CHEMISTRY COMMUNITY

Shanna Yu 1C

Ahh, thank you for this. I haven't even looked at my score for the previous one because I get too anxious thinking about it, and now I'm stressing over the upcoming one. This helps a little bit; I really appreciate it!

Shanna Yu 1C

Hi! You kind of already illustrate how internal energy and enthalpy relate to each other in your question, but I can maybe rephrase it in a way where it's more obvious: H (enthalpy)= U (internal energy) + PV (work). Enthalpy is just the sum of the internal energy of the system and the product of pre...

Shanna Yu 1C

Hi! I'm guessing you're asking about #7 from the Week 5/6 homework? Let's look at CH3OH. What I did first was take the difference between the final and initial masses, then divide that by CH3OH's molar mass to get the mass difference in mols. Since we want to know how much of the heat supplied went ...

Shanna Yu 1C

Hi! I also got a very large K value for the first part of that question. I can only imagine this means the reaction just has a high concentration of products at equilibrium. I believe that each reaction has its own K value, though this is dependent on the given temperature the reaction takes place a...

Shanna Yu 1C

Hi! I answered this question for someone else, so I'll paste it here, but I will modify it a bit to fit your specific question. Let's look at the particles inside each container. Container A has a monoatomic ideal gas, which means it's going to be more disorderly than a diatomic one, as there's mor...

Shanna Yu 1C

Hi! I answered this question for someone else, so I'll paste it here, but I will modify it a bit to fit your specific question. Let's look at the particles inside each container. Container A has a monoatomic ideal gas, which means it's going to be more disorderly than a diatomic one, as there's mor...

Shanna Yu 1C

Hi! Could this be what you're looking for? The columns have the standard enthalpies of formation for certain compounds, delta hF. https://sites.google.com/site/chempendix/thermo Oh, and just a heads up, I think a general policy is to post or include the question you're referring to in your post. Lem...

Shanna Yu 1C

Hi! I answered this question for someone else, so I'll paste it here, but I will modify it a bit to fit your specific question. Let's look at the particles inside each container. Container A has a monoatomic ideal gas, which means it's going to be more disorderly than a diatomic one, as there's more...

Shanna Yu 1C

Hi! Let's look at the particles inside each container. Container A has a monoatomic ideal gas, which means it's going to be more disorderly than a diatomic one, as there's more molecules to bounce around and do their chaotic, molecule-y thing. This automatically puts it above Container B and C, as t...

Shanna Yu 1C

Hi!

I was under the impression that there wouldn't be, as it's a holiday. Can someone confirm?

Shanna Yu 1C

Hi!

I think there are step-up sessions you can attend where you go over problems and practice concepts with some TAs! I heard those really help people.

Shanna Yu 1C

Hi! I just answered this for someone else, so I'll just copy and paste what I wrote there, if you don't mind. I'd start out by balancing that reaction's equation, then make sure you know which one the limiting reactant is. Should be zinc. You want deltaH for this reaction, which you can find by addi...

Shanna Yu 1C

Hi! I'd start out by balancing that reaction's equation, then make sure you know which one the limiting reactant is. Should be zinc. You want deltaH for this reaction, which you can find by adding up the heat of formation for the products — heat of formation for the reactants. For ZnCl2, you can fin...

Shanna Yu 1C

Hi! Did you use deltaU=q + w? q is just nCdeltaT, where n is your sample size, C is 4R (as given to you in the question), and deltaT is your change in temperature, 17.9K. Once you’ve plugged that all in, you have your q. Work done at constant volume is zero. That means we’re just doing deltaU=q+0, w...

Shanna Yu 1C

Hi! The heat required to melt ice + heat required to bring said ice to Tfinal (final temperature) = heat released to bring all 394g water to Tfinal. I used q= For each of the three components up there, with delta T written out as (Tfinal-Tinitial) so that you can isolate and solve for Tfinal. It’s b...

Shanna Yu 1C

Hi! This is kind of really messy, but I’ve attached a link to a picture of my work for that question. Since you say you’ve already read through the other posts regarding this Sapling question, I hope you’ll get what exactly it is I did, but if you don’t, lemme know and I’ll explain! https://imgur.co...

