CHEMISTRY COMMUNITY

Sami Siddiqui 1J

If you add more reactants, the reaction will favor the products and vice versa in order to satisfy Le Chatelier's Principle. As a rule of thumb, if you add more of a compound on one side of a reaction, then the reaction will want to remove that sudden stress by using that compound in reaction.

Sami Siddiqui 1J

On a quantitative level, we can see if an acid or base is weak if we look at its Ka (or Kb) value. As these values increase, the value of pKa (or pKb) decreases and the strength of the acid or base respectively increases. As the values of Ka or Kb become smaller, the pKa and pKb values increase and,...

Sami Siddiqui 1J

You can find the Ka value for this quesiton by looking at Table 6D.1. It has values specific to cations acting as acids.

Sami Siddiqui 1J

I used a graphing calculator for all my exams in Chem 14A and had no issues. Just as long as you aren't using some wacky programs on your calculator or cheating in any way, you will be fine.

Sami Siddiqui 1J

You will get points for doing both components.

Sami Siddiqui 1J

Any of the Star Wars films are a great option to be honest. I also enjoy any Disney Renaissance film whenever I need to unwind; my personal favorites are Aladdin, The Lion King, and Beauty in the Beast in that order.

Sami Siddiqui 1J

I got tripped up on this one, too. The most important thing to remember is that, when you try neglecting the term "-x" in your equilibrium expression, you have to check to make sure the % ionization is less than 5%. In this question, if you were to find the % ionization while neglecting th...

Sami Siddiqui 1J

I think the most important thing you must do to keep your motivation is, as cliched as it sounds, to save time for yourself. I understand that it may be difficult for you to do so because it seems like everyone's always doing something productive, even in the later hours of the night. Regardless, I ...

Sami Siddiqui 1J

If the K value is generally 10^-4 or smaller, we can assume the change in concentration (x) that we write in our equilibrium tables is negligible in our calculations. This will make it easier for us to make calculations for the equilibrium concentrations of our reactants and products because we won'...

Sami Siddiqui 1J

For me, I normally take notes on the content of an upcoming outline before Dr. Lavelle covers it in lecture. You don't have to understand everything when you go through it the first time because his lectures will essentially tie any loose ends that you may have regarding a particular section. Also, ...

Sami Siddiqui 1J

a) is actually endothermic and b) is exothermic. If H is negative, it is exothermic and therefore heat is released as a product. If H is positive, it is endothermic and heat is added as a reactant. When the temperature is increased, it will cause a shift towards the side with less heat. Hope this h...

Sami Siddiqui 1J

Until you're able to fully access Sapling, you can't do homework yet. However, as plenty of other people have just mentioned, there's ample time to get this fixed up so you don't lose any credit on your homework assignments. For many of us, myself included, there have been some technical issues when...

Sami Siddiqui 1J

I think knowing this is pretty much outside the scope of this class. However, if you haven't taken LS30A yet, you'll actually find out there when you start covering equilibria.

Sami Siddiqui 1J

The user control panel likely isn't functioning the way it should so I just re-registered a new alias under the same UID. Because you're using the same UID, all your previous posts will be saved, so there's nothing to worry about there.

Sami Siddiqui 1J

I can't think of an explicit chemical reaction where the equilibrium concentration of reactants is greater than products; however, I can point out a scenario in which this may be possible. Let's say you have an exothermic reaction such that: R \leftrightarrow P + heat If this reaction was at equilib...

Sami Siddiqui 1J

If you haven't seen the email already, Dr. Lavelle plans to give an extra 10 bonus points (capped at 180 pts, 100%) for all of us, so that's some good news.

Sami Siddiqui 1J

To solve this, we need to break up the salt into two components: the cation and the anion. If we did this in part a, for instance, we'd get Na+ and C6H5O-. Knowing from Dr. Lavelle's lecture that Na+ (or any Group 1 or 2 cation for that matter) does not significantly change the pH, we can simply ign...

