Final Exam

The final exams are graded, I'll enter course grades first thing tomorrow morning.  If you'd like to see your final exam, you can stop by and look at it but you can't keep it.  You should be able to see your grade in eServices some time tomorrow or Friday at the latest.  Have a great break and I'll see you next semester.

Final exam...

Less than 1 hour until the final exam starts...

Final exam info

You will have the same info on the front of your final exam as was on Exam 4.

Finding exams and keys...

A few people have been having trouble finding the exams and keys I posted yesterday.  I suspect you may be running into a cached page problem?  If you're using Chrome, try opening my webpage in an incognito window; other browsers have similar features, but I don't know what they're called.

web.mnstate.edu/bodwin
In the left panel, click on "Chem 150" under "Fall 2011"
The new page should open in the right panel.  Scroll down and all the exams and keys should be there.

Exams and keys posted

All the exams from this semester are posted, all the keys except Exam 4 are posted.  I might get to those today, but we went through most of Exam 4 in class just a few days ago.  Let me know if you have any questions.

VSEPR

Valence Shell Electron Pair Repulsion Theory!
VSEPR is the theory used to predict molecular shapes.  Because each region of electron density (lone pair, single bond, double bond, triple bond) is negatively charged, the regions of electron density repel one another as much as possible.  This repulsion dictates the shape of the molecule or polyatomic ion.  If you're having trouble visualizing these 3-dimensional shapes, try the PhET simulation we looked at in class:
http://phet.colorado.edu/en/simulation/molecule-shapes

Have a good weekend and don't forget to look at the OWL assignments that are currently posted.

Lewis Structures

As with everything, your ability to understand and draw Lewis Structures depends upon 3 equally important things.  #1 - Practice drawing Lewis Structures.  #2 - Practice drawing Lewis Structures some more.  #3 - Most importantly, everyone needs to practice drawing Lewis Structures.  We'll work through some more examples in class and (hopefully) do some practicing in class, but you really really really really need to practice them yourself.  I've posted the lab info for the experiment we're doing after break, take a peek for some more practice using Lewis Structures.

On Friday, VSEPR.  What's VSEPR?  Come to class on Friday...

Lab exam

A few people have asked about the lab exam you will have this week, specifically how to study/prepare for it.  To give everyone the same info, here's the reply I sent to someone who asked:
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Hopefully you're already prepared. ;)  I might review some of the techniques and procedures we've consistently used throughout the semester; things like error handling, graphing, different types of glassware, etc.  Because this is more of a techniques and procedures exam, it's not necessarily something that can be studied for.  As I've said, the exam is not going to be a bunch of experiment-specific detail (What was room temperature for the Al + HCl experiment?, What color was the nickel solution in the Clandestine Lab experiment?, etc).
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Let me know if you have any further questions, I'm not sure how much more detail I can go into about the lab exam, but you can always ask.

Electron configurations, etc

We've been looking at electron configurations and what we can do/predict with them.  Sizes, charges, stability, magnetism.  Don't forget about the OWL assignments that are posted.

All the nitty gritty of the electron world...

Monday and today we've been exploring the world of the electron a bit more.  The vast majority of chemistry is really a study of the electron: where are they, why are they there, where do they move, when do they move, how fast do they move.  On Monday, we looked at quantum numbers as a way to address electrons, but writing out explicit quantum numbers can be a bit ponderous, so today we looked at a shorthand way to express quantum numbers with electron configurations.  Electron configurations describe the energy levels and orbitals that are occupied (or might be occupied) in an atom or ion and provide a very useful tool for studying electrons.  Practice them.

Light!!

We've been looking at the nature of light for the last 2 days (as well as getting exams back and going through problems) and have just gotten to the point of using that light to explore the structure of atoms. Next week, the fun will be beyond measure.  On Monday, I promise there will be fire.

Have a great weekend.  Volleyball has their final home games of the year tonight and tomorrow, and football has their last home game tomorrow.

Exam tonight...