Shanna Yu 1C

Hi! I've answered this question for a couple of others, now, so I'll just copy and paste my answer here. I'll be using 6.D.15's part b as an example, if you don't mind. So for this, I think it's good to know that Cl won't be included in this whole thing, as it doesn't affect the pH. Al3+ is the (we...

Shanna Yu 1C

Hi! I've answered this question for a couple of others, now, so I'll just copy and paste my answer here. I'll be using 6.D.15's part b as an example, if you don't mind. So for this, I think it's good to know that Cl won't be included in this whole thing, as it doesn't affect the pH. Al3+ is the (wea...

Shanna Yu 1C

Hi!

I'm under the impression that when the muscles make lactic acid, the haemoglobin will react with its H+ (as it's an acid) to create HHb, which will set the O2 free, in a way. This means the concentration of HbO2- is lowered in the presence of lactic acid.

Lemme know if you need clarification!

Shanna Yu 1C

Hi!

That stands for combustion.

Shanna Yu 1C

Hi! So for this one, the reason why the answer key shows a bunch of balanced equations is because you need to do that here. The end enthalpy of something is just the final enthalpy minus the initial enthalpy, so what'd you do after balancing everything is sum up the enthalpies for your reactants, su...

Shanna Yu 1C

Hi! I got rid of the NO2 first by multiplying the first reaction by 2 so that it cancels out with the other 4NO2 in the second equation. From there, you can kinda piece together the overall reaction, so what you'd do then is add their deltaH together, just with the first deltaH multiplied by two for...

Shanna Yu 1C

Hi! From my understanding, you used the ICE table to find the Ks for both Cl2 and F2, right? The K of chlorine ends up being smaller than the K of fluorine, which means that chlorine favours its reactants more than fluorine does (products over reactants). Something that favours its reactants more me...

Shanna Yu 1C

Hi! When you see a reaction start out as one compound and end up as two or more separate components (breaking bonds), that means the reaction takes energy to happen, which means it's endothermic (taking in energy). Endothermic corresponds to a positive value. When you see a reaction that forms a com...

Shanna Yu 1C

Hi! I actually answered this for someone else, so I'll just copy and paste it here. So for this, I think it's good to know that Cl won't be included in this whole thing, as it doesn't affect the pH. Al3+ is the (weak) acid, here, as it attracts water molecules enough that said water's hydrogen atom...

Shanna Yu 1C

Hi! I feel as if I'm not quite getting your question, as to me, when you have to calculate a strong acid's dissociation in water, it's just the molarity of said strong acid (because as you said, you're assuming monoprotic and also that strong acids dissolve completely). I'm either misreading, or you...

Shanna Yu 1C

Hi! I actually answered this for someone else, so I'll just copy and paste it here. So for this, I think it's good to know that Cl won't be included in this whole thing, as it doesn't affect the pH. Al3+ is the (weak) acid, here, as it attracts water molecules enough that said water's hydrogen atoms...

Shanna Yu 1C

Hi! The person above answered your question, but I'm adding this on to provide a more complete/detailed walkthrough of what you need to do. I think percent deprotonisation is your concentration of hydronium ions over the initial concentration of the acid itself. With this concept in mind, we've got ...

Shanna Yu 1C

Hi! So for this, I think it's good to know that Cl won't be included in this whole thing, as it doesn't affect the pH. Al3+ is the (weak) acid, here, as it attracts water molecules enough that said water's hydrogen atoms break off. The reaction goes like this, I think: Al(H2O)6 3+ + H2O ---> H30 1+ ...

Shanna Yu 1C

Hi! I answered a question similar to this one last quarter. When pH equals pKa, the concentrations of the conjugate base and conjugate acid are equal, so that's one way to relate them. At any rate, you know that pH is basically a measure of the concentration of hydronium ions. pKa, a.k.a. the acid d...

Shanna Yu 1C

Jazlyn Romero 1I wrote:Thank you both so much! So helpful!!

No problem! |>

Shanna Yu 1C

Hi! I was under the impression that we say the change is negligible when the K is smaller than 10^-3. 10^-4 is also a small enough K value, but Lavelle did say in his most recent lecture that the cutoff is 10^-3. Just thought that clarification might help. On your other concern, I recall that we are...

