CHEMISTRY COMMUNITY

Zaid Bustami 1B

Boosting this too!! Thanks to Dr. Lavelle and the ENTIRE teaching team for their hard work and for making this class so memorable and awesome, hope you like the suprise (:D)

Zaid Bustami 1B

Yup that approach is correct, exactly what I'd do too. Assuming no heat escapes into the surroundings, total entropy change for the process would be equivalent to asking for the sum of the entropy changes for each iron rod. The entropy change that occurs when temperature changes is given by this equ...

Zaid Bustami 1B

Hey Nicole, Using the Nernst equation would really help in understanding what these changes would do. Also since a concentration cell by definition has an E naught equal to 0, I'm writing a simplified version without E naught. E=-\frac{RT}{nF}lnQ I'm also gonna use Ag+ as an example, so I've written...

Zaid Bustami 1B

Hey Alen! Yea, the anode side is always the oxidation side and the cathode side is always the reduction side. Note that the word "anode" refers to the electrode that is on the oxidation side, and the word "cathode" refers to the electrode that is on the reduction side. This is im...

Zaid Bustami 1B

Hey Sophia! Since this problem doesn't really provide acid or base equilibrium constants right off the bat, I'd start off with identifying which ones are acids or bases and which ones are strong or weak. a) HCl is a strong acid b) CH3NH3Cl is a salt that will dissociate into CH3NH3+ and Cl-. The chl...

Zaid Bustami 1B

Hey Arezo, If you had a reaction that violated microscopic reversibility, what would happen is the pathway of the forward process would not be the same as the pathway of the reverse process, and you could end up with different activation energies in either process. Since calculating K under the micr...

Zaid Bustami 1B

If left exposed to the air, many metals corrode over time - you probably know that rusting is an example of corrosion, and it involves the oxidation of iron in the presence of water and oxygen. Corrosion involves the oxidation of the metal, so in order to prevent corrosion you need to coat the metal...

Zaid Bustami 1B

Hey Hannah! What you have here is a concentration cell, meaning that the species in both the anode and cathode sides are identical. This is because both electrodes are made of nickel, and both solutions contain nickel ions. The tricky part is that the problem makes things a bit more complicated by s...

Zaid Bustami 1B

Hey Minahil, What you need to do is compare the activation energies of the forward and reverse processes in order to determine whether a reaction is endothermic or exothermic. It helps to first look at what activation energies look like in the forward and reverse reactions of a reaction that's endot...

Zaid Bustami 1B

Hey Emma, Your first half-reaction is pretty much perfect, the last step you need for it is to balance out the charges on either side by adding electrons. Since the left side has a total charge of -2 (there are two chlorides, each with a -1 charge), you need to add 2 electrons to the right side so t...

Zaid Bustami 1B

Basically what you've got going on in the second compartment is a precipitation reaction, where the nickel(II) nitrate reacts with the sodium hydroxide to produce an insoluble product, nickel(II) hydroxide. You know this because the Ksp value (the equilibrium constant for a dissolution reaction) lis...

Zaid Bustami 1B

Hey Minahil, Both are listed as redox couples in the problem, so even if the problem gives you La 3+ /La it doesn't necessarily tell you the reaction will for certain go from the ion to neutral La and result in reducton. Though what you did writing them both as reduction reactions is the exact first...

Zaid Bustami 1B

I'm actually having trouble understanding why the book did it they way they explained it in the answer key, and I keep getting a different answer, but I thought it could be helpful to explain how I'm looking at this problem. The first thing I did was a calibration calculation for the pH probe to fin...

Zaid Bustami 1B

Yup that's exactly what happens when you get a short. Given you have a constant voltage for your cell, resistance is inversely related to current (E = current * resistance). As resistance approaches zero (you'll get this in a circuit where you have basically no resistance, like one just containing a...

Zaid Bustami 1B

Hey Marc, Yea that's an awesome question! A lightbulb works by converting energy in the form of electricity (a current, which is essentially just moving electrons) into heat and light. As electrons pass through a filament (that thin piece of metal kind of dangling in the middle of a bulb), they enco...

