2010/10/01

Almost exam time...

We've finished up chapters 4 and 5, exam next Wednesday. Let me know if you have any questions or anything specific you'd like to review on Monday in class. I've posted problem set #4 and the answer key on my mnstate.edu page, let me know if there are problems.

Have a good weekend and good luck preparing for the exam.
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3 comments:

  1. AnonymousOctober 3, 2010 at 3:26 PM

    This is the feedback from an OWL assignment problem where we need to assign oxidation numbers to each element in the compound that I got wrong...

    The oxidation numbers for the elements in Bi(OH)3 are determined by first assigning oxidation numbers to H and O using the rules:... Hydrogen has an oxidation state of +1 in compounds.
    Oxygen has an oxidation state of -2 in compounds.
    You can then calculate the oxidation state of Bi remembering that the sum of all the oxidation numbers in Bi(OH)3 = 0.
    ...x + 3(+1) + 3(-2) = 0 ... and ... x = 3


    It says near the end that the sum of all oxidation numbers in Bi(OH)3 = 0. I do not get how it is zero. If you could either post something back or go over it in class tomorrow that would be great.

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  2. AnonymousOctober 3, 2010 at 3:36 PM

    Just kidding I figured it out.

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  3. Dr. BodwinOctober 5, 2010 at 8:18 AM

    Just in case someone else didn't figure it out...
    The sum of the oxidation numbers of all the atoms/elements in a formula is equal to the charge on that formula. In this case, Bi(OH)3 is a neutral formula so the charge is zero. This means that the oxidation number of Bi + 3 times the oxidation number of oxygen + 3 times the oxidation number of hydrogen must equal zero. Oxygen is almost always -2, hydrogen is almost always +1, so... well, the rest is in the OWL feedback above..

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