    (Original post by thomaarrss)
    is a reaction spontaneous if delta G is below or equal to zero or just below 0???
    Think below but you set it =0 as -0.00000000001 and 0 makes no difference. In practice it's the same
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    (Original post by jammypancake)
    http://filestore.aqa.org.uk/subjects...5-QP-JUN12.PDF
    2b? how to calculate the gradient?
    Take the point (460,0) and the point (700,23.5) then do change in y over change in x to get 23.5/240 to get an answer of 0.0979 which rounds to 0.098. Its hard to see but it is at 23.5 just a bad question for accuracy
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    (Original post by zombaldia)
    I meant it uses up energy. It's still endothermic
    ah ok sorry if i misread
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    (Original post by GO97)
    hey sorry your quiz is right, its the cr2o7^2-/cro4^2- ions that are orange/yellow. the hexa-aqua ions are violet, this could have cost me big time in tommorows exam haha thanks for that
    http://m.imgur.com/ulgdobg

    It's yellow and the exams consider them yellow check 2013 or the 2014 pap
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    (Original post by thomaarrss)
    is a reaction spontaneous if delta G is below or equal to zero or just below 0???
    i think it is just below 0 but for calculation purposes its classed as 0, i could be wrong though correct if i am please
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    (Original post by Chembio123)
    Only starting to revise tonight, am I doomed?
    No I think chem 5 easy. Just read over chemrevise
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    (Original post by shamk123)
    i think it is just below 0 but for calculation purposes its classed as 0, i could be wrong though correct if i am please
    It has to be below 0 and in calculations when you round it make sure you choose the rounding that gives make delta g below 0 if you round the wrong way it will be positive
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    (Original post by Bloom77)
    And how would you do that
    I normally combine them to find the ratio like you know for those maths chem questions at the end


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    1MnO4- for 5 Fe 2+
    2MnO4- for 5C2O4 2-
    3:5
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    (Original post by Chembio123)
    Only starting to revise tonight, am I doomed?
    Im the same as you i started about an hour ago after 7-8 weeks of neglecting chem5 its safe to say best option is to pull an all nighter, so i took some modafinil
    to keep me awake for the whole night, im just gonna get some sleep after the exam
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    (Original post by YoloSwagginz)
    http://m.imgur.com/ulgdobg

    It's yellow and the exams consider them yellow check 2013 or the 2014 pap
    yeah thanks, just more confusion from my side lol ive deleted that comment.
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    Can someone please explain when al(oh)6 3- is formed? All my notes say al(oh)4- but the mark scheme for Jan 2011 says Al(oh)6 3-.
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    Hey guys, not sure if we need to know this but I'm curious anyways:

    It says the units for planck's constant are J s what does the 's' stand for?
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    reading the textbook about how exothermic reactions can be spontaneous and its talking about the decrease of energy in the system allows an increase the entropy of the surroundings.... we don't need to know this much detail do we? I've never seen it come up
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    O
    (Original post by AL20xxx)
    Can someone please explain when al(oh)6 3- is formed? All my notes say al(oh)4- but the mark scheme for Jan 2011 says Al(oh)6 3-.
    Sure

    Ligands that are similar sizes have the same coordination number so

    Nh3 oh h2o and cn as Ligands will all form al(c)6

    Where cl- ions will form al(cl)4 due to it being considered "larger" think it may have to do with it having a larger charge cloud but idk cause we don't need to know its theory
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    when asked to draw a complex do we do arrows to the ligand or do the dotted lines/triangles to show the shape???
 
 
 
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