CHEMISTRY COMMUNITY

MAC 4G

I assuming that we are still having the final, lecture, and discussion in person until Lavelle says otherwise. As for take home finals, I'm not sure if this will be his chosen method, because it could also be taken online where they lock the screen. But, it would be tough to do being that our exams ...

MAC 4G

Lyndon's review packet probably won't be announced or up until much closer to the day of the review session, so most likely Wednesday.

MAC 4G

Both are correct as mentioned above, and I don't think you'd lose points for having either. Typically we haven't really left coefficients in fractions, at least on exams and tests.

MAC 4G

The final is cumulative, so it's best to go over everything. But if you want to focus more on something than anything else, I would focus on the parts of the outlines that weren't on the midterm.

MAC 4G

Lyndon's review session will be Friday March 13, 6-9pm, Moore 100

MAC 4G

The book does have it as mol.L-1.s-1 and I think the correct answer is L.mol^-1.s^-1 being it would be 1/M*s

MAC 4G

For the most part, I generally just looked at the values that corresponded the reduction equation that had 1e- being transferred.

MAC 4G

You can convert torr to atm by dividing the torr values by 760. The units will then cancel out as you solve the equation. after this do you implement the pressures into the Q equation? I keep getting the wrong overall answer so I'm not sure what I'm doing wrong. Yes, you would plug in the converted...

MAC 4G

The easiest way to do it for me, is write down what's being reduced or oxidized. From there, see if you need to add e- to the reactants for a reduction (gaining e-) or add e- to the products for a oxidation (losing e-). When you combine them and balance them out the number of e- transferred should b...

MAC 4G

I believe it's because those are the values of their concentration being there's H2 as a product and a reactant. So, in their respective places when finding Q, you'd need to include their concentration which is 1atm.

MAC 4G

When you write out the half reactions you should get that 2 e- are being transferred, so n should equal 2.

2H+ +2e- -> H2
H2 -> 2H+ + 2e-

MAC 4G

In your example, it should be ||Cl-|Pt

A trick is that solids should always be on the outermost part of your diagram, then it's gases (if they are present), and closest to the salt bridge, are your aqueous solutions.

(s)|(g)|(aq)||(aq)|(g)|(s)

MAC 4G

Typically, yes, the anode would be written on the left of the salt bridge and the cathode would be written on the right. Assuming the question provided you with the cell diagram.

MAC 4G

An example to look at would be Mn2+ being reduced to Mn and Al is being oxidized to Al3+. Here, Mn2+ is gaining electrons and Al is donating electrons. So, using Le Chatelier’s principle, an increase in concentration of Mn2+ causes equilibrium to shift the reaction towards the products, which would...

MAC 4G

I think you can look at the redox pair: Fe3+ and Fe2+. Since there are 6 of Fe3+ that means the charge is 18+. There are 6 of Fe2+ so the charge is 12+. The difference between the two is 6 electrons.

MAC 4G

You can relate Gibbs free energy to K using the following equation: ∆Gº = -RTlnK As you can see, this is not dependent on pressure or temperature. However, the max work that can be done is given by ∆G at constant pressure and temperature. Hope this helped! You said that this isnt dependent on tempe...

MAC 4G

Remember that pressure is inversely proportional to volume (Boyle's Law), so at a constant temperature (isothermal) V2/V1 = P1/P2.

MAC 4G

It would seem that you have the equation wrong or you plugged the wrong value in for the pressures. The equation should be: delta S = nRln(P1/P2) So for this problem, n=1.50 mol, R=8.314 J/K*mol, P1=20bar, and P2=5bar. When you plug in these values accordingly, you should get a positive value (17.3 ...

MAC 4G

An example to look at would be Mn2+ being reduced to Mn and Al is being oxidized to Al3+. Here, Mn2+ is gaining electrons and Al is donating electrons. So, using Le Chatelier’s principle, an increase in concentration of Mn2+ causes equilibrium to shift the reaction towards the products, which would ...

