CHEMISTRY COMMUNITY

Sophia Wendin 3L

I'm taking Chem 14BL!

Sophia Wendin 3L

I'm not 100% sure, but since it isn't in the outline and hasn't been mentioned in the lectures, I don't think we need to know about it.

Sophia Wendin 3L

Catalysts aren't used up in a reaction. When looking at steps of a reaction, the catalyst will first appear on the left/reactant side, then appear later on the product side like it has been "consumed" then "produced."

Sophia Wendin 3L

I compared the rates using k = Ae^(-Ea/RT). We don't know A, but since it will be the same for both pathways, it will cancel. I plugged in the two different activation energies for Ea then compared the two different k values. I found that the rate constant was about 6 x 10^8 bigger for the path with...

Sophia Wendin 3L

k or what I call "little k" is the rate constant of a reaction. A large k value will indicate a faster reaction. It is an important aspect in determining the a reactions rate law and rate at a given time. K or what I call "big K" is the equilibrium constant of a reaction, which ...

Sophia Wendin 3L

(b) diluting the solution at the anode would increase cell potential because adding pure water decreases the concentration of solution there. Think of the chemical equation and how increasing or decreasing concentration shifts the rate of reaction. Since it is at the anode (where the concentration ...

Sophia Wendin 3L

I try to set reasonable goals for the day and give myself breaks often. Everyday I take a longer break during lunch to watch tv with my roommates. Some days I get a lot of work done and other days I barely do anything. On days where I'm feeling super unmotivated, going outside helps me a lot.

Sophia Wendin 3L

Anthony_3C wrote:If you want a more detailed explanation, here is a great website:
https://chemistry.stackexchange.com/questions/5520/why-do-liquids-and-solids-have-constant-concentrations

Thanks! This was helpful.

Sophia Wendin 3L

The half reactions have electrons on different sides of the equations so they will cancel out when you add the two reactions together. Before adding, make sure both half reactions have the same number of electrons, and if they don't multiply one or both by a coefficient.

Sophia Wendin 3L

I like to go through the textbook section and take notes, then attempt all of the problems. When I correct my answers I pay close attention to what kinds of problems I got wrong and what common mistakes I made. For the last midterm, I decided to do the textbook problems twice to make sure I was prep...

Sophia Wendin 3L

I follow these general steps for balancing equations 1. find oxidizing and reducing agent 2. determine half reactions for each agent 3. balance everything in half reactions except H and O 4.balance O and h (acidic conditions balance O with h20 and H with h+)(basic conditions O with oh- and h with h...

Sophia Wendin 3L

Mine did that too and wouldn't change even when I refreshed the page or logged in and out. Luckily I had already attempted the problem a few times so I went back to a previous attempt and it let me type again.

Sophia Wendin 3L

Write out the half reactions and see how many electrons are used in those. For example, Au3+ + Zn ––> Au + Zn2+ (unbalanced) can be written as 2Au3+ + 6e- --> 2Au and 3Zn ––> 3Zn2+ + 6e-. The n value here would be 6!

Sophia Wendin 3L

It's easier said than done, but I try not to think about it or else I stress myself out. Going outside and talking to friends helps me get my mind off of school.

Sophia Wendin 3L

There are a few rules that you can follow. First, any neutral element that is not a part of a compound (such as O2 in this problem) will always have an oxidation state of 0. Second, oxygen and hydrogen usually have oxidation states of -2 and +1, respectively, when attached to any compound. So for t...

Sophia Wendin 3L

To heat the water to 100° the kettle also has to be at 100°, so we calculate q for both the water and kettle and add them together to get the total heat needed.

Sophia Wendin 3L

First I did qcold = -qhot to find the final temperature.

(50g)(4.18J/gC)(Tf - 20) = -(65g)(4.18J/gC)(Tf-50)

Once I had that, I used ΔS = nCln(T2/T1) for the hot and cold water.

Finally, I added the two ΔS values together to get ΔStotal.

Sophia Wendin 3L

Also, I don't know how much we'll need it, but it's probably a good idea to practice using the equation so you feel prepared for any questions on the midterm. Reading the textbook and doing the practice problems helped me understand it better.

Sophia Wendin 3L

Here we use the equation S = klnW and W is calculated using O^n or orientations^(number of particles). For (a) all molecules are in the same direction, so there is one orientation and W = 1^64. For (b), there are 4 orientations so W = 4^64. You can then plug those values in to the equation to get S.