Shanna Yu 1C

Hi!

Andrew answered your question well, as you do need an ICE table, but here’s a (kinda messy) step-by-step I did for you, just in case you need any further help!

https://imgur.com/2zMypSI

Shanna Yu 1C

Hi! I cross-checked several different sources for ya to see what was more commonly used, and the general consensus (invariably, actually) is 10^-3. I also agree with the above poster in that it's in your best interest to base your calculations off of what he says in the most recent lecture, as no on...

Shanna Yu 1C

Hi all, I'm having some trouble answering a sapling question. reactants : H2(g) + I2 (g) product: 2HI (g) Kc= 53.3 Given: 0.700 mol H2 and 0.700 mol I2 placed in 1.00 L container Question: what concentration of HI present at equilibrium? I have my ICE table correct but I think I am getting stuck wi...

Shanna Yu 1C

Hi! It took me a second to figure this one out, too, but here's how it goes (assuming you did your ICE table correctly): You have Kc=[HI]^2/([H2][I2]. What you're going to do now is express Kc in terms of x, here, where your ICE table should reflect that H2 is .3-x and I2 is also .3-x, while 2HI is ...

Shanna Yu 1C

Hi!

The thing you pointed out in part D is actually a typo; the H2 should be on the right side, alongside the D2. Your proposed alternative explanation is correct in that having H2 on the other side would justify the given answer!

Shanna Yu 1C

Hi! From experience, a lot of the stuff we learned in 14A are mostly fundamentals that we end up applying in concept/calculation to things we learn later down the line, but I think a nice one to keep in mind is the acid-base topic and concepts we touched on towards the end of 14A, as we'll definitel...

Shanna Yu 1C

Hi!

Someone can correct me if I'm wrong, but I think it's because you need to convert to the proper SI units when calculating K for stuff. For pressures, this is atm or bar. 100 kPa per 1 atm. :>

Shanna Yu 1C

I didn't take chem 14A last quarter (I took 20A), so I recently purchased the sapling course from the bookstore. Is anyone else still waiting for the code to be mailed to them? hi! from what i recall, a lot of people last quarter were experiencing delays and were also waiting for the book code, lik...

Shanna Yu 1C

hey, this is unrelated to your question since it's already been answered, but your screen name is currently your UCLA id number. your login username should be your id number, but your screen name can just be your name + discussion session instead of your id (for security reasons hahaha)

Shanna Yu 1C

Hi! Large objects with large masses have such small uncertainties in their positions that it's very, very hard to see with anything we have available to us, much less to the naked eye. It's not that it's not applicable, just that the uncertainty for larger objects is too small to be noticeable or si...

Shanna Yu 1C

Hi! I'm not sure if I'm reading your question properly, but electrons get excited when they get energy, be it from absorbing photons that come their way or collisions with other atoms/particles (transfer of kinetic energy). They are emitted when the photons match the threshold energy, which will exc...

Shanna Yu 1C

Hi! They aren't quite the same thing. When pH equals pKa, the concentrations of the conjugate base and conjugate acid are equal, so that's one way to relate them. I'm not sure if Dr. Lavelle will go over that in his final lecture tomorrow. At any rate, you know that pH is basically a measure of the ...

Shanna Yu 1C

Hi!

If I'm not mistaken, the oxygen in this compound has three electron densities: two lone pairs and one bonding site (a double bond, in this case). Because there are three, it makes sense that its hybridisation is sp2, as s(1)+p(1)+p(another 1)=3.

Shanna Yu 1C

Hi! I answered someone else who had the same question you did a while back, and here's what I said: s-character is just another way of saying the extent of sigma bond character in a hybridised structure. With that in mind, angles increase when you increase the s-character. Let me know if you need cl...

Shanna Yu 1C

Hi! We only add a hydrate if there is a hydrate in the compound. An example of this would be Na2Co3 . 10H20, where the period is more of a multiplication dot than a period. Anything with these H2Os tacked on at the end qualifies as a hydrate. In the example's case, this would be sodium carbonate dec...

