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    (Original post by JSN)
    can someone explain how to get the answer for question 3a) to d)

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

    dont understand it at all!

    A) a hydrogen electrode is 0 by definition. (All E values are relative to hydrogen)

    B) is 1.23V. (You need to remember how the O-H fuel cell works. The E value at the oxygen is 1.23 - the hydrogen electrode)

    C) that the IUPAC convention for writing cells. I suggest doing a few questions using it if you don't know how to write it. (I can't be bothered at the minute :-P)

    D) you have 2 redox half equations. The one with the most positive E value iz the reaction you don't flip. The other you make an oxidation half equation and then combine by balancing the electrons. Sounds complex how I put it but...it simplifies nicely to 2H2 + O2 ---> 2H20

    hope that makes sense
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    (Original post by Zazuwaved)
    Credit to Kev.1995

    Okay so imagine we have a solution of unknown concentration of Fe3+ ions.

    -Add a ton of Potassium Thiocyanate to it, this gives it the blood red colour.

    -Now we have a solution of red liquid of an unknown concentration.. But we DO know it contains Fe3+ bonded to thiocyanate ions!

    - Prepare a series of test tubes of different known concentrations of Fe3+ with the thiocyanate

    - Now because each tube has a different conc, they will be different intensities of red.

    - Use a colorimeter to determine the absorbance of light through these series of test tubes and draw a graph. (It'll be proportional)

    - Find where your unknown conc is on the line and there you go, the unknown concentration of Fe3+

    NB- They all need to have the same volume, easily done by adding water to them to make them up to the same volume.
    Thank you... one more question... how can you work out the formula?
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    (Original post by Gazfink)
    If I were you I would just do what the marksheme says. You can't lose marks then. Obviously it would be stupid if they marked you down, because yeah it is correct. But just do what the mark scheme says...even if It can seem illogical at times :-D
    cheers for replying! yeah Ill go with the mark scheme then
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    why are the ionic oxides basic in water but the covalent oxides are acidic in water??? and why is aluminum oxide amphoteric?
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    (Original post by Gazfink)
    A) a hydrogen electrode is 0 by definition. (All E values are relative to hydrogen)

    B) is 1.23V. (You need to remember how the O-H fuel cell works. The E value at the oxygen is 1.23 - the hydrogen electrode)

    C) that the IUPAC convention for writing cells. I suggest doing a few questions using it if you don't know how to write it. (I can't be bothered at the minute :-P)

    D) you have 2 redox half equations. The one with the most positive E value iz the reaction you don't flip. The other you make an oxidation half equation and then combine by balancing the electrons. Sounds complex how I put it but...it simplifies nicely to 2H2 + O2 ---> 2H20

    hope that makes sense
    Oh right right, i need to go over the fuel cell, thanks!
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    (Original post by lifeisgood2012)
    why are the ionic oxides basic in water but the covalent oxides are acidic in water??? and why is aluminum oxide amphoteric?
    Aluminum oxides are amphoteric as they can act as both bases and acids. They can react with both bases and acids.
    Reaction with both acids and bases is what defines amphoteric.

    Ionic oxides are basic in water as they form "hydroxides". Metal hydroxides are alkaline in solution.e.g. NaOH,, Mg(OH)2 etc.
    "Covalent oxides"or oxides containing a non-metal are acidic in water as they form acids by accepting H+ ions and becoming H3PO4, H2SO4 etc. These compounds are acidic. Thats all you need to know regarding your question.
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    Why is silicon dioxide an acidic oxide despite not being soluble in water?

    Posted from TSR Mobile
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    (Original post by frogs r everywhere)
    Aluminum oxides are amphoteric as they can act as both bases and acids. They can react with both bases and acids.
    Reaction with both acids and bases is what defines amphoteric.

    Ionic oxides are basic in water as they form "hydroxides". Metal hydroxides are alkaline in solution.e.g. NaOH,, Mg(OH)2 etc.
    "Covalent oxides"or oxides containing a non-metal are acidic in water as they form acids by accepting H+ ions and becoming H3PO4, H2SO4 etc. These compounds are acidic. Thats all you need to know regarding your question.
    hi thankks..... i understand the amphoteric and the ionic oxides but when you explained the covalent oxides i didnt get it cos acids donate H+ rite?? but here your saying they accept it???
    Could you explain that to me and also could you do an ionic equation for one of the covalent oxides reacting with water please?
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    (Original post by homeworkwiz)
    Why is silicon dioxide an acidic oxide despite not being soluble in water?