MAC 4G

I believe someone answered the question of when to use which equation really well here:
viewtopic.php?f=135&t=59535&p=224882&hilit=delta+g+delta+g+naught&sid=dbd85a72ff5163bdbee84a646ab24983#p224882

MAC 4G

I don't know if it helps or if you already know but there is a typo in the question. The Cl2 on the right side should be Cl- where it says Cl2(g) -> HClO(aq) + Cl2(g)

MAC 4G

The above response is correct in the sense that 14A-D are typically prerequisites for upper divs and your main focus should be getting those done. But, labs typically don't require a lot of time to complete the work if you stay on top of all your work and know how to best manage your time. From what...

MAC 4G

What circumstances should we use the difference values of R? The best way to figure out which value of R to use, I would look at what units you're given from the question to see what you can cancel out. Similarly, you can also look at what units you need in your final answer as to what you are look...

MAC 4G

I've found that most people end up taking both courses at the same time because it's the most manageable combination. But, of course, everyone's ability to do so will vary depending on each person. I would also consider the workload of your other classes. For the most part, labs don't require too mu...

MAC 4G

Matthew ILG 1L wrote:Is the change in enthalpy supposed to be much larger for part b than it is for part a?

Depends on what you define as much larger, assuming I calculated both correctly, part b is 510 J more than part a's enthalpy.

MAC 4G

Entropy is 0 because the question mentioned that an isochoric method was used meaning the volumes are the same therefore there is no volume change. So if you want to look at it in terms of numbers, if you had V2 as 5 and V1 as 5, then in the equation for delta S the ln(5/5) or ln(1) will equal 0. Th...

MAC 4G

I think you want to look at whether or not Delta G is negative for spontaneity that is accounted for entropy * T is larger than delta H. The equation for delta G = delta H - TdeltaS So, even if delta H and delta S are positive, increasing the temperature to make the term TdeltaS larger than delta H ...

MAC 4G

The above post is correct in how you find delta T, I found that it was easier to delta T = q/nCv,m and then use the equation Tf = delta T + Ti. Then to find enthalpy of part b, you would use the equation delta H = q + nRdeltaT and plug in all the values you were given and the delta T value you found...

MAC 4G

Because the midterm being on Wednesday, I believe you're allowed to do any of the problems. I did mine on thermodynamics.

MAC 4G

I don't know if this answers your question nor if I'm 100% right, but the way I look at the equations and figuring out which one to use is based on the information the question provides. ΔS=qrev/T is used if you were given heat that's reversible (heat can move in both directions of the reaction) and...

MAC 4G

Doing the review worksheet that Lyndon made is a good option as well, but your best bet is the homework problems. If he uses anything to make his exam questions, it'll probably be homework questions and past exams.

MAC 4G

Does anyone know when the answer key will be posted?

MAC 4G

I believe they are still there at Young 3034, at least they were earlier this week.

MAC 4G

A closed system can transfer energy (heat or work) but not matter with its surroundings.

An isolated system cannot transfer energy or matter with its surroundings.

MAC 4G

He should be finishing up outline 4 this Friday, but whatever he covers up until on Friday will be the material that is fair game for the midterm.

MAC 4G

Typically, you don't have to memorize any equations because they will all be provided on the equation sheet. But, it would be helpful to know how and when to use each equation.

MAC 4G

He'll most likely make an announcement later on this week, as we get closer to the review session.

MAC 4G

I believe that this is an error in the textbook, but it is 28J, as it is correctly written in the solutions manual.

MAC 4G

Steam has more phase changes than water, it releases more energy when it comes into contact with skin. It transfers more energy to your skin resulting in more damage. Here's a more detailed answer: viewtopic.php?t=1844

MAC 4G

Here's a more detailed answer: viewtopic.php?t=1844

MAC 4G

Can we pick it up at any time?

MAC 4G

The test will be returned most likely in your next discussion, but the grade for it may come out earlier.

MAC 4G

When I was working on this question: Air in a bicycle pump is compressed by pushing in the handle. If the inner diameter of the pump is 3.0 cm and the pump is depressed 20 cm with a pressure of 2.00 atm. Part c) What is the change in internal energy of the system? I thought that it would be equal to...

MAC 4G

I had a similar question and I'm not sure if it's right, but when I looked it up, it said that increasing the pressure, the reaction will shift towards the side with fewer moles of gas. And just the opposite for a decrease in pressure.