Sophia Wendin 3L

How can we use the equation ΔH = ΔU + PΔV (which is only true under constant pressure) for this question if we're under constant volume not constant pressure? For constant volume we can use ΔH = ΔU + ΔPV and ΔP = nRΔT/V from the ideal gas equation and when you plug that back in we have ΔH = ΔU + nR...

Sophia Wendin 3L

The textbook has the equation with initial pressure in the numerator and final in the denominator. Since PV = nRT, P and V are inversely related so that's why it's flipped.

Sophia Wendin 3L

Bethany Yang 2E wrote:someone made this doc, and this might help with trying to recall equations!
https://docs.google.com/document/d/1A-G ... sp=sharing

Thanks for sharing this! I've been struggling to keep track of all of them.

Sophia Wendin 3L

The equation sheet tells us that 1L = 1dm^3 and we can convert that to 1000 cm^3! I also remember that in one lecture we learned that 1 cc (cubic centimeter) equals 1 mL which is another way to convert.

Sophia Wendin 3L

I think the textbook answer of 90 x 10^2 is an error. The solution manual says the answer is 4.9 x 10^2 which is what I got when I solved for w = deltaU - q.

Sophia Wendin 3L

Educated by Tara Westover! I don't usually read memoirs but her story is so interesting.

Sophia Wendin 3L

I think we use 8.314J/K*mol for that equation because the mol and K cancel with n and T leaving us with J.

Sophia Wendin 3L

I know some of my past classes have referred to entropy as the "randomness or disorder" of something, which relates back to how Doc. Lavelle said in lecture that entropy is the representation of all of the possible "microstates" of a system. On a broad thought, the entropy is th...

Sophia Wendin 3L

Work is only not accounted for in calculating deltaU when we have a constant volume (because no expansion work is done). Because this problem involves a constant pressure (as opposed to constant volume), volume can change and thus we have to account for both q and w in calculating deltaU. Here are ...

Sophia Wendin 3L

My routine is to spread out the different sections on the outlines over a few days. I like to read the textbook and take notes on the main concepts then attempt the practice problems for that section. When I check my work I make sure to pay attention to which problems I get wrong and what common mis...

Sophia Wendin 3L

I'm struggling too and I'm planning to read through the textbook and take notes and then do the practice problems until I understand them better.

Sophia Wendin 3L

I think that if you do practice problems, you can identify a pattern in the way different categories of questions are phrased, and then you can match those categories to the equation they use in the answer key. Honestly, that's the best I can come up with and what I would do, but other people may h...

Sophia Wendin 3L

Hello! I'm currently taking Chem 14BL right now, and I just want to advise that this class, while not the most difficult material, is a lot of work. So if you think you are already going to have a busy schedule next quarter, you might want to consider whether you can handle this extra workload (wee...

Sophia Wendin 3L

I think the general rule is 10^-4 but it's good to check and make sure x is less than 5% of the initial value.

Sophia Wendin 3L

Thank you!

Sophia Wendin 3L

Percent protonation applies to bases and percent deprotonation applies to acids. Acids are proton donors, so we want to see what percent of the acid is deprotonated. Bases receive protons, so we want to see what percent is protonated.

Sophia Wendin 3L

I feel this way too! We all did our best and if we don't do as well as we hoped we can learn from it and hopefully be more successful on MT 2!

Sophia Wendin 3L

The reason why a larger Ka value correlates to a stronger acid is because Ka=[products]/[reactants] and the larger the Ka value is, the more products there are. When there are a lot of products, that means that a lot of the acid (reactant) dissociated which means there will be more H+ ions, making ...

Sophia Wendin 3L

Like everyone else said, Ka is for acids and Kb is for bases, and we can convert between the two because we know that Kw = 10^-14 and Ka * Kb = Kw.

Sophia Wendin 3L

The calculation should be the same from what I saw while doing the Sapling homework!

Sophia Wendin 3L

The textbook problems were super helpful for me last quarter. I would recommend doing all of them (or as many as you can/feel like) and then spend extra time on the topics you get wrong.

Sophia Wendin 3L

the reactants decrease in concentration since they are forming the products, so their X value is negative and the X value of the products are positive since they are being formed.

Sophia Wendin 3L

multiply out the denominator( (a + b)^2 = a^2 + 2ab + b^2) then cross multiply with 53.3. Move everything to one side so you can find X using the quadratic formula.