Exam 3 is tonight so we spent most of class reviewing.  On little bit that we (sort of) added...  When calculating the heat of reaction for an aqueous process, it is sometimes easier to use the net ionic equation.  There aren't always tabulated values available for every soluble salt, but the ions can be found for most elements.

See you tonight, 6pm in SL104.

emailed question...

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Hello Dr.Bodwin,

I have been studying up on the Exams that you have on your page and some of them have some things that we didn't go over in class and I was just wondering if those things would still be on our exam. The things that I am wondering about is like the quantum numbers, electron configuration, and the wavelength problems. 
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We'll be getting to that material after this exam, don't worry about it for now.

Old exam keys...

Looks like I haven't had time to put a key together for a while...

http://msumgenchem.blogspot.com/2010/10/old-exam-keys.html

Question

Question from email:
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Dr. Bodwin,

I am wondering if the test on Monday will have material from our last exam and the new material covered since then or just the enthalpy??
Thanks

-----------------
A few people have asked me this question and the best answer I can give is "yes".  Although there will not be any questions that are strictly "exam 2 questions" on this exam, you will have to know how to do things from exam 2 to answer exam 3 questions.  If you are trying to calculate the heat liberated or absorbed by a reaction, you will have to be able to write a balanced equation.  If you're trying to write a balanced equation, you will have to be able to write balanced formulas.  Many enthalpy problems are the same as all the stoichiometry problems we looked at for exam 2, the only difference is that instead of calculating grams or molarity or volume, you'll be calculating heat.  You may need to determine the limiting reagent, or percent yield, just like any other stoichiometry problem.

Other questions, let me know.

Heat, heat, heat...

Today we went through some more enthalpy/heat transfer problems.  We also went through a problem that demonstrated Hess' Law; for a multi-step process, the sum of the enthalpies for all the steps should equal the enthalpy of the whole process.  We've actually been using Hess' Law the whole time we've been looking at enthalpy, but we didn't formally call it Hess' Law.

If you have questions, email me.  I have a few other things going on this weekend, but I will do my best to post answers to the blog ASAP.  If you need to take Monday's exam at an alternate time and have not yet talked to me, please check in either by email or in person before Monday.

And most importantly, take a break or two over the weekend.  The weather is supposed to be quite nice, so take a little walk around the block for a study break.  Volleyball is home tonight and tomorrow and football is home this weekend.

Enthalpy

Wednesday in class we went through a couple more enthalpy problems/calculations.  Enthalpy of formation values are tabulated and refer to the heat transfer when 1 mol of the substance is produced from its standard state elements.  The magnitude of that heat transfer is the same whether a substance is being formed from its elements or the elements are being formed from the substance, only the direction of the heat transfer (and therefore the sign of {delta}H) changes.  This is the key to calculating enthalpy of reaction using tabulated enthalpy of formation values.

Tonight at 6pm in HA113, Tri-Beta will be hosting research night.  Faculty from Biosciences, Chemistry, and Physics will give brief descriptions of their research and be available for questions.  If you're interested in doing research, this is a good opportunity to see a variety of the projects taking place on campus.

New OWL

Oh, and there are new OWL assignments posted.  Enjoy.

Enthalpy - the heat of a process

Today we linked heat capacity to the larger idea of enthalpy.  Enthalpy is the heat transfer associated with a chemical reaction or physical process.  We'll work through a few more examples on Wednesday.

The Exam

We spent most of today going over the exam.  Exam 3 will cover a LOT of the same material, so make sure that you use your performance on Exam 2 to guide you toward the areas you need to study more.  Having graded the exam, I can identify polyatomic ions as the biggest problem most people had.  There's no real trick for polyatomic ions, you just have to memorize them.  As with anything (music, language, sports, etc), the more you practice the more automatic a thing becomes.  When you write out the formula for nitrate 100 times while you're studying, you will tend to remember the formula for nitrate.

We also did a quick heat capacity problem at the end of class, we'll get more into that next week.  Keep an eye on the schedule, Exam 3 is a week from Monday, so it's coming up quickly.  Have a good weekend.

