CHEMISTRY COMMUNITY

Jagveer 1I

I think that we have until this Sunday to complete our 50 chemistry community posts. So, by Sunday at midnight we should have at least 50 posts.

Jagveer 1I

I believe you are spot on, and also thank you that is a useful concept to keep in mind! I realized that if the cell is at equilibrium, nothing is spontaneous...it's just at equilibrium. If I said product conc at equilibrium is favored and not the forward reaction at equilibrium would that be correc...

Jagveer 1I

I believe you are spot on, and also thank you that is a useful concept to keep in mind! I realized that if the cell is at equilibrium, nothing is spontaneous...it's just at equilibrium. If I said product conc at equilibrium is favored and not the forward reaction at equilibrium would that be correc...

Jagveer 1I

The equation for Ecell is Ecell=-delta G/(nF), so if the reaction is multiplied by 2, delta G double, but the value of n also doubles, therefore they cancel each other out and Ecell is the same value as when the reaction was not multiplied by 2. The textbook also goes over this and provides examples...

Jagveer 1I

You are understanding the question correctly, we need to balance the charges on both sides of the equation. On the left side of the equation, the charge of water is 0 and the charge of S2O3 is -2, so the charge on the right side of the equation must equal -2. On the right side, the charge of 2 SO4 i...

Jagveer 1I

I am also going to do the textbook problems and go to all the review sessions that I can. Since there are so many textbook problems, I am going to focus on the topics that I am a little unsure on and only do a couple of problems from the topics that I already understand. I only went to one review se...

Jagveer 1I

From my experience in 14a, the final was similar to the midterm in its format since it had a couple of textbook problems on it and I am pretty sure that this final will be the same since on the syllabus he said that textbook problems will be included in the exams.

Jagveer 1I

To add on, the exam time for international students is from 7 am to 8:30 am and more information about regarding this can be found on Dr. Lavelle's website under the exam schedule tab.

Jagveer 1I

Chem_Mod wrote:Which question?

I think that this person is referring to part d of 6L.3. I was also confused on this.

Jagveer 1I

In order to solve this question you would need to know how to solve for oxidation numbers. For Cr2O7^2-, you know that O has an oxidation number of -2, so in this molecule O7 has a charge of -14. Since the total charge is 2-, this would mean that the charge of Cr2 is +12, and that of Cr is +6, since...

Jagveer 1I

Some TAs are also very accommodating and allow you to schedule a time to review your test that works for both you and your TA if you cannot attend their office hours, so you can also contact your TA and see if that is an option.

Jagveer 1I

https://www.youtube.com/watch?v=j0hI-a6EWWo
To add on, I found this video very helpful, he goes over the main rules for oxidation numbers and provides lots of examples to reinforce the ideas.

Jagveer 1I

GER is used for reduction so its leo the lion says ger (Gaining electron=reduction).

Jagveer 1I

I also noticed that, but I think that he just misspoke since around 30:30 he says that since it is a negative voltage it is non-spontaneous.

Jagveer 1I

To add on, since our TAs are the ones in charge of chemistry community grades, it is probably best to double check if they are okay with 50 total chemistry community points by the end of the quarter or if they want you to do at least 5 posts per week.

Jagveer 1I

The equation delta G naught=-RT*ln K and the Gibbs free energy equation is used to derive the van't Hoff equation. Topic 5J.3 goes into detail on how to derive the equation if you want to check it out for a detailed explanation.

Jagveer 1I

The wording of when the homework is due may be confusing since the work week usually ends on Friday, but homework is usually always due the Sunday following the week when it is due. As was seen last week, Dr. Lavelle is very accommodating and will extend the homework deadlines if he feels the need t...

Jagveer 1I

I have another midterm to do but I am definitely going to treat myself by watching lots of tv, taking lots of naps, and eating lots of junk.

Jagveer 1I

So for this problem I think that the solution manual is purposely exchanging the units of delta T from C to K to show us that the degree of change in temperature is the same in C and K. This is true because K=C+273.15, so if the temperature is changed by 1 C, then it is also changed by 1 K as well. ...

Jagveer 1I

This problem is mentioned in the solution manual errors located on the class website, but the right answer is -2.7 kJ/mol not -2.469.5. For Q I got 18, for delta G naught I got -12350 J/mol and then for delta G I got -2737 J/mol. So, maybe you typed something into your calculator wrong.

