CHEMISTRY COMMUNITY

jasmineculilap_3F

Intermediates are produced in a chemical reaction, but they are also used up to make a different product. Therefore, they won't be in the final equation. Catalysts won't be used up. It's just added to decrease activation energy, so it will be on the product side at the end of the equation since it's...

jasmineculilap_3F

You can add the exponents in the rate law together, use the units of the rate constant, or use graphs.

jasmineculilap_3F

It doesn't depend on reactant concentration. The graph of concentration vs. time is a negative linear line.

jasmineculilap_3F

Yes. The flow of electrons therefore goes from the anode to the cathode, so the anode is oxidized while the cathode is reduced. The anode should also be on the left.

jasmineculilap_3F

I'm planning to dorm if I'm vaccinated by then.

jasmineculilap_3F

I'm excited to meet people irl.

jasmineculilap_3F

An oxidizing reactant oxidizes something, so it gains electrons by doing so. Therefore, it's oxidation number decreases. Gain electron(s) means decreased oxidation number means it's the oxidizing reactant.

jasmineculilap_3F

Kc is the equilibrium constant for molar concentrations while K is the general equilibrium constant.

jasmineculilap_3F

You would also change the sign.

jasmineculilap_3F

I think the problem should tell us.

jasmineculilap_3F

I would go do the textbook problems.

jasmineculilap_3F

Anode is always on the left and cathode is always on the right.

jasmineculilap_3F

I would do it just in case because I think the syllabus says it's 5 posts per week not necessarily 50 overall.

jasmineculilap_3F

Yeah, you can post as much as you want.

jasmineculilap_3F

Yes. The oxidizing agent makes other molecules lose electrons by gaining them, therefore it is reduced.

jasmineculilap_3F

We should have 40 posts by now since it's the end of week 8.

jasmineculilap_3F

I like to remember OIL RIG(oxidize is loss, reducing is gain). Remember that oxidizing agent oxidizes, so it gains the electrons while the reducing agent reduces substances, so it loses electrons.

jasmineculilap_3F

There's nothing that can really be done since the test is over, so I try and take my mind off it by playing games or watching shows. Try to relax and do something you like to get less stressed.

jasmineculilap_3F

Textbook problems and attending UA sessions are probably the best way to review.

jasmineculilap_3F

It'll probably be up in a week or so like last time.

jasmineculilap_3F

I'm taking chem14bl, math 3c, ls 7b, and cluster 20cw.

jasmineculilap_3F

I think either one should work as long as you're consistent in what you're using. Just make sure you convert to whichever unit is wanted too.

jasmineculilap_3F

delta G < 0 aka negative means it's spontaneous, favoring the forward reaction (products)
delta G > 0 aka positive means not spontaneous, favoring the reverse reaction.

jasmineculilap_3F

Entropy is disorder while enthalpy is heat. Enthalpy is the heat absorbed/released in a system while entropy is the measure of disorder.

jasmineculilap_3F

If it's abt temp change then you don't really need to convert. Just pay attention to the units needed/what the question is asking for. Make sure you use the same temperature unit in calculations.

jasmineculilap_3F

I found the final to be harder than the midterms.

jasmineculilap_3F

It's exothermic because there are still bonds being formed that release more energy than the energy needed to break bonds.

jasmineculilap_3F

They're the same thing.

jasmineculilap_3F

I don't think he's posted an outline yet.

jasmineculilap_3F

I'm going to be taking it next quarter so I still remember material from 14B

jasmineculilap_3F

go to TA hours!

jasmineculilap_3F

ΔPV doesn't mean ΔP x ΔV, it means the change in quantity of (pressure x volume). If pressure is constant, then your only change in the quantity (pressure x volume) is volume, so you can say ΔPV=PΔV Oh, that makes a lot of sense. So, basically if there is a change in the product of pressure times v...

jasmineculilap_3F

Endothermic takes in heat while exothermic releases heat. Therefore, solid->liquid->gas would be endothermic since heat is required to break the bonds. Going reverse would be exothermic since forming bonds releases energy.

jasmineculilap_3F

Salts can be basic, acidic, or neutral depending on the anions/cations in the salt.

jasmineculilap_3F

I'm planning to take 14BL next quarter so I'll still remember what I learned in 14B :)

jasmineculilap_3F

kelvin bc R is in kelvin

jasmineculilap_3F

Weeks 1-3 lectures will be in the midterm. The week 3 topics are 4D.3, 4D.4, 4D.5, and 4E.3.

jasmineculilap_3F

It would have 4 SFs since we're following the same SF rules. One thing to remember though is that when taking the -log of a number, the total SFs of that number will be the number of SFs in the decimal place. In the case of pH = 7.834, for the -log(#), the # would have had 3 SFs that made 7.834 have...

jasmineculilap_3F

Yes, they're during lecture time.

jasmineculilap_3F

Yes, we're allowed to have print outs of the periodic table and formula sheet on his website.

jasmineculilap_3F

Changing pressure only affects gases, so you don't include aqueous.

