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Edexcel A2 Chemistry Exams -6CH04 (14th June) and 6CH05 (22nd June) Discussion Thread Watch

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    (Original post by Funky_Giraffe)
    For Edexcel, I've always learnt Fe2+ as green and Fe3+ as brown coloured.
    okay thanks
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    (Original post by Funky_Giraffe)
    Yeah
    Oh okay thanks
    I always went with gut feeling with these MCQ lol
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    (Original post by ST_123)
    https://a5c076379da85d2c0b501a60f1a4...0Chemistry.pdf

    Could someone please explain how to do 24di,ei and f please :/ So confused
    -thanks
    For ei, you need to remember how HBr interacts w double bonds, eg in alkenes by electrophilic addition. This is the same w this molecule and the HBr will eradicate all the C--C molecules in the same way, adding hydrogen and bromine to them instead. Hope this helped :-)
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    (Original post by Don Pedro K.)
    okay thanks
    according to the user guide fe2+ is a pale blue blue (basically turquoise by the looks of it) and fe3+ is red brown
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    (Original post by samb1234)
    according to the user guide fe2+ is a pale blue blue (basically turquoise by the looks of it) and fe3+ is red brown
    Oh right okay...The Fe3+ makes sense but I never thought Fe2+ was any shade of blue XD
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    (Original post by Don Pedro K.)
    Oh right okay...The Fe3+ makes sense but I never thought Fe2+ was any shade of blue XD
    tbh though a lot of these colours are pretty far away from the actual colour if you do the experiment anyway
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    I am never seem to figure out the half equations anyone have any tips on how to do This


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    (Original post by Supermanxxxxxx)
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    I am never seem to figure out the half equations anyone have any tips on how to do This


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    Alright well this is the technique I use for half equations. I'll show you with an example other than these two, so you can try and work it out yourself:

    So let's say we want to write the half equation for the oxidation of Cr2O72- to Cr3+.

    FIRST STEP:

    Write reactant ---> product

    Cr2O72- --->2Cr3+.

    (We have a 2 in front of the Cr3+ because there are two chromium atoms on the left so we need to balance that on the right hand side.

    SECOND STEP:

    Balance the oxygen atoms using water molecules.

    Now, we have 7 oxygen atoms on the left hand side, and none of the right hand side. So, we add 7 water molecules on the right hand side:

    Cr2O72- --->2Cr3+ + 7H2O

    THIRD STEP:

    Balance the hydrogen atoms using H+ ions.

    We now have 14 hydrogen atoms on the right hand side, and none on the left hand side. So, we add 14 H+ ions on the left hand side:
    Cr2O72- + 14H+ --->2Cr3+ + 7H2O

    FOURTH STEP:

    Balance out the charges on both sides using electrons.

    We have a combined charge of (-2 + 14) = +12 on the left hand side, and (2*3) = +6 on the left hand side. So, we need 6 electrons on the left hand side in order to get equal charges on both sides:

    Cr2O72- + 14H+ + 6e- --->2Cr3+ + 7H2O

    And you're done!

    Now, try and see if you can figure out the half-equations for your question using this method!
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    (Original post by Don Pedro K.)
    Alright well this is the technique I use for half equations. I'll show you with an example other than these two, so you can try and work it out yourself:

    So let's say we want to write the half equation for the oxidation of Cr2O72- to Cr3+.

    FIRST STEP:

    Write reactant ---> product

    Cr2O72- --->2Cr3+.

    (We have a 2 in front of the Cr3+ because there are two chromium atoms on the left so we need to balance that on the right hand side.

    SECOND STEP:

    Balance the oxygen atoms using water molecules.

    Now, we have 7 oxygen atoms on the left hand side, and none of the right hand side. So, we add 7 water molecules on the right hand side:

    Cr2O72- --->2Cr3+ + 7H2O

    THIRD STEP:

    Balance the hydrogen atoms using H+ ions.

    We now have 14 hydrogen atoms on the right hand side, and none on the left hand side. So, we add 14 H+ ions on the left hand side:
    Cr2O72- + 14H+ --->2Cr3+ + 7H2O

    FOURTH STEP:

    Balance out the charges on both sides using electrons.

    We have a combined charge of (-2 + 14) = +12 on the left hand side, and (2*3) = +6 on the left hand side. So, we need 6 electrons on the left hand side in order to get equal charges on both sides:

    Cr2O72- + 14H+ + 6e- --->2Cr3+ + 7H2O

    And you're done!

    Now, try and see if you can figure out the half-equations for your question using this method!
    Thank you so much I did it one second it wasn't hard at all


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    (Original post by Supermanxxxxxx)
    Thank you so much I did it one second it wasn't hard at all


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    Haha no problem! See, it's easy when you know the method
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    (Original post by Supermanxxxxxx)
    Thank you so much I did it one second it wasn't hard at all


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    (Original post by Don Pedro K.)
    Haha no problem! See, it's easy when you know the method
    Just so you know that method only works if done in acid, it's very slightly different in alkali
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    (Original post by samb1234)
    Just so you know that method only works if done in acid, it's very slightly different in alkali
    It is? How so?
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    (Original post by Don Pedro K.)
    It is? How so?
    easiest way is to add alkali to both sides of the equation at the end to get rid of H+ ions
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    (Original post by samb1234)
    easiest way is to add alkali to both sides of the equation at the end to get rid of H+ ions
    ah okay, thanks
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    samb1234 Can you explain how to do part ii of this question please? I didn't get it right and I don't understand the mark scheme xD

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    (Original post by Don Pedro K.)
    samb1234 Can you explain how to do part ii of this question please? I didn't get it right and I don't understand the mark scheme xD

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    You have the amount of energy transferred for that number of moles, you need to find the energy for 1 mol, convert to KJ and stick the right sign in front since the temp increased
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    (Original post by samb1234)
    You have the amount of energy transferred for that number of moles, you need to find the energy for 1 mol, convert to KJ and stick the right sign in front since the temp increased
    ahhh okay that makes sense haha! Thanks
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    samb1234 Can you check my reasoning behind the idea of water being neutral despite a decrease in pH?

    So, from what I understand, in pure water, the concentration of [OH-] and [H+] is always the same. So, if [H+] was to increase, the pH would decrease, since pH = -log([H+]), but the water still remains neutral since the [OH-] also increases.

    Is that right?
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    Hey im not sure on some Unit 4 Qs. 6b ( why is there permamnent dipole dipole interactions...is it due to the polar alcohol bond) and 6c ( the wrong answer c seems feasible to me but why is that fragments without H are unlikely) Name:  20160605_201553.jpg
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    Thanks in advance
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    (Original post by Don Pedro K.)
    samb1234 Can you check my reasoning behind the idea of water being neutral despite a decrease in pH?

    So, from what I understand, in pure water, the concentration of [OH-] and [H+] is always the same. So, if [H+] was to increase, the pH would decrease, since pH = -log([H+]), but the water still remains neutral since the [OH-] also increases.

    Is that right?
    Yes that is correct. H2O <-------> OH- + H+. When you vary the temp the equil will increase/decrease the conc of oh- and h+ equally. A nice way of doing pH is to say that since conc of oh is same as h+ you can say h+ =root Kw so pH = -logrootKw
 
 
 
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