MAC 4G

It doesn't really matter being that it's connected to your UID (I think), so you should get the credit. Last quarter, I received full credit and my username isn't my full name. And I believe that Lavelle mentioned that having the discussion listed is specifically for TAs to see what questions their ...

MAC 4G

To approximate you would want a K value less that 10^-3, but if you aren't sure whether or not it meets the cut off, then I suggest using the quadratic formula just to be safe.

MAC 4G

For this homework question, early in the chapter there was a table that provides all of the Kb values. That's what you would use to determine x in your ICE box, which you could then use to figure out pOH and then pH.

MAC 4G

Generally, you would use the smallest number of sig figs you were given in a problem. For example, if you used a ICE box to find that x is 4.22*10^-5, then you're pH or pOH should have 3 sig figs (2 numbers after the decimal) like 4.38.

MAC 4G

It's hard to say what will be on the test, but generally, anything that shows up as a homework problem in the Chemical Equilibrium and Acid and Bases sections are fair game and could be on the test.

MAC 4G

Last quarter the tests were all pretty much free response, and there will be 7-8 questions.

MAC 4G

And another thing to look out for, even if the question doesn't specify whether or not it wants Kc or Kp look at the given equation. If the reactants and products have "(g)" next to them, you'll look to find Kp because they're gases and you would use the partial pressures for it.

MAC 4G

I believe that based on the information you are given, like whether the products or the reactants are increasing or decreasing, would determine whether the change is positive or negative. Something to keep in mind: -If you are given initial concentrations and all of them are nonzero, calculate the r...

MAC 4G

He'll let us know for sure, but it's all the material covered up until the Friday before the Test. So, for test 1, it will most likely be all the information presented up until this Friday, January 17.

MAC 4G

Do we not have to calculate anything? This question doesn't require you to calculate anything, it just wants you to take the given information and see if you understand that if K at 700K is smaller, then it would lead you to the conclusion that there are more reactant present. Therefore, there woul...

MAC 4G

I believe so, since the pressure of O2 will increase as the reverse reaction is favored shifting .

MAC 4G

My best guess is that if they don't give you an initial concentration, then it's implied that it's 0. But, I'm not sure what the actual reason is. Sorry, I can't give you a definitive answer but hopefully someone else will have the answer, and you can ask a TA or Lavelle. And if you get the right an...

MAC 4G

I wrote my answer for part a as K=[C 2 H 4 Cl 2 ] 2 [H 2 O] 2 /[C 2 H 4 ] 2 [O 2 ][HCl] 4 but the textbook has it in terms of partial pressures because they are gases. This is also how we had to write it when doing the modules. If I do not write it in terms of partial pressures on the exams or home...

MAC 4G

K doesn't have any units because we are using activity values. He used the example of finding Gibbs free energy change or ΔG0 = −RTln(K). I don't quite remember what the units are but I'm sure you can look them up and I believe we'll be covering this later on in the course. But in relation to the qu...

MAC 4G

I believe you did the problem right, at least that's how I did it. I believe you want the positive number because negative concentrations are unphysical, or in other words, you want a positive number to make chemical sense.

MAC 4G

"Bars" is a pressure unit so I believe you would be able to simply plug in however many bars you have to all of the molecules. But, if you need a conversion to or from atm then you'd use 1 atm = 1.01325 bar.

Not sure if this answers your question, sorry in advance if it doesn't.

MAC 4G

The general rule is that nonmetal oxides are acidic, metal oxides are basic, and semimetal oxides are amphoteric. Try drawing out lewis structures and seeing if the compound can give, receive, or both give and receive lone pairs. If it can do both then its amphoteric. Here's an explanation for part ...

MAC 4G

It was just posted the key word is Marshmallow.

MAC 4G

Because Bronsted acids are defined to be the molecules to that donate a proton (H atom), NH4+ would be the Bronsted acid here as it is the one that donated a proton to HCO3- and became NH3. Lewis acids are defined to be the lone pair acceptors so NH4+ would be considered a Lewis acid since it is po...