Zaid Bustami 1B

First off identify your oxidation and reduction half-reactions. Because the left side represents the anode and the right side the cathode, we know the left side represents the oxidation reaction and the right side the reduction reaction. Now just write down the half-reactions, balance them, and comb...

Zaid Bustami 1B

Hey Crystal! The way sapling does it is by using the first method we learned to use to balance equations, where you just change the coefficients of the species. I found it easier to do this in terms of separating into half-reactions and then balancing it that way. When writing your half-reactions, s...

Zaid Bustami 1B

Hey Tanya! Keep in mind that a coulomb (C) is a unit of charge rather than a unit of energy, and so a coulomb of charge in and of itself doesn't really do work. In lecture Prof. Lavelle said that one volt (V) is equal to one joule per coulomb. This tells us that when you have a voltage of one volt, ...

Zaid Bustami 1B

Hey Yasmina! Before we do anything though it's worth noting that the platinum listed is the material of the electrode itself, and is used because none of the species that actually take part in the redox reactions are conductive solids. You can just ignore it for this problem. So the first thing you ...

Zaid Bustami 1B

Hey Harrold, Yea it would be a good way to think about potential as something's electron-pulling ability, but I would describe the oxidation and reduction reactions themselves as having these potentials rather than the electrodes. Keep in mind that anodes and cathodes are often a product/reactant of...

Zaid Bustami 1B

Hey Samantha! Yea you could definitely use the Van't Hoff equation to calculate your values, but you also need to keep in mind the process would be different. What you'd do first is calculate your K at standard conditions by using the "plain" Van't Hoff equation. You'd use the table the pr...

Zaid Bustami 1B

Hey Caelin, Yea there's a final exam listed if you add it to your class planner. It's Monday, June 7 from 6:30pm-9:30pm. If you add all your classes to your class planner each class will have its final exam info listed with it, so it's also nice if you're trying to compare final dates. Best of luck!...

Zaid Bustami 1B

Hey Stella, When approaching problems with a lot of numerical info like these, it helps to translate everything in the problem into a set of mathematical expressions: ΔS o vap @100 o C = 109J/Kmol C P liquid = 75.3J/Kmol C P vapor = 33.6J/Lmol ΔS o vap @85 o C = ? Based on those given values it's go...

Zaid Bustami 1B

Hey Quinton! My guess would be some sort of calculus equation. If you remember from the integral equations that we used for w and for ΔS, when we evaluated the integrals we were left with algebraic expressions involving natural log. w=-\int_{V_{1}}^{V_{2}}PdV =-\int_{V_{1}}^{V_{2}}\frac{nRT}{V}dV = ...

Zaid Bustami 1B

Hey Jeffrey, Yea you've got the right idea. At the most basic level the difference between a reversible and an irreversible process in terms of q and w is that for a reversible process, q flows into the system as the system does work, whereas for an irreversible process q flows into the system after...

Zaid Bustami 1B

Hey Cooper, Molar enthalpy of vaporization is the amount of energy it takes to vaporize one mole of a substance. So in order to calculate it from the data, you need to divide the amount of heat that each system absorbed divided by the moles of substance in each system that were vaporized. You're giv...

Zaid Bustami 1B

Hey Lilyana, The first thing you wanna do for part b is calculate the ΔU for the reaction of 1.4g of CO. ΔU = q V = -q calorimeter = -C calorimeter ΔT ΔU = -(3)(22.799-22.113) = -2.058kJ What you just calculated is ΔU for 1.4 g of CO. To calculate ΔU in terms of energy per mol of CO, essentially wha...

Zaid Bustami 1B

A bomb calorimeter is a type of calorimeter that is used to measure the heat released/absorbed by a reaction at constant volume. Here's a diagram of one: calorimeter.jpg The basic idea is that the reactants are placed into a sealed vessel that's surrounded by water. When a reaction is run, the therm...

