CHEMISTRY COMMUNITY

Aydin Karatas 1F

BOOST!!! You need to see this Dr. Lavelle!

Aydin Karatas 1F

I'm taking a break from writing my soc paper to tell you, Dr. Lavelle, that I could not be more appreciative of you. Chem 14A and 14B would not have been the same without your amazing energy and commitment to teaching. Your music was awesome, the UA sessions were a life saver, and the only thing tha...

Aydin Karatas 1F

By their definition, the anode is being oxidized and the cathode is being reduced. So, yes.

Aydin Karatas 1F

I'm definitely planning on dorming! A weird side effect of quarantine those is I'm pretty fine with having a single now even though I so wanted a traditional dorm with roomies senior year of high school.

Aydin Karatas 1F

The way I do it is by looking at an element and seeing what happens to its charge on the other side of the reaction. Then, I can double-check with the other element.

Aydin Karatas 1F

Angelica Soriano 3L wrote:My favorite song at the moment is an anime opening but it's so good!!
Shogeki - Yuko Ando
https://open.spotify.com/track/5QwAdWCn ... zzNnCQbYsg

YOOO! mad respect to this

Aydin Karatas 1F

By Your Side by Nito and Gengar Musique is a banger (both the vocals and guitar). Also, new Day6 album is coming out next month so I'm kinda hyped for that.

Aydin Karatas 1F

Since OH- has to do with the disassociation of water, our solvent, we can include it in our final rate law because we can think of it as not a true intermediate. It comes from our solvent rather than the actual reactants.

Aydin Karatas 1F

From Kate's final review, we clarified that even though [OH^-] is an intermediate, it's included because it's a result of the disassociation of water, which is our solvent. Therefore, we can include it and don't need to consider it as a true intermediate like the other intermediates in this reaction.

Aydin Karatas 1F

Yes! Since the catalyst brings the Ea down for the whole reaction (you can see this when you graph a catalyzed vs. uncatalyzed reaction), both the forward and reverse rates will increase.

Aydin Karatas 1F

Following your steps, since RT is in the denominator for Ea/RT=ln(k/A), you should multiply the value you got for ln(k/A) by RT rather than divide by RT (so RT's cancels out on the left side of my equation). Also, do be mindful of any negatives that appear in your calculations but it should all end ...

Aydin Karatas 1F

The main thing I would recommend is to take it slow and be methodical because there's a lot of calculations going on. These were the values I was given. I want to be able to solve for N, M, and L individually in rate=k[A]^{N}_{0}[B]^{M}_{0}[C]^{L}_{0} . The only variable where I can immediately do t...

Aydin Karatas 1F

For a diagram (either picture or cell notation diagrams) the anode will be on the left and the cathode will be on the right. Another sanity check is the cathode should be the oxidizing agent (i.e. being reduction) while the anode is the reducing agent (i.e. being oxidized).

Aydin Karatas 1F

The overall order of a reaction is found by summing the orders of the elementary reactions.

Aydin Karatas 1F

pretty bare-bones skeleton structure if you ask me

Aydin Karatas 1F

When an oxidizing agent oxidizes a target molecule, the oxidizing agent is reduced in the process. Therefore, a strong oxidizing agent should have a high reduction potential and a poor oxidizing agent will have a low reduction potential. These values should be given in the problem or on a table.

Aydin Karatas 1F

I took typed-notes for fall quarter to test it out but I prefer handwritten. Investing in an iPad for this specific purpose was worth it in my opinion. I feel much more comfortable with handwritten notes because of the ability to draw pictures, arrows, and other symbols with ease.

Aydin Karatas 1F

Ter- is a prefix for "3". Termolecular reactions involve the collision of 3 species. Probability-wise, this is a rare occurrence so they are not as common as unimolecular and bimolecular reactions.

Aydin Karatas 1F

I'll be in CHEM 14BL, MATH 33B, LS 7B, and AN N EA 10W. I ready for this quarter to be over but this class was pretty fun and wholesome.

Aydin Karatas 1F

This was definitely the coolest event I've participated in over Zoom! No questions about it!

Aydin Karatas 1F

I'm a freshman and I can't wait to actually be in person. Seeing some of my friends who are going to UCSD (where dorming hasn't been completely destroyed) having fun with their roomies just makes me want to be on campus even more.

Aydin Karatas 1F

If you write out all the steps of a reaction and add them together for the overall reaction, intermediates should cancel out because they exist in both the forward and reverse reactions (i.e. they are products of one step and reactants of another step). Therefore, for the overall reaction rate, inte...