Jagveer 1I

For this question, you need to use the equation delta S=Cv * ln(T2/T1). However, we cannot use the Cv given in the problem directly since we need the heat capacity for the number of moles of gas. In order to do that, as you were doing, we need to use PV=nRT to solve for n, the number of moles. Since...

Jagveer 1I

If a system is at thermal equilibrium, the temperature of the system and the surroundings is the same, so there is no net movement of thermal energy between the system and surroundings. So, if the temperature of the system and surroundings are the same, then you know that thermal equilibrium has bee...

Jagveer 1I

To add on, in an irreversible pathway expansion occurs rapidly, so the temperature is dropped and therefore it is not an isothermal pathway. On the other hand, the reversible pathway is slow, so it is at thermal equilibrium with its surroundings and therefore the temperature does not change.

Jagveer 1I

For this problem, we need to note that entropy is a state function, so we can add up entropies from many steps to get the total entropy. To solve this problem we need to break the problem into 3 steps: heating the liquid from 34 C to 100 C; allowing it to vaporize; cooling the vapor back to 34 C. We...

Jagveer 1I

For this problem, we need to note that entropy is a state function, so we can add up entropies from many steps to get the total entropy. To solve this problem we need to break the problem into 3 steps: heating the liquid from 28 C to 100 C; allowing it to vaporize; cooling the vapor back to 28 C. Yo...

Jagveer 1I

Yeah, he talks about Cp and Cv in lecture #9.

Jagveer 1I

For this question they want us to find the final temperature, so we will need to use the equation that you described to find it. We need to first calculate the moles of ice and then multiply that by the delta H fus of water and this calculation gives us the heat required to melt the ice, so the enth...

Jagveer 1I

To use delta H fus, we need to first calculate the moles of ice and then multiply that by the delta H fus of water and this calculation gives us the heat required to melt the ice. The way that you would set up your equation would: enthalpy of fusion of water, which is the value that was just calcula...

Jagveer 1I

For this question we need to first find out how much heat is released by compound A, so we need to convert the mass to moles and then multiply that by the heat of combustion to get the kJ of heat released, I got 35.58 kJ using your numbers. Now we have to use C=q/delta T to find the heat capacity of...

Jagveer 1I

You are correct in regards to the equation being delta U= q+w, so w=-P * delta V, which is why the overall equation was delta U= delta H -P * delta V. The w is not being subtracted, it is just a negative value.

Jagveer 1I

Yes, you are correct. For the delta H we just do the same manipulation that we did to the chemical equation, if we multiply the chemical equation by 3 then we multiply the delta H by 3 as well. As you stated, for K, when a chemical is manipulated then K is changed by K^manipulation. If the chemical ...

Jagveer 1I

Yes, you are correct, topic 6E also goes over this in detail and a very good example is 6E.1. Since it is a part of the outline, it may also be covered in the test so it will be good to go over it. Hope this helps.

Jagveer 1I

I think that there will be certain answer choices that are not the correct answer, but are very close to the answer. For example, maybe providing the [H3O+] instead of [OH-]. Not completely sure how this will work but this was how I thought that it will go.

Jagveer 1I

Yes, Professor Lavelle just sent out an email that has instructions attached that says that we need a primary device to do the test on and a secondary device for zoom. So, it is the same format as Chem 14a.
https://lavelle.chem.ucla.edu/wp-conten ... dterm1.pdf

Jagveer 1I

It is something different, on CCLE, there is a MIDTERM 1 section on the left hand side above Week 1 and that has all of the links for the midterm. You just use the one for your lecture and discussion section, I think that the TAs will have all of their discussion sections in one zoom link.

Jagveer 1I

To add on, for 14A, Dr. Lavelle modeled the questions on the midterms and final after the textbook problems mentioned in the outline. So, that is a good outline of what is expected for the exam. The outlines are on his class website and they are listed under learning outcomes. As already stated, the...

Jagveer 1I

The textbook uses K to refer to Kp, and Kc is used to specifically refer to equilibrium constants calculated using concentration. So, although K can be used to represent either Kc or Kp, the book only uses it for Kp.

Jagveer 1I

Some UA workshops go over past midterm problems as well, so that may help with what is expected of us for the midterm. I find the problems in the outline the most helpful because Dr. Lavelle based his midterm and final exams off of those textbook problems last quarter and some questions were the exa...