Zaid Bustami 1B

Hey Samantha! Because everything in your calculations is gonna be in terms of the heat gained/lost by the calorimeter, you actually don't need to consider mass. Remember for any calorimetry problem, the calibration step is to calculate the heat capacity of the calorimeter. It isn't really useful to ...

Zaid Bustami 1B

Hey Blake! Anytime I get overwhelmed with a bunch of info I find it really helpful to convert everything in the question into mathematical expressions. Based on this problem, what we're given is this: m ice = m water = 150g ΔT ice = ΔT water = +5 o C t ice = 10.5 hour t water = 0.5 hour It's also he...

Zaid Bustami 1B

Hey Minahil, The reason that the q value you use to solve for heat capacity of the calorimeter is positive is because the calorimeter is absorbing the heat that the calibrating reaction releases. In terms of the reaction, q is negative because the rxn releases 3.5kJ of heat, but in terms of the calo...

Zaid Bustami 1B

Hey Arezo, When the system experiences a sudden increase in volume due to a large pressure difference between the system and the surroundings, the work that the system is doing is only on the initial pressure of the surroundings. So for an irreversible expansion, you only need to consider one pressu...

Zaid Bustami 1B

Hey Brandon, The most important thing you want to do is find out what type of expansion is occurring in each step of each path. You want to keep in mind that the total work for a path is the sum of the work for each individual step. For path A, there is only one step which is a reversible expansion....

Zaid Bustami 1B

Hey Eve, The reason why you're seeing this is that path A and path B are not just identical reversible and irreversible expansions, respectively. Path A involves a single step: a reversible expansion from 4.45L to 8.78L at 298K and 2.68 atm. However, path B involves two steps: a decrease in the syst...

Zaid Bustami 1B

Hey Mackenzie! If you remember from Prof. Lavelle's last lecture he told us that the formula you need to use to calculate work for a reversible process is this integral: w=-\int_{V_{1}}^{V_{2}}PdV In order to actually solve this integral, P has to represent some function in terms of V because it mus...

Zaid Bustami 1B

Hey Crystal, A really important concept to keep in mind is that for an energy transfer between two things, the energy lost by one is equal and opposite to the energy gained by the other (we're assuming that energy transfer is ideal and energy does not escape into the surroundings, like the problem s...

Zaid Bustami 1B

Yea it seems like it's counterintuitive. I'm not entirely sure what the answer is, but I'll give it my best shot with what we've covered already from class. I think it would help to think of this in terms of the concepts we learned about from heat capacity in today's lecture. ΔH is analogous to heat...

Zaid Bustami 1B

Hey Austin, Looking at this problem and the information that's given, I would say the first thing to think about is what method of calculating ΔH is going to be easiest. In this case it's to use standard enthalpies of formation. Now the thing the problem wants you to find isn't the enthalpy change f...

Zaid Bustami 1B

Hey Emma, Yea I never thought about the possibility of that happening, that's a really interesting question. Because resonance is verified experimentally, and of course bond enthalpies are determined experimentally, what I'd think would happen is that the book would provide a bond enthalpy for the b...

Zaid Bustami 1B

Hey Austin, Yea so when you're looking at these equations for \Delta H^{o}_{c} you want to keep in mind that they all imply that they are referring to the enthalpy change when only 1 mole of the substance is being combusted. Because of this rule, fraction coefficients are allowed and are actually ne...

Zaid Bustami 1B

So this is an instance where it's too close to call just by looking at your K a . We know that our definition for a small K has an upper limit of 1x 10 -3 , and since your given K value is really close it's better to stick with a more precise method of judging wither the assumption is valid. To save...

Zaid Bustami 1B

The logic here is exactly the same with bases, though I think it would be easier to rethink real quick about what the comparison between pK a and pH really means. -When the substance you add to a solution has a pK a that is greater than the pH of the solution, it means that the substance is a weaker...

Zaid Bustami 1B

Yea what you want to do for these is ALWAYS start with the species that is given. If you look at the ionization/protonation reactions for NH 3 and NH 4 + , you'll notice they involve different products: NH 3 + H 2 O ⇌ NH 4 + + OH - NH 4 + + H 2 O ⇌ NH 3 + H 3 O + You'll notice too that they're not e...