Shanna Yu 1C

Hi!

s-character is just another way of saying the extent of sigma bond character in a hybridised structure. With that in mind, angles increase when you increase the s-character.

Let me know if you need clarification/elaboration! I can explain further.

Shanna Yu 1C

Hi! Someone can come by and correct me if I'm wrong, but I believe the porphyrin ligand in this case is made up of the nitrogen you mentioned, except they're also all connected by carbon atoms, to simplify things (it's got bridges n' stuff that aren't super relevant to this question). They're kind o...

Shanna Yu 1C

Hi! I believe a good way to tell is if there are any lone pairs present, as the strong repulsions they have with each other and with other atoms in the molecule are usually the reason why bond angles end up being less than the standard 109.5, 120, etc.,. We can take H 2 O as an example. If it didn't...

Shanna Yu 1C

Hi! I don't think transition metals ever have a negative oxidation number (or even an oxidation number of 0) when we're dealing in terms of naming compounds with transition metals, so they never have negative charges. Negative ions with negative oxidation states don't need Roman numerals, and transi...

Shanna Yu 1C

Chem_Mod wrote:Hexacyanoferrate(II) would be correct.

Aah. Now my reply looks redundant, as Doctor Lavelle beat me to it by a few seconds, hahaha. Please ignore mine.

Shanna Yu 1C

Hi!

I think it's most commonly referred to as ferrocyanide, but for a name that reflects its oxidation number and other conventions we learned of, I believe it's hexacyanoferrate(II).

Shanna Yu 1C

Hi! The Roman numeral simply reflects the oxidation number/charge of the metal ion in question. You can look up videos if you're unsure of how to find the oxidation number of an atom, but I'll give an example here, too (this is taken directly from one of the lectures): Let's say we have [Fe(CN) 6 ] ...

Shanna Yu 1C

Hi! I believe this works for very nearly everything: Find the atom you want. Then count the number of atoms that are connected to it. Then count the number of lone pairs that are on it. Add those two together. 4=sp3 3=sp2 2=sp 1 is hydrogen, likely. Alternatively, if you know the molecular geometry:...

Shanna Yu 1C

Hi! Someone can correct me if I'm wrong, but I believe it typically has to do with the idea that lone pairs want to be as far away from each other as possible, so whatever arrangement allows for the greatest distance apart possible will be what they settle into. Having all three lone pairs on the eq...

Shanna Yu 1C

Hi! Ozone's central oxygen atom actually has three regions of electron density—one double bond, one single bond, and a lone pair on the central atom. With this in mind, its molecular geometry is actually bent, which kind of looks like an obtuse angle, for lack of a better comparison. This is consist...

Shanna Yu 1C

Hi!

Someone can correct me if I'm wrong, but I believe combustion problems typically operate under the assumption that oxygen is in excess unless it's stated otherwise.

Shanna Yu 1C

Hi! A region of electron density can be defined as things like bonds (double or triple bonds count as just one region of electron density) and lone pairs. With this in mind, let's take a simpler molecule, like carbonate, or CO3^2-. When drawn out, it has three bonds—two single bonds and one double b...

Shanna Yu 1C

Hi!

Ideally, we'd want the central atom to be as stable as possible. Stable=formal charge of zero, so it'd be unfavourable to have charges on the central atom, even more so negative charges because the central atom is typically the least electronegative one of the bunch.

Shanna Yu 1C

Hi! Ionic compounds have high melting points. This is due to the really strong attraction between cations and anions. That being said, Mg does have a 2+ charge, which is higher than Na's 1+, but I think it helps to keep in mind that Mg is bonding with two Cl atoms, which means it's 2+ matching up wi...

Shanna Yu 1C

Hi! Without a specific example, I'm afraid I'm going to have to speak in more general terms, but I believe what you need to know is electronegativity. We figure out which end of the molecule the negative pole is by asking ourselves which of the atoms/ends is more electronegative than the other, as t...

Shanna Yu 1C

Hi! Someone can correct me if I'm wrong, but I think a big part of the reason behind this effect is because (typically for heavier p-block elements) the ns2 electrons that make up the inert pair aren't quite as well-shielded from the pull of the atom's nucleus as the p-electrons are (ineffective shi...

