CHEMISTRY COMMUNITY

Emma Chang 1G

I would try to memorize the strong acids and bases from this list, and then everything else is either a weak acid or base. Hope this helps!

https://sites.google.com/site/chempendi ... cids-bases

Emma Chang 1G

That makes sense, I forgot to account for resonance. Thank you so much!

Emma Chang 1G

Thank you so much, I needed this! Good luck guys, we're almost there!

Emma Chang 1G

I usually look up lo-fi music on Youtube. I find that it works well as background noise without being too distracting!

Emma Chang 1G

Place the following molecules or ions in order of decreasing bond length: The SO bond in SO2, SO3, SO3^2- The answer was SO3^2- > SO2 = SO3 (SO2 and SO3 bond length are about the same) When I drew out the Lewis structures for all three of these, I thought all of them had only double bonds in their m...

Emma Chang 1G

I wasn't able to attend Dr. Lavelle's Zoom Review on Wednesday, and I was confused about the third slide from the bottom. https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/Review_Zoom.pdf I was wondering if anyone could tell me how we determine that there are eight H-bonding sites wi...

Emma Chang 1G

I'm going to try to review all the textbook questions we've done throughout the quarter to get more practice. I'll probably go through the outlines and make sure I understand the bullet points, and then I'll read through my textbook and lecture notes to hopefully refresh the material. I'm planning o...

Emma Chang 1G

On Dr. Lavelle's website, the toolbox isn't listed as one of the sheets we are allowed to use on the final. I would probably memorize them just to be safe!

Emma Chang 1G

Hi I was wondering if anyone could explain to me how you would go about naming [Co (NH3)3Cl3] I thought it would be triamminetrichlorocobalt(II) I think Dr. Lavelle addressed something similar to this in a post from a few years ago. In that example, the overall charge of the complex was -1, since C...

Emma Chang 1G

I don't think we'll get credit necessarily, but my TA asked us to fill them out since it would be really helpful for them, especially since everything is a little different because we're online.

Emma Chang 1G

I think we use OH2 since O is the one binding to the central atom, but I think either is fine since the final is multiple choice.

Emma Chang 1G

All single bonds are sigma bonds, and both double and triple bonds each have one sigma bond. The remaining bonds left over for double and triple bonds are pi bonds. A double bond has one sigma bond and one pi bond, and a triple bond has one sigma bond and two pi bonds. Hope this helps!

Emma Chang 1G

For the final I'm going to really make sure I know how to do all the textbook problems we've been given, especially the trickier ones. If I can, I'll go through the video modules from the first few topics again since we haven't touched on that material in a long time. I'm also going to try to attend...

Emma Chang 1G

I try to exercise in the morning to get the day started, but I've really been enjoying walking my dog while listening to music during my study breaks!

Emma Chang 1G

I think the elements starting in the p-block of the third period and all the elements after can have an expanded octet, and I believe phosphorus, sulfur and chlorine are some fairly common ones that can accommodate more than eight e-. Hope this helps!

Emma Chang 1G

I think the quantities of each atom in you drawing are correct, but I believe Dr. Lavelle said that carbon cannot have an expanded octet and almost always forms four bonds. You would have to draw the carbon atom bonded to three of the hydrogen atoms and the one oxygen atom, and the remaining hydroge...

Emma Chang 1G

We learned that chromium and copper were two exceptions for electron configurations, but I think Dr. Lavelle said that the elements in the same group as those two also follow that exception. I believe having a 4d shell that is half full or completely full makes it more stable, so you would pull an e...

Emma Chang 1G

Ionization energy increases as you go across a period and decreases as you go down a group. As you go across the period the number of protons in each element increases, which draws the electrons in more tightly and makes them more difficult to remove. As you go down a group, the valence electrons ar...

Emma Chang 1G

I believe a coordinate covalent bond occurs when the Lewis base contributes both electrons to form a bond with the Lewis acid. The Lewis acid needs two electrons to complete its octet, so the Lewis base supplies both of these electrons.The Lewis base is the electron donor and the Lewis acid is the e...

