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    (Original post by lyricalvibe)
    Are you OCR B? My teachers said its quite likely this'll happen as its the last year for the spec, are you planning on looking at old spec stuff?

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    Oh no OCR A. OCR B seems more interesting, wish we did it. Really? They're probably right but who knows.
    I was just thinking since it's the last official year for this type of A level- maybe they might do so. I mean we still have resit exams in 2016 specialised for our year only but I doubt that they would want many students to resit the paper next year so maybe the paper this year papers wouldn't be that hard or surprising. Now, this isn't exactly a good thing as it could lead to high grade boundaries. Ah, it's 50/50. All of us would get As only in my dreams lol Anything could happen this year.
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    (Original post by blueberry389)
    when you say sth is being oxidised or reduced, do you say the element of the compound or the compound?
    e.g. the S in SO2 is reduced / SO2 is reduced?

    thank you!
    The sulfur element is being reduced because you're talking about the oxidation state of sulfur only.
    SO2 - O = -2 x 2 = -4 so S = +4
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    When I am talking about acidified potassium dichromate (IV) orange to green chromium (III) ions would I be right in saying the potassium dichromate is REDUCED to chromium ions?
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    Hi guys, I have produced a glossary of terms for F321 mod-1 (atoms and reactions) if you want to have a look. Thanks.

    http://www.thestudentroom.co.uk/g/re...tions_glossary
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    Hi

    I'm a tiny bit confused about the investigation for the ease of decomposition for Carbonates.

    I understand the procedure.

    But why are we using this apparatus? What are we observing? How long it takes for the lime water to become cloudy, due to the CO2 formed in the equation:

    MCO3 --> MgO + CO2

    Is the experiment for all carbonates?
    I thought aqueous Ca(CO3) is lime water.. So why are we using this?

    Lol ok maybe not a 'tiny' bit confused
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    (Original post by Dinaa)
    Hi

    I'm a tiny bit confused about the investigation for the ease of decomposition for Carbonates.

    I understand the procedure.

    But why are we using this apparatus? What are we observing? How long it takes for the lime water to become cloudy, due to the CO2 formed in the equation:

    MCO3 --> MgO + CO2

    Is the experiment for all carbonates?
    I thought aqueous Ca(CO3) is lime water.. So why are we using this?

    Lol ok maybe not a 'tiny' bit confused
    Yes it is how long it takes for the lime water to become cloudy to compare the time taken for decomposition between two carbonates.
    Ca(OH)2 (aq) is lime water. When CO2 is passed through it reacts like this Ca(OH)2 (aq) + CO2 --> CaCO3 (s) + H2O. CaCO3 is the white precipitate.


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    (Original post by samwillettsxxx)
    When I am talking about acidified potassium dichromate (IV) orange to green chromium (III) ions would I be right in saying the potassium dichromate is REDUCED to chromium ions?
    Yah. The compound has been reduced, therefore reduction has occurred. We don't worry too much about the potassium, we treat it as a spectator ion.

    I can see why you'd be hesitant to call it reduction, because that would imply that everything in the chromate (inc. the oxygen) is being reduced. It would be more accurate to say that the chromium in the chromate is being reduced, but saying that the 'chromate has been reduced' is still deemed more than acceptable.

    Source: Bossed A2 Chem.
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    (Original post by C0balt)
    Yes it is how long it takes for the lime water to become cloudy to compare the time taken for decomposition between two carbonates.
    Ca(OH)2 (aq) is lime water. When CO2 is passed through it reacts like this Ca(OH)2 (aq) + CO2 --> CaCO3 (s) + H2O. CaCO3 is the white precipitate.


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    That's a superb explanation!

    Thank you again :blushing:
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    (Original post by Dinaa)
    That's a superb explanation!

    Thank you again :blushing:
    Np! Happy to help


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    Anyone having a tough time getting their head around organic chemistry :no: (ocr a)

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    I have been researching to try and find the correct way to convert a number from mol/dm^3 to g/dm^3 but I am really struggling. Which is the correct way to do it accurately?
    Could you provide an exam so I can see how you do it.
    thanks
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    (Original post by KeilahDeere)
    Anyone having a tough time getting their head around organic chemistry :no: (ocr a)

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    Yes, me !
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    (Original post by locket0511)
    I have been researching to try and find the correct way to convert a number from mol/dm^3 to g/dm^3 but I am really struggling. Which is the correct way to do it accurately?
    Could you provide an exam so I can see how you do it.
    thanks
    multiply the moldm-3 by the mr(relative molar mass) of whatever it is to get it to gdm-3
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    When cracking takes place, a large number of products are formed.Suggest why a large number of products are formed.
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    can't do chem yay
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    (Original post by ThatGuyRik)
    multiply the moldm-3 by the mr(relative molar mass) of whatever it is to get it to gdm-3

    so is this correct?

    A sample of vinegar contains 0.1 mol/dm3 ethanoic acid. What is its concentration in g/dm3? ( Mr, of ethanoic acid is 60)
    concentration in g/dm3 = concentration in g/dm3 × Mr
    concentration = 0.1 × 60 = 6 g/dm3
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    (Original post by Kadak)
    When cracking takes place, a large number of products are formed.Suggest why a large number of products are formed.
    Large carbon chain breaks into many smaller carbon chains?

    Assuming from what i remember last year.
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    (Original post by locket0511)
    I have been researching to try and find the correct way to convert a number from mol/dm^3 to g/dm^3 but I am really struggling. Which is the correct way to do it accurately?
    Could you provide an exam so I can see how you do it.
    thanks
    From the amount I know (at As level), you simply multiply the concentration (mol dm-3) by the Mr of the substance (in g mol-1) to get the concentration in g dm-3. There may well be more advanced/accurate ways of doing it.
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    (Original post by _NMcC_)
    From the amount I know (at As level), you simply multiply the concentration (mol dm-3) by the Mr of the substance (in g mol-1) to get the concentration in g dm-3. There may well be more advanced/accurate ways of doing it.
    Thanks a lot!!!
    so is this correct?

    A sample of vinegar contains 0.1 mol/dm3 ethanoic acid. What is its concentration in g/dm3? ( Mr, of ethanoic acid is 60)
    concentration in g/dm3 = concentration in g/dm3 × Mr
    concentration = 0.1 × 60 = 6 g/dm3
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    (Original post by locket0511)
    Thanks a lot!!!
    so is this correct?

    A sample of vinegar contains 0.1 mol/dm3 ethanoic acid. What is its concentration in g/dm3? ( Mr, of ethanoic acid is 60)
    concentration in g/dm3 = concentration in g/dm3 × Mr
    concentration = 0.1 × 60 = 6 g/dm3
    Yes that looks fine. I'm getting the same answer. It's acceptable to have the units either as g/dm3 or g dm-3
 
 
 
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