Sophia Wendin 3L

The total pressure is the sum of all the partial pressures of a system. The problem gives you the Kp and the initial concentration of PCl5, so you can set up an ICE box to find an equation for x, and then plug in x to find each of the partial pressures of PCl5, PCl3, and Cl2. Add them all together ...

Sophia Wendin 3L

Clueless and Heathers are my comfort movies and New Girl and Skins are my comfort shows. There are a bunch of other movies/shows that I like to rewatch over and over again but these are my most-watched.

Sophia Wendin 3L

I made an ICE table and found the initial concentration of SO3 by dividing .680 mol/3.5L. The initial concentrations for SO2 and O2 are 0. Then I put in the change, which is -2x for SO3, +2x for SO2 and +x for O2, and since we are given the amount of O2 in moles at equilibrium, we can find x by divi...

Sophia Wendin 3L

I've been trying to spread my studying out instead of cramming right before and I think that has helped. Also, while doing the textbook problems, I focused on what I got wrong so I didn't make the same mistakes during the midterm or final.

Sophia Wendin 3L

Concentrations can't be negative so pick the X value that makes it so the concentrations of the products and reactants aren't positive.

Sophia Wendin 3L

They are optional but they are good practice! They were helpful for me last quarter before the midterms/final.

Sophia Wendin 3L

Similar to what others said above, an easy way to remember this is by thinking of heat as either a "product" or "reactant" in the reaction. If the reaction is endothermic, heat is a "reactant," so increasing temperature shifts the reaction to the right and increases K....

Sophia Wendin 3L

Some reactants are being turned into products so the reactant concentration will decrease (-X) from it's initial concentration of 3 and the product concentrations will increase (+X) from their initial concentration of 0.

Sophia Wendin 3L

Hi! An inert gas is just a gas that, when added to the equilibrium system, won't cause a chemical reaction. When an inert gas is added to the sealed reaction vessel, it does increase the pressure but because the volume of the reaction vessel itself doesn't change, the volume of the gases contained ...

Sophia Wendin 3L

I had the same question! The Lewis structure I drew makes it seem like it's polar and when I looked it up a bunch of websites said it is polar so I think it might be a mistake.

Sophia Wendin 3L

A Bronsted acid is a proton (H+) donor and a Bronsted base is a proton acceptor. In the textbook problem HNO3 donated a proton to HPO4 so HNO3 is the bronsted acid and HPO4 is the base.

Sophia Wendin 3L

There's a helpful list in section 9C (table 9C.1) of the textbook with a list of common ligands that also tells you if the ligand is monodentate, bidentate, tridentate, or hexadentate. Using this, count up the number of sites ligands are attached to the metal and that's the coordination number.

Sophia Wendin 3L

Hi! Dr. Lavelle mentioned in lecture that both the -ido and -o are the same thing, except -ido is a newer notation and -o is the one we will mainly use in this course. So technically, both hexacyanoferrate(II) and hexacyanidoferrate(II) are correct. If the ligand ends with -ide, you change the endi...

Sophia Wendin 3L

I was confused about this too! There are five regions of electron density, and three of those are lone pairs. Since a trigonal bipyramidal shape has three equatorial positions and two axial positions, it makes sense for the two atoms to go on the axial positions (making their bond angle 180) and the...

Sophia Wendin 3L

The UA worksheets are helpful. I also like to go through my notes and the textbook and rewrite the main concepts in an organized way.

Sophia Wendin 3L

Claire_Kim_2F wrote:https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/NamingCoordinationCompounds.pdf here is the link for anyone who needs it!

Thank you!

Sophia Wendin 3L

Katie Lam 2G wrote:I read the final exam information on Dr. Lavelle's website, and he said it will be approximately weighted to the amount of time covered in the syllabus. He wrote that quantum was the biggest section, so we should expect more questions from that section. Hope this helps!

Thank you! This was helpful

Sophia Wendin 3L

I'm definitely a little stressed about getting ready for finals!

Sophia Wendin 3L

Sophia Hu 1C wrote:We know H2O and NH3 is neutral. All halogens have a typical oxidation state of -1. Since there are two Chlorines then there is two -1 charges. However, it is all neutral which means Co must have a +2 charge.

I was confused about this too! Thank you!

Sophia Wendin 3L

Elements in the p-block of period 3 or later have empty d-block orbitals that can hold electrons so they can have more than 8 electrons in their outer shell. S, P, and Cl are some common atoms that can have expanded octets. This confused me at first too!