Zaid Bustami 1B

Hey Samantha! For this problem you'd actually need to refer to table 6D.1 in chapter 6D of the textbook. I know the problem doesn't say it, but tables 6C.1 and 6C.2 don't contain any information on those highly charged cations that we know produce acidic solutions. Table 6D.1 lists the K a for Al(H ...

Zaid Bustami 1B

Hi Alexandra! For this problem, you're going to need to use an ICE box. It might seem like in this scenario you're supposed to use a different method but keep in mind that ICE boxes works for any problem where you have K and the initial concentrations of all your species, even if you have nonzero pr...

Zaid Bustami 1B

Hey Eliana! Basically the idea is that if you have a sample of pure water, you know that due to autoprotolysis it will have a hydronium ion concentration of 1x10 -7 M. Here's the autoprotolysis of water reaction just for your reference: 2H 2 O ⇌ H 3 O + + OH - , where K = K w = 1x10 -14 . Let's say ...

Zaid Bustami 1B

Hey Jake, Based on high school chem I think it's definitely possible to have to deal with a cubic equation when you have a large K value, but like was said earlier maybe not in this class since Prof. Lavelle never went over it. Just thought it could be helpful to explain how it works. For this scene...

Zaid Bustami 1B

Hey Siwa! Wanted to mention real quick that the K c value the table lists at 298K is incorrect, and it should be 6.1 x 10 -3 rather than 6.1 x 10 23 . Check out https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=49&t=70852&sid=00c68c80630373633d338553da5437c3 for more info about that. Also ...

Zaid Bustami 1B

Hey Minahil! Just to add to Samantha's awesome explanation, for any problems like this there's a rule that states that if you multiply all stoichiometric coefficients by a certain number, the K for that equation gets raised to that number. If you're given this equation: 1 / 3 N 2 + H 2 ⇌ 2 / 3 NH 3 ...

Zaid Bustami 1B

Hey Minjoo! I would say from looking at the table that the numbers on the left are K p values and then obviously the ones on the right are K c values. What the book does I think is they omit the "p" subscript on the K for the first column because using K by itself implies that it correspon...

Zaid Bustami 1B

Hey Bella, This could be an error in the textbook itself actually. There's a pattern in the K values in the table where they increase as temperature is increased, and so I'd expect the same pattern in the K c values listed. It would make more sense if the K c for the reaction N 2 O 4 <->2NO 2 at 298...

Zaid Bustami 1B

Hey Elizabeth, Correct me if I'm overlooking something but I think this could be the textbook's fault. The reaction they list is 2HD + H 2 <-> D 2 , which is doesn't make sense because there are no regular hydrogens present on the right side of the reaction (I say regular because D stands for deuter...

Zaid Bustami 1B

Hi Dana! HF is not a strong acid because fluorine is a very small atom. Although the electronegativity difference in this molecule is larger than for HBr and HI, which are strong acids, the small length of the bond makes for a very strong H-F bond. The short length of the bond overcomes any extra we...

Zaid Bustami 1B

I think I answered something similar in another topic right here, might be helpful to look at too https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=57&t=69400&sid=f39e3f22a505137e8fd07240f943ce1c . Basically the idea is that for things that completely dissolve in water, writing either both...

Zaid Bustami 1B

Hey Shannon! De Broglie's hypothesis tells us that everything with mass (matter) has a dual particle-like and wave-like property. Based on that, look at each answer choice: a. This experiment revealed how atoms are largely made of empty space between the nucleus and the orbiting electrons. It tells ...

Zaid Bustami 1B

Yea to add on all atoms with 2 or more regions of electron density around them need to be hybridized in order to properly distribute their valence electrons about themselves in a way that each of them can be used to form a bond/lone pair and in a way that allows the proper bond angles. Any species w...

Zaid Bustami 1B

Hi Eileen, So I think what you're referring to is whether or not an acid will deprotonate depending on its conditions. In the lecture, Prof. Lavelle said that an acid will deprotonate if the pH of its environment (i.e. the solution it's in) is greater than its pKa. In other words, an acid that is mo...