    Posted from TSR Mobile
    Si02 can react with the OH- ions( Base) so acts as a BL acid.
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    Could someone please explain how the moles for question 6d(i) is 3:5 for june 2012 paper? I don't understand.

    http://filestore.aqa.org.uk/subjects...5-QP-JUN12.PDF
    http://filestore.aqa.org.uk/subjects...W-MS-JUN12.PDF
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    (Original post by starfish232)
    Could someone please explain how the moles for question 6d(i) is 3:5 for june 2012 paper? I don't understand.

    http://filestore.aqa.org.uk/subjects...5-QP-JUN12.PDF
    http://filestore.aqa.org.uk/subjects...W-MS-JUN12.PDF
    Ok so you're given all the equations, and the bottom 2 you're given, try to think of them as one; because if you look i the description, it is FeC2O4.2H20 so their equations must be 1:1 together, when compared to the top.
    So you must balance 5 electrons against the 3 electrons of the other 2 bottom equations. So to get that you can 3x the top and the bottom 2 can be 5x for both, giving 15 electrons each. Therefore when you combine the half equations you get 3 MnO4- + 5 FeC2O4.2H20; 3:5.
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    (Original post by lifeisgood2012)
    hi thankks..... i understand the amphoteric and the ionic oxides but when you explained the covalent oxides i didnt get it cos acids donate H+ rite?? but here your saying they accept it???
    Could you explain that to me and also could you do an ionic equation for one of the covalent oxides reacting with water please?
    Lets consider a covalent molecular structure such as SO2.

    When SO2 reacts with water, an acidic compound is produced....H2SO4 (sulfuric acid). You don't really need to know the mechanism of this, but all you need to know is:

    " If non-metal-oxide react with water, the product is acidic. If a metal-oxide reacts with water, the product is alkaline"

    Regarding your question of accepting H+ ions, well yes: a non-metal oxide reacts with H2O which consists of H+ and OH- ions.

    An example may well be:

    P4O10(s) + 6 H2O(l) => 4 H3PO4(aq)

    Do ask if you still don't understand.
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    (Original post by frogs r everywhere)
    Lets consider a covalent molecular structure such as SO2.

    When SO2 reacts with water, an acidic compound is produced....H2SO4 (sulfuric acid). You don't really need to know the mechanism of this, but all you need to know is:

    " If non-metal-oxide react with water, the product is acidic. If a metal-oxide reacts with water, the product is alkaline"

    Regarding your question of accepting H+ ions, well yes: a non-metal oxide reacts with H2O which consists of H+ and OH- ions.

    An example may well be:

    P4O10(s) + 6 H2O(l) => 4 H3PO4(aq)

    Do ask if you still don't understand.
    thank you
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    Could someone please help me on answering question 4c and 5cii... for both of the questions i keep getting the vanadium half equation wrong... how do i know which one is the correct one? the questions and the answers are attached below
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  1. File Type: rtf questions.rtf (102.8 KB, 473 views)
  2. File Type: rtf asnwers.rtf (62.5 KB, 77 views)
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    Is silver chemistry required for the module??????? like the equations and stuff?
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    (Original post by homeworkwiz)
    Why is silicon dioxide an acidic oxide despite not being soluble in water?

    Posted from TSR Mobile
    SiO2 is insoluble in water due to its macromolecular structure. However, it acts as a Lewis Acid. This property allows it to react with NaOH
    SiO2 + 2NaOH -> Na2SiO3 + H2O

    Q 2g
    http://filestore.aqa.org.uk/subjects...W-QP-JUN10.PDF
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    (Original post by lifeisgood2012)
    Is silver chemistry required for the module??????? like the equations and stuff?
    Not really. However, you can be asked it as the Chem5 paper may contain synoptic questions from Chem2.
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    (Original post by lifeisgood2012)
    Could someone please help me on answering question 4c and 5cii... for both of the questions i keep getting the vanadium half equation wrong... how do i know which one is the correct one? the questions and the answers are attached below
    Yep.

    For 5cii)

    It asks:
    (ii) Identify all the species in the table which could convert I–aq) into I2(aq) but which could not convert Br–(aq) into Br2(l).

    If you look at the table it gives us, the species we need to give need to have an E value higher than 0.54 but it need to be lower than 1.07
    They can only be Fe3+ an NO3-.


    For 4c)

    V2+undergoes a series of oxidations. Initially when it mixes with MnO4- ions it changes into V3+
    But, the table also shows that V3+ oxidises into (VO)2+ (third half- equation).
    (VO)2+ can be further oxidised, according to the table into (VO2)+ (sixth half equation in the table)

    (VO2)+ is the final product of this reaction

    Its oxidation state is +3.
    The half equation can therefore be formed.

    If you don't understand how I got this, do ask.
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    (Original post by frogs r everywhere)
    Yep.

    For 5cii)

    It asks:
    (ii) Identify all the species in the table which could convert I–aq) into I2(aq) but which could not convert Br–(aq) into Br2(l).

    If you look at the table it gives us, the species we need to give need to have an E value higher than 0.54 but it need to be lower than 1.07
    They can only be Fe3+ an NO3-.


    For 4c)

    V2+undergoes a series of oxidations. Initially when it mixes with MnO4- ions it changes into V3+
    But, the table also shows that V3+ oxidises into (VO)2+ (third half- equation).
    (VO)2+ can be further oxidised, according to the table into (VO2)+ (sixth half equation in the table)

    (VO2)+ is the final product of this reaction

    Its oxidation state is +3.
    The half equation can therefore be formed.

    If you don't understand how I got this, do ask.
    hi thank you... you helped me sooo much..... so to confirm for question 4c.... the element with the highest oxidation state will be formed?
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    anyone got any tips on how to memorise colour changes?
 
 
 
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