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OCR A2 CHEMISTRY F324 and F325- 14th and 22nd June 2016- OFFICIAL THREAD watch

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    Please can someone explain why you multiply the solar energy by 6? Im really confused
    Please dont say becausr theres 6 moles of co2 reacting

    DeltaH= -q/n
    To figure out moles you do, -q/deltaH
    I dont understand why you need to multiply the solar energy by 6.
    Does it somehow imply that the solar energy is only for 1 mole of co2?Attachment 554471554473
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    (Original post by asunny465)
    Please can someone explain why you multiply the solar energy by 6? Im really confused
    Please dont say becausr theres 6 moles of co2 reacting

    DeltaH= -q/n
    To figure out moles you do, -q/deltaH
    I dont understand why you need to multiply the solar energy by 6.
    Does it somehow imply that the solar energy is only for 1 mole of co2?Attachment 554471554473
    glucose for photosynthesis
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    One thing that's always confused me is, if you got an octahedral complex with 2 monodentates and 2 bidentates, then in the trans form, is there still optical isomerism? I know the cis form definitely does but since this molecule is symmetrical, it doesn't seem to make sense to be considered a 'separate' optical isomer.
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    (Original post by Serine Soul)
    What oxidation state of iron can we associate with red?
    Ignore that read it wrong
    What does VI mean?
    I don't think the red colour is what's useful to deduce the formula. The VI is the critical information. It tells you about the oxidation state of Fe. VI means +6. Plugging this in and considering the overall -2 charge should help you work it out.

    If the question were to say IV or V, you can deduce that Fe would be +4 or +5 oxidation state respectively.
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    (Original post by KB_97)
    Attachment 554457
    Can someone help me with this question please?
    FeO42-

    because Fe has to be +6 charge
    Oxygen is usually -2 and the only way to make the overall charge at 2- is to to (4x-2) + 6
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    Do we have to know about half neutralisation?
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    (Original post by megafidget)
    One thing that's always confused me is, if you got an octahedral complex with 2 monodentates and 2 bidentates, then in the trans form, is there still optical isomerism? I know the cis form definitely does but since this molecule is symmetrical, it doesn't seem to make sense to be considered a 'separate' optical isomer.
    Yes - you can get an optical isomer. It's literally a mirror image of the trans!
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    (Original post by VMD100)
    Nope
    http://hyperphysics.phy-astr.gsu.edu...electrode.html

    This diagram is clearer
    Attachment 554315
    The reactions we are doing generate electrical energy, not require it.
    Do we need to know anything about electrolytic cells or do we need to know both?

    Posted from TSR Mobile

    No optical isomers of the trans form.
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    (Original post by Ezexamsalltheway)
    HI , guys how many of the transition metal aqueous ion colours do we actually need to know. all of them? thought i should better start my revision now :P rip
    Copper
    (Blue) Cu2+ or [Cu(H2O)6]2+
    (Light blue) Cu(OH)2 (s)
    (Yellow) [CuCl4]2- (this one looks green if you don't have excess Cl-
    (Deep blue) [Cu(NH3)4(H2O)2]2+
    Spoiler:
    Show
    (White) CuI (S)

    Cobalt
    (Pink) Co2+
    (Blue) [CoCl4]2-
    (Blue) Co(OH)2
    Spoiler:
    Show
    (Purple) Cis [Co(NH3)4Cl2]2+
    (Green) Trans [Co(NH3)4Cl2]2+

    Iron
    (Pale Green) Fe2+
    (Green) Fe(OH)2
    (Yellow) Fe3+
    (Brown) Fe(OH)3

    Chromium (as in alcohol oxidation reactions)
    (Green) Cr3+
    (Orange) Cr2O72-

    Manganese
    (Purple) MnO4-
    (Pink, colourless) Mn2+

    And the colour changes with thiosulfate and iodine in the clock reaction. Not transition elements, but useful to know.
    Spoiler:
    Show
    You use an oxidising agent to do 2I- -> I2 + 2e- which is light brown due to I2, then react I2 with S2O32- to make S4O62- and I-. Add starch when the second step goes a "pale straw" colour due to the lack of I2, it then turns blue black. Stop the clock when the blue-black disappears and goes colourless (when all the I2 has reacted).
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    is 2nd equation so charges on right = charges on left? https://gyazo.com/7d6e8331e15fc94e20faa04b0f006cef do i have to do that with all ionic equations?
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    (Original post by RayMasterio)
    Yes - you can get an optical isomer. It's literally a mirror image of the trans!
    But that's the thing, the mirror image of the trans will look exactly the same as the trans...., I guess they will rotate plane-polarised light in different directions so they wouldn't be the same molecule but the molecules appear to occupy the same spaces (from the diagrams at least).
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    (Original post by lai812matthew)
    that's what i write
    do you have to be specific and say two, what if you say bacteria produced one and lab produces many i.e. more than one, do you have to specify 2
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    How do you know if an indicator is suitable for a certain acid-base titration?
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    (Original post by ImNervous)
    Do we need to know anything about electrolytic cells or do we need to know both?

    Posted from TSR Mobile

    No optical isomers of the trans form.
    No we don't need to know about electrolyte cells, yeh that's what I thought.
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    (Original post by Roses98)
    I don't think the red colour is what's useful to deduce the formula. The VI is the critical information. It tells you about the oxidation state of Fe. VI means +6. Plugging this in and considering the overall -2 charge should help you work it out.

    If the question were to say IV or V, you can deduce that Fe would be +4 or +5 oxidation state respectively.
    Yeah I know that.

    I was asking the OP what the VI stood for? To check they understand?

    I hadn't realised that they had given roman numerals when I first saw the question
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    (Original post by lai812matthew)
    like there is only [Fe(NH3)4(H2O)2]n+ complexes but not [Fe(NH3)6]n+ complexes.
    Is that just memory or should we know why that doesn't occur?


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    (Original post by M.Branson98)
    How do you know if an indicator is suitable for a certain acid-base titration?
    If the peak of the acid is high then there's a strong base, if the bottom curve is low there's a strong acid.

    Indicators will have their range given so you just have to see if it falls in the range right?


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    (Original post by mechanism)
    Do we have to know about half neutralisation?
    what's that??
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    I'm so nervous for this exam. I just can't to the calculations towards the end of the paper. I'm so tired I keep making stupid mistakes. Bye bye a*
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    (Original post by tcameron)
    what's that??
    To do with acid and bases where HA=A-.
 
 
 
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