Exam Monday

We reviewed in class on Friday. A couple things people are having trouble with:

1. Changing {delta}H when you change a reaction. Let's look at an example, consider the reaction A-->B with {delta}H = 25kJ/mol. Positive {delta}H means this reaction is endothermic, so we have to add 25kJ of energy to convert each mol of A into a mol of B. If we reverse the order of the reaction, B-->A, the conversion of a mol of B into a mol of A will release 25kJ of heat, making it exothermic with {delta}H = -25kJ/mol. What if we triple the reaction, 3A-->3B? For each mol of A that is converted to a mol of B, the reaction still requires 25kJ of heat, but now the reaction as written is converting 3 mols of A into 3 mols of B, so {delta}H = 3(25kJ/mol) = 75kJ/mol.

2. Heat capacity and {delta}H for a phase change. When a substance freezes, melts, boils or condenses, it releases or absorbs heat without a change in temperature. This means that if I have a pot full of water on the stove and I start to boil it, no matter how hot I make the burner the boiling water will be 100 degC until all of it has boiled away. If you're looking at a problem that goes through a phase change and changes temperature, you can break the problem down into a couple smaller problems. For changing temperature without changing phase, it's a heat capacity problem; for changes in phase, it's {delta}H of phase change problem.

Remember, if you're looking at old exams we have not talked about quantum numbers or electron configuration or wavelengths yet, that's coming up in Chapter 7.

If you're in the mood for a study break, today at 4pm the MSUM Dragon volleyball team will be Digging for a Cure to raise money for cancer treatment and research. Wear purple and cheer on your Dragons as they crush Upper Iowa. Purple T-shirts will be sold at the game for $10.

Let me know if you have questions, I will answer to the blog.

Fuels and enthalpy...

Today we looked at the enthalpy involved in the combustion of a variety of fuels. "Combustion" or "burning" means reacting a fuel with oxygen to produce carbon dioxide and water for hydrocarbon fuels. We will look over the numbers a little more on Wednesday.

For EVERYONE in lab this week, you may have noticed that there is no experimental procedure included in your lab manual. The procedure is posted on my mnstate.edu website under the Chem 150L link. This is for everyone in my labs AND Dr. Marasinghe's labs.

There's a new MC assignment posted, due Sunday. See you all in class on Wednesday.

Back to work...

Welcome back from the long weekend. I handed back exams today, if you weren't in class I will bring the exams to class on Friday.

We finished looking at the enthalpy of reaction problem that we started last week, on Friday we will look a little more closely at quantifying heat exchange.

As I mentioned in class today, don't forget that advising starts this week. If you have any questions about advising, choosing a major, or classes for next semester, let me know, I'm happy to try to answer them.

I've updated the SI times (at left...). I will also be putting up a new Masting Chemistry assignment either tonight or tomorrow morning, the MC system is down for scheduled maintenance right now.

Seminar Speaker

Tomorrow (Friday, October 9th) at 1:30pm the Department of Chemistry will host Dr. Darrell Eyman from the University of Iowa for a seminar. Dr. Eyman's talk is entitled "Grafted Active Site Catalysts: Synthesis, Characterization and Applications". The seminar will be held in SL118.

Extra exam question...

I did not give your exams back today because I wanted to give everyone a chance to earn a few more points. In class, I handed out an extra exam question worth up to 20 additional points. If you were not in class today, stop by my office and pick one up. I will also have them in lab tomorrow. This extra question is due by noon on Friday.

Today in class we started talking about thermodynamics and heat transfer, including specific heat capacity and how to calculate energy transfer from temperature change and vice versa. We stopped in the middle of a problem, we'll continue it on Friday.

Next Wednesday from 8am until noon there will be a representative from the University of Minnesota College of Pharmacy. If you are interested in a career in pharmacy or just want some more information, stop by and check out the info. He will be in HA405. UMinn's College of Pharmacy is a very highly ranked program (3rd in the nation the last time I checked...) and there are a number of MSUM alumni who are currently attending the college.

Last questions?

How can you look at a molecular formula and know if it's a strong acid or not? like in the question 4 from fall 2007, which of the following the the strongest acid? KOH CHLO4 HC2H3O2 H20 or NH3

We didn't specifically talk about identifying strong acids or bases this semester, so this type of question will not be on the exam. Strong acids and strong bases are those which "completely" ionize in solution. For bases it's a little easier because the strong bases are those that are soluble, so they follow the solubility rules for hydroxides (alkali metal hydroxides and soluble alkali earth metal hydroxides are strong bases). For acids, it's probably a matter of memorizing the strong acids (perchloric, sulfuric hydrochloric, hydrobromic, hydroiodic, nitric) and all the rest of the acids are weak. But again, we didn't specifically address this so it won't be on the exam.

