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If I earn a "D" grade in 10B, do I have to drop 10BL (lab)?
Yes, you may continue in 10BL. You are not requirewd to drop it.
Hi Prof. I asked earlier about using TI82 calculators on the midterm and you said that there was not enough time to check for any programs, so I was wondering if your decision changed because of the extended time for the final??? Thanks!
No. I am sorry, but I will have to stand by the decision to allow only non-programmable calculators.
Professor Knobler, Where can we find the two examples you posted on the web? Thank You.
The problems are lited under "Problem Sets"
Prof. I have a very hard time understanding your lectures. You only use symbols when explain and not actual numbers. It is very confusing to myself and many others. We would like it if you were more clear with your explainations. After all we cannot fit all the students that want to learn in the first four rows. Thank You Jonathan X
A key to inderstanding science is to work with symbols first and then put in numbers. I try to make my explanations as clear as I can -- if they are not clear to you then you should either ask questions in class or ask me or your TA's questions in office hours.
hi professor, i have a few questions- for 18.1 and others like it- why do you have to subtract from the rate at t=0? (this is how they do it in the solution book) also, for 18.79 can you assume that you are right that the order of c is 0 to find a and b because c is never held constant. if so, that would only work if c was to the zero order right? hope you can understand this, thanks, erin dowling
I assume that you are referring to a problem like 18.5 not 18.1--there's no subtraction of anything in 18.1. The rate is expressed as the change in the concentration over the change in time. To get the change in concentration you must subtract the initial concentration. (If this is not the question you've asked, contact me again.) No, you must not assume that the reaction is zeroth order with respect to C. Compare experiments 1 and 4. You'll see that the concentrations of A and B are the same -- only the concentration of C has been changed. You can therefore obtain the order with respect to C from them and then use it with the other experiments to obtain the order with respect to the other reactants.
Prof.Knobler, I believe I speak for everyone when I say that not being allowed to use our Graphing calculators (TI's) causes us to get confused and more anxious during the test. I know that the reason why you wouldn't let us use the programmable calculators is that you think that we'll put some program or some notes there, but that's what the 8.5'x11' sheets are for right? So, if we could show to you or to the TA's that our calculators have NO programs on them, would you let us use them?
The problem with asking the TA's to check the programmable calculators is that there we must delay the start of the exam until all of the calculators are checked. It is not just the information stored -- wehich is equivalent to the "crib" sheet, but the possibility of programming certain calculations. The policy about the use of calculators has been known to you for over three weeks -- in that period of time you should be able to become familiar with the workings of a simple calculator. I would change the policy if I were assured that everyone in the class had access to a programmable calculator. Until that time those people with programmable calculators have a potential advantage over those who don't and I must be certain that there is a level playing field.
Professor, if our discussion is after Wednesday are we suppose to turn in both homework #4 and #5 at Wednesday's lecture or just homework #4? Thanks, Deshanett C.
No, only the 4th homework set. You do not have to turn in the 5th set.
Good evening Professor: I have three questions for you: 1)when did you assign 18-124 for week 3 homework? 2)when will the answers for 18-110 and 18-114 be available? 3)In Atkins-Jones, example 18.3 (pg680), we are asked to find the reaction order in respect to individual reactants and to the overall reaction. I understand how to determine the order for BrO3(-) and Br(-), but I don't see how an answer for H+ was deciphered. I don't see how reaction 2 and 4 differ by a factor of 1.5? What did they do in order to find a ratio of 1.5? How can we spot a second order reactant?
1. I announced in class April 13 that there was an error in the 3rd week's homework assignment and that 114 (which was already assugned) was replaced by 124. 2. I will ask that it be put on VOH today. 3. The [H+] in experiment 2 was 0.10 and it was 0.15 in experiment 4. The concentrations of the Bro3- and Br- were the same in 2 and 4, so only the [H+] changed -- it increased by 0.15/0.10 = 1.5. The ratio of the rates in the two experiments was 5.4/2.4 = 2.25. Since 1.5^2 = 2..25, the reaction must be second order in [H+].
Professor, I know you gave us the median for the midterm in class today, but I was wondering if you have a rough idea of what the standard deviation is for this test. Could you please post it? Thank you.
The standard deviation doesn't tell you much when the distribution is not Gaussian, i.e. when it is not symmetric around the mean. What is a more meaningful measure is how you stand in the class. The distribution of grades will be posted on VOH and you'll be able to determine your standing.
What will be covered on the Final Exam? Will it be cumulative,or will it just cover things after the midterm?
The final exam will be cumulative but there will be extra emphasis on the last two weeks of the course.
Prof. Knobler, On HW 3, CH.17 Prob 50, the rxn for Ag|Ag+ is supposed to be switched to match that written on the table on the back of the book. Since you swithched it, why did you not switch the sign to negative as well? Your equation has 1.23-0.8. Why is it not 1.23-(-0.8)=1.23+0.8?