Sami Siddiqui 1J

We can calculate pH if we're given the value of H+ concentration: -log(H+) = pH We can calculate pOH if we're given the value of OH- concentration: -log(OH-) = pOH The relationship between acid strength, the Ka, and pKa is very simple. The higher the acid strength, the higher the Ka and the lower th...

Sami Siddiqui 1J

Honestly, the Organic Chemistry Tutor is your best bet. I would also recommend Khan Academy as well. I used it for my AP Chem exam a couple years back and it worked very well.

Sami Siddiqui 1J

Strong acids are able to completely deprotonate, which means that all of the acid molecules can release a proton when making contact with a base in solution. Weak acids do not completely deprotonate, which means only a fraction of the molecules will actually give off protons to bases in solution. St...

Sami Siddiqui 1J

I noticed in the two midterms we've completed that most questions are, indeed, modeled on the homework questions. If you do those, you'll be set for sure.

Sami Siddiqui 1J

We will probably have to memorize the names of the ligands provided on the sheet. It's also advisable that you know which ones are polydentate.

Sami Siddiqui 1J

Hi! I simply count the number of the ligands and that gives you the coordination number. I do this, too. I also consider the fact that some ligands are polydentate, which means that a given ligand atom may be able to attach to more than one site. There isn't really a methodical way of noting when o...

Sami Siddiqui 1J

Memorize the ligand names and really just memorize the blue blurb in the textbook that goes over the process of writing formulas and naming compounds step-by-step.

Sami Siddiqui 1J

The central atom in any coordination compound is always going to be the transition metal cation. To find the coordination number, you just look at the ligands that are next to the metal in the bracketed formula and count them. While counting, you must be cautious of the fact that some of the ligands...

Sami Siddiqui 1J

Complex ions, as far as I know, are literally just coordination compounds but with charges. An example of this would be a hexaaquacopper(II) ion, which is listed in the textbook. So, in general, coordination compounds can be classified as neutral complexes or complex ions with a certain charge.

Sami Siddiqui 1J

Thank you so much, Hannah! You came in clutch with this resource for sure.

Sami Siddiqui 1J

I came back to this thread just to say that Star Wars: The Clone Wars is the move. For anyone with Disney+ and at least a slight interest in sci fi narratives, this series is perfect.

Sami Siddiqui 1J

For me, I usually designate a decent amount of time to take study breaks from doing a ridiculous amount of practice questions. It's very helpful to realize that, in the end, stressing about the test isn't going to do much good for you physically nor will it likely end up bolstering your performance....

Sami Siddiqui 1J

Annika Tamaki 3L wrote:On Lavelle's website, he has a list of ligand names, so if we are supposed to memorize anything, it is likely his list.

This is a very smart way of seeing it. I'm probably going to do this myself.

Sami Siddiqui 1J

Dr. Lavelle addressed this exact issue in today's lecture. In this class among many chemistry courses we'll be taking in the future, we'll be replacing the -ide suffix with -o, so chloride becomes chloro. Chlorido is technically correct by IUPAC standards, but we're just not using it here.

Sami Siddiqui 1J

There isn't a specific place to find these answers; instead, you must contact your TA to discuss your answers. I'm also fairly certain that Dr. Lavelle is going to cover the most commonly missed questions as our finals loom closer.

Sami Siddiqui 1J

I'm feeling great right now, thanks for asking! I've honed in my focus this quarter and look to be finishing strong heading into our upcoming finals. I hope everyone is doing well!

Sami Siddiqui 1J

The points update every so often to reflect the number of points we should've earned by a given checkpoint. So, for instance, people who completed at least 35 posts by the end of Week 7 should be seeing 35 pts under their grade. Of course, by the end of the quarter, we should have at least 50 posts,...

Sami Siddiqui 1J

Yes, this would be the case because of the lone pair. Because the lone pair would make the dipole vectors uneven, we would get a polar molecule.

Sami Siddiqui 1J

The final is going to be held over the weekend. I would need to double check the date but I'm pretty sure it's December 13th.