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.

----Question----

On the Spring 2010 exam, for number 17, it says to redo experiment 3 at 16.31 degrees, the rate is 7.53x10-4. How do you calculate the new k for this problem. The answer is 7.17x10-5

----Answer----

The new k is calculated from the concentrations used in experiment 3 and the new rate. When the temperature of a reaction changes, the mechanism doesn't change (over small temperature changes) so the rate law expression is the same, you can plug everything in and use the same orders you determined at the original temperature.

----Question----

I'm having trouble with #14 on the Spring 2008 exam. How do you solve it?

14. You have found the following value in a table of equilibrium constants at 25ºC:

Cu2+(aq) + 4 NH3(aq) 􀀧 [Cu(NH3)4]2+(aq) Kc = 1.7x10-13

What is the equilibrium constant for the reaction:

2 [Cu(NH3)4]2+(aq) 􀀧 2 Cu2+(aq) + 8 NH3(aq)

----Answer----

This is similar to the lead bromide question... To get from the first equilibrium reaction to the second, we have to: 1)reverse the reaction; 2)multiply by 2. To convert the equilibrium constant, that means we have to: 1)take the inverse; 2)raise it to the 2nd power. So:

K{new} = ( 1 / (1.7E-13))^2 = 3.46E25

----Question----

17. You have found the following value in a table of equilibrium constants at 25ºC:

PbBr2(s) 􀀧 Pb+2(aq) + 2 Br-(aq) Kc = 6.29x10-6

What is the equilibrium constant for the reaction:

½ Pb+2(aq) + Br-(aq) 􀀧 ½ PbBr2(s)

----Answer----

To get the second reaction, the first reaction must be: 1)reversed; 2)multiplied by 0.5. Therefore, the equilibrium constant must be: 1)inverted; 2)raised to the 0.5 power. So the "new" equilibrium constant value is:

( 1 / 6.29E-6 )^0.5 = 399

----Question-----

On the exam -summer 2007 (exam 2) #9, I wasn't sure how you got the second grams (16.246 g / 331.208) because in the periodic table it said pb is 207.2, so what do I have to do to get it to 331.208?

----Answer----

The source of the lead(II) is lead(II) nitrate. When you're looking at equilibrium problems (or kinetics problems, or any stoichiometry problems) you can usually just use the net ionic equation, but you need to include spectator ions when you're weighing out reactants.

----Question-----

On exam - fall 2004 (Exam 1), I wasn't sure how to calculate #12. For #14 I converted it to grams then to the concentration but I don't seem to be getting the right answer either.

----Answer----

That's an old exam... #12 has an error, so there is not correct answer listed, but here's how to do it. The rate of oxygen consumption is:

0.433mols/8.5oL/min = 0.0509 M/min oxygen consumed

From the balanced chemical equation, for every 5 mols of oxygen consumed in the reaction 4 moles of ammonia is consumed, so the rate of ammonia consumption is:

(0.0509 M/min oxygen)(4 moles ammonia / 5 moles oxygen) = 0.0408M/min

Last minute questions...

----Question----

I can't figure out the ppm question when it asks what is the concentration in ppm of a solution made by dissolving 14.18mg of dinitrotoluene (182.13g/mol) in 4.00L of water. I thought you took .01481g/182.13 than you took that answer divided by 4kg and than take that times 1 million but its not working out for me.

----Answer----

"ppm" is (by convention) usually thought of as a mass-mass unit, so it's part of the mass fraction family.

ppm (dinitrotoluene, DNT) = [ {mass DNT} / {total mass of sample} ] * 10^6 =

[ {0.01418g DNT} / {4000g + 0.01418g} ] * 10^6 = 3.55ppm DNT

NOTE: because the volume of water only known to 3 sig figs and is so large compared to the mass of DNT, the denominator simplifies to just 4000g.

----Question----

You don't need to know molar mass of a substance to find what: molarity, molality, normality, mole fraction, or mass percent?