MAC 4G

I think the best thing is to do as many homework problems, if not all, as you can. He likes to pull questions from the homework as well as make questions that resembles the homework problems. Reading the textbook is also a good idea to help you understand the information more in depth. I found that ...

MAC 4G

I was wondering the same thing, so I looked for it. Lyndon did have a practice final that only covers the material after the midterm being that he covered everything else for the midterm. So, if he does he'll probably announce it during Wednesday's lecture.

MAC 4G

How to find Conjugate acid or base of a given compound? To find the conjugate acid: Bases form conjugate acids. So in order for Reactant X to produce a conjugate acid it must behave like a base. So what a base does is accept H+. There you go just add H+ to X. (part a) To find the conjugate base: Aci...

MAC 4G

Hard to say because depending on which TA is grading that question, if it shows up on the exam, will determine whether or not they except it. So, it would be something to ask your TA about or Lavelle. But, the safest bet would be to write how the textbook has it. Sorry I can't give you a more defini...

MAC 4G

1) Draw the Lewis structures of the molecules 2) Determine which bonds have the greatest difference in electronegativity (for example, C-Cl has a greater difference than C-H) 3) Draw the dipole moments for those bonds of greatest electronegativity difference. 4)Compare the directions of the dipole m...

MAC 4G

viewtopic.php?p=2547&sid=0d36d6a0a83ae99de2b5592ede5e800f#p2547

Here's a detailed answer.

MAC 4G

I'm not 100% sure, but my guess is that we'll be learning up until the last day of lecture, or at least until the Wednesday before the final.

MAC 4G

CH2O has a trigonal planar shape because it does not have any lone pairs on the central atom, making the angles 120 degrees. Also, there are 3 regions that carbon is bonding with, making the hybridization sp2. But, carbon can form 4 bonds because it has 4 valence electrons so that leaves a remaining...

MAC 4G

The general trend is that it increases towards Helium. In other words, it increases from left to right in a periodic table and up a column. Some extra info that might help: If two atoms have electronegativity values that are extremely different, they won’t share electrons between them. The atom with...

MAC 4G

The sigma bond is stronger than the pi bond because in the sigma bond, there is a direct overlap of the orbitals. Where as in the pi bond, the overlap is side wise, which reduces the area of overlapping.

MAC 4G

I don't believe that they'd expect us to have the electronegativity values memorized, but definitely understanding the trend and how it's affected by the atomic number and the distance at which its valence electrons reside from the charged nucleus.

MAC 4G

It's hard to say which ones will be on the test and the exam, but I'd say definitely any shape that Lavelle has shown us in lecture as well as any that appear in the homework. But everything on the chart below would be a good thing to memorize, just to be safe. https://www.google.com/url?sa=i&so...

MAC 4G

A sigma bond is a bond that forms when one electron from each of the two orbitals with interacts end-to-end. Overlapping just means it's in phase, or the orbitals are sharing the same space.

MAC 4G

My best guess would be all the material after the midterm up until this Friday's lecture, at least that's how it went for the first test. But I'm sure lavelle will mention it at some point this week, but if not, then our TAs I think will know by the end of this week. So we can ask them.

MAC 4G

Either way is fine if you get to the correct answer, but on an exam I think it's better to write out using the formula because in doing so, I believe it'll get you some partial credit if you were to make a mistake when drawing a Lewis structure or a resonance hybrid.

MAC 4G

Based on how the midterm was, I'll be going over each component of the equations and how you would go about finding missing parts and whatever the question is asking for. And then doing as many of the homework problems as possible.

MAC 4G

For the full electron configuration for Nitrogen in the ground state, it would be up to 2p^3. You would do this if it asks for the configuration of a specific element and/or ion. At least that's how it has been for the homework and the midterm so far. But, conceptually I think it's important to know...

MAC 4G

I believe it is only material that he's taught from after the midterm to the end of this week, but I'm not sure. But, I'm just going based on how it was for the first test. I'm sure he will bring it up sometime this week, if not, you can double check with your TA.