Sami Siddiqui 1J

I'm not exactly aware of how this grading curve works. Can someone please clarify because I thought the points we earned individually was the only thing that was being factored into grading.

Sami Siddiqui 1J

To best answer this question, consider the following circumstance: If we had a molecule where the central atom had 3 regions of electron density, then we know that the hybridization notation is sp2. This means that one s and two p orbitals were mixed. Because a p-subshell contains 3 orbitals, that m...

Sami Siddiqui 1J

The convention doesn't really matter in this case. I usually just say sp3d because it sounds better and I've seen this notation more frequently in textbooks.

Sami Siddiqui 1J

I'm pretty sure what you said should work fine. I do think it's necessary to mention all this though. Oxidation numbers are typically distinct from the concept of formal charge. It must be kept in mind that formal charges exaggerate "covalent character" by delegating one electron to each a...

Sami Siddiqui 1J

Hi Brendan, I optimized my success on these midterms by doing the following things: 1. Attending UA workshop sessions. They are incredibly useful for additional practice and to clarify any little misconceptions. 2. Reading through notes I've taken in my textbook so I don't have to sit through and re...

Sami Siddiqui 1J

The final is definitely cumulative, so plan accordingly!

Sami Siddiqui 1J

I can confirm that I got an answer of 10 electrons, as Victor had said earlier. There were two single bonds and three lone pairs in the central chlorine atom, so 5 * 2 = 10 electrons surrounding that chlorine atom. It can be confusing given that we associate one electron for a bond if we were to cal...

Sami Siddiqui 1J

The only possibility I could consider for part b in which we can see a linear geometry with lone pairs is if we had an octahedral electronic arrangement with 4 lone pairs. That way, we'd be left with two atoms to be oriented directly across from each other to minimize the amount of repulsion. Edit: ...

Sami Siddiqui 1J

I wasn't so sure either, but I'm sure this past Chemistry Community thread may be able to answer your question:

viewtopic.php?t=37869

Sami Siddiqui 1J

As a final clarification, the molecule as a whole is nonpolar, but the bonds themselves are polar. If you need a further explanation of this, I actually found a past Chemistry Community thread on this exact subject:

viewtopic.php?t=23561

Sami Siddiqui 1J

I thought Si has only 2 double bonds, one with each O, which would give it a formal charge of 0. Why does it have a lone pair? I realized that I assumed Si to have 6 valence electrons instead of 4. That is my mistake. The Lewis structure would be linear. If that is the case, I'm actually not quite ...

Sami Siddiqui 1J

Dr. Lavelle mentioned that the quantum number component and Outline 3 in its entirety is to be focused on for this midterm. Anything else would simply be unnecessary.

Sami Siddiqui 1J

The lectures will probably be asynchronous. Because of this, I heard that if you get on the waitlist for 14B, you can still get admitted into the class. If this is true, then a lot of us can still clinch 14B even with the second pass for enrollment.

Sami Siddiqui 1J

The Lewis structure of SiO2 is such that there are two Si-O double bonds and one lone pair with the Si atom. Because of that lone pair pushing the other two regions of electron density closer to each other, the shape of the molecule isn't exactly symmetrical. As a result, there will be a permanent d...

Sami Siddiqui 1J

Enrolling in Chem 14B and BL in the same quarter is an option that some people take. However, for other students such as myself, the more ideal option may be to take BL in a separate quarter. You know your schedule better than anyone else, so whichever option you take is your call on what you think ...

Sami Siddiqui 1J

What everybody above said is correct. If I may add, polar molecules will experience permanent dipole-dipole interactions due to the uneven electron distribution within each molecule, which leads to partial charges and so forth. Nonpolar molecules, because the electron distributions are even, do not ...

Sami Siddiqui 1J

Additional stability refers to the fact that we can change the Lewis structure such that the formal charges for each of the atoms becomes closer to 0. We can usually do this by extending the octet for central atoms in the third period or larger. If you need a specific example in which the extended o...

