CHEMISTRY COMMUNITY

Taylor K 2D

Usually Ka is associated with acids and Kb with bases. If you need to convert between the two, just use the equation Ka x Kb = Kw = 1x10^-14

Taylor K 2D

Get lots of sleep and don't psych yourself out - you got this!!! Also don't forget to change your clocks!!!

Taylor K 2D

That all makes sense to me! But for some reason the textbook says that "A thermodynamically stable compound is a compound with a negative standard Gibbs free energy of formation (water is an example)."

Taylor K 2D

Hi can someone tell me how to know if something is thermodynamically stable and kinetically stable? I saw another post on this but every response said something different so now I'm even more confused.

Taylor K 2D

I don't believe we explicitly went over this. I guess you could get it by deriving the lnk= -Ea/RT +lnA, but I would just try to remember the Arrhenius equation by itself so you don't have to try to derive it.

Taylor K 2D

I was already a little nervous for the final and then I remembered that it is going to be on daylight savings so we're gonna lose an hour of sleep the night before/morning of the final ;_;. At least it's pi day which means discounted pizza and pie. How are you guys planning to get your stress out th...

Taylor K 2D

I love sushi, pasta, and waffles!

Taylor K 2D

I try my best to space out my school work with self care/free time. I know that I get burnout a lot more when I study for long blocks of time, especially when it is all one subject. I find I do a lot better when I divide study time and space it out/alternate between subjects more frequently.

Taylor K 2D

Going to workshops and doing the textbook problems is what really helped me!!

Taylor K 2D

I believe at equilibrium E naught equals zero since K=1 and ln(1) is zero.

Taylor K 2D

I know that there will be a lot of office and ta hours and I think he might do what he did last semester with the last lecture being review but I'm not sure. He will probably send out an email soon!

Taylor K 2D

I think we still have access to chem community (I think the website is actually open to the general public) but I'm just not sure if we will be able to post on it anymore?

Taylor K 2D

I usually try to do them my self first, but for really difficult ones, I will read the solution and see if it helps me to understand the problem better.

Taylor K 2D

I like workshops for the extra practice and step-ups when I'm struggling!

Taylor K 2D

Can someone please help me with this? I don't understand what I am doing wrong?

Taylor K 2D

Does anyone know if we can review our Midterms in another TA's office hours? Like if we can't make the time that our own TA has office hours, can we go to a different one's?

Taylor K 2D

I watched WandaVision and caught up on the Bachelor :)

Taylor K 2D

Isabelle Hales 1J wrote:Hi! Going off of this, is the melting point also found at equilibrium (deltaG = 0)?

I would think so! I'm pretty sure that at the temp for all phase changes (melting point, boiling point, etc) that G=0!!

Taylor K 2D

Conceptually: when there is no change in volume, there is no expansion or compression, and therefor no work being done. Mathematically: for the equation w = -P∆V, and so if the change in volume is zero, then the work done will also be zero. for the equation w = -nRT ln V2/V1, the natural log of V2/V...

Taylor K 2D

A state function is independent of the path or phase changes taken to arrive at the final destination. In other words, it only really cares about the initial and final values.

Taylor K 2D

It honestly depends on which units you need in your final answers. I just keep my equation sheet next to me to help with that!

Taylor K 2D

I would just email your TA to make sure you can stop by during their office hours and then you should be able to review your midterm :)

Taylor K 2D

The d stands for "derivative," but I think that professor Lavelle just used that to explain how he was deriving his equations :)

Taylor K 2D

The "reverse" of the natural log (ln x) is e^x, so I just raised both sides to look like e^lnk = e^ (the rest of the equation). The e will cancel out with the ln k so you'll end up with k = e^ (the rest of the equation). I hope this helps!

Taylor K 2D

The first thing that I did was try to convert watts to kilojoules (1W = 1J/s). And then the Hvap = energy of the heater/change in mole. I calculated the change in mole by subtracting the final mass from the initial and then converting from grams to mole for each compound. Once you find the Hvap for ...