jasmineculilap_3F

I understand the relationship between Q and K, but can someone tell me why/when we would want to measure the Q/when it would be important? You could measure Q in order to figure out what direction the reaction is going towards. When Q<K, the reaction favors the forward reaction/products and if Q>K,...

jasmineculilap_3F

I like Studio Ghibli movies and Spiderman: Into the Spiderverse!

jasmineculilap_3F

When Q < K, the reaction will favor the forward reaction, so there will be more products formed. When Q > K, the reverse reaction is favored with more reactants forming

jasmineculilap_3F

A shift to the left means that reactants are favored while a shift to the right favors products.

jasmineculilap_3F

You want to use the inverse of Kc when solving for the Kc of the reverse reaction.

jasmineculilap_3F

Kelvin. You can convert celsius to kelvin with K = C + 273.15.

jasmineculilap_3F

You only count gases since the question is asking abt pressure.

jasmineculilap_3F

I found the assigned textbook problems on the syllabus to be really helpful in practicing my understanding. Also, try to space out your studying instead of a long intensive study session.

jasmineculilap_3F

In an email, Professor Lavelle said 5 posts per week = 5 pts per week. I think the deadline is on sundays, but different TAs might have a different rules.

jasmineculilap_3F

yeah, solids and liquids aren't included because their change in concentration is very small.

jasmineculilap_3F

Tetrahedral shapes are AX4, so there are only 4 bonded atoms while square planar is AX4E2, so it has 2 more lone pairs.

jasmineculilap_3F

Ka is the acidity constant and pKa is the -log of Ka. Ka = [H+][A-]/[HA]. Both can be used to find out how strong an acid is. A higher Ka means a lower pKa and stronger acid while a lower Ka means a higher pKa and weaker acid.

jasmineculilap_3F

Equatorial are on the same horizontal plane as the central atom while axial are above/below that plane.

jasmineculilap_3F

You use ate when overall charge is negative.

jasmineculilap_3F

I remember that Lyman is around 100nm while Balmer is around 400-600nm, but knowing the energy levels is important for equations.

jasmineculilap_3F

I think lecture videos should be working now.

jasmineculilap_3F

There's ion-ion, ion-dipole, dipole-dipole, dipole-induced dipole, and induced dipole-induced dipole. Ion-ion is between 2 ions (Na+ and Cl-). Ion-dipole is between an ion and something that's not an ion but has ionic characteristic (Cl and H2O). Dipole-dipole is between two non ions but they still ...

jasmineculilap_3F

I think knowing the ligands might be something to memorize since I think we can only have the periodic table and formula sheet.

jasmineculilap_3F

My lecture videos are working right now, maybe try checking again or refreshing?

jasmineculilap_3F

pH = -log[H+]. I don't think we need to how it was derived.

jasmineculilap_3F

Lone pairs affect molecular shape. They have more repulsion and change bond angles.

jasmineculilap_3F

I also feel like it's going pretty well, but I'm learning new material now so I'm slightly nervous for that and for finals.

jasmineculilap_3F

Don't forget to take breaks, so you don't get tired. Also, try not to be in a distracting environment (I avoid the living room bc the TV is on). I tend to study at night since that is when I feel the most focused personally. Taking notes and having a set goal for what I want to have studied also hel...

jasmineculilap_3F

jasmineculilap_3F

I'm going to solve the textbook problems assigned to us, and I'm probably going to attend some workshops as well throughout weeks 9 and 10 to avoid cramming.

jasmineculilap_3F

Look at electronegativity. A smaller electronegativity difference means more covalent character.

jasmineculilap_3F

You need 5 posts per week, so 35 posts are needed.

jasmineculilap_3F

Week 7 and 8 are in the same sapling assignment, so it's due next sunday.

jasmineculilap_3F

I think we'll get them by this week since it took abt a week for midterm 1.

jasmineculilap_3F

lecture 22!

jasmineculilap_3F

C should have a formal charge of -2. Valence - free - bonding/2 = 4 - 4 - 4/2 = -2.

jasmineculilap_3F

It's negative due to electron-electron repulsion. Adding the first electron makes group 16 elements an anion, so the second electron affinity is negative since the electron is trying to go into an already electron dense space, causing them to repel and need energy to add it.

jasmineculilap_3F

The electronegative N and O and the slightly positive Hs can all hydrogen bond. N has 1 lone pair, so it can form 1 H bond (2 Ns = 2 H bonds) while O has 2 lone pairs, so it can form 2 H bonds. Each H (4 Hs in urea) can have one H bond, so the total becomes 8 bonds.

jasmineculilap_3F

it's lectures 10-18.

jasmineculilap_3F

N can't have an expanded octet since it's in period 2.

jasmineculilap_3F

Lone pairs are the electrons that aren't in a covalent bond, so they aren't shared between atoms.

jasmineculilap_3F

You can find VE quickly by counting which group it's in. There are 8 groups from left to right when you don't include the d-block/transition metals, and the group it's in corresponds to the element's valence electrons.