For winter 2006 number 12, I took (1.62M)(50.00mL) = C2(500.0mL) and figured out C2, but I don't get the same answer as the test. Is there another step I'm missing?

That question asks for the concentration of potassium ions, not the concentration of potassium carbonate. The concentration of potassium carbonate is 0.162M, but for every mol of potassium carbonate that is dissolved there are 2 mols of potassium ions dissolved, so the concentration of potassium ions in the solution is twice the concentration of potassium carbonate, 0.324M.

Email question...

For exam 2a from fall 2008 # 13, where or how do you get n2 = (14.5mols)(0.875) = 12.7mols. I get where the 14.5 mols is coming from, but where is the 0.875 come from? and how do I get to that point?

The problem states that 12.5% of the gas is lost to a leak as the balloon rises, so 87.5% of the gas remains in the balloon.

Email questions...

There are a bunch, I'll answer them one at a time...

I have some questions on the old exams. One is balancing equations. I know we've talked about that alot in class, but I'm still kind of confused. I understand how to balance the numbers out when we're given the equation, but I get confused when starting from scratch. For example, number 7 for fall 2006 exam: Magnesium hydroxide solution + Lead (IV) nitrate solution --> Lead (IV) hydroxide + Magnesium nitrate. I know we have to know the polyatomic ions, but does the charge have anything to do with the numbers behind of the element? Why is hydroxide OH2 in the answer, when it's a OH- polyatomic ion? Does that have to do with balancing?

For cations that have ambiguous charge, the oxidation state is given by Roman numerals after the name. Something like sodium is pretty much always +1, but the transition metals and main group metals (like lead, tin, etc) can have a number of different stable charges, so these are specified. For this question, we need 2 hydroxides in the formula of magnesium hydroxide and 4 hydroxides in the formula of lead(IV) hydroxide. Hopefully it's "(OH)_2_" in the answer and not "OH2"...

Another question I have is number 11, winter exam 2006: How many grams of hydrogen are required to make 34.061g of ammonia by the following reaction? xH2(g)+ yN2(g) --> z NH3(g). I have no idea how to do this problem. Do you have to do something with mole ratio?

Yes, you need the mol ratio. First, balance the equation. Once you have correct numbers for x/y/z, then convert34.061g of ammonia into mols, use the mol ratio (x/z in this case) to convert mols of ammonia to mols of hydrogen, then use the molar mass of hydrogen to convert to grams.

How do you do concentration problems like numbers 12: 50.00mL of a 1.62 M potassium carbonate solution is diluted to 500.0mL. What is the concentration of potassium ions in the resulting solution, [K+]? and 13: what is the concentration of a perchloric acid stock solution if 21.53 mL of 1.054M Mg(OH)2(aq)is required to titrate 15.00mL of HClO4(aq) to the equivalence point in the following reaction?: a HClO4 (aq) + b Mg(OH)2 (aq) --> c H2O(aq) + Mg(ClO4)2(aq).on winter 2006 exam?

Hmm, this is a 2-fer. When you are diluting a solution of known concentration, use the formula C1V1 = C2V2 where C's are concentrations and V's are volumes. In this case, plugging in numbers gives:
(1.62M)(50.00mL) = C2(500.0mL)

The second one is a titration problem, which is just a specific type of stoichiometry problem. Write a balanced chemical equation, convert 21.53mL of 1.054M Mg(OH)2(aq) to mols, use the ratio from the balanced equation to convert mols Mg(OH)2 to mols HClO4, then use the given volume to convert mols HClO4 to concentration (mols/L).

Number 14, winter exam 2006: 50,00mL of 1.119M Co(NO3)3 (aq) is combined with 60.00mL 1.821 M Na2Co (aq). 2.946g of precipitate is recovered from this reaction. I understand parts a and b, but I don't understand c: What is the percent yield of this product?

Percent yield is the actual yield divided by the theoretical yield time 100%. Actual yield is the amount you collect or "recover" from the reaction, theoretical yield is the maximum possible amount you could produce if you use all of the limiting reagent to make product.

