CHEMISTRY COMMUNITY

Abhi Vempati 2H

It seems the same as 14A, and I think the only thing that might change is how he adjusts the grades based on the class's performance this quarter.

Abhi Vempati 2H

I have heard his final covers pretty much all the topics in equal amounts, so I would expect about 1-2 questions on kinetics. Hope this helps!

Abhi Vempati 2H

You have to look at the activation energies (not the rate constants). If the fwd reaction has a lower activation energy than the rev reaction, you know that there is a smaller "hump" going forward and a larger one going backward. If you draw the reaction profile, this corresponds to an exo...

Abhi Vempati 2H

If you read the section above Example 7C.1, the book talks about steady-state approximation, which Professor Lavelle said we would not be doing. I think we will not be expected to know this, but I would just confirm after class or during office hours in order to make sure. If you do so, would it be ...

Abhi Vempati 2H

K is actually measured by the activities of all the reactants, which have no units. Using concentrations is technically an approximation technique, and in order to account for this, we remove the units. This was covered in one of the equilibrium video lectures on his website. Hope this helps!

Abhi Vempati 2H

Isothermal means a constant temperature, so we can use the formula ΔS = nR ln(V 2 /V 1 ). Since entropy is a state function, only the initial and final states matter, not the pathway taken to get there. This means that for entropy change of the system, there is no difference between reversible and i...

Abhi Vempati 2H

Going over the outlines that Professor Lavelle posts on his website helped me last quarter as well. Lyndon's review session and doing the end of chapter textbook problems were also nice, as some of those problems were very similar to/exactly the same as what was on the actual final.

Abhi Vempati 2H

You can assume/make it a pseudo-first order reaction by making the initial concentration of one reactant extremely large. When we do this, we can essentially say that the change in reactant concentration during the reaction is so tiny that it is insignificant, i.e. that reaction concentration remain...

Abhi Vempati 2H

Adding on to what @Amanda Lin 2I said, an overall reaction can be a sum of several elementary steps, which is helpful for kinetic analysis. Hope this helps!

Abhi Vempati 2H

As @705302428 mentioned, you would only see species in the denominator if they have a negative order. Professor Lavelle didn't cover this in lecture, so I don't think we'll be expected to know this. Hope this helps!

Abhi Vempati 2H

Like everyone is saying, I wouldn't worry too much about 4 or more molecules colliding, and if you know how to identify uni-, bi-, and termolecular reactions, you should be fine. Hope this helps!

Abhi Vempati 2H

you use it to find the equilibrium constant K. It defines the relationship between activation energy, the equilibrium constant, temperature, and A which is the frequency factor. How do we get A or the frequency factor? I'm assuming that it's experimentally determined and will probably be given to u...

Abhi Vempati 2H

For a concentration cell, the species in the anode and the species in the cathode are the exact same. In a general galvanic cell, they can be different. Hope this helps!

Abhi Vempati 2H

Pages 538-543 in the textbook were really helpful for me, and it explains each step clearly and gives examples to follow along with. Hope this helps!

Abhi Vempati 2H

It may also be helpful to email your TA and clarify what's cool with them. Hope this helps!

Abhi Vempati 2H

Like everyone is saying, nitrate diffuses through the porous disk to balance the charges on both sides. The actual reaction that is occurring is for Ag⁺(aq) and Ag(s), but NO₃^- is important for the concentration cell as well.

Abhi Vempati 2H

In addition to what everyone is saying, there was a specific example done in lecture. In his example last Friday, Professor Lavelle showed a cell with copper and iron in it. In the copper half reaction, Cu(s) was present, so we can use the solid itself as an electrode. However, in the iron half reac...

Abhi Vempati 2H

In addition to what @Brian J Cheng 1I mentioned, the ion concentration inside the glass meter is different form the ion concentration in the solution, creating an electrochemical gradient, and therefore, a e^- current.

Abhi Vempati 2H

In addition to what @Jasmine Fendi 1D mentioned, the anode is where the oxidation takes place and the cathode is where reduction occurs. Hope this helps!

Abhi Vempati 2H

E^o is the cell potential under standard conditions, and E is the cell potential at any other condition. Hope this helps!

Abhi Vempati 2H

We use an inert conductor when one of the redox couples does not have a solid in it. In his example on Friday, Professor Lavelle showed a cell with copper and iron in it. In the copper half reaction, Cu(s) was present, so we can use the solid itself as an electrode. However, in the iron half reactio...

Abhi Vempati 2H

Yes, the salt bridge prevents a buildup of charge in the galvanic cell by letting ions from one electrode flow to the other side. Professor Lavelle mentioned the flow of Cl^- ions in the galvanic cell example on Friday, so referring to that might help.