Sami Siddiqui 1J

Grades are something I intentionally distract myself from thinking of because they can be stressful. For those of you who didn't perform as well on the first midterm as you would have liked, I would just say that stressing about potential outcomes of failure really isn't going to do anything to help...

Sami Siddiqui 1J

I haven't seen the option for subtitles on Bruin Cast. However, if you feel the need to, you can just increase the speed of the prerecorded footage to 1.25x speed. I usually go at this speed if I'm already comfortable with the content and need to get through the explanations faster while making sure...

Sami Siddiqui 1J

Here are some things I consider while doing a Lewis structure: 1. Make the element with the lowest ionization energy the central atom. 2. Make other elements in the Lewis structure symmetrical about the central atom. 3. To start, assign any two neighboring atoms a single bond and assign any lone pai...

Sami Siddiqui 1J

Cations are much more common as you move closer to the left of the periodic table, otherwise known as the s-block. This is because the ionization energy for removing electrons decreases as you move to the left of a given period (and also as you go down a group). Because of this concept, metals in th...

Sami Siddiqui 1J

I would say yes, it's a great idea to remember what some of the names of the polyatomic ions entail in terms of molecular formula. It'll just make it easier to do all the questions involving Lewis structures. I know it'll take a while to remember all of those polyatomic formulas, but once you know t...

Sami Siddiqui 1J

I really love Parks and Rec on Netflix; in my honest opinion, it's a way funnier version of The Office. I'm also patiently waiting for season 4 of Stranger Things, so there's that.

Sami Siddiqui 1J

The number on the top right corner of the brackets is used to denote the charge of the entire polyatomic species you're focusing on. If you were to calculate the formal charge of each of the atoms in the Lewis structure, you'll notice that the net formal charges will be equal to the charge of the en...

Sami Siddiqui 1J

I assume you have a grip on how to get each of these numbers based on the value of n you're given and the subshell you want to focus on, so for the purpose of simplicity, I'm just going to state what each quantum number means in the shortest way possible: Principal quantum number (n): This value per...

Sami Siddiqui 1J

Partial credit will very likely be given for questions that asked for multiple answers within each answer choice. I would say the last couple of questions would be where this idea of partial credit really comes into the picture; I remember those questions specifically weighing more in terms of point...

Sami Siddiqui 1J

As of me posting this, I can also confirm that the homework for weeks 5-6 is still not available. Just do the exercises that the Unit Outline recommends in the meantime if you need any practice.

Sami Siddiqui 1J

As many others have already stated, the concept "4s before 3d" just means that electrons are filled in the 4s orbital before the 3d orbital. This is because, initially, an empty 3d orbital is inherently associated with a higher energy state than 4s, which would entail that electrons fill t...

Sami Siddiqui 1J

We can tell if an electron is excited by looking at the electron configuration in a couple ways: 1. The electron is elevated to a higher shell by absorbing a photon. This would be reflected in the electron configuration by writing out the configuration as is while writing the subshell in which the e...

Sami Siddiqui 1J

To solve for this question, consider one thing when it comes to ejecting electrons: energy of the incoming photon = energy needed to remove electron (work function) + kinetic energy If we want to maximize the wavelength, we would need to set the kinetic energy equal to 0 because we want to know what...

Sami Siddiqui 1J

To further clarify Rich's comments, We normally set hv equal to the work function under the circumstances he specified because of the law of conservation of energy. To say this in a form simpler than the equation we use for these cases, think of it this way: energy of incoming photon = energy needed...

Sami Siddiqui 1J

In general, when it comes to this topic, all you need to recall is that squaring the wavefunction phi (any function for this, really) will present the probability density of finding an electron in a given region of the electron cloud. I can only infer that squaring phi (which is the height of the wa...

Sami Siddiqui 1J

I can confirm what Courtney said here. The format of the Rydberg equation really won't matter too much, so long as you're using it properly. Just use what you're most comfortable using and you'll do great!

Sami Siddiqui 1J

For all the homework and additional practice questions I've done, I used 3.29e15 as Rydberg's constant and it worked like a charm.