Shanna Yu 1C

Hi! I'm afraid I may not be fully understanding your question, but from what I know of Coulomb's law and electrostatic potential... You're correct in the charge and distance between charges bit, but that concept and looking at effective nuclear charge aren't mutually exclusive as your question impli...

Shanna Yu 1C

No problem! :>

Shanna Yu 1C

I forgot to note that the textbook says the answer is 6 valence electrons. I'm assuming the textbook assumes that the student doesn't know about the half-filled orbital for stability thing, as Co3+ tends to be a bit of an exception. If we were doing it conventionally (i.e. following the set rules),...

Shanna Yu 1C

Hi! SO2 is a polar molecule. You can tell because of its molecular geometry and the idea that it has lone pairs on the central atom (S), but I'm not sure if we've gotten to VSEPR theory and all that just yet. That being said, because it's a polar molecule, the interaction between two SO2 molecules i...

Shanna Yu 1C

Hi!

If I'm not mistaken, the electrons that are taken out are as follows:

One electron from 4s2 (half-filled orbital is more stable)

Two electrons from 3d7

This gives an electron configuration of [Ar] 4s13d5.

Shanna Yu 1C

Hi! I answered this earlier under your other question relating to Q4, stating that it was connected to this question. I explained both this and the aforementioned other question, so I'll just repeat my answer here! The experimentally determined bond lengths imply that the bonds within the resonance ...

Shanna Yu 1C

Hi! This question is based off of the one before it. The experimentally determined bond lengths imply that the bonds within the resonance structure contain both C(double bond)N and C(double bond)O character. You can tell, because the C—O bond length 128pm falls closer to the length for C(double bond...

Shanna Yu 1C

Hi! If you're familiar with how each periodic trend relates to each other, then the only one you really need to remember is electronegativity. If you remember that fluorine is the most electronegative element (i.e., electronegativity increases going up and to the right), then you can figure out all ...

Shanna Yu 1C

Hi! As you may know, fluorine only needs one more electron to have a complete octet. In simpler terms, this means it really, really wants to get the one electron that it needs to complete said octet, which means it's pretty electronegative. Now, as to why it happens to be the most electronegative ou...

Shanna Yu 1C

Hi! I don't think we've actually covered this material yet, but here we go! A Lewis acid accepts electrons in a reaction. It gets said electrons from a Lewis base, which donates electrons. With this in mind, let's take a look at your first one: Cl- + AlCl3 ---> AlCl4- If we just look at the reaction...

Shanna Yu 1C

Hi! When it comes down to formal charges, the reason we want a low value (preferably 0, a.k.a no formal charge at all) in Lewis dot diagrams is because a 0 indicates a stable molecule, and molecules tend to want to stabilise, which is why the structures that are being resonated between typically hav...

Shanna Yu 1C

I agree with most of the others in the audio-visual topics and the textbook problems! I've only been to one workshop so far and that was pretty okay, but the Sapling questions as well as the homework on the syllabus are what help me feel more prepared for things. :>

Shanna Yu 1C

Hi! They're related. The atomic spectrum/spectra is/are the frequencies of electromagnetic radiation that's emitted or absorbed when an electron makes a transition between energy levels. An electron emits energy when it's returning to original/lower energy levels and takes in energy when it's jumpin...

Shanna Yu 1C

This is super helpful! Thank you so much.

Shanna Yu 1C

Hi! As energy levels increase, the pull that the protons at the nucleus (centre of the atom) has on the electrons within those levels decreases. It does have to do with effective nuclear charge, as you said—the farther away the electron is from the atom, the smaller effective nuclear charge it feels...

Shanna Yu 1C

Hi! The way I figure out electron configs is by looking at the periodic table. This diagram should help: https://guidancecorner.com/wp-content/uploads/2018/05/electron-configuration-and-the-modern-periodic-table.png The way he wrote the last part is the same as if he were to write 2p^2, but he said ...

Shanna Yu 1C

Hi!