Taylor K 2D

I was a little confused on this as well but I think that because it's asking about the entropy of the liquid samples that the order would be reversed, so the molecules with the highest entropy in gaseous form would have the lowest entropy in liquid form? Again I am not completely sure :/

Taylor K 2D

I'm confused about this also. I wish he had covered this in lecture :/

Taylor K 2D

I was wondering this too but I think it is because the system is at a constant volume, so the volume of the samples is not really something we need to take into account. Additionally, since we aren't asked to find the moles or molarity, and are already given the temperature, we don't need the moles ...

Taylor K 2D

I just click "your posts" and use the number at the top.

Taylor K 2D

To find q requires knowing the molar heat capacity at constant volume. Since CO2 is a linear molecule (if you draw out the lewis structure it shows that CO2 has a linear shape), then the Cv for CO2 is 5/2 R. Use that value for Cv and plug it into your equation for q. For the second part, since the s...

Taylor K 2D

I was able to find q, but I am having a really hard time finding U. Can someone please help?

Taylor K 2D

I am not sure but I remember someone saying hopefully by the end of the week?

Taylor K 2D

I would think that if we aren't given temperature (and aren't being asked to find it) then we can assume room temp or maybe STP? That's a really good question.

Taylor K 2D

I tried to find the q of ice and the q of water, and then use ΔHfus+qice=−qwater to solve for final temp. I hope that helps!

Taylor K 2D

I have been wondering this too!

Taylor K 2D

I think it's due next Sunday!!

Taylor K 2D

Does anyone know up until what content the midterm will cover? Do you think it will cover up until today's lecture?

Taylor K 2D

Im not sure! It looks like you got the same final temperature as the textbook did!

Taylor K 2D

You use the invert of Kc when using the reverse reaction, and you use Kc^n when multiplying the entire equation by the same coefficient.

Taylor K 2D

I think Professor Lavelle said he wouldn't give us a cubic equation that has a K larger than 1x10^-4 since we did not learn how to do that and it's a lot of extra work.

Taylor K 2D

I think it's because as a gas, the steam or vapor has a higher enthalpy than water and thus a higher internal energy

Taylor K 2D

I think that for these problems you need to use the Kb and Ka tables (Table 6d1 I believe). It says it right before the 6D questions but I also didn't see it at first.

Taylor K 2D

The conjugate base of a weak acid will be a strong base.

Taylor K 2D

The best way to calculate this is the same way you would calculate Kc: Q = [products]/[reactants], or in this case, Q = [NH3]^2/[N2][H2]^3 and then you can just plug in the concentrations.

Taylor K 2D

Based on previous lectures, it seems we only really need to know the six strong acids (HCl, HBr, HNO3, H2SO4, HI, and HClO4), and the general weak ones that he keeps mentioning (acetic acid, H2SO3, HF, HNO2). For bases, I think we just need to know that group one and two oxides/hydroxides are strong...

Taylor K 2D

I am also having this issue. I think that the solution manual has a typo.

Taylor K 2D

For problems like these, you first have to look at all of the molecules that make up the products and reactants, in this case that is NH3, 3I2, N2, and 6HI. Then, you can look at the one step equations given and make note of the ones that contain the molecules you got from step one. Then, from the o...

Taylor K 2D

I am struggling with this one too.

Taylor K 2D

I believe so!

Taylor K 2D

Qp = (P product)/(P reactant)

So for this problem, QP = (PNO2)^2/(PN2O4)

QP= (.5)^2/(.5)

Taylor K 2D

First, convert from mol to mol/L, so just divide your given mol values by the 5 L. Then, you can set up your I.C.E. or equilibrium table.

Taylor K 2D

When going from moles to atoms/molecules, you would divide by Avogadro’s number, and when going from atoms/molecules to moles, you would multiply by Avogadro’s number.

Taylor K 2D

That’s a great question. I don’t know if there would be any conjugate pairs because this reaction is between a strong acid and a strong base.

Taylor K 2D

Usually, I either do the same exact thing or the complete opposite: stay up super late studying cause I’m so stressed out and then wake up even more stressed, but this time I’m watching holiday movies and I’m gonna try to go to bed early lol.

Taylor K 2D

I think a favorable force just refers to the force that acts as the main intermolecular/bonding force. So in NH3, I think hydrogen would be the more favorable force because it's the strongest IMF for NH3. I hope that helps!