Zaid Bustami 1B

Hey Aina! Just like for carbon, the orbitals in N from NH3 have to be hybridized so that the NH3 molecule can have the proper shape. Although NH3 is a molecule in which the N is only bonded to three atoms, which is theoretically possible without hybridization because nitrogen has 3 p orbitals with u...

Zaid Bustami 1B

Hey Hana! Keep in mind that in order to calculate pH/pOH you need the concentration of the strong acid/base that you're dealing with. So, to solve these problems this is the value that you're trying to get to before you do the logarithmic calculation for pH/pOH. (d) For this one you're dealing with ...

Zaid Bustami 1B

Hi Brett, Thought it would also be helpful to give some examples of structures that would be impacted by pH. So organic structures often contain atoms that allow them to act as acids or bases. A lot of organic molecules often contain a nitrogen with a lone pair on it, and under normal pH conditions ...

Zaid Bustami 1B

Hey Sofia! Yea this goes back to how reactions taking place in water (in an aqueous environment) are written. Keep in mind that all these acid/base reactions need to be taking place in an aqueous environment for them to even work. The equation you wrote is showing the correct reaction, but you have ...

Zaid Bustami 1B

Hey Ian, Yea it's a bit counterintuitive because like Kailani said there are two regions of overlap in a pi bond, one up above and one down below. Keep in mind though that a pi bond is only made up of two overlapping p orbitals. Since p orbitals have two lobes, it looks like you have two pairs of ov...

Zaid Bustami 1B

Hey Jeffrey! To the best of my knowledge pi bonding can only happen between two p orbitals overlapping side by side, and not hybrid orbitals. If you look closely at molecules with double or triple bonds you'll notice that all of them have at least one unhybridized p orbital left over which is used t...

Zaid Bustami 1B

I would say hybridization happens only in circumstances where there are two or more regions of electron density around an atom. Like you said, for H2 since each H has only one region of electron density around it (the H-H bond), then there really wouldn't be anything to hybridize because only 1 s or...

Zaid Bustami 1B

Hey Kayko! I think it's just the geometry of the carbonate ion that allows it to do that. Since the two oxygens with formal charge -1 are relatively close to one another, a TM can form either one or two coordinate covalent bonds with the carbonate. Even if the carbonate has enough lone pairs to form...

Zaid Bustami 1B

I think the way to think about it is to prioritize giving the most electronegative atom an octet. If you use this approach to problems I think it will work (at least it works both with the ClO and OH radicals), but I'm not sure as to when electronegativity stops becoming a factor or what happens whe...

Zaid Bustami 1B

Hey Stuti! Yea you're absolutely right. p orbitals in an atom are oriented 90 degrees from one another, so in molecules like ammonia or methane the orbitals have to hybridize in order for the 109.5 degree bond angles to be possible. I have a picture here that visualizes what happens when carbon orbi...

Zaid Bustami 1B

Hey Ben! Yea, I think a good example would be ethyne (C2H2). I think any molecule where there are two central atoms each with two regions of electron density around them would all have two sp hybridized orbitals. So this would make nitrogen gas (N2) a good example too, and any others you can think o...

Zaid Bustami 1B

Hey Alvin! This goes back to when we talked about formal charge. I've drawn here what ClO2+ would look like with either one double bond or two double bonds: ClO2+.jpg When looking at the atoms' formal charges, it is clear that having only one double bond is more unstable because there are more forma...

Zaid Bustami 1B

Hello! I agree I don't think knowing nomenclature is required for this class, but I think it would be helpful going forward to be familiar with them. You're right those prefixes need to be paid attention to - they tell you how many of each atom is present in each molecule. This is the nomenclature s...

Zaid Bustami 1B

Hey Jaylin! What I find helpful to remember all the shapes is to memorize the shapes that are formed when no lone pairs are present (i.e. linear, trigonal planar, tetrahedral, trigonal bipyramidal, and octahedral) since these tend to be a bit easier to remember and they simply involve spacing out at...

