CHEMISTRY COMMUNITY

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It is cumulative, so everything is fair game.

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The TAs do not even see the final before its released. It is all created by professor Lavelle

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Clara Cho 2K wrote:Do you think we will still meet for discussion?

No discussions or classes of any sort will meet in person!

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At equilibrium does deltaG=0? If so why? Thank you!

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the reaction is correct either way. You can multiply by 2 if you want.

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it could possibly be heavy on thermodynamics and kinetics. That would be my guess. We were heavily tested on acids and bases already on the midterm.

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Do we include liquids like H20 in our rate laws? Or does the concentration of H20 not matter for the rate law? thank you

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Hello, I was wondering if anyone knew any helpful links to help understand pre-equilibrium that was mentioned in the text book 7C.3 and in lecture this Friday. How do we go about solving these questions??

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Can an intermediate always be in the rate law expression? Or is there certain instances where it cannot be placed in the rate law expression and has to be replaced with something other. Thank you !

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the rate limiting step, is the step that matches the rate law that was originally given. That would be your slow step.

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What does it mean to be a reducing or oxidizing reagent? I am also having problem understanding this true or false question: " A reaction with a negative standard reduction potential favors oxidation of the reactant" it is true. What does it mean to have a 'negative standard reduction pote...

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The Nernst equation, can be used for various things in electrochemistry. It can be used to calculate the concentration of a substance, maybe the pH, the cell potential, standard cell potential etc... However, i think the dead give away would be when they give you concentrations and/or pH's.

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Gibbs free energy under conditions of constant temperature and pressure! your delta G must be negative because that is the energy free to do work!

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Test 2 will cover all of electrochemistry and the second page of thermodynamics.

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you must use Kelvin because of the units in the gas constant R

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you should use K when the problem is at equilibrium, and Q when the problem is not specifically indicated to be at equilibrium.

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The salt bridge is there to balance out the charges to keep the electron flow moving within the cell.

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When working with a cell diagram, the left must have the anode (oxidation) and the right must have the cathode (reduction).

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yes the solutions manual is correct. Make sure you are doubt checking your negatives. When you take the ln of your concentrations it should be a negative value. Then you should multiply that negative value with (-RT/nF) which makes a positive .

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Specifically which videos from Dr. Lavelles website did you watch for help? Thank you

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another way you can think about it, that is personally easier for me is to use: Ereduction + Eoxidation = Ecell

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Hello everyone, I have a question on part A of this question because I do not exactly know what to do with redox reactions when they are at equilibrium. Are there specific rules? Or strategies when doing these? Thank you!

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An oxidizing agent is the substance that causes the oxidation in another substance. Common oxidizing agents include oxygen, hydrogen peroxide and halogens. A reducing agent is a substance that causes another substance to reduce. So to identify an oxidizing agent, simply look at the oxidation number...

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how do you identify in redox reactions the reducing and oxidized reagents? I having trouble with this, does anyone have any tips?

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Yes you should be able to cancel out the electrons when you combine both half equations.

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I had a question regarding when a reaction is spontaneous or not. I thought a reaction is only spontaneous when deltaG is negative. However in this question deltaG is positive and the solutions manual says it is spontaneous. Is there any exceptions to when a reaction is spontaneous or not ? Thank you.

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GFolk_1D wrote:Hi! Will there still be UA review sessions today despite the holiday?

Because of the holiday there will not be a UA session this week!

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Hi! To answer your question, when we talk about which molecule has more positional or residual entropy we have to take into account of how many microstates the molecule can take because molecules that have more microstates will have more positional entropy. When comparing these two molecules it is ...

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Hello everyone, I have a question on this specific homework problem. I do not get exactly why COF2 can be more disordered compared to BF3 just because of the different atoms it composes. It there a real explanation for this ?

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Yes the calculator should be fine!

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For this question, I am confused on why you add the the energy to melt the ice into the MCAT equation? I've never seen a question use this concept and just confused on it.