Jagveer 1I

To add on, steam causes more severe burns when it comes in contact with our skin compared to when water at the same temperature as the steam comes in contact with our skin because, as seen in the diagram provided in lecture, more energy is released by the steam, around 48 KJ/mol, when it contacts ou...

Jagveer 1I

I think that the lecture will be posted at 11, he did not say anything about there being no lecture today.

Jagveer 1I

On the chem 14 b website there is a solution manual errors link and that states that the there is a typo for 6A. 19 part c and that 3.1 x 10^-3 mol/L should be used. I included the link to the solution manual errors below. https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Solution_Ma...

Jagveer 1I

The initial concentrations of the reactants are given, so we are able to make an ice table for this problem. Using the table, at equilibrium [SO2]=0.522-2x, [O2]=0.633-x, and [SO3]=+2x. Since K is very small we can simplify x when solving for it and this gives us x being equal to 4.94 x 10^-6. We th...

Jagveer 1I

Yes, I think that he was just showing us how we could solve them, but we won't have to for this class.

Jagveer 1I

Yes you are correct. Since the equation for K is [Products]/[Reactants], very small and very large values tell us whether products or reactants are favored. When K is very small, the denominator, Reactants, is much larger than the numerator, Products. Therefore the reaction lies to the left. The opp...

Jagveer 1I

I think that only Dr. Lavelle or the TAs can upvote responses.

Jagveer 1I

To add on, K is the equilibrium constant and Q is the reaction quotient. K has a specific value and is calculated using either the concentrations or partial pressures at equilibrium. On the other hand, Q can be calculated at any time during the reaction, not just at the equilibrium.

Jagveer 1I

The moles are given in mmol and the volume is 0.5L, so the concentration needs to be calculated with that information accounted for. I got [SO3]=2x10^-4, [SO2]=0.0024, [O2]=0.001. After using those values I got the answer of 6.9. Hope this helps.

Jagveer 1I

Some of the problems on the outline need you to calculate the molar concentration and since the exams are similar to those problems, we may need to calculate it on our own.

Jagveer 1I

I think that if K is less than 10^-4, then the calculation of x can be simplified. But, this maximum value of K changes when talking about acids and bases.

Jagveer 1I

The homework is due Sunday of week 2, there is a trial that you can use for Sapling to complete the homework. I used it last quarter and it saves the work that you complete during the trial once you have the access code. To complete the homework, Sapling is required.

Jagveer 1I

I also got the same answers!

Jagveer 1I

In topic 6a of the book, it states that metals typically form basic oxides and nonmetals typically form acidic oxides and the elements on the diagonal border between metals and nonmetals have acidic and basic character, they form amphoteric oxides. So, I use the periodic table to determine whether s...

Jagveer 1I

I also think that it is classified as a strong acid since Dr. Lavelle did not specifically tell us that we should classify it as a weak acid.

Jagveer 1I

When the pH is less than the pKa, the acid will not protonate, but when the pH is greater than the pKa, the acid will protonate. So, for any ionization to happen the pH has to be higher than the pKa. Hope this helps.

Jagveer 1I

I am having the same problem. Even when I connect to the UCLA VPN I still get either the error message or buffering, so I am not sure what the problem is either since connecting to the VPN usually fixes these problems for me.

Jagveer 1I

To add on, for the molecule to rotate about the double bond, the π-bond would need to break and then reform. It is not possible for a π-bond to remain intact and rotate.

Jagveer 1I

Both UA and step-up sessions are great ways to get extra help and practice, but step-up sessions are more detailed and slower paced than UA sessions. If you need more help in understanding a concept, step-up sessions are recommended since they go step by step through explanations.

Jagveer 1I

Yes you are correct that prefixes are not taken into account when looking at alphabetical order, so we can ignore tetra and di. Since it is ammine and aqua and m comes before q in the alphabet, ammine is before aqua. Hope this helps.

Jagveer 1I

In the textbook it was mentioned that for a molecule to rotate about the double bond, the π-bond would need to first break and then re-form. So, I believe that this is the only way that the central atom can rotate with π-bonds.

Jagveer 1I

Just to add on, a polydentate is a ligand that can attatch to more than one binding site simultaneously and the prefix used is bis-, tris-, or tetrakis- to define that polydentate. An example of a polydentate is bidentate ethylenediamine, where each end of the molecule has a nitrogen atom with a lon...