Because you never switch the signs of the reduction potentials. The definition of Eo is Eo(cathode) -Eo (anode) . You never need to change the signs,
Professor Knobler, perhaps I am blind, but could you tell me where to locate the old exam on voh? I am having trouble locating it. Thank you very much.
It is not on VOH.
Dr. Knobler, I was wondering if its possible to post answers to 17.82 and 17.114. Also on the 3rd homework assignment it say that problem 17.114 is assigned and it was already assigned on the 2nd homework. You posted the answer for 17.124, so is that the problem that is assigned rather than 17.114 again. Thank you.
Yes, I announced in class that 17.124 was the problem that should be done in the 3rd homework set. Apparently 17.82 ans 17.114 were not scanned in. I will have it done but it takes several hours to get things posted.
Dr. Knobler, I have a question about the solution posted for problem H-29. I think you used the value of delta H of formation of H2O2 (l) rather than H20. The value I obtained for ln K(310)/K(298) was -.872 the same as on the answer key that Carlos gave us.Yet, when I calculate e to the - .872 I get .42, so the value for K(310) I get is 4.2 x 10 to the -15. But, the answer on Carlos' sheet is 2.4 x 10 to the -14. Can you please give me the correct answer. Thank you, Desirée Salazar
You are right. I misread the table and used the wrong valur of deltaH. The solution is otherwise correct.
i was just wondering how you would find the volume in self-test 17.14A on page 659. do we know the Molarity of a given gas at 25C and 1.00atm? thanks
V= nRT/P
professor knobler, i have a question similar to martin's. on page 642 the book says that E>0 is spontaneous and in example 17.5 goes on to say that the sum of the standard potentials should be positive. however, doesn't the example in toolbox 17.2 (pg. 645) contradict this statement because here the sum of the standard cell potentials is negative. how do we know when the standard cell potential is okay to be negative and when it should be positive? thanks, mia sugi
Identify what is being oxidized (the anode) and what is being reduced (the cathode) for the overall reaction that you have written. Then calculate Eo = Eo(cathode) - Eo (anode) If this is positive the reaction is spontaneous, if it is negative the reaction is not spontaneous.
Prof. Knobler: I printed up the anwers to Homework 3 and noticed that on problem #50 in Chapter 17, for the value RT I thought we were to use the values at standard conditions, so I used .0257. The solution used the number .060. Can you explain why this is? Thank you Kristine Puich
You are talking not about RT but RT/F. The book gives the value of RT/F. I have given you the value of 2.303RT/F which one uses ith the log to the base 10 rather than the natural log (ln).
Dr. Knobler, I wa wondering if on the exam we will me given the name of a compound or the chemical formula. Thanks.
I will give you chemical formulas.
Hi professor Knobler, Today in lecture you said that we should never change the sign of the standard potential of a particular half reaction when calculating the standard potential of a cell; however, in the textbook (pg 642) it states that when combining two half reactions we should flip one and reverse the sign of its standard potential if necessary. Please clarify. Thanks.
The book is misleading. Read page 636 -- the first equation on the page -- and the statement just above it, which is precisely what I have told you -- and is correct.
Professor Knobler, I was wondering if you recommend doing all of the questions in the handout in preparation for the midterm on friday. Also, what type of questions are going to be on the midterm? will they be multiple choice? Thanks alot. Shyama Kamat
There will be no multiple choice questions; they will all be numerical problems similar to those on the homeworks. Yes, it would be helpful for you to do as many problems as possible, not just the ones assigned.
I was wondering if it would be possible to change the review session to 5-7pm because a bunch of us have a lab until 5:00. I know that personally I need as much help as I can get, and I don't want to miss out on any review session. Thanks.
Sorry, I can't change it to 5-7 because I have another engagement.
Professor, I am having trouble solving problem 26 on the handout.I am confused about what Q means and how it is found. I know that it is equal to K but I'm unsure how to slove for it. Also for number 29 on the handout I could not find an equation that related K to enthalpy. Should I solve for G and then convert to H using a different equation. Please steer me toward an equation that I can use. Thank you.
The reaction quotient has the same form as an equilibrium constant but it is not the equilibrium constant. You should read Section 13.6 on page 485 of Atkins-Jones, which defines Q and discusses its meaning. K is not related to the enthalpy but its temperature dependence is. Read the section in the handout at the bottom of page 435.
Professor Knobler, In lecture you said that the midterm will cover everything through this coming Wednesday. So, to prepare, should we have the homework questions for week 3 completed-- or is that for the final? Thank you-- Maya
You should complete the homework for week 3.
Professor Knobler, Could you please release the answers to the problems on the hand out that were not assigned for those of us who would like to practice more problems. Thanks Mike
If you would like to check the answers to these problems, please consilt with me or with your TA's.