Aydin Karatas 1F

E^o is measured at 1 atm, 298 K, and 1 M. It is used to find cell potential under nonstandard condition, E, with $E=E^{o}-\frac{RT}{nF}ln(Q)$ .

Aydin Karatas 1F

n refers to the number of e- being transferred. You can find this once you balance the equation (make sure the charges are balanced as well).

Aydin Karatas 1F

I recently watched Ride Your Wave and would recommend it. The final scene lives in my head RENT FREE.

Aydin Karatas 1F

Looking at the equation $E=E^{o}-\frac{RT}{nF}ln(Q)$ , we know E^o will stay the same, so it won't result in a change of E. Both n and ln(Q) double, so this doubling would "cancel out".

Aydin Karatas 1F

The problem should explicitly state some of the initial states of the species (s, l, g, aq). When adding H⁺ or OH^-, those ions should be aq, and H₂O as a product will most likely be l.

Aydin Karatas 1F

V (reduction potential) refers to how likely a molecule wants to be reduced. Oxidizing agents oxidize a target molecule by removing electrons from the target molecule, meaning the oxidizing agent takes on those electrons and is reduced in the process. So a higher V means a stronger oxidizing agent.

Aydin Karatas 1F

Another mnemonic that is helpful when looking at an unlabeled diagram is r eduction happens on the r ight. I really liked the anode mnemonic that someone else said but another way of thinking about the same process is cathode as a "t" in it, which looks like a positive sign. This means ele...

Aydin Karatas 1F

not all heroes wear capes :')

Aydin Karatas 1F

If we look at the equation for \Delta G at standard condition, we get \Delta G^{o}=-RTln(K) . For \Delta G to be negative, thus meaning the reaction favors the products, we need ln(K) to produce a positive number. When K > 1, ln(K) < 0. When 0 < K < 1, ln(K) > 0. Sinc...

Aydin Karatas 1F

Larger, more complex molecules (more particles, more possible energy states) will have high entropy values.

Aydin Karatas 1F

Higher entropy means higher disorder and more possible arrangements. The entropy of gases > the entropy of liquids > the entropy of solids. However, when looking at whether a molecule is likely to form, we also need to take into account enthalpy and the temperature of the reaction to see if the form...

Aydin Karatas 1F

Enthalpy, Gibbs Free Energy, heat capacity (not specific heat capacity), heat, and internal energy are dependent on the amount of substance.

Aydin Karatas 1F

From what I understand, all reversible reactions are isothermal because the energy lost by the system doing work is at such a slow rate that heat from the surroundings replenishes the energy deficit. However, if a reaction is isothermal, this does not necessarily mean it will be reversible. There ar...

Aydin Karatas 1F

From what I understand, water in ice form has a fairly strong crystalline structure which prevents a lot of movement of the individual water molecules. However, liquid water molecules have much more freedom in movement. Even though liquid water might be denser, the individual molecules can possibly ...

Aydin Karatas 1F

In Lec 13, we focused on \Delta S in an isothermal system undergoing reversible expansion (T was constant but V was changing). However, if T was changing, we would refer to the equation in Lec 14 where \Delta S=nCln(\frac{T_{2}}{T_{1}}) . For C, we would use a C P value if the pressure was c...

Aydin Karatas 1F

Going from higher energy states to lower energy states will decrease disorder in the system but the released energy from the reaction will increase disorder in the surroundings. So yes, the entropy of a gas will decrease when it transitions into a liquid state.

Aydin Karatas 1F

Since these values are taken from the perspective of the system, yes. A negative work value means the system has to "exert" itself and do work on the surroundings. A positive work value means the surroundings are doing work on the system.

Aydin Karatas 1F

The pressure is inversely related to volume. Low pressure results in a higher volume, which means more space for molecules to be in. With more possible positions for a molecule to be present, the entropy increases.

Aydin Karatas 1F

Looking at the problems from Sapling, $\Delta S$ is generally J.K^-1 while $\Delta H$ is in kJ. If you want you $\Delta G$ value to display as kJ, remember to dividing the entropy values by 1000. The main thing is to maintain consistency.

Aydin Karatas 1F

For this problem, we will assume that all heat lost by the iron is gained by the water: $q_{water}=-q_{iron}$ . From here, calculate $q=mC\Delta T$ for both sides of the equation, with $\Delta T_{water}=T_{f}-25.0^{o}$ and $\Delta T_{iron}=T_{f}-94.7^{o}$ . Isolate and solve for T_f.