Abhi Vempati 2H

When he showed the midterm average, Professor Lavelle started playing a song, and it was really catchy and helped me smile through the pain. Now the tune's stuck in my head and I don't know what it is – can someone please tell me the name of the song so I can bump it after getting destroyed by Test ...

Abhi Vempati 2H

Since entropy is a state function and the path between initial and final states doesn't matter, the entropy of the system is nRln( \frac{V_{2}}{V_{1}} ) for both reversible and irreversible expansion. However, as @Samuel G Rivera - Discussion 4I mentioned, the main difference is what happens to the ...

Abhi Vempati 2H

Count the moles of gas on each side by counting the coefficients, which will help you determine the correct way equilibrium shifts.

Abhi Vempati 2H

Adding on to what was said above, in constant pressure systems, we typically have to account for expansion work, but in constant volume systems, we don't (the gas can't expand if there's no volume change). Hope this helps!

Abhi Vempati 2H

STP, which is standard temperature and pressure, is 1 atm and 273 K (273.15, to be specific).

This is often confused with standard state/standard conditions, which is 298 K, This is also referred to as room temperature.

Hope this helps!

Abhi Vempati 2H

He doesn't curve tests, but he said that he looks at the point distributions at the end of the quarter and adjusts grades as necessary. Last quarter for 14A, I think it helped people by a couple points, but I wouldn't bank on it to save a grade. Hope this helps!

Abhi Vempati 2H

You can use both interchangeably. According to my TA (Dane), the per mol part simply clarifies that we are calculating the energy absorbed/release if the entire reaction occurred 1 mol times. Hope this helps!

Abhi Vempati 2H

What about when we are calculating the change in internal energy in a system under constant pressure? Work is involved, do we always use the negative sign in the expression for work? We don't just put the negative sign because work is involved; rather, we add the negative sign because the system is...

Abhi Vempati 2H

I also found Lyndon and Matthew's 3 hour session to be super helpful last quarter. It's on Monday, February 10 this time from 6-9 P.M., and they give a super helpful worksheet and talk through the problem solving process, so I highly recommend attending if you have the time!

Abhi Vempati 2H

We may cover something in lecture this week that involves integrals, so I would review the equation sheet and integral rules just in case. Hope this helps!

Abhi Vempati 2H

Since we are also incorporating a lot of thermochemistry and First Law concepts into our lectures for the Second and Third Laws, I'm pretty sure it would be fine to have some problems from the previous unit for Week 5 HW. However, I would include some problems from the current unit as well. Hope thi...

Abhi Vempati 2H

Divide by the number of moles, i.e. amount of substance.

Abhi Vempati 2H

Adding on to what @Zoya Mulji 1K said, if we're given a table of standard enthalpies of formation, the problem will want us to use enthalpies of product minus enthalpies of reactants. Most of the time, it should be pretty clear what method the problem wants you to use. Hope this helps!

Abhi Vempati 2H

A measured bond enthalpy is the average value of a bond that was calculated from many different molecules, so it is less accurate. A C–H bond demonstrates this perfectly. When we look at CH 4 , C 2 H 2 , and CHCl 3 , the bond enthalpy for each C–H bond will be different for each molecule. So, in ord...

Abhi Vempati 2H

Adding on to what @Goyama_2A mentioned, understanding the basics of the theorem will help when deriving the formulas mentioned later in the chapter. Hope this helps!

Abhi Vempati 2H

If you go through the acid-base examples that we did in class (on a Friday, I believe), there was a problem that had a similar structure to #5. This should help clarify the steps as well. Hope this helps!

Abhi Vempati 2H

Adding on to @KeyaV1C, you calculate [concentration of deprotonated species] with an ICE table and the K value. Hope this helps!

Abhi Vempati 2H

In order to vaporize a substance, you need to break all the IMFs holding the sample together (because a gas, for our purposes, exhibits no IMFs). This takes a lot of energy compared to melting (where you only need to break a few IMFs), which is why the enthalpy of vaporization is so high. Hope this ...

Abhi Vempati 2H

Adding on to what @Hui Qiao Wu 1I mentioned, the lower the pk_a of an acid, the higher the pK_b of its conjugate base. This is because of the equation pk_a + pK_b = 14. Hope this helps!

Abhi Vempati 2H

Yes, the reverse reactions (condensation, freezing, and deposition) would be exothermic because they release energy, forming a more stable phase of matter.