Jagveer 1I

Yes, the decimal point makes the first one 2 sig figs and the lack of it makes the second one one sig fig.

Jagveer 1I

Yes, generally the stronger the intermolecular forces are, the higher the surface tension is.

Jagveer 1I

I found this information in the textbook under topic 2D. The electronegativity is the average of the ionization energy (I) and electron affinity (E_a) of the element. So, the formula given was X=(1/2)(I+E_a). Hope this helps.

Jagveer 1I

I think that we will have a lecture on Friday since we had lecture on Veteran's day, but I am not completely sure on that. For discussion sections, it is probably best to email your TA and check in with them to see what their plans are.

Jagveer 1I

CH2Cl2 is polar because Cl is more electronegative than H, therefore pulling the electrons towards it. Since the electrons are not shared equally by the atoms, there is a dipole moment that does not cancel and therefore the molecule is polar. On the other hand, CH4 is nonpolar because all of the bon...

Jagveer 1I

Professor Lavelle has not taught us about any polarizability trends, but the general rule is that the bigger the molecule, the bigger the polarizability. So, I would assume that the polarizability trend would be similar to the trend of atomic radius, decreasing across a period and increasing down a ...

Jagveer 1I

Yes you are correct. The electrons in the s-orbital are very close to the nucleus, which is why they are able to penetrate through the inner shells. On the other hand, electrons in the p-orbital do not penetrate the nucleus as much as electrons in the s-orbital because they have orbital angular mome...

Jagveer 1I

Polarizability is the ability to which a species of electrons can be distorted. If they can be easily distorted, they have a higher polarizability. Polarizability depends on the size of the molecule and therefore the number of electrons. The less tightly the electrons are held, the more polarizable ...

Jagveer 1I

I have recently started watching Superstore and it is very funny and reminded me of The Office.

Jagveer 1I

There is no designated review session, but worksheets and other resources have been posted that are helpful review materials.

Jagveer 1I

We are not expected to know that. The questions on the midterm are going to be modeled after the problems given in the syllabus and 1D#5 was not on the syllabus, so I do not think that we will be quizzed on it.

Jagveer 1I

What should you do is no one laughs at your chemistry jokes?
Keep telling them until you get a reaction.

Jagveer 1I

I have lots of distractions at home, so the best way that I can study from home is to completely isolate myself. I take only the things I need for schoolwork and head to my room and do my work. I also occasionally take breaks so that I do not burn myself out.

Jagveer 1I

I am also worried about my grade and I am not feeling as confident for this midterm, but I am constantly calculating my grade and this kind of destressed me. The final will probably be the most stressful since all of the other grades are final and our total grade solely depends on how we perform on ...

Jagveer 1I

I always panic whenever I click the start button, so the first thing I do is take a deep breath and then I try to calmly proceed and slow down so that I do not miss something or misread the question. Whenever I do not understand a question I move on and come back to it when I feel ready.

Jagveer 1I

Is purchasing a hard copy of the textbook worth the $190? I have difficulty viewing the textbook well on Sapling's site and like have a physical copy, but it is expensive and if it helps with the exams it would be worth it. I have only used the e-book and it has been working for me, but if you woul...

Jagveer 1I

I don't think that we have to draw the resonance hybrid for compounds, but we should understand what resonance hybrids are and how resonance structures relate to resonance hybrids. We should draw the resonance structures of compounds because they are easier to draw since we do not have to account fo...

Jagveer 1I

Samin Kabir wrote:When do we get the midterms back?

I have heard that the scores will be released this Thursday, but I do not know if we will be able to see our responses for the midterm.

Jagveer 1I

AlyssaMaynard1C wrote:Will each question be all or nothing for points? Or how could points be awarded if the answer was almost there?

My TA said that partial credit would be given on some questions, but I am not too sure about how many questions have the possibility of partial credit.

Jagveer 1I

lauren_tran_3J wrote:What is the day/time in PST when participation is counted? Thank you!

Participation is counted on Sunday at 11:59 pm PST.

Jagveer 1I

The [Ar] is the closest noble gas to Cr that has a lower energy than Cr. He did that so that you do not have to write out the electron configuration up to [Ar], so it was just to save time and make the process easier. He just showed us a different way to write the electron configuration for Cr and h...