It will not help them. I interpreted this as a bit of a joke question, but someone can come by and correct me if I'm wrong.

edit: To clarify, I don't mean that your question itself is a joke. I was referring to the problem in the module. :P

Shanna Yu 1C

Hi! The quantity of photons doesn't matter in this case, where light is behaving like particles. Increasing the number of photons only increases the light's intensity, which isn't directly related to the variables the question gives you. It's kinda like in the photoelectric effect, where increasing...

Shanna Yu 1C

Hi!

Here's what I did:

v=R(1/2^2 - 1/3^2), where R=3.29x10^15

Then I plugged v into wavelength=C/v to get the wavelength in metres.

Then I multiplied that by 10^9 to get it in nanometres.

Lemme know if you have any questions!

Shanna Yu 1C

Hi! So when you think about how many transitions are in between the n=6 shell and the ground level (n=1), there's five, right? So that's the number of spectral lines. So now you want the wavelengths for a transition between n=6 to n=1 (the lower wavelength) and n=6 to n=5 (the higher wavelength). Yo...

Shanna Yu 1C

Hi! The quantity of photons doesn't matter in this case, where light is behaving like particles. Increasing the number of photons only increases the light's intensity, which isn't directly related to the variables the question gives you. It's kinda like in the photoelectric effect, where increasing ...

Shanna Yu 1C

Hi! Use the wavelength you're given in DeBroglie's equation, where wavelength(electron)=h/p or h/(mv). Solve for v by plugging in the mass of an electron and Planck's constant. Then plug in the velocity you just solved for into E(electron)=1/2mv^2 to solve for E(electron), again plugging in the mass...

Shanna Yu 1C

Hi!

I ran the same calculations you did:

E=hv, E=(6.626x10^-34)(1.28x10^15), E=8.48128x10^-19

Then I plugged it into E(photon)=(work function) + KE:

(work function)=(8.48128x10^-19)-(3x10^-19), (work function)=5.48128x10^19 J. Is this the answer you're looking for?

Shanna Yu 1C

Hi! I think a good keyword here is wavelength, which is what part a asks you for. De Broglie's formula is wavelength=h/p or h/mv, so finding the wavelength using this formula is the fastest, most efficient way instead of using an equation for kinetic energy. The way I see it, if you're asked for wav...

Shanna Yu 1C

Hi!

I think you're correct, the answer is D.

Shanna Yu 1C

Hi!

The mass of an electron is 9.109x10^-31 kg. It goes where m is in Ek=1/2mv^2. Hope that helps!

Shanna Yu 1C

I found a link for Discussion 3C!

https://groupme.com/join_group/63065999/Dx0k7JYq

Shanna Yu 1C

Hi!

I did a search for Groupme discussion links on both the Discord server and the Lecture 2 Groupme and don't think a Groupme for 2I exists yet.

Shanna Yu 1C

Hi! While I didn't exactly need or use what you're asking about to solve this question, I don't think we're expected to know them specifically (personally, I looked their molar masses up). If you're wondering where the H on the product side comes from, though, propyls and propanes are compounds comp...

Shanna Yu 1C

Hi! So first, you wanna figure out what the molar mass of the metal is (M). You can do this by subtracting the hydroxide's molar mass from the given molar mass (74.10 g/mol). This should give you 39.94 g/mol, which again, is the molar mass of the metal (M) in the compound. If you look at the periodi...

Shanna Yu 1C

Hey, so when they ask for 75% yield, they’re asking for 75% of the theoretical yield, which you can solve for by running the given quantity of the limiting reagent from part b through the reaction to solve for grams of product (CO2, in this case). So whatever quantity of CO2 you get from that, multi...

Shanna Yu 1C

Hey, so I did a mass search for 2F links on both the Discord and the Groupme for lecture 2, but it doesn’t seem like a Groupme for 2F exists, yet?

Here’s the link to lecture 2’s Groupme if you haven’t joined, yet. Maybe you can ask someone there?

https://groupme.com/join_group/62528879/AM6bvZV6

Shanna Yu 1C

I wrote this out step-by-step for you because I feel like it’s difficult to explain through words! I hope you can read it, and if you have any questions, lemme know.

(Can’t figure out how to post images here, so I uploaded my work onto imgur for ya.)

https://imgur.com/G0o49Zg

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