Abhi Vempati 2H

While a substance is undergoing sublimation, all the energy (from the heat) is being used to break the IMFs that hold the solid together. Therefore, the temperature (which is the average kinetic energy of the molecules) does not change. During sublimation, the heat doesn't make the molecules move fa...

Abhi Vempati 2H

This link on the Chem 14B website (titled "Everything you want to know about Significant Figures") was really helpful for me: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/EVERYTHING_YOU_WANTED_TO_KNOW_ABOUT_SF.pdf The second page has the rules for logarithms, and I hop...

Abhi Vempati 2H

Professor Lavelle said that everything up until the end of Friday was fair game for the test, and since we had a buffer example in class that day, there's a chance that it may be on the test. Hope this helps!

Abhi Vempati 2H

Like @asannajust_1J mentioned, there might be an email sent out, but if not, I would just drop by on Tuesday or later to check it out. Hope this helps!

Abhi Vempati 2H

I believe the correct answer is 3.2 x 10^-15 M.

In the solutions manual, the [H₃O⁺] = 3.1 x 10^-3 M, but in our book, it's just 3.1 M, leading to a different answer. Hope this helps!

Abhi Vempati 2H

Professor Lavelle mentioned at the end of class on Friday that it was only chemical equilibrium + acids and bases, so basically everything we covered in the first two weeks. Hope this helps!

Abhi Vempati 2H

When the book says the "activity", it simply means the concentration/partial pressures. As stated in lecture, we approximate the activity by using the concentrations or partial pressures of the species in the reaction. So, all we need to do is write the expression for K like you would norm...

Abhi Vempati 2H

We use reactivity to calculate K, and concentrations/partial pressures are just an approximation of the reactivities. Reactivities have no units, so K will therefore not have any units. This was covered in the modules and lectures in further detail. Hope this helps!

Abhi Vempati 2H

After asking this question in office hours yesterday, Professor Lavelle said that if the problem simply asks you to write the expression for K for a reaction with only gases, you can use K p . However, he did make it explicitly clear that K is just a general term for either K c or K p , so it doesn'...

Abhi Vempati 2H

Yes, you need to take the ΔH into account. Since ΔH < 0, the forward reaction is exothermic and the reverse reaction is endothermic. In other words, the forward reaction gives off heat and the reverse reaction uses up heat. Since heat is added, Le Chatelier's Principle states that the system will tr...

Abhi Vempati 2H

You can make this assumption whenever K < 10 -4 , as stated in Professor Lavelle's modules. Since the value of K is extremely small, we know that only a very, very tiny amount of product will be formed. In other words, only a very, very tiny amount of reactant is used up. So, when subtracting x (the...

Abhi Vempati 2H

As everyone is mentioning, it would probably be useful to know the amphoteric oxides themselves, which is something you have to memorize. It's in the textbook (page 450), and he may ask us questions about them on the final. Hope this helps!

Abhi Vempati 2H

i. SO₂ (aq) + H₂O (l) $\rightarrow$ H₂SO₃ (aq)

ii. CO₂ (aq) + H₂O (l) $\rightarrow$ H₂CO₃ (aq)

I believe these are the two reactions we need to know for acid rain. Hope this helps!

Abhi Vempati 2H

I think Week 9 HW should include at least some coordination compound problems (Week 9 was mainly when we covered this topic), but I also did some acid-base problems for Week 9. Week 10 should definitely be acids and bases, though. Hope this helps!

Abhi Vempati 2H

viewtopic.php?f=47&t=54235&p=200584&hilit=cisplatin&sid=a21d2330bf22d2b242277ad29b7fb59a#p200584

Hope this helps!

Abhi Vempati 2H

When a ligand is polydentate and you have more than one of these ligands, you use bis-, tris-, etc. This is done because some polydentate ligands already have the prefixes di-, tri-, etc. in them (ex: di ethylene tri amine). By using bis-, tris-, and the alternate prefixes, you can clearly show how ...

Abhi Vempati 2H

I had my discussion on Wednesday and got my test back, so I'm assuming that everyone who had discussion on Monday, Tuesday, and Wednesday received their tests. If you have discussion on Thursday or Friday, you'll most likely get it back in discussion next week or after a lecture (either Monday or We...

Abhi Vempati 2H

Most TAs are collecting Week 9 and Week 10 HW together during their discussions next week. As @NRobbins_1K also mentioned, Professor Lavelle sent out an email saying that we would have the option of turning in Week 9 HW during Week 10 discussion.

Abhi Vempati 2H

The final will be cumulative, but it will probably focus a bit more on the second half of the quarter because the first half was tested extensively during the midterm. However, I would still study everything carefully to be prepared. Hope this helps!

