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    (Original post by melony852)
    Yeah thats right! You just need the understanding of these cells to apply in questions which may be related.
    thank you! what kind of understanding? lol like basically do you we just need to know that when they're being re charged the equations go the other way?

    and to answer your question above, yeah sadly it is
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    (Original post by melony852)
    I'm getting a little confused now.....
    So is the paper synoptic or not? This is the first time i've heard its synoptic...???
    They can ask anything from CHEM1 and CHEM2 but CHEM4 won't come up. It's brought in things like electronegativities in periodicity in the past.
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    (Original post by louise_234)
    why when drawing the conventional representation we dont include water or h+ ions?
    It's assumed that we can produce half-equations from the other species.
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    do we have to actually need to know the reactions for batteries? or just how they work and adv and disadv?
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    thanks.. for the helpp.. yaay.. i can forget about unit 4 ...
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    (Original post by jimmy_95)
    Can somebody please explain to me why for questions asking to explain entropy change sometimes the mark scheme refers to an increase in moles and other times to an increase in the number of particles. I just want to know which one I should say, particles or moles, or if there is a condition for using each one.
    Either will get you the mark, essentially the same thing. Btw in the mark scheme it says mol not moles. Mol is short for molecules whereas moles is that avagadro bull
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    Can someone give me a step by step working out with the answer of Q 8 (C) on AQA Chem 5 (January 2010) paper?

    Im really struggling on it.

    Thanks.


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    Hi, noticed under entropy and all that in the spec it says, increasing disorder can be illustrated by 'chemical change e.g dissolution, evolution of CO2 from hydrocarbonates with acid' wondering how that increases disorder?
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    (Original post by JSN)
    do we have to actually need to know the reactions for batteries? or just how they work and adv and disadv?
    + can some explain why the reactions with en will be different for 8ei and ii

    http://filestore.aqa.org.uk/subjects...5-QP-JAN12.PDF
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    (Original post by homeworkwiz)
    Either will get you the mark, essentially the same thing. Btw in the mark scheme it says mol not moles. Mol is short for molecules whereas moles is that avagadro bull
    Thanks.
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    (Original post by idunnotbh)
    hey guys random question here but can anyone confirm this for me please...

    we do NOT have to learn the half equations and overall equations of re chargeable batteries and non re chargeable batteries, for example for lead acid batteries or zinc/carbon like in the text book?

    thank you
    The specification says
    be able to use given electrode data to deduce the reactions occurring in non-rechargeable and rechargeable cells and to deduce the e.m.f. of a cell
    So they'll give you the half equations and you might be asked to write the overall equation but you don't have to remember anything except which way the reaction will go
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    For complexes, why is the charge on say [Cr(NH3)6] 3+ and not 3-? As when you have [Cr(OH)6] it's 3- and don't -OH and -NH3 both have a charge of -1?
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    (Original post by FM1994)
    Can someone give me a step by step working out with the answer of Q 8 (C) on AQA Chem 5 (January 2010) paper?

    Im really struggling on it.

    Thanks.


    Posted from TSR Mobile
    It helps if you start by writing equations for what's going on

    Phosphorus (V) oxide reacting with water:
    P4O10 + 6H2O \rightarrow 4H3PO4
    Phosphoric acid reacting with sodium hydroxide:
    H3PO4 + 3NaOH \rightarrow Na3PO4 + 3H2O

    Work out the number of moles of NaOH which react:
    n = MV / 1000
    n = (0.5 x 21.2) / 1000 = 0.0106 mol

    The ratio of H3PO4 to NaOH is 1:3, or in other words, for every 3 moles of NaOH which react, 1 mole of H3PO4 reacts. So the number of moles of H3PO4 is 0.0106 / 3 = 0.0035333... mol

    This is the number of moles in 25cm3, but you have 25,000dm3. Since 1dm3 = 1000cm3, 25,000dm3 = 25,000,000cm3

    25,000,000cm3 / 25cm3 = 1,000,000

    So the number of moles in 25,000dm3 is 1,000,000 x 0.0035333... = 3533.333... mol

    From the first equation you can see 1 mole of P4O10 reacts to form 4 moles of H3PO4, so the original number of moles of P4O10 in the tank must be 3533.333... / 4 = 883.333... mol

    Then you can use the equation m = n x Mr to work out the mass of P4O10 you have in the tank, where the Mr is 284

    m = 883.333... x 284 = 250,866.666... g

    To convert this into kg, just divide by 1000. The final answer should then be 251kg to 3 significant figures