Aydin Karatas 1F

I went to UA sessions (really good worksheets, questions, and explanations provided!) and also reviewed my discussion section problems to see I I could do them on my own.

Aydin Karatas 1F

The side that starts with the higher temperature will have the negative because it will release heat as it reaches equilibrium with the other substance.

Aydin Karatas 1F

I found the UA sessions INCREDIBLY helpful, especially with bridging some gaps in knowledge because they came up with good questions. They're all really helpful UAs, but shoutout to Justin :). As said before, being able to do questions in an efficient manner does help with timing and not being stres...

Aydin Karatas 1F

Looking at the equation, $\Delta U=q+w$ , we can figure out that w=0 since there is no change in volume. A change in volume indicates that work was either done by or on the system.

Aydin Karatas 1F

From what I understand, the mathematical relationship is $C_{v}=C_{p}-R$ , so what you wrote would make sense as $\frac{7}{2}R-\frac{2}{2}R=\frac{5}{2}R$ . So nice (educated) guess :)

Aydin Karatas 1F

To understand this, we can look at the equation $\Delta U=q+w$ . Remember that w, whether the reaction is reversible or not, is dependent on volume change: (1) $w=-nRTln(\frac{V_{2}}{V_{1}})$ , (2) $w=-P\Delta V$ . If the volume doesn't change, w=0, so $\Delta U=q$ .

Aydin Karatas 1F

Work is the amount of energy needed to move something, which can be seen in the equation w=Fd . If you take this equation and F=PA , we can derive the equation for work on an irreversible reaction: w=-P\Delta V. . The negative is there because this equation looks at work from the perspective of the ...

Aydin Karatas 1F

Specific heat capacities should be given in the problem. The equation sheet also gives various heat capacities for water at different physical states.

Aydin Karatas 1F

Which wor function you'll need depends on if the reaction is reversible or not. Reversible reactions occur when a change in pressure happens at a very slow rate (the problem might also mention that the reaction is reversible). In this case, we'll use w=-nRTln(\frac{V_{2}}{V_{1}}) . A reactio...

Aydin Karatas 1F

Kandyce Lance 3E wrote:For the constant pressure, why do we multiple 7/2?

The problem just tells us that the constant‑pressure molar specific heat for CO2(g) is equal to 7R/2. So we need to multiply the gas constant by 7/2 to get the correct constant‑pressure molar specific heat.

Aydin Karatas 1F

If a system is insulated, then no heat transfer can occur. This would be a trait on an isolated system where neither matter nor energy can be exchanged with the environment. A closed system is "isolated" in the sense that no matter can enter or exit the system, but energy can still be tran...

Aydin Karatas 1F

One recommendation I have to make work/school time seem less exhausting to periodically take a short break to watch an episode of a show you are watching. A general rule you could follow is 40 minutes of studying followed up with one ~20-minute episode. Rinse and repeat. Doing this can also motivate...

Aydin Karatas 1F

Since pKa is -log(Ka), Ka would be 10^-(pKa). This process would also work for going between pKb and Kb, pH and [H₃O⁺], and pOH and [OH^-].

Aydin Karatas 1F

If the temperature isn't given, I believe we just assume it's at the standard 25^oC.

Aydin Karatas 1F

For determining which gas constant to use, look at the values given in the problem and what the unit of measurement the answer should be in. The correct gas constant will cancel out all variables present and leave you with just the unit of measurement of the answer.

Aydin Karatas 1F

If a reaction requires energy, such as going from a low energy state to a high energy state (ex. solid to liquid), it is endothermic. If a reaction is going from a high energy state to a low energy state (liquid to solid), it will release energy and be exothermic.

Aydin Karatas 1F

A phase change is a general term for the various processes (sublimation, vaporization, melting/fusion, etc.) where a substance changes its state.

Aydin Karatas 1F

Another test to see if the approximation is acceptable is to check (after approximation) x < 5% of the initial concentration where x = the final concentration (so taking the ionization/protonation %).

Aydin Karatas 1F

Hess's Law states that enthalpy is a state function. This means that values of enthalpy can be added/subtracted because state properties are not dependent on the path taken to obtain a state.

Aydin Karatas 1F

If a concentration is given as the initial, that molecule will likely decrease. So use -x for that molecule and all other molecules on that side of the reaction. Molecules on the other side of the reaction will have a +x and have an initial concentration of 0.

Aydin Karatas 1F

Look's good to me!

Aydin Karatas 1F

When water is the solvent, it's ignored. When it is involved in the reaction (ex. being in gas form), it is included in calculations.

