CHEMISTRY COMMUNITY

AmyHo2K

My thinking is that the slowest step is the rate determining step because if the slowest step is not completed, the reaction can't be completed no matter how fast the first step is.

AmyHo2K

I think we leave the catalyst out of the overall rate law, but I could be wrong. My thinking is that we don't include it because it's not being consumed in the reaction.

AmyHo2K

I enjoyed the topics from 14B more personally, especially kinetics and chemical equilibrium.

AmyHo2K

I am going to UA review sessions, doing textbook problems, and reviewing the problems we did for hw on sapling. Sapling is nice because it provides an explanation on how to do the problems.

AmyHo2K

The final is going to be cumulative and if it's anything like 14A, the number of questions for each topic would depend on how much time we spent on it.

AmyHo2K

The average rate is the rate over the total time of the reaction and the reaction rate is the rate of the reaction at specific points (think of derivatives).

AmyHo2K

I'm a part of Association of Chinese Americans (ACA) and SEACLEAR, but I'm hoping to join more organizations in the Fall.

AmyHo2K

I'm pretty sure the question has to tell us because the result from it being in an acidic solution would be different from what you would get if it was in a basic solution.

AmyHo2K

I feel like I need more context for this question, but depending on the question, you could be asked to calculate k. Kate's week 9 worksheet had a question regarding that if you're interested.

AmyHo2K

I think we would be given reaction data to determine which order of reaction it would be. The first order reaction plots ln[A] vs time and leaves us with a straight line with a negative slope and the second order reaction plots 1/[A] vs time, leaving us with a straight line with a positive slope. Wh...

AmyHo2K

The moles are the moles of electrons being transferred. Make sure to put in the number from the balanced reaction.

AmyHo2K

If retyping it in does not work, try logging out and logging back in. I heard other people had the same problem and that fixed it.

AmyHo2K

We add OH- to balance basic solutions and I think it's coefficient is the same as the one we used for H2O on the other side of the equation.

AmyHo2K

K is only used when the reaction is at equilibrium and Q is when it's not. Be sure not to confuse it with each other because it could get you wildly different answers.

AmyHo2K

How come when doing log(Q), we are allowed to do log (partial pressure/molarity)? Since one of them is in atm and one is in mol/V (Molarity), shouldn't you have to convert one or the other so that the units are the same? I saw on another post that it's because pressure and concentration are in stan...

AmyHo2K

Standard conditions are the conditions where temp is at 25C and pressure is at atm, etc. They will have different equations and would therefore get you to different answers.

AmyHo2K

They should be pretty similar unless your TA switches sections with another TA because of a possible time conflict.

AmyHo2K

Yes, I think so. I remember doing problems on it around the end of section 5.

AmyHo2K

Hi, I believe that only gases work for this law.

AmyHo2K

I would recommend going to workshops and step ups for their worksheet, but if you're not able to, a lot of UA's have folders with their worksheets or you can just ask someone to send the worksheet to you.

AmyHo2K

Spontaneity depends on a negative deltaG value so increasing the Gibbs free energy would not lead it to be spontaneous unless the absolute value of deltaH is still smaller than the absolute values of TdeltaS.

AmyHo2K

They can't be negative. To my understanding, it is because you can't take away work and energy to increase the temperature.

AmyHo2K

ΔG° is Gibbs free energy under standard conditions which is in a system where it's 25 degrees C and 1 atm, while ΔG is not under the standard conditions.

AmyHo2K

I'm planning to take Chem BL, LS 7A, and Stats 13. Hope I get them though since Chem BL and Stats 13 are both impacted.

AmyHo2K

I am planning to take 14bl in the fall to get lab experience and I also think getting through the labs in person might be more time efficient and easier. However, I heard the workload for the virtual class is lighter.

AmyHo2K

I am pretty sure that the letter grade cutoffs are the same standard ones listed on the UCLA websites.

AmyHo2K

I have the same issue focusing during test time too, with siblings and family in the house. I ended up moving rooms to create more of a testing environment.

AmyHo2K

If you do 50 by the end of the quarter, you should be fine. If you get 25 by the end of the week, I think your grade should turn to a 25/25. If not, you can probably ask your TA about it.

AmyHo2K

I feel like UA sessions and doing textbook problems help a lot. Also, if you're confused on a specific concept, you can look up Youtube videos on it. I really like The Organic Chemistry Tutor.

AmyHo2K

You can go to any TA's office hours and ask if they can review it with you. I think you ask them any week, not just this one.

AmyHo2K

It usually depends on the week, but I usually go to 3 to 4 UA sessions a week and try to do textbook problems every weekend so I would say 16 hours is pretty accurate. You might not need 16 though. It depends on skill set and I find that doing a lot of practice problems help me so that's how I study.