Taylor K 2D

I think s-character has to do with the ratio of s to p orbitals in hybridization? So an sp hybrid orbital would have more s character than an sp3 hybrid orbital. I found this chem community post from 2011 that might help more.

viewtopic.php?t=1255

Taylor K 2D

This is a really good question. I would assume that it could be monodentate because of its shape and the angles between the two single-bonded oxygens preventing it from successfully bonding to the same metal but I honestly don't know. I found this link from a 2014 chem community post on the same top...

Taylor K 2D

Yes! I got this one too and it took me longer than I'd care to admit lol. The name of that compound is Tetraamminedichlorocobalt(III) chloride. I got this using the rules Professor Lavelle mentioned in class. Tetra because there are four ammonias and ammine from the chart on the chem 14 A website. D...

Taylor K 2D

This is a good question. I don't know if there a specific way to determine the bonding capacity for all transition metals. I know that it's hard to pinpoint and exact number of valence electrons a TM refers, as we've seen with their ability to hold more than a full octet. From what we've done in lec...

Taylor K 2D

Well it depends. If you remember earlier in the quarter, when we first reviewed unhybridized orbitals, we learned that p orbitals have 3 different orientations-- px, py, and pz. Each p orbital was on a different axis. The reason the unhybridized p orbital was perpendicular rather than in plane was ...

Taylor K 2D

Hi! Just want to point out the number of bonds and number of ligands are not the same thing. The coordination number is the amount of bonds to the central transition metal atom, but that will not necessarily be the number of ligands. After all, polydentates can form multiple bonds; for example, one...

Taylor K 2D

That's a really good question. Chloride is a diatomic molecule but in one of the sapling questions I got I had to say dichloride and not just chloride so I am not entirely sure. I hope someone else can clear this up because I had a similar question. You might need to say di- regardless of whether or...

Taylor K 2D

I'm not entirely sure, but my guess would be that repulsion would be less, as there is less magnitude of negative charge to repel other electrons attached to atoms. I would agree with this. With one less electron, the repulsion will probably be less as the charge cloud is smaller than if there were...

Taylor K 2D

Can someone explain why the structure for H2CCCCH2 would not by cyclical/have the carbons form a square with one hydrogen attached to each carbon instead of the 4 carbons being in a line?

Taylor K 2D

There are two linear structures, just like there are two different "bent" structures. Molecules with 5 charge clouds, 2 bonds, and 3 lone pairs are also considered linear structures. As the others mentioned, this structure is also considered linear because the lone pairs repelling the bond...

Taylor K 2D

I'm a little confused on your question, but I don't believe this would affect bond angles. In our lewis structures, the surrounding atoms often have lone pairs, otherwise, they wouldn't really be a region of electron density. I hope this helps!

Taylor K 2D

I was wondering this too.

Taylor K 2D

Yes! All molecules experience London Dispersion Forces!

Taylor K 2D

What did silver say to gold at the bar?

"Ay you (Au), get outta here!"

Taylor K 2D

Two chemists walk into a bar. One tells the bartender, "I'll have an H2O." The other says, "I'll have an H2O too!" The second chemist dies.

Taylor K 2D

That's a good question. I would assume that resonance structures exist to demonstrate the possibility of different ways the compound can exist, and then the formal charges help us figure out the probability of those structures and their contribution to the resonance hybrid and the actual bond length...

Taylor K 2D

The lone pairs are drawn that way also because the two hydrogen bonds cannot be directly across from each other, because the repulsion from the two lone pairs would be too strong. The only stable way to place the lone pairs is "next" to each other.

Taylor K 2D

For things like this, I usually also guess and check, or try and calculate the formal charge because that helps me identify where the issues with my structures are.

Taylor K 2D

The lewis structure of AsF3 reveals that there is one lone pair on the central atom, As, with three single bonds. This shows that AsF3 is a polar molecule (its not symmetrical and there is a clear dipole). AsF5, on the other hand, is nonpolar, as revealed by its symmetrical lewis structure. Polar mo...