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for both parts you want to use q=ncAT. However, since the question says to treat it as an ideal gas, for the heat capacities the calculation will be different. For part a: for C you calculate it using c=(5/2*R). While for part b, you have to use c=(3/2*R). R being equal to 8.314

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I would check the constants and equations sheet from your previous test 1, that is what you will receive on the midterm. So whatever that is not on there will be what you have to memorize. I would make sure to know everything from lecture and the homework.

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Zoya Mulji 1K wrote:Did we go over the equation for work done by expansion when P is not constant? Is that the same work equation with the integral included?

I am pretty sure we have not went over that.

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for part B of this homework question, I keep calculating a different answer than what is on the solutions manual. For my calculation for work i am getting W= -3.18 , while the solutions manual is w= -326. I do not get why that is the solution on the manual if all our work is the same. Is this an err...

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On this specific homework question, the questions compares two final work values and asks which one is doing more work.
One answer was w=-122 and the other was w=-344. Why is the answer that is more negative (-344) doing more work? Even though it is a more negative answer?

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It is because Enthalpy only depends on initial and final values which make the reaction a state function.

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Kp is used with partial pressures like gases such as ATM. But Kc is used when you have things like aq concentrations.

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It will take place week 6. February 12th

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Ka and Kb values can also be given to strong acids and bases they will just be excessively big I think

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We wil be covering Outline 4, so I am sure anything from that section would be okay!

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Exothermic indicates that the heat is in the products, and you can treat heat as a molecule in a sense. So if the heat is increased then you can see that the reaction will shift to the left

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Within your ICE table, you should only include aq solutions and gases. If H2O is presented as a gas then you should still put it in the ICE table.

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Every question can possibly be different, so you can necessarily pinpoint down one equation. It just depends on the question and units.

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when your K value is less than 10^-3 you can assume that it is too small to affect your concentration.

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I wondered the same thing. I just thought the solutions manual to be wrong and wrote the equation as HClO2 + H2O ⇌ ClO2 + H3O+ After this, I calculated the concentration of H3O+ from the pH value. Now I'm trying to solve for the Ka but I was wondering, is 0.10 M of HClO the equilibrium concentratio...

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This question has created some confusion for me just now. HClO2 is a weak acid however, it seems to completely ionize in the solutions manual. A weak acid should contain a reverse reaction correct? Can somebody please explain why this question is worked out this way?

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you would get the ((change in concentration)/(initial concentration) )x100% and that is how you calculate it.

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A buffer solution is a reaction with contains a weak acid and its conjugate weak base in a reaction, as far as how much do we need to know about them? I am not exactly sure. I would check the learning outcomes on the website.

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If the moles are the same, then there will be no shift in the reaction.

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Last Monday's session Lyndon said he will not be having his session today due to the holiday.

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Yes, the 5% rule applies to both acid and base and chemical equilibria. They are essentially the same process.

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yes! just make sure everything is in SI units and they are all the SAME. ICE tables are so useful.

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Lavelle will post the young hall peer learning session on the website when they are ready. They'll probably cover recent material, it just depends on the UA!

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It will be during week 3 in discussion, Lavelle will probably go over everything that will be on the quiz friday in lecture.

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Most likely on friday professor will let us know everything we need to know for test 1.

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Q will also be unitless since it is calculated the same way as K. Just make sure you are using the correct units for molarity.

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I know CaO is a base, however is it a strong or weak base?

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You just have to assume the charges of the molecules or atoms. For cyanide you have to know that the charge is negative one so you would add the "O". Chlorine most likely also has a negative one charge. You have to learn these charges it is not really specifically given.

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Im pretty sure that has to do with the size difference between F and I. HI is a much bigger molecule than HF which leads to it being more acidic.

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I would just depend on the homework problems. Get all of them down and then begin asking your TA for a more specific layout of what you need to know. Homework problems are really helpful.

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For the final, would it be necessary to know how to draw out the hybridization diagrams?

