CHEMISTRY COMMUNITY

Liam Bertrand 3

Thank you Dr Lavelle, Reid, and all the UAs for a great quarter!

Liam Bertrand 3

I didn't think that the final was easier or harder than any of the midterms, pretty much a similar difficulty.

Liam Bertrand 3

TYSM!

Liam Bertrand 3

Peaky Blinders is an 11/10

Liam Bertrand 3

I feel so burnt out. I've lost motivation for schoolwork, class, exercising, really everything. I'm at the point where I just want the quarter to end rather than feeling motivated to do my best.

Liam Bertrand 3

The overall reaction should not include a catalyst.

Liam Bertrand 3

I'm going to review sessions, reviewing discussion slides (which have textbook problems), and reviewing sapling.

Liam Bertrand 3

it will be cumulative

Liam Bertrand 3

Kinetic stability has to do with the speed of the reaction. Slower reactions mean higher kinetic stability.

Liam Bertrand 3

Its hard for me to stay motivated when I'm just sitting at my desk, I miss walking to campus and sitting in a lecture hall a lot more than I thought I would.

Liam Bertrand 3

I remember Dr. Lavelle using 10^-3 as sort of the cutoff, but I also heard in a review session last night people saying smaller numbers. I used 10^-3 for the first midterm. I think for questions on the final it will be very obvious if you can approximate or not.

Liam Bertrand 3

rende burritos after practice are a big miss

Liam Bertrand 3

Q shouldn't be affected, because an increase in water will still mean the same proportion of reactants and products.

Liam Bertrand 3

Enthalpy involves heat released, while entropy is the number of states a system can be in.

Liam Bertrand 3

sushi!

Liam Bertrand 3

Higher reduction potential = better oxidizing agent

Liam Bertrand 3

I went on Youtube and watched a few videos and then did some practice problems and I feel pretty confident now.

Liam Bertrand 3

I agree with the two other responses

Liam Bertrand 3

Use the charges given and the oxidation numbers to figure out how many e-'s to add

Liam Bertrand 3

Its probably just that one question, if we were to get a standards Gibbs free energy calculation question on a test you would be given the values for reactants and products and could use the normal equation.

Liam Bertrand 3

A little bit of Peaky Blinders for me!

Liam Bertrand 3

Cell potential is not a state function. Usually state functions have a delta value in their equations

Liam Bertrand 3

How is 14bl compared to 14b in terms of content? Are they super related

Liam Bertrand 3

How did the partial pressures in bars end up cancelling? I tried converting bars to joules but it isn't on our equations sheet. Do we not have to do that?

Liam Bertrand 3

Degeneracy is the number of arrangements a molecule can have. I know its a bit confusing since its W and work is w.

Liam Bertrand 3

Prof Lavelle has always just said if work is done on the system w is positive and if the system does work then w is negative

Liam Bertrand 3

Does anyone know how the second part would be formatted? I did it the same way as the top part and sapling keeps rejecting my answer as well. I tried a big number and I tried scientific notation.

Liam Bertrand 3

Thanks for the info, sapling is really annoying sometimes where it doesn't take certain answers even though they are right. I was doing the same thing you did and was super frustrated because I knew I wasn't doing anything wrong but couldn't get sapling to accept my answer.

Liam Bertrand 3

Delta G is negative means more products, and a positive delta G means more reactants.

Liam Bertrand 3

Thank you so much, I struggled for a while with this problem. I was using q=mC(deltaT). Is there a reason why you wouldn't use this equation for this problem? Probably something I missed as to why not to use that formula.

Liam Bertrand 3

Conservation of Energy!

Liam Bertrand 3

No, this is not the case. In fact, when pressure is constant, the volume usually changes and when volume is constant the pressure usually changes. This can be obtained from the ideal gas law.

Liam Bertrand 3

I go to workshops often and review the hw, but I definitely need to do more textbook problems for practice

Liam Bertrand 3

I try and schedule my days out really well, so that I know exactly what I have to get done and it helps me feel better knowing I got everything done that I wanted to for a day.

Liam Bertrand 3

I think of it kind of like a flow chart from solid->liquid->gas. going to the right on that sequence requires energy (endo) and going to the left releases energy (exo).

Liam Bertrand 3

Definitely guessed on 2, and didn't feel great about 1 other question. Maybe professor Lavelle will curve or if not we have another midterm and the final to go!

