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Chemistry OCR Salters B! revision help!

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    Can someone help me please! Iam really confused, as to how much of the green industry do we need to know? How much detail? the exam is just a week away now :/
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    (Original post by sharon800)
    Can someone help me please! Iam really confused, as to how much of the green industry do we need to know? How much detail? the exam is just a week away now :/
    I suggest you read the syllabus as that will tell you exactly how much you need to know. It's on the OCR website.
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    (Original post by clownfish)
    I suggest you read the syllabus as that will tell you exactly how much you need to know. It's on the OCR website.


    explain the term electronegativity; recall qualitatively the
    electronegativity trends in the Periodic Table; use relative
    electronegativity values to predict bond polarity in a
    covalent bond; decide whether a molecule is polar or
    non-polar from its shape and the polarity of its bonds;

    What topic is this? Do you know any websites that teach you this? How do you know whether a molecule has a dipole or not?
    please help me! Thank you
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    (Original post by sharon800)
    explain the term electronegativity; recall qualitatively the
    electronegativity trends in the Periodic Table; use relative
    electronegativity values to predict bond polarity in a
    covalent bond; decide whether a molecule is polar or
    non-polar from its shape and the polarity of its bonds;

    What topic is this? Do you know any websites that teach you this? How do you know whether a molecule has a dipole or not?
    please help me! Thank you
    The topic is electronegativity, if you don't have any notes on it take a look at page 8-9 on here: http://www.knockhardy.org.uk/sci_htm_files/08sandb.pdf
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    (Original post by clownfish)
    The topic is electronegativity, if you don't have any notes on it take a look at page 8-9 on here: http://www.knockhardy.org.uk/sci_htm_files/08sandb.pdf
    Hi, can you help me to understand this concept! I just dont get it!

    CFC's have a higher ozone depleting potential, and so the HFC'S are used instead....but, HFC'S contain c-f bond which is stronger, so it is surely more stable! and has less reactivity, so DOES NOT get broken down in the stratosphere, beacuse high energy is required to break the C-F bond? I was doing one of the past papers, and it says " HFC'S are already broken down in the trophosphere?" HOW??

    please help me!
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    (Original post by sharon800)
    Hi, can you help me to understand this concept! I just dont get it!

    CFC's have a higher ozone depleting potential, and so the HFC'S are used instead....but, HFC'S contain c-f bond which is stronger, so it is surely more stable! and has less reactivity, so DOES NOT get broken down in the stratosphere, beacuse high energy is required to break the C-F bond? I was doing one of the past papers, and it says " HFC'S are already broken down in the trophosphere?" HOW??

    please help me!
    The HFCs react with OH radicals in the troposphere forming water. This involves the breaking of the C-H bond, not the C-F bond: books.google.co.uk/books?id=NN5S0_3dEvkC&pg=PA289&l pg=PA289&dq=hfcs+broken+down+in+ troposphere&source=bl&ots=W4un13 Iwi3&sig=NG4rUfP-_U86ypllfA9nUkbdEa0&hl=en&sa=X&e i=5I6yT469IYP28gO2qIGICQ&ved=0CF UQ6AEwAw#v=onepage&q=hfcs broken down in troposphere&f=false

    also see Jan11 Q5d (and it's marking scheme).
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    (Original post by clownfish)
    The HFCs react with OH radicals in the troposphere forming water. This involves the breaking of the C-H bond, not the C-F bond: books.google.co.uk/books?id=NN5S0_3dEvkC&pg=PA289&l pg=PA289&dq=hfcs+broken+down+in+ troposphere&source=bl&ots=W4un13 Iwi3&sig=NG4rUfP-_U86ypllfA9nUkbdEa0&hl=en&sa=X&e i=5I6yT469IYP28gO2qIGICQ&ved=0CF UQ6AEwAw#v=onepage&q=hfcs broken down in troposphere&f=false

    also see Jan11 Q5d (and it's marking scheme).
    ah i get it now, so its the C-H bond in the HFC'S Which get broken, because they have lower bond polarity? and low bond strength...
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    (Original post by clownfish)
    The HFCs react with OH radicals in the troposphere forming water. This involves the breaking of the C-H bond, not the C-F bond: books.google.co.uk/books?id=NN5S0_3dEvkC&pg=PA289&l pg=PA289&dq=hfcs+broken+down+in+ troposphere&source=bl&ots=W4un13 Iwi3&sig=NG4rUfP-_U86ypllfA9nUkbdEa0&hl=en&sa=X&e i=5I6yT469IYP28gO2qIGICQ&ved=0CF UQ6AEwAw#v=onepage&q=hfcs broken down in troposphere&f=false

    also see Jan11 Q5d (and it's marking scheme).
    I have a question, so i was thinking if you can check it and see i've got the right uidea...

    describe how rates of the reaction changes with high temperatures.
    Rates of reaction increases with high temperature
    so more molecules have the minimum kinetic energy( or at least the activation energy) , so successful more collisions per unit time.

    Is this correct? Am i missing anything?
    Thank you
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    (Original post by sharon800)
    I have a question, so i was thinking if you can check it and see i've got the right uidea...

    describe how rates of the reaction changes with high temperatures.
    Rates of reaction increases with high temperature
    so more molecules have the minimum kinetic energy( or at least the activation energy) , so successful more collisions per unit time.

    Is this correct? Am i missing anything?
    Thank you
    This is right. I would say:

    At a higher temperature more molecules have energy greater than the activation energy, therefore there are more successful collisions so the rate will increase.
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    thank you so much for your time and help!
    with catalysts is it the same idea : so
    catalysts work by providing an alternative pathway with lower activation energy, so more molecules have energy greater than the activation energy, so more successful collision, and increase rate of reaction
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    (Original post by sharon800)
    thank you so much for your time and help!
    with catalysts is it the same idea : so
    catalysts work by providing an alternative pathway with lower activation energy, so more molecules have energy greater than the activation energy, so more successful collision, and increase rate of reaction

    Perfect
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    (Original post by clownfish)
    Perfect
    yay!
    thank you!!!

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