MAC 4G

Do you know if Cr and Cu are the exceptions to this rule because their 3d orbitals are filled first? Thank you![/quote] Cr and Cu are are exceptions in which they don't have a full 4s shell. One of the e- is moved into the 3d shell. So Cr: [Ar]3d^5 4s^1 and Cu: [Ar]3d^10 4s^1. My TA said you will ne...

MAC 4G

Sorry it should be [Ar]3d^5 4s^2

MAC 4G

If you find its electron configuration, you'll find that it's [Ar]4s^2 3d^5. You can see here that there are 2 in the s and 5 in the d, so 2+5 = 7 valence e-

MAC 4G

From the homework questions, it seems like we do. I'm worried that we do? As for hw questions, I think the reason they do this is because we have access to look them up. But, I don't think you'll have to worry about it for the exam being that they can't expect us to know the exact compound name for...

MAC 4G

My TA said that if it is a compound we haven't seen much of, they will provide it for us. But, he said it would be unlikely for us to have to know compounds names. The only ones that they'll expect you to know are compounds like CO2, H20, NH3, etc. The ones we've been seeing in review problems and o...

MAC 4G

Just some extra information for this type of question, if they give you a percent yield that isn't 100%, you have to use the formula % yield = actual/theoretical to find the theoretical as the starting value of grams. Then, you'd follow the process that was explained above. For example: 50g of Si3N4...

MAC 4G

I think all you need to know how to calculate different parts using the equation and know conceptually what it's used for and all of its parts.

MAC 4G

I don't know the answer for sure, but I would go based on whether or not it was used in one of the homework questions, then there's a chance it could be on the exam. And if you have some time to learn, I don't think it would hurt to go over it. Sorry this isn't a definite answer.

MAC 4G

I don't believe you can find the answers to Test #1 online, you'd have to either find someone who got all of the answers correct or ask specifically here. And someone will explain it.

MAC 4G

The midterm is up to Concept 2D, but only the parts that Lavelle teaches up to this Friday, November 1. I believe that he mentioned next week he will still be discussing parts of 2D, so the midterm will only be 2D up to this Friday's lecture.

MAC 4G

The explanation above is right. The question is asking for the charge of the ion when it gives or receives e- to become stable. In this case, it would be easier for Ga to lose and give away its 3 valence e- rather than looking to gain 5.

MAC 4G

Going off the explanation above, when you are drawing Lewis structures it's best to think about the octet rule, in which every element has 8 e- in the outer most shell (through sharing valence e- in this case). But, there are a few exceptions (the slides mention the exceptions as H, He, Li, and Be) ...

MAC 4G

It'll be like Test 1 where all of the formulas as well as a periodic table will be provided with the exam, you will just have to know how to use them.

MAC 4G

You can also go to your studylist from my.ucla and on the right side it says grades right under class website.

MAC 4G

Also, using the kinetic energy of the ejected e- in part a and the work function (threshold) you found in part b being it is the minimum required energy to eject the e-. Then to find the energy of the incident light: You find in part a) the kinetic energy of the ejected e-: 5.909x10^-18 J. E - 1.658...

MAC 4G

In part C, you use that 1/2mv^2 to find the kinetic energy of the electron ejected. But, for the energy of the incoming photon, you have to add that energy to the energy threshold implying that the energy of the incoming photon was enough to overcome the threshold energy. It must have an energy grea...

MAC 4G

I also realized that there were some mistakes in my answers. a)ns^1 b)ns^2np^3 c)(n-1)d^5ns^2 d)(n-1)d^10ns^1 I believe that the d orbital comes first being that orbitals are filled in the lower energy orbitals first, then fill higher energy orbitals. For instance, the 3d orbitals have a slightly hi...

MAC 4G

(a) The alkali metals are all the metals in Group 1A, belong in the s-block, and have 1 valence electron each. Therefore, depending on which shell the element is in, the configuration will be ns^1. (b) Group 15 elements are in the p-block with 15 valence electrons each, 2 in the s sublevel, 10 in th...

MAC 4G

After you find the electron configuration, you need to look at the orbital which comes last in the configuration that is being observed. It may be - s,p,d or f . The ‘s’ orbital contains max of 2 electrons whereas max electrons in p,d and f are 6,10 and 14. If the last orbital contains electrons les...

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