On exam fall 2006, number 9 is assigning oxidation numbers to each element: AgNO3. Ag is +1, N is +5 and O is -2. How are we suppose to know that? N's charge is -3, why is it's oxidation number +5? Also, how do you know the charge of transition elements?

If nitrogen were just some random nitrogen ion, we'd probably expect it to have a charge of -3, in that case it would be a "nitride". In this case, nitrogen is part of the polyatomic nitrate ion. The sum of all the oxidation numbers of the atoms in a polyatomic ion is equal to the charge of the polyatomic ion. For nitrate, we'd expect the oxygens to have oxidation numbers of -2, so:
(Ox# nitrogen) + 3(Ox# oxygen) = (charge of nitrate)
(Ox# nitrogen) + 3(-2) = (-1)
(Ox# nitrogen) = +5
For transition and main group metals (see one of the answers above), the Ox# will either be given as a Roman numeral, or it will have to be determined from a given formula. In this example, the charge of nitrate is -1, so if the given formula is "AgNO3", then the silver must have a charge of +1.

I hope this helps, I'll probably check in again a little later...

Email question...

I am having problems figuring out the questions that give volume, Temp, and Pressure. Like for example number 10 on fall 08 exam 2a, a 2.65 L steel tank contains an ideal gas at 15.83 degrees C and 1.15 atm. what is the temp if the pressure changes to 1.48 atm? I tried using the formula P1V1/T1 = P2V2/T2 and I'm having no luck getting an answer. How do I figure this problem?

That approach is close, but if this is a rigid steel tank then the volume is probably not going to change. From the comparative form of the ideal gas law, V and n cancel (because they don't change), so we're left with P1/T1 = P2/T2. Plugging in values from the problem:

(1.15atm) / (288.98K) = (1.48atm) / T2

T2 = 371.90K = 98.75 degC

Problem set key

I've also posted a key for the problem set we did in class, it's on my Fall 2009 Chem 150 page.

Keys posted

The keys for last year's Exam 2 are posted on my web page under "Previous Gen Chem I". If you have other questions, email them, I'll be responding to the blog tomorrow morning.

If you're in need of a study break, the volleyball team is at home this weekend, 7pm tonight and 4pm tomorrow afternoon.

Good luck studying.

Gases

On Monday we started talking about gases and gas laws including kinetic-molecular theory of gases, the "simple" gas laws, and finally the ideal gas law. We'll look at gases again on Wednesday and probably start reviewing for the exam next Monday.

There's a new MC, due Sunday.

I got a question in email about the Redox assignment in MC. It's a little terminology issue that can cause some confusion so let me try and handle it here. If a substance is oxidized it loses electrons. (That's the "OIL" part of "OIL RIG" or the "LEO" part of "LEO GER".) Because oxidation and reduction cannot take place independently (if you have one, then you must have the other), you could say that if substance A loses electrons and is oxidized, then it will cause something else, let's say substance B, to be reduced. So if A is oxidized, then it is acting as a reducing agent toward B. Stated simply:

If something is oxidized, then it is a reducing agent or reductant.

If something is reduced, then it is an oxidizing agent or an oxidant.

These definitions are all about the cause-and-effect relationship between oxidation and reduction.

If you have any specific questions/topics you'd like to review for the exam, let me know.

Stoichiometry practice...

Friday in class we looked at a problem set to practice stoichiometry. It's copied below. There is a new MasteringChemistry assignment posted, due Friday. Next week we will look at gas laws and continue with stoichiometry. Your next exam is October 5th.

Chem 150 – Fall 2009 – Problem Set #2

You are studying the reaction of 1.132M potassium phosphate solution with 1.275M barium nitrate solution.

1. Write a balanced chemical equation for this reaction.

a. How many grams of precipitate could you make if you completely react 125.0mL of the potassium phosphate solution?

b. How many grams of precipitate could you make if you completely react 175.0mL of the barium nitrate solution?

c. How many mL of the barium nitrate solution is required to react completely with 125.0mL of the potassium phosphate solution?

c. How many mL of the potassium phosphate solution is required to react completely with 175.0mL of the barium nitrate solution?

d. What is the theoretical yield of precipitate (in grams) if you react 125.0mL of the potassium phosphate solution with 175.0mL of the barium nitrate solution?

e. What is the limiting reagent in part d? How many moles of the excess reagent remain after the reaction is complete?

f. Write the balanced net ionic equation for this chemical process.