Liam Bertrand 3

Temperature is the main one

Liam Bertrand 3

You can only assume x is insignificant if the Ka/Kb value is 1x10^-3 times smaller. Basically when Ka or Kb is really really small you can disregard the x

Liam Bertrand 3

You would still take into account the exponents.

Liam Bertrand 3

Energy is always needed to break bonds and released when forming bonds

Liam Bertrand 3

If water is the main solvent it is not included.

Liam Bertrand 3

coefficients!

Liam Bertrand 3

HW and textbook problems are the way to go.

Liam Bertrand 3

Once you get x, plug it into the "equilibrium" section of the ICE box. Also, check if you need to convert moles to M by dividing by the Liters.

Liam Bertrand 3

You just have to identify the base and the acid, which can be hard. Best way to learn those is thru practice.

Liam Bertrand 3

There is so much water, and so little [H30+] and [OH-] are produced in comparison, that the concentration of water basically stays the same. The concentration change is so many decimal points out that it is negligible

Liam Bertrand 3

Go to workshops constantly (I try for 2 a week) not just before tests. It really helps

Liam Bertrand 3

When you have log10(a)=c, to get to exponential form it would be 10^c= a. Thus, since the pka is c in this example, raise 10 to whatever the pka is and it should give you ka. Someone might want to double check me on that one though because its been a while since I've done that kind of math

Liam Bertrand 3

K will stay the same, but Q will need to be calculated again and then you can compare the K and Q to determine reaction direction.

Liam Bertrand 3

I think that most of the time we'll be given larger K values, so that we have to use the quadratic equation.

Liam Bertrand 3

Textbook problems and reading, step up sessions, rereading lecture notes are what I've been doing. Hopefully by starting week 1 I'll be better prepared for the midterms and final

Liam Bertrand 3

This should always be in kelvin, and I think the kelvin/celsius conversion formula is on our formula sheet, in case its needed on an exam.

Liam Bertrand 3

Yes, Q just applies to a reaction that is not at equilibrium, while K is at equilibrium. You can compare the two values to determine which direction the reaction is going.

Liam Bertrand 3

The UA sessions are something that I didn't utilize until the final but its something that I'll go to every week in 14b

Liam Bertrand 3

The last two questions were complete guesses for me, 3 minutes per question was way shorter than I thought it would be

Liam Bertrand 3

I'm going to just flip through my 14a notebook occasionally, specifically on sections that were difficult for me in the class

Liam Bertrand 3

A wedge means it is coming towards you (picture straight out of the paper). And the dash means that it is going directly away from you (straight behind the paper). The wedge and the dash are in the same plane, just in opposite directions. Its a 2d representation of where the atoms or molecules are i...

Liam Bertrand 3

They are just two separate ways of explaining what is happening. Two different POVs If you will.

Liam Bertrand 3

The easiest way to determine this is by the length of the bond. Longer bond = weaker bond. In your example, carbon-oxygen bond would be weaker because it is slightly longer than a C-N bond. There are of course other factors but this is the one that professor keeps mentioning the most, so I would ass...

Liam Bertrand 3

Lewis structures that do not have ideal formal charges would not be incorrect, but rather "less ideal" or more unlikely to occur naturally.

Liam Bertrand 3

Drawing out the lewis structure with dipole arrows is the best way to figure this out, because you can see if they cancel.

Liam Bertrand 3

I would assume it will probably be proctored on respondus like the midterms.

Liam Bertrand 3

Lone pairs repel each other as well as other atoms +bonds. This is why, in say NH3, the lone pair "on top" of Nitrogen in the lewis structure pushes the N-H bond angles away from it.

Liam Bertrand 3

The triple bond will just contain 1 sigma and 2 pi bonds, I think that's all the detail we need to know.

Liam Bertrand 3

Think of benzene, where there are 3 double bonds, however, each carbon could theoretically form a double bond. Furthermore, the pi bonds are all in the same plane, allowing a sort of highway where electrons can move freely. Thus at any point the pi bond could be between any two of the carbons in ben...

Liam Bertrand 3

To add on to everyone else, the easiest way to visualize the resonance structures is to draw the lewis structure. Then, consider where 2 electrons (usually a lone pair or double bond) can be moved to still contribute to a correct structure. Formal charge can then be used to decide which resonance st...

Liam Bertrand 3

^^ I agree, looking into the formula for molecules you already know how to draw is a good way to start.

Liam Bertrand 3

Hopefully y'all are right, I missed a week as well.