Abhi Vempati 2H

Would the pKa and Ka values be given or would we have to calculate that ourselves? You only need one value to make the comparison, so if you're given either the K a or pK a , you can tell which one is the stronger acid. However, sometimes you may not be given either of these values. If this is the ...

Abhi Vempati 2H

In addition to naming, Professor Lavelle also could ask us about the shape and coordination number of the coordination complex. He may ask us which ligands are polydentate, and as @Vincent Leong 1A mentioned, he could also ask us about their biological importance.

Abhi Vempati 2H

I took my test today and nothing related to intermolecular repulsions was there, so you should be good.

Abhi Vempati 2H

To expand upon @Brittney Hun 1K and clarify a couple of things, both CHF 3 and CHI 3 exhibit dipole-dipole forces, so this doesn't matter in determining the melting/boiling points. What matters, therefore, is the LDFs of each molecule. CHI 3 is a larger molecule/has a higher molar mass, and this cor...

Abhi Vempati 2H

Like @Caitlyn Tran 1G mentioned, there's no specific reason or time where you need to write the ammonia backwards. It's just a clarification thing, so you won't need to memorize when you would need to write a molecule backwards.

Abhi Vempati 2H

After asking my TA and also taking the test, I can confidently say that no, we won't need to know it. Hope this helps!

Abhi Vempati 2H

HO^- is more tightly bound because the oxygen has a smaller atomic radius, so the nuclear forces of attraction are much stronger and pull the electrons more tightly. The high electronegativity of oxygen also leads to this conclusion.

Abhi Vempati 2H

If two different atoms are covalently bonded together, the electrons are shared unequally. The more electronegative atom pulls all the electrons closer to itself, making the atom partially negative. Since the electrons are being pulled away most of the time from the less electronegative atom, the le...

Abhi Vempati 2H

Yes, it is extremely unstable for radicals to exist, which is why they're highly reactive and don't last for a long time. For example the CH₃ molecule is a radical, and it easily reacts with another H atom to form CH₄, a much more stable molecule.

Abhi Vempati 2H

I'm pretty sure it's just something you have to memorize. For some molecules it's pretty intuitive (ex: since CO 2 is a linear molecule, the bond angle is 180 degrees), but once you get to the more complicated molecules, you just have to memorize them. There's a chart in the textbook/online to help ...

Abhi Vempati 2H

This model wouldn't be able to hydrogen bond even if a carbon had a lone pair. This is because for hydrogen bonding, 1) the H needs to be bonded to a highly electronegative atom, specifically N, O, or F and 2) the N, O, or F has to have a lone pair. Since C 2 H 4 doesn't have an H bonded to an N, O,...

Abhi Vempati 2H

Test 2 will most likely cover intermolecular forces, VSEPR and Lewis structures, and whatever we learn during lecture on Friday/Monday. There is a chance that it could be a cumulative test, so I would just ask the TAs to be sure. Hope this helps!

Abhi Vempati 2H

For future reference, the test and exam schedule is on the Class Website. It's a link titled "Test and Exam Schedule" and it's pretty easy to find. Hope this helps!

Abhi Vempati 2H

He just announced in the 2 P.M. lecture that we're getting our midterms back on Wednesday during our lectures. Good luck!

Abhi Vempati 2H

@Cassandra_1K If you're unsure, you can ask your TA for clarification and I'm sure they'd be happy to answer. Chemistry Community is also a great resource!

Abhi Vempati 2H

Like @Daniel Honeychurch 1B mentioned, for Chem 14A, we'll be given the experimental values of the dissociation energies. Professor Lavelle mentioned that we will be calculating dissociation energies in Chem 14B, though.

Abhi Vempati 2H

The +/- charge doesn't really depend on the central atom. Instead, it depends on the electronegativity of the atoms. You want the most electronegative atom to have a - charge on in, so for something like PO 4 3- , it wouldn't be the P that has a negative formal charge, but rather the oxygens around ...

Abhi Vempati 2H

As everyone has mentioned, there is probably a way to find the exact bond length (in angstroms) based on its strength or vice versa, but I believe that's beyond the scope of this course. I think the main thing we need to know is the relationship between bond length and strength – the longer the bond...

Abhi Vempati 2H

The reference sheet is probably going to be exactly the same as the one we got for Test 1, so you could look at that for reference while preparing. Hope this helps!

Abhi Vempati 2H

As @AGulati_4A mentioned, drawing the "slightly negative" and "slightly positive" symbols along with the vectors indicate a dipole moment. The vector should start from the slightly positive atom and point toward the slightly negative atom.