AmyHo2K

You can find the bond enthalpies using the formula others have mentioned. I find it really helpful to draw out the Lewis structures first and then count how many there are each of each bond then add or subtract the enthalpies using the table given.

AmyHo2K

We definitely don't need to memorize them but we should be able to find it though other values given such as Kw, pka, and pkb.

AmyHo2K

Once you download it, it should be good to go. Make sure to check if it's fully updated before your exam!

AmyHo2K

Bond enthalpies should be provided to us. I went to a couple of UA workshops where they discussed questions like this and they provided the table in the question.

AmyHo2K

I think he will send out an email soon about what would be on the test because that's what he did in 14A. Also, I heard that week 4 material shouldn't be on the test.

AmyHo2K

Kelly Singh wrote:
Alara Aygen 3K wrote:Doing the problems mentioned in the outlines and attending the UA workshops really help!

Which UA workshops are most helpful?

I normally go to Kate's or Michaels' workshops. Justin is also super helpful, but I haven't been attending his this quarter because of a time conflict.

AmyHo2K

When the partial pressure of the reactants increase, then the system will shift forward and backwards when the pressure of the products increase to balance itself. Some people pretend the pressures are on a scale, but it's not a super accurate way of thinking about it.

AmyHo2K

I think I remember hearing from my TA that the content for the first midterm cut off would be tomorrow.

AmyHo2K

Because pKa + pKb = pKw, and at 25 celcius and pKw = 14, pKw is pertained for both.

AmyHo2K

Endothermic reactions require heat so adding heat can cause it to overcome the activation energy. The equilibrium would then shift towards the left or to the products' side.

AmyHo2K

We don't exclude H20 when it's a gas. We only exclude it when it's a liquid or solid.

AmyHo2K

Madeline Louie 1B wrote:I understand why water is neglected when its in excess, but will we be told it is in excess or do we assume it is?

I'm pretty sure that we just assume water is in excess.

AmyHo2K

Hi, I also realized I needed to do better at studying so something I am trying to do is to do all the textbook problems on the outlines after we went over all the content during class.

AmyHo2K

Hi. The midterms are on Jan. 29th and Feb. 19th during class time.

AmyHo2K

You are switching the reactants and products of the forward reaction, and when you put this into the K equation, you would end up with the reciprocal of the first recation.

AmyHo2K

M is molarity or moles per liter.

AmyHo2K

It really depends on you and how much you understand the material. I usually like to go 3 to 4 times a week which may seem like a lot, but they're always extremely helpful, especially before exams.

AmyHo2K

The second letter is just to describe that the reactant quotient is measuring. Qc is used for molar concentration and Qp for partial pressure.

AmyHo2K

In lecture today, Professor Lavelle said since their values are so similar, we would be approximating 1 bar as 1 atm.

AmyHo2K

I had the same question and the responses were helpful. It makes sense why they're not require to have an octet.

AmyHo2K

Electronegativity is lower down a group so there would be more of difference between Ba and Br.

AmyHo2K

Yes, I would start by getting familiar with common strong and weak acids and bases.

AmyHo2K

Having the additional Oxygen would make the acid more stable because the charge would be spread out more and therefore stronger.

AmyHo2K

Yes, and I'm not exactly sure. Maybe it's less strong than the other strong acids so others might consider it a weak acid.

AmyHo2K

Strong bases normally have metals from the first two groups.

AmyHo2K

It might be a good idea to memorize it because knowing whether common household items are bases or acids was one of the objectives in the outline.

AmyHo2K

It would just be chloride. You name the ions outside the brackets normally with no prefixes.

AmyHo2K

I'm ready for this quarter to be over and for Winter Break. Otherwise, a little stressed about finals.

AmyHo2K

I get really bad test anxiety too and I found that doing jumping jacks helps a little with getting rid of the jitters.

AmyHo2K

To find the coordination number, I look to see how many ligand bonds are formed with the transition metal.

AmyHo2K

I'm pretty sure we have to memorize it. The ones we are expected to know are on a worksheet on Lavelle's website.

AmyHo2K

I think there is a bigger range for passing and if the class average is low enough, the grades would be adjusted.

AmyHo2K

I had the same question and this helped. I do wonder if it's curved this year though because the class averages seem pretty high.

AmyHo2K

It should be cumulative.

AmyHo2K

I feel the same way and it gets worse when I think about how badly it might drop my GPA. :(

AmyHo2K

It should be all multiple choices. Does anyone know if it's going to be all or nothing again?

AmyHo2K

This made me realized I'm not the only one worrying about grades. Also, I am so lost about where my grade even stands.

AmyHo2K

I had the same question and the responses were super helpful!

AmyHo2K

We look at intermolecular forces because we want to see the interactions between the molecules.

AmyHo2K

I usually start out by drawing the Lewis structure to see how everything is bonded then seeing what kind of interactions each atom has with another.