Zaid Bustami 1B

The seesaw bond angles would be <90 and <120 (as well as <180 if you're comparing the atoms at the top and the bottom). This is due to the fact that the remaining atoms experience greater repulsion from the lone pair than they would if the lone pair was an atom. Thus, the bond angles normally observ...

Zaid Bustami 1B

Just wanted to let you know that your answer is correct, b is nonpolar and d is polar. The answers from the textbook solution manual say that is the correct answer as well, so I don't know if whatever it is you're accessing for the correct answers has the right answers listed. Hope this helps :)

Zaid Bustami 1B

Hey David, Yea it's kind of weird because for the textbook problems, it seemed like formal charge wasn't what determined which atom has the unpaired electron. However, from what I've seen it seems like the most electronegative atom will want the full octet, so the less electronegative species will l...

Zaid Bustami 1B

Keep in mind that the name of the shape comes from the arrangement of the atoms themselves, and not the arrangement of the regions of electron density (i.e. bonds and lone pairs). So even though SO32- does have a "tetrahedral" arrangement of electron density regions, you would name its sha...

Zaid Bustami 1B

Hey Noelle! Just wanted to add to the explanations. It's difficult to know exactly which reactants will donate a lone pair without drawing their lewis structures first. So, the first step is to draw out their structures, which I've shown here: Lewis Acid-Base Rxn.jpg By definition, a Lewis acid is a...

Zaid Bustami 1B

I'm doing some course planning and I was just wondering about how 14B and 14BL were designed to be taken. I've heard from some that deciding when to take each class depends on how much of a workload you can take, whereas others have said that 14BL applies concepts from 14B and it might be better to ...

Zaid Bustami 1B

Hey Cecilia, I would say for this problem you're going to need to use electronegativity. You're right in saying that you can use electronegativity to compare covalent bonds and polarizability to compare ionic bonds. It's just that in this case you're actually looking at a set of compounds with mixed...

Zaid Bustami 1B

Hey Vivian! 1. I asked Prof. Lavelle about this too, and he said that determining exactly which electrons are valence electrons for d-block elements can be inconsistent. What he told me is that if you define valence electrons as the electrons occupying the outermost shell, then the number of valence...

Zaid Bustami 1B

Hey Jalaila, Adding on, I just wanted to say something about why resonance hybrids exist. It's more stable to have a more evenly distributed electron density than to have differences in electron density throughout a molecule. Think about a dam. Sure it's possible to maintain a gradient in the water ...

Zaid Bustami 1B

Hey Thomas! This comes down to the electronegativity differences, since we're dealing with covalently-bonded atoms. The greater the difference, the more ionic in character the bond is because an electron can be more easily transferred vs shared when the electronegativity difference is higher (the mo...

Zaid Bustami 1B

Hey guys, So for part c, the question is asking how many valence electrons manganese has. How is it that Mn has 7 valence electrons and not 2, since the 3d electrons are in a lower energy shell than the 4s electrons? If valence electrons are defined as electrons in the highest energy shell, why are ...

Zaid Bustami 1B

I know we're probably gonna be going over this soon but I thought it would be helpful to explain a bit. So the strength of a bond tells you how much energy you need to break it. This energy (which you might have come across while researching) is called the bond enthalpy, so bonds with greater bond e...

Zaid Bustami 1B

Also Jeffrey I just wanted to add, when you're writing the electron configurations, you follow the order as indicated by the periodic table until Calcium, which has 20 protons. In calcium, 4s is filled rather than 3d because an electron with a 4s state in Ca has lower energy than an electron with a ...

Zaid Bustami 1B

Wait correct me if I'm wrong but isn't affinity positive when energy is released from an atom gaining an electron (since the calculation for affinity is initial energy of the atom minus its final energy)? So Alexandra wouldn't electron affinities can be negative if energy is absorbed and positive wh...

Zaid Bustami 1B

Hey Jaden, Just wanted to add, electron affinity is a measure of how favorable it is for an atom to gain an electron. If you remember the terms endergonic and exergonic (probably from bio), these would also apply to electron affinity. A positive affinity indicates that energy is released when an ele...