Sami Siddiqui 1J

What David said is probably the most we'll need to know for the midterm. I also attended a UA workshop session today and found that this concept is not normally tested to a great extent, so don't sweat it if some parts of it may appear confusing. Nonetheless, you should still hone in your knowledge ...

Sami Siddiqui 1J

The negative sign in Bohr's equation is important because it indicates one thing: the energy of an electron more closely bound to the nucleus is less than that of an electron farther away from the nucleus. As you get further away from the nucleus, the energy slowly becomes less negative (increases) ...

Sami Siddiqui 1J

Personally, I just did them all for general practice. For the upcoming midterm, however, I placed a much greater emphasis on questions not involving quantum numbers or electron configurations (aka the material that Dr. Lavelle said wouldn't be tested this week). Other than that, you should be fine.

Sami Siddiqui 1J

Never mind, I didn't scroll down far enough on the document. It, indeed, shows Mg as the limiting reactant.

Sami Siddiqui 1J

The LR for this problem would be Mg because the weight of 3 Mg's are greater than the weight of B2O3. The book seems to also use Mg as the LR so you may have done your calculations wrong. In the textbook answer document provided on Sapling, I notice that it uses B2O3 as the limiting reactant. Using...

Sami Siddiqui 1J

The book seems to use B2O3 as the LR because it uses that reactant to solve for the mass of boron produced. Problem: Solid boron can be extracted from solid boron oxide by reaction with magnesium metal at a high temperature. A second product is solid magnesium oxide. (a) Write a balanced equation f...

Sami Siddiqui 1J

On the most fundamental level, it's important to consider that the derivation leads us to the mathematical assumption that the wavelength of an object (e.g., electrons) is inversely proportional to the momentum of this object. By setting the equations E = pc (which is the essential equivalent to E =...

Sami Siddiqui 1J

We multiply by two because that would most ideally represent the spread of a given value. For example, say that you can throw a baseball at 95 mph +- 5 mph. The range of values that your throw can take on are between 90 and 100, so your uncertainty in velocity wouldn't just be 5; in fact, it would b...

Sami Siddiqui 1J

To answer your question Margia, We need to address two specific things. The first thing to consider is that the p, d, and f orbitals have a nodal point at the center of their wavefunctions. What this means is that the electron density is 0 at a very close distance to the nucleus for these types of o...

Sami Siddiqui 1J

To address your question, we couldn't really ever know if an electron is in a specific subshell because all electron orbitals in one-electron atoms are degenerate. In other words, the energies of the orbitals are identical to one another in a given energy level (e.g., n = 5 as you've mentioned in yo...

Sami Siddiqui 1J

To address your last question Chris, you may occasionally calculate a negative value for the frequency from the Rydberg equation and that would cause you to calculate a negative energy if you kept the sign. However, because negative frequencies don't exist, we simply ignore that sign and, as such, w...

Sami Siddiqui 1J

Usually, we can tell if a wavelength is detectable if the actual number you came up with is smaller than the size of an atomic bond, which is basically in picometers. If you start going smaller than that, we really get into some gray area in that regard. As a rule of thumb, if you're working with ma...

Sami Siddiqui 1J

Also, to address your second question (which I haven't clearly addressed in my first response), wavelength is almost always interpreted as the peak-to-peak distance for essentially any wave you can think of. I can't think of instances when this doesn't hold true, but if someone does know of these ci...

Sami Siddiqui 1J

To add onto what Kelly said, I would first like to address the experiment that she was referring to. According to Dr. Lavelle, when an electron beam is passed through a crystal, a diffraction pattern will occur. Because only waves can interfere both constructively and destructively to form diffracti...

Sami Siddiqui 1J

Hey Sami, thanks for your reply and the article, it's a good one to come back to. I actually rewatched Lavelle's lecture and he said that wavelengths that weren't absorbed actually pass through the atom. Still trying to see how this information works with the basics physics info that materials abso...