2. These questions deal with concentrations of the solutions used above:

a. What is the concentration of potassium ions in the 1.132M stock solution? Phosphate ions?

b. What is the concentration of barium ions in the 1.434M stock solution? Nitrate ions?

c. How many grams of potassium phosphate are present in 125.0mL of the 1.132M stock solution? How many grams of barium nitrate are present in 175.0mL of the 1.275M stock solution?

d. If you dissolved the mass of potassium phosphate in part c in enough water to make 300.0mL of solution, what would the concentration be? If you dissolved the mass of barium nitrate in part c in enough water to make 300.0mL of solution, what would the concentration be?

e. If you dilute 125.0mL of the potassium phosphate stock solution to a total volume of 300.0mL, what will be the “new” concentration of potassium phosphate? If you dilute 175.0mL of the barium nitrate stock solution to a total volume of 300.0mL, what will be the “new” concentration of barium nitrate?

Things that go boom...

Today we looked at a couple more reactions including redox. The key to redox reactions is being able to figure out oxidation numbers for substances from their chemical formulas and it takes practice.

Practice? Yes indeed! There's a new Mastering Chemistry assignment, due Tuesday.

Introducing...Stoichiometry!!

Today we looked at the relationship between mols of reactants and products, stoichiometry. For any stoichiometry problems, there are a logical set of steps to follow:

1. Balance the equation.

2. Convert the quantity you know into mols.

3. Convert mols of what you know to "mols of interest" using the coefficients (ratios) in the balanced equation.

4. Convert "mols of interest" into a useful number. That useful number might be a mass or a volume or a number of other things, it all depends upon the problem.

We looked at a precipitation reaction and an acid-base reaction today, we'll look at a couple other reactions on Wednesday.

There is a new MasteringChemistry assignment posted, due Sept. 28th.

More on recognizing reaction types

Friday we talked about a couple classes of reactions and how to recognize them. We looked at precipitation reactions which required use to develop solubility rules to identify ionic compounds which we wouldn't expect to be soluble in water. Next, we looked at some molecule-forming reactions, starting with acid-base reactions, so we talked about acids and bases which react to form a salt and water (a molecule). We also looked at some gas-forming reactions, those that form a gas molecule when they react. We will continue looking at classes of reactions Monday, and there will be a new MC assignment on Monday.

Balancing equations...

Today I handed back exams, the average was around 125/150. For those of you who didn't do as well as you might have liked or missed the exam, remember that your low exam score is dropped in determining your final score for the course, so whether it's an exam that you miss or just the lowest score out of your 4 exams, it will not impact your course score directly.

We started talking about balancing reactions today. Your textbook introduces balancing reactions in Chapter 3 and the really gets into it in Chapter 4. We introduced things today and will really get into it on Friday. See you then. (Or tomorrow if you're in one of my labs...)

Speaking of labs, if you are in my lab make sure you remember to look at the online pre-lab and take the quiz on D2L.

Exam #1, Fall 2008

I posted a key for last year's exam if you'd like to check your answers.

www.mnstate.edu/bodwin, in the left panel click on "Gen Chem I" under the "Previous" heading.

Review...

Today in class we reviewed for Monday's exam. We went over the questions from the in-class problem set, the answers were: cobalt has a +3 charge, the alloy is 85% iron.

If you have other questions, let me know and I will answer them to the blog. Don't forget about your MasteringChemistry assignments.

Exam reminders:

- At least 1 empty seat between everyone

- No graphing calculators, no cell phone calculators, and turn off your phones

See you all Monday.

Naming organics

An email question...

what do the numbers in front of the name mean? (for instance 2 pentene or 4 nonene)

The numbers indicate the location of the functional group. Many (most?) organics exist as chains of carbon atoms so if we want to draw a correct picture of the molecule based only on the name, it's necessary to specify where on the chain to put functional groups. For alkenes, the molecule contains a double bond, so if we start with pentane:

CH3-CH2-CH2-CH2-CH3

and want to make one of those bonds a double bond (and remove 2 hydrogens) to form pentene, we could either make the first bond in the chain a double or the second, so we need to specify whether it is 1-pentene or 2-pentene. The same is true for other functional groups, 1-propanol has the -OH group on the first carbon, 2-propanol has the -OH group on the second carbon.

This comes up in your MasteringChemistry assignment, but it is not a super-important distinction for us right now. The chemical formula of 1-propanol is C3H7OH, and the chemical formula of 2-propanol is C3H7OH, so for purposes of calculating mass or moles they can be treated the same. When we start looking more closely at properties of molecules, we will start to see differences between them, but that's a few chapters away at this point.