----Answer----

The only reason you would need the molar mass would be to calculate moles of the substance, so anything that includes moles will require the molar mass. Because mass percent is just a ratio of masses, you can calculate it without knowing the molar mass of the substances involved. This is related to the previous problem above.

Questions...

A few email questions came in. I may be able to answer a few more of these tonight, but by about 8 or 9pm I'll be offline until morning, so if you have questions, get them in sooner rather than later.

--Question--------------

I just had a question for the freezing point depression problems..for example on the Spring 2009 exam 1b #13, as far as the calculations where do you get the #mols particles/mols ? In that problem it says (2 mols particles/mol LiNO3) how do you get that part?

--Answer--------------

Quite a few people are having trouble with this. The number of particles per solute (the van't Hoff Factor) is a measure of how many pieces each formula unit of solute breaks into when it dissolves. For molecular solutes, the solute formula is a single piece in solution; for example, when a sugar molecule dissolves in water, it's still a single sugar molecule, it doesn't break down into individual carbon, hydrogen and oxygen atoms. When ionic solutes dissolve in water, they break down into the ions that make up the formula. In the LiNO3 example given above, when LiNO3 dissolves in water it does NOT float around as LiNO3 units in solution, it breaks down into lithium ions in solution and nitrate ions in solution, so for every 1 LiNO3 unit, there are 2 particles in solution.

--Question--------------

I am not really understanding the formula for the ppm problems, for ex. (Winter 2006, Exam 1), you have times 10^6, how did you get to the sixth power?

--Answer--------------

10^6 is a million. When you're converting a fraction (mass fraction or volume fraction) to ppm {or ppm(v)}, you have to multiply by a million, 10^6.

--Question--------------

I wanted to make sure I did my work right for the mol fraction questions. If the question asks, what is the mol fraction of sugar in a saturated aqueous sugar solution at 25 C. (solubility of sugar in water is 211.4 g/ 100 mL.

Would it be 211.4 - 180.1548 (total grams of sugar) / 180.1548 = .1744

--Answer--------------

This is essentially a unit conversion problem. The given solubility means that 211.4g of sugar will dissolve in 100mL of water. The mols of sugar is:

211.4g / 180.1548g/mol = 1.173mols sugar

Mols of water in the system:

(100mL)(1g/1mL) / 18.015g/mol = 5.55mols water

So the mol fraction of sugar in this solution is:

(mols of sugar) / (total mols) = (1.173mols sugar) / (1.173mols sugar + 5.55mols water) = 0.1745

--Question--------------

I know we did our lab on molar mass today, but I am not sure how to calculate question #16 on exam 1a (spring 2009). It has atm added to it.

16. A newly discovered protein has been isolated from seeds of a tropical plant and needs to be characterized. A total of 0.137g of this protein was dissolved in enough water to produce 2.00mL of solution. At 31.68°C the osmotic pressure produced by the solution was 0.134atm. What is the molar mass of the protein? (20pts)

--Answer--------------

In lab we were looking at one colligative property (freezing point depression) to determine the molar mass of an unknown. This problem is looking at a different colligative property (osmotic pressure) to determine the mass of an unknown. The expression for osmotic pressure is:

P = MRTi

OK, we have to make an assumption here, we're going to assume that the protein is a single particle in solution, which makes the van't Hoff Factor (i) equal to 1. After that, we can start plugging in the info from the problem.

0.134atm = M(0.08206 L.atm/mol.K)(31.68+273.15K)(1)

M = 0.0053569 mols protein/L solution

We've made 2.00mL of solution so:

(0.0053569 mols protein/L solution)(0.00200L solution) = 0.000010714mols of protein

So the molar mass of the protein is:

(0.137g protein)/(0.000010714mols of protein) = 12800g/mol

Exam #1 next Friday

We've gone through Chapters 10 (Gases) and 11(Solids and Liquids), we'll look at Chapter 15 (Solutions) next. The first exam is next Friday, let me know when you have questions, I'll post answers here.