AmyHo2K

They can be expanded past 10 when they're in the f block.

AmyHo2K

HilaGelfer_3I wrote:Hi!

I attached the link with the rules to finding oxidation numbers that sapling created. I hope this helps :)
https://sites.google.com/site/chempendix/oxidation

I had the same question and the link really helped me understand some of the rules better. Thank you!

AmyHo2K

In AsF3, there are dipole-dipole interactions which increases the boiling temperature. AsF5 have no dipole-dipole interaction so the boiling temperature is lower.

AmyHo2K

To find the oxidation number of ClO4- you can use a simple equation. Assuming oxygen will have a (-2) oxidation number and knowing the total charge of the molecule is (-1) you can write the equation Cl+4(-2)=-1. Solving for Cl gets you Cl=7, which means the oxidation number for Cl is +7. I apologiz...

AmyHo2K

Nitrogen is only in period 2 so it's not able to have an expanded octet.

AmyHo2K

It's because Hydrogen can only bond with Nitrogen, Oxygen, or Fluorine for it to be a Hydrogen bond. In this case, it's Oxygen.

AmyHo2K

The oxidation number is 7. 2 is from the O, and 5 is from Cl. The structure that's most plausible is determined by which structure do Cl have the largest formal charge in.

AmyHo2K

He doesn't say specifically so I would just assume that you take your total points you got, divide by 100, and the percent would translate to "normal" grading methods.

AmyHo2K

I had the same question because I wasn't sure when Professor Lavelle went over this in his lectures, and this helped a lot.

AmyHo2K

I had the same question and I hope our TAs would go over the problems. I have a good idea on what I got wrong, but want to know to be sure and to better prepare for the next midterm and final.

AmyHo2K

As others had said, Group 13 elements don't require an octet since they would require too many electrons to get one which is unrealistic.

AmyHo2K

Lewis acids can accept electrons and Lewis bases donate electrons. To determine whether an ion or molecule is one or the other, you can look at the valence shells of the ion and molecule.

AmyHo2K

I was confused about the same thing. Thanks for asking this question.

AmyHo2K

I used to watch Netflix, but stopped because I would get hooked and fall behind on schoolwork. Now, I just call my friends and see how they're doing.

AmyHo2K

For my notes I write cursive v for frequency and normal v for velocity. The equation sheet does a similar thing where v for frequency is italicized.

AmyHo2K

Yah. I was wondering the same thing. Does anyone know? I think that the partial points were to be given in questions that had multiple parts (i.e. questions that are divided into parts a, b, c, etc.) but I don't remember any of the questions on the midterm being like that so I doubt that we will ge...

AmyHo2K

Sad to hear that there won't be any more audio-visual learning topics. They were super helpful especially for studying before tests.

AmyHo2K

For Lyman, it means the final level of energy would be 1 and for Balmer it means the final level of energy is 2.

AmyHo2K

I would review the modules because I personally found it really helpful. If rereading helps you sure, but I think rewatching lectures and doing the practice problems might be a better way to use your time.

AmyHo2K

This helped a lot. I was under the impression that all gas molecules had the 2 in it.

AmyHo2K

Studying has been hard for me too. I usually watch the lectures, take notes on them, go to workshop, and then attempt the homework. If I'm still lost, I go to step-up. They also give out worksheets which was super helpful.

AmyHo2K

I wouldn't worry about it. Dr.Lavelle said it was just something to help us understand the concepts we would later learn.

AmyHo2K

Hi! So the energy of the incident light is equal to the sum of the kinetic energy and work function. The work function is given to you (9.83*10^-19J). So you solve for the kinetic energy which is 1/2mv^2. After plugging in your velocity into that equation you should add that plus the work function ...

AmyHo2K

I attended a workshop and left early so I'm not really sure what to do for the next steps of this problem. An electron from zinc has a velocity 4.0×10^6 m/s. If 9.83×10^19 J of energy are required to remove an electron from a zinc atom, what was the frequency of light incident onto atom? So far, I o...

AmyHo2K

Hi, I recently got an Ipad and so far, I think it's better than paper notes. I use Notability to take notes and it's working well for me. For reading classes, you can import readings into the app and write in the margins. I also use it during Step-Up sessions and workshops when they would give us wo...

AmyHo2K

Hi, you don't need to sign up for them. You just click onto the zoom link during the session. There's no worksheet you need to print out beforehand. They usually post a google doc link in the zoom chat when you get there.

AmyHo2K

No, because electrons want to go back to their stable state which is their ground state.

AmyHo2K

Generally, we would need to use Avogadro's number when a question ends with "..remove one electron.." because you're trying to find the needed energy for a single electron instead of a mole of electrons.

AmyHo2K

I would say to do a couple and see how well you do on them. If you feel like you're struggling, you could do more textbook and worksheet problems.

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