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OCR Chemistry F322~ 4th June 2013~ AS Chemistry Watch

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    Can someone help me with this question? Jan 2011, 3 b) (iii)
    Why is the answer 23 and not 12? Its really confusing me

    http://www.ocr.org.uk/Images/65360-question-paper-unit-f322-chains-energy-and-resources.pdf
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    (Original post by Beths_7)
    Can someone help me with this question? Jan 2011, 3 b) (iii)
    Why is the answer 23 and not 12? Its really confusing me

    http://www.ocr.org.uk/Images/65360-question-paper-unit-f322-chains-energy-and-resources.pdf
    The answer to (i) is 2310 but it needs to be converted to KJ so that will be 2.3

    the answer to (ii) is 0.200 moles

    (iii) is asking for the enthalpy change of reaction, so all you do is divide the answer to (i) 2.3 by the answer to (ii) 0.200 moles, then make sure to times this by 2 ))

    this comes out as 23
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    (Original post by Beths_7)
    Can someone help me with this question? Jan 2011, 3 b) (iii)
    Why is the answer 23 and not 12? Its really confusing me

    http://www.ocr.org.uk/Images/65360-question-paper-unit-f322-chains-energy-and-resources.pdf
    Ah!
    I done this paper, yesterday evening...

    Right, so!?
    She equation that you must use - and I will share in written form - is: Q (which you calculated through mCAT) divided by the one marker question in the mole (which I believe was 0.2mol)

    Then, I believe it is twice the amount due to the moles of Ammonia Thiocyanate = 2 due to the stoichiometric ration.

    = 23.1 kJmol^-1.

    To two significant figures, 23 works out just fine!
    And, then it must always be a negative value with the enthalpy of combustion - as I am sure you know.

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    [QUOTE=ellie2996;42900592]The answer to (i) is 2310 but it needs to be converted to KJ so that will be 2.3

    the answer to (ii) is 0.200 moles

    (iii) is asking for the enthalpy change of reaction, so all you do is divide the answer to (i) 2.3 by the answer to (ii) 0.200 moles, then make sure to times this by 2 ))

    this comes out as 23[/QUOTE

    why times 2?
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    [QUOTE=Smko;42900666]
    (Original post by ellie2996)
    The answer to (i) is 2310 but it needs to be converted to KJ so that will be 2.3

    the answer to (ii) is 0.200 moles

    (iii) is asking for the enthalpy change of reaction, so all you do is divide the answer to (i) 2.3 by the answer to (ii) 0.200 moles, then make sure to times this by 2 ))

    this comes out as 23[/QUOTE

    why times 2?
    because of the stoichiometric ration
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    [QUOTE=ellie2996;42900678]
    (Original post by Smko)

    because of the stoichiometric ration
    I see! Thanks!! Good luck on Tuesday!
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    (Original post by Ché.)
    Ah!
    I done this paper, yesterday evening...

    Right, so!?
    She equation that you must use - and I will share in written form - is: Q (which you calculated through mCAT) divided by the one marker question in the mole (which I believe was 0.2mol)

    Then, I believe it is twice the amount due to the moles of Ammonia Thiocyanate = 2 due to the stoichiometric ration.

    = 23.1 kJmol^-1.

    To two significant figures, 23 works out just fine!
    And, then it must always be a negative value with the enthalpy of combustion - as I am sure you know.

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Views: 139
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    When ever you are finding out the answer to these questions, and when the ask you for the number of moles of something, are you always finding one mole of the substance?


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    [QUOTE=ellie2996;42900678]
    (Original post by Smko)

    because of the stoichiometric ration
    Okay I just didn't times it by two...I don't understand, why do you have to times by two? Is it because there's two moles of NH4SCN?
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    [QUOTE=Beths_7;42900705]
    (Original post by ellie2996)

    Okay I just didn't times it by two...I don't understand, why do you have to times by two? Is it because there's two moles of NH4SCN?
    yep
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    [QUOTE=ellie2996;42900710]
    (Original post by Beths_7)

    yep
    Okay thank you! I thought that enthalpy change of reaction was for one mole though...:confused:
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    I got Bio Unit 2 Edexcel exam tomorrow and then Chemistry the next day... WHY?!
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    (Original post by Jimmy20002012)
    When ever you are finding out the answer to these questions, and when the ask you for the number of moles of something, are you always finding one mole of the substance?


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    Typically, you would do one mole without even checking the stoichiometry of the equation provided.
    Most, if not all except this one, supplies an equation with one mole per reactant at least.
    If that makes sense?

    That question was a little awkward hence why I had to check the question again to figure out if that was needed; the knowledge from F321 was creeping on me!
    LOL.


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    (Original post by scorpio22)
    thats an actually really good question.And one i find asking myself.
    There is some areas such as mechanisms: electrophillic addition and nucleophillic sub.. learn the dipoles,movement of electrons etc
    just learn it because it will come up somehow

    boltzmann:temp or catalyst guranteed to come up
    catalyst convertor process almost always comes up.

    carbon capture almost comes up.
    polymers;alcohol

    I think this exam is about attention to detail, there is so many areas to make mistakes like writing incomplete instead of complete combustion.
    drawing full displayed instead of sketal, significant figures. when it requests
    There are also so many areas to get marks:

    one thing that will definitely come up is definitions and enthalpy calculations.thats one i would encourgae to understand.
    how to calculate formation from combustion
    combustion from formation
    enthalpy reaction
    the reverse reaction activation etc
    how to calculate formation from combustion
    combustion from formation
    enthalpy reaction
    the reverse reaction activation I don't really know how to do that are there any questions I can do to practise on that please


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    Can anyone help me with working out enthalpy changes

    Enthalpy change of:

    reaction when given formation values
    reaction when given combustion values
    reaction when given average bond enthalpies
    formation when given reaction values
    formation when given combustion values
    formation when given average bond enthalpies

    I never know what to do, either products - reactants or reactants - products
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    Anybody got Jan 2013 paper ?
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    (Original post by Joey952)
    Anybody got Jan 2013 paper ?
    http://www.scribd.com/doc/120653070/...Jan-2013-Paper
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    Anyone got any tips on how to learn the mechanisms? I just can't seem to get my head around them but they're worth so many marks!


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    Completely confused myself is an exothermic reaction bond breaking or bond making???
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    (Original post by ellie2996)
    Completely confused myself is an exothermic reaction bond breaking or bond making???
    Endothermic = bond breaking (+ ve)
    Exothermic = bond making (-ve)



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    i was just wondering..if our ISA was about oxidation of alcohols will that be in the exam??
    thanks
 
 
 
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