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.
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...
ReplyDeleteThe 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.
Just kidding I figured it out.
ReplyDeleteJust in case someone else didn't figure it out...
ReplyDeleteThe 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..