I am currently enrolled in Section 1F Tuesday at 1:00-1:50 and I personally don't like that times slot because of scheduling problems. I would like to transfer in to Section 1K on Thursday at 11:00 but it's already full. Is it possible for me to get enrolled in the section? Thanks!!
Sorry, but you can't transfer. You can transfer to a full section only if there is someone who would like to make a switch with you.
hello dr. knobler. i am confused with how to solve number 26 on the handout. my problem for number 26 is what does 27.2% dissocation mean, what does the reaction equation look like? please help stear me in the right direction. do we get to use a sheet of paper with all the equations on them for the midterm and final?
The reaction is the reverse of the one we discussed yesterday: N2O4 ---> 2 NO2 27.2% dissociation means that if we started with 1000 molecules of N2O4, 272 would dissociate producing 454 NO2 molecules and leaving 728 molecules of N2O4 undissociated. You are told that the total pressure is 1 atm, which means that the partial pressure of N2O4 and the partial pressure of NO2 must add up to 1 atm. Given these pieces of information you can calculate the equilibrium constant for the reaction. I have not yet decided if you will be allowed to have a sheet of notes with you. If you have a sheet of notes, then I will provide no equations -- only constants and data. If you don't have notes, I will supply a number of equations -- some students prefer one form of test, others prefer the other.
Hi Professor, I was having trouble with problem number 20-6 in the handout. I was able to figure out all the parts except for the w and delta U. Can you please show or explain how to work those sections out on VOH? Thank You Very Much!
I won't give you the answer but I'll steer you into the right direction. Think carefully about the path in part (a) -- it is a path at constant external pressure (1 atm) and along the path the pressure of the system also remains constant at1 atm. What do you know about P,V work when a system is expanding or is compressed at constant pressure? (Check your lecture notes and Atkins-Jones -- you could look up work in the index if you don't remember where it is discussed.) Once you have w you should be able to calculate delta U because you already have q.
We have several questions from the handout. The first question in on problem 14. Would the entropy of ethanol at 50*C be the same as the entropy of vap. of ethanol at 50*C? Does delta H of vap. change with temp? When calculating delta S, is that supposed to be the difference between the entropy of vap. at 50*C and the entropy of vap. at 78*C? If we have other questions, we'll submit 'em later. Thanks. --Kasey Crettol and Ashley Dragoman
1. No, the entropy of ethanol at 50 deg C is not its entropy of vaporization. The entropy of vaporization is the difference beteen the entropy of the vapor and that of the liquid. 2. Yes, the entropy of vaporization changes with temperature. It decreases as the temperature is increased. 3.The delta S that you are asked to calculate is the difference between the entropies of the vapor and the liquid AT 50 DEG C, NOT the difference between the entropy of the vapor at 78 deg C and that of the liquid at 50 deg C.
I am just wondering if I ahould memorize all the equations you have thus far given us and will give us in the future. Are you going to give us a conversion sheet with some equations of none for the final exam? The reason I am asking is so that for this first week I can take a proper approach to the material, I certainly will memorize all equations if required.
There is no need for you to memorize all the equations. It is important, however, that you know in detail just when equations are applicable, e.g. if they apply only to an ideal gas, or only at constant temperature or pressure. I would expect that you know some basic equations, which should occur by repeated use: delta U = q + w, delta H = delta U + delta(PV), delta G = delta H -T delta S, q = c delta T, delta s = q rev/T.
Hi Professor Knobler, I was doing question 20 from chapter 16 and I couldn't find the standard molar entropy for S8, from part a. Could you tell me what this value is or the reason it is not included in any of the tables? thanks Kevin Claudeanos
The standard molar entropy for S8 is that for rhombic S, 31.80 J/Kmol. Unfortunately it is not identified until page 779 in Chapter 20.
Professor K, I'm slightly confused on the relationship between potential energy and distance between molecules. The text states (pg.590)that "squashing" atoms close together raises their potential energy. This makes sense to me because raising potential energy corresponds to lowering kinetic energy (so molecules move less). But in the handout (p418)) it states that a tendency for potential energy to be minimized corresponds to condensation. Doesn't condensation require less molecular movement (thus less distance between them)which would correspond to "maximizing" (not minimizing) potential energy to achieve the condensation? Am I reading/undersatnding this wrong?
Molecules do not only repel each other at short distances but they attract each other at longer distances. The attraction and repulsion just balance at one distance, which is a minimum (most negative value) of the potential energy. Thus, molecules at long distances can lower their energy by coming closer together. This is the energy that is given up when a gas condenses to a liquid. Unfortunately there is no plot of the potential energy against distance in Atkins. You can find it in most other books. If, for example, you check Principles of Modern Chemistry by Oxtoby and Nachtrieb, which you can find in the library, you will find such a curve on page 119. If you stop by at office hours I can show it to you as well.
test!
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