Sophia Wendin 3L

Compounds with double or triple bonds (meaning there's at least one pi bond) and resonance structures have delocalized pi bonds because the bonds can have multiple positions in the compound.

Sophia Wendin 3L

The lone pairs have stronger repulsion and therefore want to stay as far away from each other as possible. Occupying the equatorial positions allows for this, as the axial positions would still require one pair to take up an equatorial position, creating a smaller angle between the lone pairs. At t...

Sophia Wendin 3L

I find the textbook problems super helpful, but before I start a section I like to look through the textbook and my notes and write down some of the key concepts for that topic. Once I'm done, I check my answers and figure out what topics I struggle with and need to focus on the most.

Sophia Wendin 3L

My notes are a lot more organized now and I like to divide them up by topic instead of putting as much information as possible onto a page. I also am learning to study over a few days instead of cramming the night before.

Sophia Wendin 3L

I am slightly confused about drawing lewis structures, and specifically the octet rule. Do elements with 1 or 2 valence electrons have to fill an octet? For example, when drawing out a compound that contains the element Na (which has one valence electron), is it ever possible for it to fill the oct...

Sophia Wendin 3L

The textbook problems on the outline help me the most because I can see what topic I need to focus on based on which problems I get wrong.

Sophia Wendin 3L

Megan Singer 1F wrote:
I thought the atom with the lowest ionization energy/lowest electronegativity goes in the middle

Yes! The atom with the lowest ionization energy does go in the middle. For example, C is in the middle in CCl4 and C has a lower ionization energy than Cl.

Sophia Wendin 3L

Angelica Soriano 1F wrote:Professor Lavelle just came out and said that winter enrollment in all his classes will increase! :D I linked it down below.

viewtopic.php?f=157&t=67870

Thank you!! I was super stressed about this.

Sophia Wendin 3L

When the cation is small with a high charge (high polarizing power), it can distort a large anions (high polarizability) electrons really well so there is a shared region between the two ions. This shared region gives the ionic bond covalent character. In a covalent bond, atoms share electrons, so a...

Sophia Wendin 3L

I solved this problem by adding the 2 electrons back into the configuration so I could tell what metal M was more easily. For a) the non-ionic e- configuration would be [Ar] 3d7 4s2, which is Co. For b) the e- configuration would be [Ar] 3d6 4s2 so the metal is Fe.

Sophia Wendin 3L

Elements in row 3 or higher (specifically p-block elements) can have expanded octets because they can utilize their d-orbitals when bonding. Group 13 elements (like B and Al) typically don't have complete octets because they would need to gain 5 e- which is a lot. However these compounds do have a ...

Sophia Wendin 3L

Studying at home is hard for me too. I like to study or do work for a little while, like an hour, then take a break and do something else for 15 or 20 minutes. I think my breaks are most effective when they don't include more screen time, so usually I'll go outside or make a snack or talk to my room...

Sophia Wendin 3L

The sum of oxidation numbers for each element must add up to the overall charge of the molecule (so in this case, it should equal -1). Oxygen has an oxidation of -2, and since there are four oxygen atoms, the sum of oxidation states for oxygen is -8. Now, you can pick an oxidation state for the chl...

Sophia Wendin 3L

This confused me too. Since the experimental bond lengths were in between the values for single and double bonds for both carbon-oxygen and carbon-nitrogen bonding, I answered that there is ample C=N and ample C=O bond character.

Sophia Wendin 3L

Hannah Montana is my go to show but I also recently watched The Haunting of Hill House and The Haunting of Bly Manor and I highly recommend.

Sophia Wendin 3L

I try not to get too stuck on difficult problems so I'll make not of questions I'm unsure about and revisit them later. I'm also a lot less nervous when I feel confident going into the test, so I like to do a lot of practice problems until I'm at the point where I can correctly answer most of them a...

Sophia Wendin 3L

Electronegativity increases going right on the periodic table because the number of protons in the nucleus increases, the electronegativity or attraction will increase. The nucleus brings in the electrons closer together because of the increase in attraction. Electronegativity will then decrease go...

Sophia Wendin 3L

For me, practice problems are super helpful and I pay close attention to where I make mistakes so I can watch out for that on tests. I also like to rewrite my notes because it helps me remember some of the key concepts from the lectures.