Aydin Karatas 1F

Use the initial value given for a molecule. This initial will most likely decrease during the reaction, so you know it will be negative. Based on which side of the equation it is on, make sure all other molecules on the same side as the molecule with the initial value will decrease and all species o...

Aydin Karatas 1F

Le Chatelier's principle would tell us that the reaction would shift to the right because it wants to return to equilibrium. If there is an increase in the concentration of reactants, more product must be formed to maintain the same K value at that temperature.

Aydin Karatas 1F

Kw can be used for problems regarding the concentrations of acids and bases and pH. So yea, but it can be used in related scenarios.

Aydin Karatas 1F

All the K's are equilibrium constants. Kc denotes the equilibrium constant when using concentrations. Kp denotes the equilibrium constant when using partial pressures of gases. Keq is can denote the equilibrium constant of an equation. Q is the reaction quotient, which might not be an equilibrium. I...

Aydin Karatas 1F

The stronger acid will have a comparatively smaller pKa.

Aydin Karatas 1F

Concentration should be given.
If not, moles of the P/R and the volume of the solution might be given. In this case, $Concentration=\frac{n}{V}$ .

Aydin Karatas 1F

That part of the lecture was talking about how you can convert between partial pressures and concentrations for gases. PV=nRT is the ideal gas law. If we rearrange it to P=\frac{nRT}{V} , we notice that n/V is just concentration ( M=\frac{mol}{L} ). Therefore, if we ever need to convert partial pres...

Aydin Karatas 1F

Any solvents, liquids, and solids are not included in concentration calculations. So, unless the water is in gas form, its concentration is not used because it either doesn't have a concentration (such as when it is a liquid or solid precipitate) or has a negligible concentration (such as when it is...

Aydin Karatas 1F

To add to what's already been said, for gases, a pressure increase can be caused by either a decrease in volume or the addition of inert gases. If the volume decreases, the side of the reaction with more moles has is favored because its concentration will increase faster. If the pressure increase is...

Aydin Karatas 1F

P represents pressure and can be measured in atm, bar, or Torr. V represents volume and needs to be in L. n represents the number of moles. R is the gas constant and can differ depending on which unit of pressure you are using. The constants and equation sheet for this class has all of them, so you ...

Aydin Karatas 1F

In the week 9, Monday lecture, Dr. Lavelle explained that the porphyrin ring (a chelating ligand structure) fills up 4 of these sites, creating the heme structure. The fifth site is occupied by histidine, an amino acid in the myoglobin. The sixth site is what oxygen would bind to so that oxygen can ...

Aydin Karatas 1F

Lillian wrote:I'm going to sleep until my organs fail, obviously.

how can i learn this skill

Aydin Karatas 1F

Because transition metals can have multiple oxidation states, the charge of a d-block atom will depend on the molecule it's in. For example, in [AgCl 2 ] - we know that the charges of all the atoms of Ag and Cl must add up to 1-. Since the charge of a chloride ion is 1-, gold must have an oxidation ...

Aydin Karatas 1F

Practice and watching videos are a great way to "consume" the needed information. But to really get it down, I try to teach myself or someone else the concept to fully "digest" it. So I'd recommend not just reviewing material, but also testing if you could explain it without the ...

Aydin Karatas 1F

Just to clarify, will we be expected to calculate equilibrium constants or just have a basic understanding of them? I believe that equilibrium constants are discussed further in Chem 14B. A general understanding of how and why they work should do but the basics of actually calculate them aren't too...

Aydin Karatas 1F

To add on to what's been said, weak acids will combine with water, specifically H+ ions, leaving behind OH- in solution, thus resulting in a basic solution. Weak bases will combine with water, specifically OH-, leaving behind H+/H 3 O+ in solution, thus resulting in an acidic solution. This is all b...

Aydin Karatas 1F

I like listening to calming music when studying, especially Lo-Fi. I have heard the opposite for some people (listen to intense music or no music at all). I know the final is in 2 days but it would be helpful to find what works for you.

Aydin Karatas 1F

Transition metals are those listed in the d-block. According to the IUPAC, they're atoms with incomplete d sub-shell or can give rise to cations with an incomplete d sub-shell. For this class, elements of the first row of the d-block are the only transition metals to familiarize yourself with (parti...

Aydin Karatas 1F

To find the coordination number, you need to find out how many times each ligand can bind with the central metal. For example, the N on NH 3 has a lone pair that can be donated. Since it has one lone pair on one atom, it can only have one coordinate covalent bond with the central atom. Therefore, (N...