Emma Chang 1G

You should have 35! I think the five posts for each week are due at midnight on Sunday of that week!

Emma Chang 1G

The boiling point is the temperature at which a substance changes from a liquid phase to gas phase, and the melting point is the temperature at which a substance changes from a solid phase to a liquid phase. I believe both follow the same trends in relation to IMF. Boiling and melting points are hig...

Emma Chang 1G

Since the midterm is multiple choice, I'm assuming that we won't have to draw out Lewis structures like we did for the Sapling hw. My guess is that we'll have to identify the one Lewis structure out of a few structures that has the lowest total formal charge, since this one would generally be the mo...

Emma Chang 1G

This is great! I also wasn't totally clear on identifying Lewis acids and bases so this is definitely helpful.Thank you so much!

Emma Chang 1G

Yes I think it should say SO2 instead of H2S. I said it was still dipole-dipole since the slightly positive S end of one molecule is attracted to the slightly negative O end of the other molecule.

Emma Chang 1G

I think the electronegativity difference between C and H is not big enough for the molecule to be considered polar. Since CH4 is non polar, London (dispersion) forces are the only ones that apply. Hope this helps!

Emma Chang 1G

I'm sure a lot of us are in the same boat now! It's definitely stressful when each question carries a lot of weight, but all we can do is try our best! I found that the workshops and step-up sessions help a lot! In the last few days leading up to midterm 1 I tried to review all the lectures notes, d...

Emma Chang 1G

I believe the formatting will be the same! Dr. Lavelle's email said that the questions are generally shorter and are worth fewer points, so I don't think that the number of questions will necessarily be the same.

Emma Chang 1G

1/s is 1 Hz (Hertz, or cycle per second), and the units for Joules are kg.m^2.s^-2.

Emma Chang 1G

Hi! I find the that having things to look forward to after sitting through a lecture or getting a good amount of homework done is pretty helpful for me. Usually that'll be a snack or some time to work on a hobby I enjoy. I also have a hard time studying on a bed or a couch since it's easy to fall as...

Emma Chang 1G

I think there are higher sub shells but we don't necessarily need to know them for this class. We probably just need to know the s, p, d, and f subshells, but even so I heard that we will deal with f subshells less than the other three. Hope this helps!

Emma Chang 1G

I'm think it might depend on your TA, mine didn't mention anything about not having discussion but he also doesn't have any discussion sections on Wednesday. I don't think Dr. Lavelle has sent out an email yet, but hopefully we'll have some clarification soon!

Emma Chang 1G

I don't think we have to draw double bonds shorter than single bonds since Lewis structures are simplified models. I think it's fine to draw them with the same lengths as long as we understand conceptually that double bonds are shorter than single bonds!

Emma Chang 1G

The energy of the incoming photon must be greater than or equal to the work function in order to eject electrons. Therefore, the energy of the incoming photon must be at least 4.540×10−19 J. You would use this in the equation E = (hc)/λ, or rearrange it to λ = (hc)/E, to solve for the minimum wavele...

Emma Chang 1G

I'm not sure but I thought that n1 refers to the final level that the electron returns to, while n2 is the initial level. I don't necessarily think that you would always subtract the higher level from the lower one; if the electron absorbs energy and is excited to a higher level, n1 would be higher ...

Emma Chang 1G

I'm not entirely sure but I think it's a manipulated version of Einstein's equation E = mc^2. You could rewrite this as E = mcc, and since c is the speed of light (velocity constant) you could write it as E = mvv. Since momentum (p) = mass (m) x velocity (v), you could write E = (mv)v as E = pv. Goi...

Emma Chang 1G

I believe R = 3.28984 x 10^15 is used for the equation v = R[(1/n1^2)-(1/n2^2)]. I think they get 1.0966 x 10^7 when they set the equation equal to 1/λ instead of v. Since v = c/λ, you could write the equation as c/λ= R[(1/n1^2)-(1/n2^2)], and then you would divide c over to get 1/λ= (R/c)[(1/n1^2)-...