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How I look at it when i am determining such things like monodentate or polydentate is the charges. On CO3^2- there are two oxygens with a negative charge and one without. Therefore, there can either be two places for a transition metal to attach since they are both negatively charged. Or the transit...

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I was wondering what the chemical formula for ethylenediaminetetraacetato (edta) would be?

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305416361 wrote:I think either one is fine, since we were taught both

Sounds good! Thanks

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gbernal1C wrote:Lavelle said to use the one that has the asterisk next to it but he also said in his email that either or is fine.

Thank you!

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When naming coordination compounds, do we use the original names or the new "IUPAC" convention naming?

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yes . Everything from week 1-10.

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These "induced-dipoles" are referring to the repulsions of the electrons in an element. They are instantaneous and all always happening . It is the weakest IMF but is involved with every element.

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Dipole- Dipole is a stronger IMF than LDF so that is why AsF3 has the higher boiling point. Yes AsF5 has higher LDF but it does not matter due to the Dipole-Dipole.

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Yes, we will be having lecture . Doctor Lavelle sent an email about it a couple days ago.

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Elizabeth Johnson 1I wrote:
105311039 wrote:It will cover material from the midterm and on!

ah wait so are you saying that the material on the midterm is fair game?

No, sorry like material from after the midterm and on!

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VSEPR stands for Valence Shell Electron Pair Repulsion theory. I do not think we need to memorize it but it would not hurt to actually know it.

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Hybridization is not on test #2 however, sigma and pi bonds are!

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TAs would be the best to head to first!

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Yes it should be treated as a whole region of electron density. It would still cause repulsion, however, I do not think it would be the same amount as a complete lone pair.

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Yes they are the same thing! However, Professor Lavelle uses Bent so I would stick to that!

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I do not think for now it matters what side the single electron goes on since we have not discussed shape yet.

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Yes! Draw every possibility of the bond. However, there will always be one "best" structure!

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In lecture on friday, Lavelle said that we will get it back on wednesday!

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The two exceptions are copper and chromium, the rest of the group under those two elements.

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Dipole occurs when two atoms are bonded, and they have a large difference in electronegativity. The most electronegative atom will pull the electrons more to one side of the bond causing a "dipole".

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It will cover material from the midterm and on!

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I think they will specifically ask for resonance structures when they want them.

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Yes, 1 midterm and we will have another test and a final.

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Formal charge helps you determine whether you have drawn the best structure possible, but you should not have to necessarily write it out unless it is asked.

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Double bonds are closer because they have a stronger bond, so they pull the elements closer together.

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It is more stable to have 10 electrons in the D orbital, rather than 9 in the D orbital and 2 in the S orbital, therefore you move one electron to the D orbital to make it as stable as possible.

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Both sections for this week should be okay!

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Can someone explain to me why the quantum numbers for 1E.9 part B are not possible? I understand part A and C, but something about B is really throwing me off. Thanks so much! - For this question 1e.9 part B is not possible because of if the Ml value. Since l is equal to 0, Ml can only also equal 0...

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How many orbitals can have the following quantum numbers in an atom: (a) n=2, l=1 (b) n=4, l=2, ml= -2 (c) n=2 (d) n=3, l=2, ml= +1? For (b) and (c), how does the magnetic quantum number tell you how many orbitals there are? I don't understand how to find the number of orbitals with ml. -Ml specifi...

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The electron spin is all determined by the quantum number Ms. There are two possible values for Ms which are 1/2 and -1/2. If Ms=1/2 then the electron is spinning upward, however if the Ms value is -1/2 the electron is spinning downward.

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Yes! The 4s electrons are removed before the 3d electrons!

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This is due to Hund's rule, which was mentioned by Dr. Lavelle in lecture.

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Its the distance between two touching atoms and their nuclei.

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There can only be two electron spins, which depends on the Ms quantum number.

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It should be the first equation!

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Yes! The speed of light is constant! And will never change.

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Possibly in your discussion this week!

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Yes, its recommended to watch them, but not required!

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