Sophia Wendin 3L

I was wondering this too. I think I'll probably try enroll during my second pass and hopefully it works out!

Sophia Wendin 3L

Anions usually have a bigger radius than cations/the parent element because there's more electron repulsion, and cations have smaller ionic radii because there's less repulsion and Rb+ would have no electrons in the n=5 energy level.

Sophia Wendin 3L

I struggle with remembering size patterns but it helps me to think of what happens when moving across a period vs down a group. Across a period, nuclear charge increases and e- are pulled in more so atomic radius increases. Down a group, the number of energy levels increases so atomic radius decreas...

Sophia Wendin 3L

I struggle with this too! It helps me to think of why the radius would increase or decrease, like when going across a period, the nuclear charge increases so the electrons are more tightly bound and the radius is smaller. When going down a group, n increases so the additional energy shells means the...

Sophia Wendin 3L

I like going on walks to take a break from staring at a screen all day!

Sophia Wendin 3L

I plugged in the mass of an electron for m, which is 9.11 x 10^-31 kg to get the electron's energy.

Sophia Wendin 3L

This confused me too! There's a pdf on the Chem 14A website called solution manual errors and it says the correct answer for this problem is 6.7 x 10-37 m which is what I got when I used 10 m/s for Δv.

Sophia Wendin 3L

I like to rewrite some of my notes/key concepts and do a bunch of practice problems from the textbook, sapling, audio-visual modules, etc. It helps me to pay close attention to what problems I get wrong or common mistakes I make so I don't make the same mistakes during the midterm.

Sophia Wendin 3L

E = hc/wavelength so wavelength and energy are inversely proportional. A smaller wavelength means higher energy so that's why the Lyman series has n=1. The Balmer series wavelengths are longer than the Lyman series, so they start at n=2 instead of n=1.

Sophia Wendin 3L

- For (a) n=2 and l=1 means we are looking at the 2p subshell. Since l=1, ml can be -1, 0, and 1, so the 2p subshell can have 3 orientations or 3 orbitals - For (b) n=4 and l=2, which is the 4d subshell, and we are given ml = -2, so there is only 1 possible orientation, so 1 possible orbital. - For ...

Sophia Wendin 3L

Since we are only given n, the answer is all of the orbitals in the n=2 energy level. So the answer would be 4 since there is 1 2s orbital and 3 for 2p!

Sophia Wendin 3L

So is delta v equal to the change in velocity, essentially the uncertainty they give us in the problem? Also, I was under the impression that when we were using uncertainties like ± 5.0 m/s we had to double it for it to be equal to delta v. I thought this because in this case the answer could be an...

Sophia Wendin 3L

For me, it helps to convert the wavelength into a frequency. If the frequency is bigger than the speed of light, which it would be in this case, then it probably isn't possible to measure the de Broglie wavelength.

Sophia Wendin 3L

The mass of an electron is 9.11 x 10^-31 kg, so use that and the given velocity to find the kinetic energy!

Sophia Wendin 3L

I was confused about this too, but first I used the given wavelength to solve for frequency so I could use the equation frequency = -R [1/n1^2 - 1/n2^2]. The textbook tells us that for the Lyman series (the lines in the UV region), n1 = 1, so we only need to solve for n2 since n1 is given.

Sophia Wendin 3L

Yes, both constructive and destructive interference create diffraction patterns. The diffraction patterns look different depending on which type of interference it is.

Sophia Wendin 3L

Hi! I think the balanced equation for this problem is actually C14H18N2O5 + 16O2 –> 14CO2 + 9H2O + N2. In that case, O2 would actually be the limiting reactant because 1 mol of C14H18N2O5 needs 16 moles O2 to react, and .0068 mol C14H18N2O5 * 16 is .1088 moles which is more than the amount of O2 (.0...

Sophia Wendin 3L

I usually wait until the end of the problem to apply significant figure rules so my answer is as accurate as possible. When I rounded at each step I noticed that my answers were a little bit off from the given answers.

Sophia Wendin 3L

It depends on the problem! When multiplying or dividing, use as many sig figs as the number with the least amount of sig figs. When adding and subtracting, round to the lowest number of decimal places.

Sophia Wendin 3L

Hey Guys, in regards to sapling HW. I was finally able to grab a hold on the access code. Are the HW problems in the syllabus, are the problems submitted on sapling? The textbook problems in the syllabus are highly recommended for review but the actual homework is on the Sapling website under the a...

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