Aydin Karatas 1F

SO₄ is capable of having 1 coordinate covalent bond with our central metal, Co. NH₃ is capable of having 1 as well. (NH₃)₅ therefore contributes 5 to the coordination number. This means that Co has 6 coordinate bonds to other atoms/molecules, thus a coordination number of 6.

Hope this helps.

Aydin Karatas 1F

Always start by identifying the charge of the atoms/molecules directly bound to the metal. Then compare that to the overall charge of the larger complex, because the charge of the metal (oxidation number) plus the charge of the other atoms/molecules should equal the overall charge. In the case where...

Aydin Karatas 1F

(en) is short for ethylenediamine (H2NCH2CH2NH2). It is a bidentate ligand (from the two nitrogens) and contributes 2 to the coordination number of the central atom.

Hope this helps.

Aydin Karatas 1F

AsO 4 3- is nonpolar because it has resonance structures. In theory, any of the O could be double-bonded, so any 3 O could have that 1- charge. These electrons are delocalized and could exist on any of the oxygens at any specific time. This, in combination with the symmetrical arrangement of O aroun...

Aydin Karatas 1F

Same atom connectivity refers to compounds arranging their atoms in the same way. This makes sense for resonance structures because atom arrangement is the same, only the electrons are moving. Between each resonance structure, the overall charge stays the same but FC per atom might differ between di...

Aydin Karatas 1F

When determining the shape, identify all regions of electron density around the central atom. These would include bonded atoms (it doesn't matter what type of bond because, in this case, they are all considered regions of electron density) and lone pairs. From there, you can figure out the most opti...

Aydin Karatas 1F

To add on to what Jeffrey said, lone pairs will slightly distort bond angles because they occupy a larger volume of electron density than a normal bond. In a normal bond, shared electrons are "confined" in-between two positively charged nuclei. Lone pairs, on the other hand, are not constr...

Aydin Karatas 1F

To add on to what's been said, keep in mind resonance structures can also contribute to symmetry and nonpolar character. For example, CO 3 2- is nonpolar. Although it seems there is a highly electronegative pull between the two O 1- because there are resonance and a symmetrical structure, the molecu...

Aydin Karatas 1F

Yes. Molecules want to be stable, so the resonance structure with the lowest energy will be preferred, for it is most stable.

Aydin Karatas 1F

The outermost electrons will be the electrons that are most affected by another atom of high electronegativity. Since the inner electrons are shielding the outermost electrons for the full nuclear attraction of the nucleus, they are the easiest to remove.

Aydin Karatas 1F

Looking at Dr. Lavelle's lecture from 10/28/2020, Oxygen (radius = 68 pm) does have a smaller radius than nitrogen (radius = 75 pm). However, there seems to be a couple of anomalies, such as Mg being larger than Na; Pb being larger than Tl. I don't think these exceptions are too big of a deal though...

Aydin Karatas 1F

Expanded octets can occur in p-block elements in row 3 or below. This is because they have an unoccupied d-orbital that can accept more elections than the conventional octet rule. The most common ones you'll see are Sulfur, Phosphorus, and Chlorine.

Aydin Karatas 1F

Intermolecular forces because it is the force between or "inter" molecules are responsible for holding materials together. Melting and boiling would cause these forces to weaken. Thus, we go from solid --> liquid --> gas.

Aydin Karatas 1F

Nitrogen will follow the octet rule. The main exceptions to keep in mind are H and He might only have 2 valence electrons; Li and Be might only have 4 valence electrons; B might only have 6 valence electrons; and P, S, and Cl can have octets larger than 8 due to having an empty d-orbital. Hope this ...

Aydin Karatas 1F

Hydrogen bonding does not occur between CH 3 CHO molecules because there are no positively charged hydrogens present. H must be covalently bonded to an N, O, or F in its own molecule to hydrogen bond with neighboring molecules. Dipole-dipole interactions are still possible since the oxygen is slight...

Aydin Karatas 1F

On top of the fact that N, O, and F are very electronegative, which does cause them to have a slightly negative charge, for H bonding to occur, the H must also be slightly positive. H in methane can't hydrogen bond because there isn't a sufficient electrostatic difference between C and H, but H in w...

Aydin Karatas 1F

The wording of this question also confused me at first. I'll use the carbon-nitrogen bond as an example to explain what this means. In this question, overwhelming, such as in "overwhelming C-N bond character", means that the size of the actual carbon-nitrogen bond in the question was much,...

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