Emma Chang 1G

Hi! This is what I know so far: 1) The midterm is taken during your regular discussion section with your TA 2) I think Dr. Lavelle recommends an external webcam, but I would ask your TA. Some of them are letting students use their phones to go on zoom while they take the test on our computer. Throug...

Emma Chang 1G

I think you multiplied the other side by delta p, but both sides should be divided by delta p. So the equation should be Δx ≥ h/(4π x Δp), or
Δx ≥ (6.626 x 10^-34 J.s)/(4π x 80.0 kg.m.s^-1). Hope this helps!

Emma Chang 1G

I think Dr. Lavelle sent out an email saying he wants us to print them out before the test, and your TA will have to check them before the midterm starts to make sure no one wrote anything on the paper beforehand.

Emma Chang 1G

I'm not sure but I got 1.8666 x 10^-9 m? I used 810 nm to calculate the energy of the incoming photon and converted the 1.1 ev to joules for the work function, then subtracted the work function from the energy of the incoming photon to get the kinetic energy of the electron. I then used Ek = 1/2 mv^...

Emma Chang 1G

The given 2.50×10^16 Hz is equal to the frequency of the work function, or the minimum frequency needed in order to eject an electron. The energy of the incoming photon (E=hv) must be, at the minimum, equal to the work function in order to eject an electron. You would plug 2.50×10^16 Hz (v) into the...

Emma Chang 1G

I keep all of the digits in my calculator until the end, and then I would round that final answer to the right amount of sig figs. My TA said that sig figs won't be a major issue, in that the multiple choice answers should be different enough that the right answer should be clear (ex. the options sh...

Emma Chang 1G

In the photoelectric effect, I think electrons are ejected as long as the energy of an individual photon is greater than or equal to the threshold energy, or the amount of energy needed to eject an electron from the metal surface. In an atomic spectroscopy experiment, the electron is excited when th...

Emma Chang 1G

When an electron has been ejected, it has no interaction with the nucleus, so its energy is said to be zero. The electron can later reattach back to the atom and descend energy levels as it emits energy. Since the reference point is zero and the electron emits energy as it moves to lower energy leve...

Emma Chang 1G

When an electron has been ejected, it no longer reacts with the nucleus. I believe the energy of the electron at this point is said to be zero. The electron can later attach back to the atom and move down energy levels, which means it is losing energy. Since the reference point is zero and the elect...

Emma Chang 1G

I think when an electron is excited, it jumps up in energy levels (such as jumping from the n=1 level to the n=2 level). This occurs when the energy of the photon matches the difference in energy between levels. When an electron is ejected, it is removed from the atom completely, and the electron no...

Emma Chang 1G

When you increase the intensity of the light, you increase the number of photons emitted. The number of photons emitted and the number electrons ejected is a 1:1 interaction, as long as each individual photon has enough energy to eject one electron. Increasing the number of photons emitted will also...

Emma Chang 1G

I think in the lecture he said centi, milli, micro, and nano would be used pretty often in chemistry, but the other ones would be good to know as well!

Emma Chang 1G

I think so! You would convert the number of moles into atoms, and then multiply the number of atoms by 2 times the given radius in order to get the total distance. You have 6.022 x 10^23 Ag atoms (since 1 mol = 6.022 x 10^23 formula units). Then you multiply this by the diameter 288pm (2 x the radiu...

Emma Chang 1G

I would first convert moles to the number of atoms. Since you are given 1.00 mol, you know you have 6.022 x 10^23 atoms (since 1 mol = 6.022 x 10^23 formula units). Since they give you the radius of the atom, you multiply this number by two to get the diameter, which is 288 pm. You then multiply 6.0...

Emma Chang 1G

I believe he said stoichiometric coefficients must be the lowest possible whole number, given that the equation is still balanced. If all of the stoichiometric coefficients are divisible by a common factor at the end, it would probably be beneficial to simplify the equation to make future calculatio...

Emma Chang 1G

I believe he said in his example of 2Na + 2H2O --> 2NaOH + H2 that chemists will more commonly refer to the reactants and products of a chemical equation in terms of moles, rather than the elements themselves. For example, he referred to 2Na as 2 moles of Na rather than 2 Na atoms. He just rephrased...

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