Liam Bertrand 3

Larger atoms do mean weaker bonds. However, there are other factors that influence a boiling point, so those are probably the reason why the boiling points are what they are.

Liam Bertrand 3

Last midterm I had respondus and the midterm on my laptop and then zoom on my phone and this setup worked pretty well.

Liam Bertrand 3

To add on to what everyone else said, partial charges are less intense (why they're called partial), can generally fluctuate a bit as the electrons move around, and sometimes only occur at specific moments. A formal charge will not change because it refers to the total number of electrons, and is no...

Liam Bertrand 3

For MT1 I reviewed HW problems, recommended textbook problems, and UA worksheets and this worked really well for me.

Liam Bertrand 3

I'm assuming on tests the comparisons will be directly to elements in the same row (i.e. further right has larger radii) or same column (further down = larger radii). Trying to determine the radii in a diagonal fashion is usually too difficult, that would probably be given to us

Liam Bertrand 3

I don't think hydrogen will ever be the central atom. Even though its on the left side of the periodic table I was taught in high school to treat it as sort of its own thing.

Liam Bertrand 3

This depends on if the bonds are double or single bonds, if resonance structures exists where the type of bond changes, or if there were differences in the two elements (rather than being a N-N bond).

Liam Bertrand 3

Lone pairs determine resonance. Try drawing the structure out and using arrows to show where electrons can be moved to form double bonds or lone pairs. This is how I learned resonance in high school

Liam Bertrand 3

When ions form a bond and come together it releases energy, it then requires energy to break that bond again.

Liam Bertrand 3

As long as you can make general electronegativity assumptions by looking at a periodic table I think that's good enough

Liam Bertrand 3

Agreed ^^

Liam Bertrand 3

The brackets and charge on the outside of the diagram has to do with the overall charge of the molecule. However, partial positive and negative charges can be attributed to each atom in the molecule.

Liam Bertrand 3

I believe that when there is a resonance structure, there is no way to confirm actual bond lengths, since the bond and electrons can move freely. Thus the experimentally observed average is generally used

Liam Bertrand 3

Noble gasses usually aren't considered. Since they have a full octet, they don't naturally accept electrons due to the energy it would take to accept. However, it can happen in labs and special situations

Liam Bertrand 3

I would definitely just put them in, its confusing if they are not there. They also give important info as to what kind of bonds and what elements that the atom in question can bond to.

Liam Bertrand 3

Adding on, covalent bonds are between two nonmetals, while ionic bonds are one metal and a nonmetal.

Liam Bertrand 3

I believe if there are 2 in the orbital they are always paired, each electron having a different spin direction.

Liam Bertrand 3

I would assume that it's the next two outlines as long as the remainder of outline 2

Liam Bertrand 3

I don't think so. I'm focusing on remembering all of the one's that aren't specifically given (like what units make up a Joule, etc.)

Liam Bertrand 3

I believe fundamentals will be on the midterm.

Liam Bertrand 3

If you are talking about for the midterm, the equation will be given, so I don't think we have to memorize as long as you know what the variables stand for. As for how he derived it, I doubt we need to know exact steps, maybe just know the general equations that he used.

Liam Bertrand 3

To add onto everyone else, also as you go higher up in atomic numbers on the periodic table the number of orbitals increases (more electrons) and the letters do as well. There's a specific order and the letters don't always go in order.

Liam Bertrand 3

Yea, I'm feeling the same. But I'm doing a couple every night to prepare for the midterm.

Liam Bertrand 3

Black body is totally hypothetical. Black cloth does absorb more light than say a white cloth, but there is no real thing that can emit and absorb all wavelengths of light.

Liam Bertrand 3

I think rewriting notes and going thru practice problems, and then quizzing myself on topics with a friend Is going to be my strategy

Liam Bertrand 3

I think they are given to us during the exam anyway.

Liam Bertrand 3

I believe it means the energy of the photon. That's what the equation is solving for.

Liam Bertrand 3

The equations are given, so as long as we know what each variable is and when to use the equation I don't think we need to know the names

Liam Bertrand 3

Totally agree. It's key to expose yourself to as many types of problems as possible before the test.

Liam Bertrand 3

Thanks everyone, I struggled with this too and this thread helped a ton. Thanks to everyone

Liam Bertrand 3

What's the dullest element?
Bohrium!

Liam Bertrand 3

When you get the number of moles, you want to divide by the smallest number to get (hopefully) whole numbers for the empirical formula

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