    Hope this helps! If you need anything explained just quote me
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    (Original post by phems)
    For complexes, why is the charge on say [Cr(NH3)6] 3+ and not 3-? As when you have [Cr(OH)6] it's 3- and don't -OH and -NH3 both have a charge of -1?
    No, NH3 is ammonia, it doesn't have a charge
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    Has anyone done the exam style paper in the back of the nelson thrones textbook? It seems quite weird and synoptic...
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    (Original post by homeworkwiz)
    Either will get you the mark, essentially the same thing. Btw in the mark scheme it says mol not moles. Mol is short for molecules whereas moles is that avagadro bull
    http://filestore.aqa.org.uk/subjects...W-MS-JUN11.PDF

    Question 2b states 'mol' but then penalises 'molecules'. I've always known mol to mean mole?
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    (Original post by crc290)
    It helps if you start by writing equations for what's going on

    Phosphorus (V) oxide reacting with water:
    P4O10 + 6H2O \rightarrow 4H3PO4
    Phosphoric acid reacting with sodium hydroxide:
    H3PO4 + 3NaOH \rightarrow Na3PO4 + 3H2O

    Work out the number of moles of NaOH which react:
    n = MV / 1000
    n = (0.5 x 21.2) / 1000 = 0.0106 mol

    The ratio of H3PO4 to NaOH is 1:3, or in other words, for every 3 moles of NaOH which react, 1 mole of H3PO4 reacts. So the number of moles of H3PO4 is 0.0106 / 3 = 0.0035333... mol

    This is the number of moles in 25cm3, but you have 25,000dm3. Since 1dm3 = 1000cm3, 25,000dm3 = 25,000,000cm3

    25,000,000cm3 / 25cm3 = 1,000,000

    So the number of moles in 25,000dm3 is 1,000,000 x 0.0035333... = 3533.333... mol

    From the first equation you can see 1 mole of P4O10 reacts to form 4 moles of H3PO4, so the original number of moles of P4O10 in the tank must be 3533.333... / 4 = 883.333... mol

    Then you can use the equation m = n x Mr to work out the mass of P4O10 you have in the tank, where the Mr is 284

    m = 883.333... x 284 = 250,866.666... g

    To convert this into kg, just divide by 1000. The final answer should then be 251kg to 3 significant figures

    Hope this helps! If you need anything explained just quote me
    Thanks, this has helped me a lot.


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    Does anyone have the January 2013 chem 5 mark scheme????


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    (Original post by JenniS)
    Hi, noticed under entropy and all that in the spec it says, increasing disorder can be illustrated by 'chemical change e.g dissolution, evolution of CO2 from hydrocarbonates with acid' wondering how that increases disorder?
    The easiest way to think about entropy for these is as a measure of disorder. When in a solid the molecules are highly ordered, and therefore the entropy is low; in a liquid they are less ordered and so entropy is slightly higher; and in a gas the molecules are rather disordered and so the entropy is relatively high. So, for these examples, dissolution is an example of increasing entropy as the molecules are going from a solid lattice to in solution and so are becoming more disordered; and for evolution of CO2 you're producing a gas (very disordered) from a solid (more ordered) so entropy is increasing.
    Hope this helps
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    (Original post by crc290)
    It helps if you start by writing equations for what's going on

    Phosphorus (V) oxide reacting with water:
    P4O10 + 6H2O \rightarrow 4H3PO4
    Phosphoric acid reacting with sodium hydroxide:
    H3PO4 + 3NaOH \rightarrow Na3PO4 + 3H2O

    Work out the number of moles of NaOH which react:
    n = MV / 1000
    n = (0.5 x 21.2) / 1000 = 0.0106 mol

    The ratio of H3PO4 to NaOH is 1:3, or in other words, for every 3 moles of NaOH which react, 1 mole of H3PO4 reacts. So the number of moles of H3PO4 is 0.0106 / 3 = 0.0035333... mol

    This is the number of moles in 25cm3, but you have 25,000dm3. Since 1dm3 = 1000cm3, 25,000dm3 = 25,000,000cm3

    25,000,000cm3 / 25cm3 = 1,000,000

    So the number of moles in 25,000dm3 is 1,000,000 x 0.0035333... = 3533.333... mol

    From the first equation you can see 1 mole of P4O10 reacts to form 4 moles of H3PO4, so the original number of moles of P4O10 in the tank must be 3533.333... / 4 = 883.333... mol

    Then you can use the equation m = n x Mr to work out the mass of P4O10 you have in the tank, where the Mr is 284

    m = 883.333... x 284 = 250,866.666... g

    To convert this into kg, just divide by 1000. The final answer should then be 251kg to 3 significant figures

    Hope this helps! If you need anything explained just quote me
    cheers you helped me out on the exact same question I was struggling on
 
 
 
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