Abhi Vempati 2H

Like @Sebastian Lee 1H said, you probably won't need to memorize the exact color, but memorizing the wavelengths for different types of light (UV, visible, X-ray, etc.) would be helpful.

Abhi Vempati 2H

Elements with electrons in the 3p subshell can expand their octets because they have empty 3d orbitals, allowing them to have more than 8 electrons. This is true for the p block in Period 3 and for all elements below Period 3, and we typically see expanded octets occurring in the halogen and noble g...

Abhi Vempati 2H

Hertz (or s^-1) is simply the unit for frequency, just like the kilogram is the unit for mass. Every time it asks for the frequency, your response will be with the unit Hz (or MHz, GHz, etc).

Abhi Vempati 2H

@Katherine Wu: Ga is a weird one because it's in Column/Family 3 of the periodic table, meaning it tends to form a 3+ charge. It confused me at first, but then I realized there's two ways to look at this: 1. Look at the family Ga is in. Ga is in Family 13 (or 3A). Then, I looked at the elements abov...

Abhi Vempati 2H

As Professor Lavelle mentioned during lecture, the values of ml and what orbitals they correspond to are somewhat arbitrary. Typically, if ml = -1, 0, 1 for the p subshell, we make px = -1, py = 0, and pz = 1 because it's convenient and intuitively makes sense. However, I don't think the numbers nee...

Abhi Vempati 2H

The subshells specifically refer to s, p, d, and f, which is what the angular momentum number gives you. Once you get to the magnetic quantum number, you're talking about orbitals, NOT subshells. So, for n = 4, l = 0, 1, 2, and 3, which correspond to s, p, d, and f. The answer is therefore 4s, 4p, 4...

Abhi Vempati 2H

Yes – once a sufficient threshold energy is reached (by increasing frequency), the amplitude of the wave does matter. Increasing the amplitude proportionally increases the number of electrons ejected. So, if you double the amplitude, you double the number or electrons ejected, and so on. Hope this h...

Abhi Vempati 2H

Does anyone know if there is a way to calculate our grades ourselves? Is it point system? The grading system is point-based, but as Professor Lavelle states on the syllabus, letter grades aren't assigned until the very end. I don't think there's a way to know what letter you have, but you can estim...

Abhi Vempati 2H

Since the material we're covering in class still relates to The Quantum World, yes, you can still do problems from this unit.

Abhi Vempati 2H

Like MTanikella_1C mentioned, the video modules, specifically the post-assessments, are pretty useful because they ask both conceptual questions and questions that require you to apply a formula. Reviewing them once or twice should help!

Abhi Vempati 2H

The Schrodinger wave function was covered briefly in the modules, but I believe that for topics we are now doing in class (quantum numbers, electron configurations, Lewis dot diagrams, etc), there are no modules for it on the Chem 14A website.

Abhi Vempati 2H

Adding onto this, the syllabus mentioned that one homework problem would be on the midterm exam, so doing all of these problems and reviewing them would probably help.

Abhi Vempati 2H

@Keerthana Sivathasan 4G
Yeah, my TA said today that we'd get it back during our discussion this week.

Abhi Vempati 2H

Yes, the light simply passes through the object. Professor Lavelle mentioned this briefly in class.

Abhi Vempati 2H

As @Akshay Chellappa 1H mentioned, the numerator term for the equation has the units J * s. Although J is kg * m^2 * s^-2, the J will be multiplied by the s to cancel out part of the s^-2, making the numerator kg * m^2 * s^-1 (Note the s^-1 instead of the s^-2). Then, the overall fraction will be (k...

Abhi Vempati 2H

@Matt Sanruk 4F: I believe Professor Lavelle and his TAs have probably figured a system out. Although I don't know how it works exactly, he's been using this system for quite a while and I trust that he knows what he's doing. I think as long as we do the 5 posts by every Sunday, we should be fine.

Abhi Vempati 2H

As many people here are stating, electrons were determined to have wave-like properties due to the diffraction patterns created during the double slit and screen experiment. It was known that diffraction patterns were a key characteristic of waves, so when the experiment resulted in diffraction patt...

Abhi Vempati 2H

Like @Rebecca Remple 1J said, we should probably getting them back by the next discussion or by the end of next week (according to the TAs). Worst case, we get them during discussion in Week 4.

Abhi Vempati 2H

Like @megan3j said, there will always be carbon dioxide and water as products in combustion reactions. There will also always be oxygen (diatomic form) as a reactant. There may also be some additional elements in the combustion reaction (e.g. nitrogen). However, for these problems, you probably will...

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