Taylor K 2D

Im not completely sure, but I believe the 5s sub shell comes before the 4d sub shell only when the 5s sub shell is incomplete and the 4d sub shell has no electrons. If the 5s sub shell is complete and there is electrons in the 4d subshell, then the 4d subshell will be placed and filled before the 5s...

Taylor K 2D

Great question! Ba-Br does have a greater EN difference than Be-Br because Ba is less electronegative. As you pointed out, EN increases going up and across the periodic table. Br is much more electronegative than Ba and Be. Be is more electronegative than Ba. This means that the greatest difference ...

Taylor K 2D

London (dispersion) forces were referred to as Van Der Waals forces and induced dipole-induced dipole forces in lecture. I believe that Professor Lavelle usually refers to them as induced dipole-induced dipole forces because that helps clarify and remind exactly what those forces do. These forces ex...

Taylor K 2D

I understand that the Midterm will cover material from the last 8 topics in Outline 2 to the end of Outline 3. In outline 2, it is mentioned that we will need to be able to "Describe the interpretation of atomic orbitals in terms of probability". In the textbook section 1D.4, there is an o...

Taylor K 2D

Catherine Bubser 2C wrote:Does is matter the order in which these orbitals are written as long as the n number are in order?

It does matter the order in which they are written because different orbitals have different energy levels so electron configuration should be arrange in a specific order.

Taylor K 2D

Yup! That looks right to me!

Taylor K 2D

Does anyone know Lavelle's grade breakdown/rubric, i.e. how many points equal an A/A-/B/etc.?

Taylor K 2D

I loved Gravity Falls as a kid and still do now. Some great shows I've watched recently are New Girl, Sherlock, and Bojack Horseman. I also plan on watching Brooklyn 99. Omg I love all of those shows (except Bojack Horseman cause I haven't seen that one yet). I also love this show called Psych, but...

Taylor K 2D

wait wow I feel stupid now. Thank you lol

Taylor K 2D

In Monday's lecture, Professor Lavelle mentioned that bond length was measured in amperes. I thought amperes were a measure of energy, so why are they used as the units for a measurement of length?

Taylor K 2D

when converting to moles from number of atoms or vice versa, you would need to use Avogadro's constant.

Taylor K 2D

Yup! It also helps to know which energy levels correspond with which series/radiation (Lyman= uv = n=1 and Blamer=visible=n=2).

Taylor K 2D

Here are some chem memes I made in my high school chem class lol

Taylor K 2D

These are memes I made when I took chem in high school lol.

Taylor K 2D

You can only simplify to the previous noble gas (last column in the periodic table). This is because the noble gases have their respective orbitals and sub shells completely filled. You cannot simplify using any other elements. For example, if you were writing the electron configuration for Aluminum...

Taylor K 2D

In Monday’s lecture, Professor Lavelle talked about how for multi-electron atoms after atomic # 20, 4s is higher energy than 3d but I’m still confused as to why. Could someone maybe clarify that please?

Taylor K 2D

"ℓ" is the quantum number that represents the angular momentum of an electron. It is used to identify which orbital the electron is in (via shape). In this case, n=6 and ℓ=0 because in an s-orbital, ℓ=0. I hope that helps!

Taylor K 2D

We can't really know for sure unless the professor posts the correct 19 questions :(

Taylor K 2D

Ok that makes more sense thank you so much!

Taylor K 2D

Does anyone know if we are allowed to have scratch paper on the midterm?

Taylor K 2D

In lecture today, Professor Lavelle mentioned this equation briefly:

ψ²≡ e⁻ρ distribution

And I am very confused. I don’t really know what this means. Can someone please explain this to me?

Taylor K 2D

I think its 1/lambda

Taylor K 2D

I believe it is because kg is the fundamental SI unit for mass. But I'm not sure.

Taylor K 2D

A black body is something that absorbs all light, but Lavelle said we didn't need to know that.

Taylor K 2D

I'm probably overthinking this, but I don't really understand how to determine the number of spectral lines based on given energy levels.

Taylor K 2D

Will we need to memorize which wavelengths belong where on the spectrum/ which types of light correspond with which wavelength? If not do you think we'll be given access to this during the test?

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