Zaid Bustami 1B

Also the thing about Boron is that it has a pretty high ionization energy for its three valence electrons, so it'll typically share electrons rather than give three of its electrons up. Now the thing is it will form molecules where it has an oxidation state of 3+, but won't necessarily exist as an i...

Zaid Bustami 1B

Hey Anirudh, So the thing is, once you remove an electron from an atom, the amount of e- repulsion decreases, so the electrons are not as easy to remove (the atom becomes smaller). Even though this is the case that doesn't mean the degree of increase for the second ionization energy is the same for ...

Zaid Bustami 1B

Hi Rita, To the best of my knowledge, when chemists make solutions, there's a lot of room for error when adding dry solute. Some solute can get stuck to the scoopula you use to scoop up the solute, some can get stuck on the sides of the beaker, some can get spilled while transferring the solute. To ...

Zaid Bustami 1B

Hey Talia! A very important technique used to determine the structure of molecules (in this case the spike proteins on SARS-CoV-2, the coronavirus which causes COVID-19) is known as x-ray crystallography, which Dr. Lavelle mentioned in his lecture. It's exactly what it sounds like: a beam of x-rays ...

Zaid Bustami 1B

Hey Alex, Yea I think you're right. The probability of finding an electron in a specific point is a specific value associated with one location. You can think of probability density as a gradient where darker regions represent a greater probability density (you're more likely to find electrons there...

Zaid Bustami 1B

Hi Anna! So if you think about Heisenberg uncertainty, you know that uncertainty in position is inversely proportional to uncertainty in momentum. If an electron were to stand still, its uncertainty in velocity would be so low that the uncertainty in position would be larger than the size of the box...

Zaid Bustami 1B

Hey Megan! Just wanted to give a more detailed explanation of the solution. So for part b, what you're looking at is a problem relating to the photoelectric effect, and how much energy in a photon is needed to eject an electron. The question says that no electrons are ejected until a frequency of 2....

Zaid Bustami 1B

Hi Jaylin! So the problem says to use a 1-dimensional box (which is basically a line) to represent the atom because this simplifies the calculation of the uncertainty in x ( \Delta x ). You know in real life atoms aren't lines, and instead you have to take into account the volume of space an electro...

Zaid Bustami 1B

Hey Brett! Adding on, from what Professor Lavelle said these transitions happen by chance. Some electrons may release the full amount of energy to return to the ground state, whereas others may not and only go down to n=2 or higher levels. Not entirely sure why this is true though. One explanation t...

Zaid Bustami 1B

Hey Jaden! Your answer's correct - the wavelength is 1.64x10^-38m. It's hard to give a precise answer as to what is the lowest measurement we can detect since it all depends on what kinds of technology we have. It was only in the last decade that scientists claimed to nail down an accurate measureme...

Zaid Bustami 1B

Hi Anika! These patterns really only apply to H since they were the result of an empirical analysis of its atomic spectrum. You'll see a similar trend (increasing wavelength with increasing lower energy level) in other atoms if they only had ONE electron, making them similar in structure to H. Howev...

Zaid Bustami 1B

Hey Quinton! Yeah it's pretty cool to think about how it is that scientists were able to depict a property of e- and atoms. Don't know if there's modern technology that's changed how spectroscopy is done, but at least from its origins I know that the fingerprint is simpler than you'd expect. The ide...

Zaid Bustami 1B

Hey Nicole, Just to reaffirm the energy of a photon has to be GREATER THAN or equal to the work function in order for it to eject the e-. If it was less than, the energy of the photon would not be sufficient to eject the e- (i.e. E photon < \phi ) and no matter how many photons are shone onto the e-...

Zaid Bustami 1B

Hey Karen! I actually did go to a workshop yesterday, basically what happens is UA brings in a prepared worksheet and as a class you work through every problem. The UA gives you time to solve each question, then after that they go over it and you can ask questions. The problems tend to be harder tha...

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