Jagveer 1I

The amount of spectral lines corresponds to the amount of transition that an electron goes through. If it is at the n=4 energy level, then it can have three spectral lines since it can transition from that energy level to the 3, 2, or 1 energy levels. So, each transition produces a spectral line.

Jagveer 1I

From what my TA told me, the midterm is going to resemble the textbook questions, so I do not think that questions will be specific to the lectures. The lectures cover what the book has, but they contain more examples and Professor Lavelle clarifies what we are expected to know and what we do not ne...

Jagveer 1I

I also tried to un-randomize sapling and I literally clicked all the buttons, but I couldn't figure it out, so maybe implementing this change is on the teacher's end.

Jagveer 1I

3d has less energy than 4s. But, when there are elements that are not in the 3d block and only are in 4s, so calcium and potassium, then we do not write 3d, but when the element is in the 3d block, then we do write the 3d first and then 4s after. Hopefully this helps, but my wording may be a little ...

Jagveer 1I

I am also struggling with this topic in photoelectric effect, is there more review in the book? or is the sapling homework a better resource? The photoelectric effect is mentioned in the book under topic 1B, but I found the audio-visual topics for the photoelectric effect very helpful. This is loca...

Jagveer 1I

Does anyone know if we'll also be given the wavelengths corresponding to the different regions of the EM spectrum, or will we need to just know it? I think that we might need to remember the wavelengths corresponding to visible light, but that's about it. However, it might be helpful to know the or...

Jagveer 1I

Photons do not have a rest mass, but they do have momentum. So the De Broglie equation does work for photons since they have a momentum and that is the requirement for the equation to apply.

Jagveer 1I

He was just trying to show that from E=mc^2, you could get E=pc since p=mv and v=c as you stated, so E=m*c*c, or E=pc.

Jagveer 1I

Is the standing circular wave model that shows waves out of phase (picture in lecture slides) inaccurate? Yes, it is inaccurate because that model shows what it would look like if you tried to model the electron in between energy levels, so you will get a circular standing wave that is not stationa...

Jagveer 1I

I think that we will be provided with the frequencies/wavelengths, but we should probably know the wavelengths of visible light.

Jagveer 1I

For the wave model, the wave's amplitude represents the intensity of the wave, so according to this model, a wave with a higher amplitude has more energy to eject electrons. However, the photoelectric effect does not support this wave model and supports the particle model instead, where light is mad...

Jagveer 1I

In today's lecture professor Lavelle said that if the energy of the photon is not greater than or equal to the threshold energy, then the metal still heats up, but no electrons are emitted. So some of the photon's energy is converted to heat energy.

Jagveer 1I

Yes, for example, doubling the intensity of the light is doubling the number of photons that make up the light. However, doubling the intensity of the light does not change the energy of the photon, it just changes the number of photons. You are correct about what changing the wavelength means.

Jagveer 1I

The kinetic energy can be zero if the energy of the photon exactly equals the threshold energy. This would make it so that the electron is emitted, but there is no excess energy, or kinetic energy. Hopefully this helps.

Jagveer 1I

I also had to search up the formula for the structure and I think that we will be provided with the formula because of where we are in the class right now, but I am not completely sure if we would need to know the formulas in the future for this class.

Jagveer 1I

So for this question they are providing you with the threshold energy, 3.607 x 10^-19 J, and the kinetic energy, 0 J. So using these givens we could solve for the energy of the photon by using the equation E(photon)-E(threshold energy)=kinetic energy, which in this case is just 3.607 x 10^-19. Now y...

Jagveer 1I

Yes, the light is acting like it is made up of individual photons instead of acting like a wave and it acts this way for both ineffective and effective light. Also, not all photons are equal, the amount of energy that each individual photon has depends on the light source. The light sources with sho...

Jagveer 1I

That is a great question. In the photoelectric experiment they observed that increasing the intensity of the light did not eject electrons and they concluded that light was not acting like a wave. Instead it was acting like it was made of individual photons, where the individual photon's energy must...

Jagveer 1I

Hi Jamie, I feel the same way and I am gradually getting used to the format of this website and one thing that helped me skip a couple of steps when I want to click on a recent post is to go over to the right side of the home page where the last post is and click on the little icon that has a small ...

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