Volleyball tonight - Diggin for a Cure!

Tonight the MSUM volleyball team will be Diggin for a Cure to raise funds for and awareness of cancer research and treatment.  Come out and support Dragon Volleyball as they crush the Golden Eagles of UMinn-Crookston.

Exam Results

Exam 2 is graded, I'll give it back in class tomorrow.  The scores were not good.  A few people did quite well, but many did very poorly leading to an average just under 50%.  We'll spend some time talking about this tomorrow in class.

With renewed vigor...

Today in class we started looking at thermochemistry, the ways that heat and chemical processes interact.  This is a small part of the larger field of thermodynamics.  We looked at some of the foundational energy things (types, units, transfer) and just got to heat capacity.  On Friday we'll get exams back and dig a little deeper into heat capacity.

Exam 2 results

OK, not exactly results, I haven't graded the exam yet, but it seems like a lot of people struggled with this exam.  This is typically the most challenging material in Gen Chem I so it's not unusual for this exam to have lower scores.  I've posted a poll, let me know if you've identified specific problems in your approach to class.  The poll is 100% anonymous and is not monitored or moderated in any way, so it's not definitive scientific data, but if there's a consistent problem identified I might be able to do some things to help.

EXAM TONIGHT

Don't forget that we are in a different room and a different building for the exam this evening.  We will be in CB111, the Center for Business.

Chem Club Tutoring Schedule

I've posted the Chem Club Tutoring schedule in a panel to the left (<-- that way <--).  Take advantage of this service and let me (or them) know what's working or not working with the schedule or tutors.

Monday's in-class problem

I'll get answers posted for Monday's in-class problem some time today or tomorrow.

Lab Hand-In

Yes, there is a hand in assignments due for last week's lab.  The assignment should be turned in to the mailboxes outside HA103.  Make sure you put your assignment in the right box (by Lab Assistants) or you will not get credit for the assignment.

OWL question - Roots

There have been a few people asking about the "Roots" question in OWL and I think the confusion may be coming from the way it's worded.  It sounds as if you will get a very large or very small number as an answer that will require scientific notation to answer, but the 3-4 times I've tried the question, my correct answer has always been easily expressed without scientific notation, answers like "4.293" and "0.271".

Simplest help = just evaluate the mathematical expression given in the problem and type in the answer you get.

First day!

Welcome to Fall 2011! Check here for class info, answers to emailed questions, and other randomness.

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For question #9 on exam 3 from this summer, how does the balanced equation that you give on the answer key match up with the stoichiometry that is also given? In the equation there are 2NaOH(aq) but in the stoich problem under it, there are 2mol KOH. I guess I am not understanding that relationship.

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Typo. It should be KOH in the equation.

One more...

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I have a question on #5b on the practice test. You posted the answer as

H3AsO4(aq) + 2 KOH(aq)  2 H2O(l) + K3HAsO4(aq)
(0.02500L H3AsO4(aq)) (0.127 M H3AsO4(aq)) (2mol KOH / 1mol H3AsO4) ( 1/0.03868L KOH(aq)) = 0.164M KOH(aq)

I am confused at where the .127 M H3AsO4 came from.

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The 0.127M came from the first part of the question. In 5b, you're using the arsenic acid solution you determined the concentration of in 5a to titrate a new KOH(aq) solution that has an unknown concentration.

Email questions...

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1. For question #14 on exam 1 from this summer, how is the "i" value 3? Also, do you have to add 100 to get the bp for these types of problems, but if it were a fp question would you subtract the value from zero?
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When potassium sulfate dissolves in water, it forms 2 potassium ions and 1 sulfate ion for a total of 3 particles. Remember, when you're calculating these numbers, you are most often calculating a change in freezing point or boiling point. Boiling point is elevated in solution, so the change you calculate is above the boiling point of the pure solvent; freezing point is the opposite direction.

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2. For question #18 on exam 1 from this summer, I understand that the order of [CH3I] is 1st and that [F2] is 0. But, if a concentration is 0 order, does that make it not part of the rate law expression? I just don't see why the [F2] isn't part of the problem after you say that its 0 order on the answer key to the test. Say if it was 1st or 2nd order, how would the problem be different?
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We could explicitly include a "[F2]^0" term in the rest of the problem, but any number raised to the zero power is equal to 1. Since we're only multiplying and dividing, including an extra term that's the equivalent of "1" will not affect the answer. If it was 1st or 2nd order, we would have to include that term. This would impact the units of "k" as well as changing the numerical value.

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3. How many questions will be on the exam? and how much time will we have on the test? Will it mostly be problems to work through or will there be some multiple choice as well?

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30-40 questions, you will have the full 2 hour 10 minute class period. This exam will have a slightly higher proportion of multiple choice questions than most of your previous exams, maybe over half multiple choice. Although there are more questions, the questions will be similar to the type of questions you've seen on previous exams.

Email me any other questions, I'll be checking in throughout the rest of the day and this evening. I will plan to be in my office (HA407H) by 7am tomorrow, if you have other questions you can stop in early. Good luck.

Questions

A couple email questions...

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Hello professor Bodwin,
I have two questions about the material for exam four. When assigning oxidation numbers, would diatomic ions have a charge of 0, like I2? And on problem set #10 number 3, for the second reaction, on the answer key you have that 3 e- needed to be added to both sides of the reduction half-rxn. I am confused why it is added to both sides and not just the reactant side.

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In diatomic molecules like I2, each iodine has an oxidation number of 0. This is iodine in its neutral, uncombined {with other elements} form. All the elements that are diatomic molecules (H2, N2, O2, halogens) are oxidation number zero when they are their uncombined diatomic molecule.

On problem set #10... oops, that's a mistake. I was copying and pasting reactions and I must have forgotten to delete some of those electrons. It should be:

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Reduction half-rxn: 2( 3 e^- + Cr³⁺(aq) ó Cr(s) )

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Let me know if there are more questions.

PS6 and E2 keys...

Blanks and keys are posted on my mnstate.edu page.

I won't be in my office this afternoon, I have an appointment off campus. Email if you have any questions.

Error in PS#05 Key

There was an error in the key I posted yesterday for PS#05, it should be corrected in the version that is online now. In the first problem, there are 3 gas particles on the reactant side forming 2 gas particles on the product side. The error said 2 and 2. Sorry about that, please let me know when you find errors (or things you might think are errors) so I can correct them.

PS#05 and key posted...

On my mnstate.edu page.

If you have questions, I'll be in my office (HA407H) tomorrow morning, probably until the early afternoon. You can also email, I'll answer questions here.

Have a good weekend.

Keys posted

The keys from Exam 1 and Problem Set #4 are posted on my mnstate.edu page.

http://www.mnstate.edu/bodwin/

And for today's problem set...

http://www.mnstate.edu/bodwin/courses/genchem2x/c210nps2k.pdf

Problem Set 1...

Here's the key for today's problem set.

http://www.mnstate.edu/bodwin/courses/genchem2x/c210nps1k.pdf

Make sure you try to work through the problem before checking the key, if you just read through the key you will think you understand the problems, but when faced with a blank page you might not know where to start.