Sami Siddiqui 1J

keV is basically kilo electron volts, so 1 keV is 1000 eV. The conversion for 1 eV to joules (SI unit of energy) is to multiply the number of eV by 1.6022e-19 and from there you have the amount of energy in joules that you can use in the context of the question.

Sami Siddiqui 1J

I believe that the module also discussed the presence of small holes that the wave would travel through to create diffraction patterns? Would someone be able to explain that portion of diffraction, and also how this would cause constructive or destructive interference? Thanks! I'm no physics whiz o...

Sami Siddiqui 1J

While reviewing physics over the summer, I recall learning about electromagnetic waves and why they are configured the way they are. Although the quantitative side of things are quite blurry right now, I do recall the fact that an electric field can sustain a magnetic field, which can in turn sustai...

Sami Siddiqui 1J

I would think the same way as well. If light of a certain wavelength cannot be absorbed by, let's say electrons of a given metallic surface, I'd assume that the light would reflect off the surface. Because these wavelengths reflected come in a different variety, your eyes will likely process these w...

Sami Siddiqui 1J

When it comes to studying, I do plenty of the things I've done in high school. What I normally do is pre-read (and pre-note if possible) the textbook to get a basic understanding of the content prior to a given lecture. Then, once lecture comes around, I would much more easily absorb the content bec...

Sami Siddiqui 1J

To answer your question, Uyenvy, we should consider one thing: c = v * wavelength This equation in its most fundamental form shows that the frequency and wavelength are inversely proportional; if one increases, the other must decrease. Applying this logic to your question, if we want to eject electr...

Sami Siddiqui 1J

A short answer to the question would be "no." A long answer to the question would be this: When considering the ejection of electrons, we should keep this fundamental equation in mind: energy of photon - work function (threshold energy) = kinetic energy of electron In order for an electron...

Sami Siddiqui 1J

As Kailani mentioned, the Rydberg Equation would be used to find the frequency of the emitted photon. To add on this discussion, I'll show the steps in finding this frequency. Step 1: Assign your known and unknown values: The Rydberg equation is as follows: v= -R[1/(n1)^2 - 1/(n2)^2] where R = 3.29e...

Sami Siddiqui 1J

I had similar teachings when I was younger but now I can safely comment that the radio waves produced by the microwave are not really harmful at all. As you've mentioned, the waves have a relatively long wavelength, so each photon (if we're addressing light as particles) won't carry much energy from...

Sami Siddiqui 1J

To figure this out, the book actually has a figure that distinguishes which specific wavelength ranges correspond to a specific type of wave. If you go Focus 1A and look for Figure 1A.9, it will provide all the classifications you need. Side note: it is helpful to remember that 700 nm to 400 nm is t...

Sami Siddiqui 1J

I wouldn't worry too much about sig figs while going through the steps in your calculations. If we do consider those during intermediate steps of our calculations, we would very likely lose accuracy in our answer. It's best to consider sig figs at the end.

Sami Siddiqui 1J

If I'm not mistaken, 2-butanone was shown using an organic chem structure, which looks like a bunch of zigzag lines put together. As a general piece of advice, each end/corner of those lines represents a carbon atom. Also recall that carbon can only make four different bonds, so the rest of the bond...

Sami Siddiqui 1J

This question is tricky, but I feel it's conceptual more than anything. Each time you double the volume, the concentration is halved. If you do this 90 times, then the concentration will decrease so much so that the number of molecules you'll find in the diluted solution essentially becomes next to ...

Sami Siddiqui 1J

The molecular formula can be the same as the empirical formula. However, you have to find the empirical formula and its molar mass to check if the ratio to the molecular formula's molar mass is 1. Then, the empirical formula and molecular formula are one and the same. Yes, Jeffrey is absolutely rig...

Sami Siddiqui 1J

That's actually very interesting! I never thought of it that way until you mentioned it. I would have to agree that the intervals you provided represent precision and the percent confidence levels would be the accuracy of a given value falling in the intervals.

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