jasmineculilap_3F

I'm taking LS7A and Chem14A at the same time right now. I think LS7A has more work because of launchpad, but it's manageable as long as you don't procrastinate a lot.

jasmineculilap_3F

Double bonds have 2 shared pairs of electrons while single bonds only share 1 pair, so the greater number of electrons results in a stronger attraction to the nuclei, shomaking double bonds shorter.

jasmineculilap_3F

When there are multiple resonance structures, formal charge can be used to find the most stable structure. The most stable have a formal charge that is or is closest to 0.

jasmineculilap_3F

You can use heisenberg's uncertainty principle where ΔxΔp >= h/4π. Δp = mΔv, so you can substitute that in and rearrange the equation for Δv which becomes Δv = h/(4πmΔx). Then, you can just plug in electron mass and Δx which is given to you.

jasmineculilap_3F

Ionization energy decreases from top to bottom since the valence electrons are in shells that are further away from the nucleus and held less tightly, making them easier to remove. Ionization energy increases from left to right since nuclear charge is also increasing, making it more difficult to rem...

jasmineculilap_3F

Kinetic energy is .5mv^2. You can use the constants sheet to get the mass of an electron, so you just need to find its velocity. You can use de Broglie's equation where λ = h / mv and rearrange it to get v = h / mλ and plug it into .5mv^2. Then, you just need to convert from Joules (since KE has uni...

jasmineculilap_3F

Avogadro's number means that 1 mole (mol) is equal to 6.0221 * 10^23 particles. Particles can refer to molecules/atoms/formula units. You usually use it when you need to convert moles to atoms/molecules. For example, the mass on the periodic table is molar mass so you would divide by avogadro's numb...

jasmineculilap_3F

You can use the de Broglie equation, λ = h/p which becomes λ = h/mv. You already have velocity, and you can get fluorine's mass by converting its molar mass using avogadro's number to go from grams/mol to grams/molecule. Make sure you find molar mass of F2 when converting since it's a diatomic molec...

jasmineculilap_3F

I got D. I used ΔxΔp≥h/4Π which can also be written as Δxmv≥h/4Π (p = mv) and manipulated it to get v. V = (h/4Π) / (Δxm). We were given that position (x) was 1% of the radius, so I found x by calculating .01 * (.05 * 10^-9) <-since you need to convert to meters. Mass of an electron is a constant, a...

jasmineculilap_3F

You don't need to find the mass percentage. You're right that you find the molar mass of the molecules, but afterwards just use molar ratios. Divide the given masses (make sure they're in grams) by molar mass to get the moles of the molecule, then use the molar ratio between the atom and the molecul...

jasmineculilap_3F

Since ΔP = M x ΔV, you can divide both sides by M to isolate ΔV, so ΔV= ΔP / M.

jasmineculilap_3F

We were sent an email with a periodic table sheet and constants/equations we can print out and use.

jasmineculilap_3F

Speed of light in a vacuum is constant where c = 3.00 * 10^8. c can also be found by c = λ * v, so frequency and wavelength are inversely related.

jasmineculilap_3F

I've been reading through the textbook sections listed on the syllabus and solving the problems as well.

jasmineculilap_3F

I think so. On his website, it said Midterm 1 is all the material through week 3 which is up to Q21 on Sapling and Topic 1D on the textbook.

jasmineculilap_3F

Since we're looking for the longest wavelength possible, this means that work function should be the same as the photon's energy (there won't be excess energy/kinetic energy from the electrons because that would mean that it wouldn't be the longest wavelength possible). We know the formula E = hv gi...

jasmineculilap_3F

An electron has a much smaller mass than protons and neutrons which are about the same. An electron's mass is basically negligible while the other 2 have masses of around 1 amu/dalton.

jasmineculilap_3F

I'm using the fundamentals/problems posted on the syllabus as study materials. I take notes on the fundamentals, watch the audio-visual modules, and take notes on the lectures.

jasmineculilap_3F

You can use the molecular formulas to find the molar mass, but I just searched it up and got molar mass of butanone on the internet. You would only need to know the formulas for the 2-butanone and 3-methyl-3-hexanol since the other product is in excess. To approach the problem, first use the density...

jasmineculilap_3F

Wavelength is measured in meters so since you know how many wavelengths are in 1 meter from the problem, you can do wavelength = 1/1,650,763.73 and use that answer to find the region it falls under. Then for the photon question, you can use the formula E = hv.

jasmineculilap_3F

Once you find the empirical formula's molar mass, you can divide the mass of the molecular formula by the empirical formula to get the ratio. You can then use that number and multiply each element's subscript/number of atoms in the empirical formula by that number to get the molecular formula.

jasmineculilap_3F

Try and note down what variables are being used in the problem. If there are 2 molarities or 2 volumes, you probably need to use M1V1 = M2V2. Another way to tell is that if a solution has been diluted/water was added, you probably need to use the same equation since number of moles (n) stays the sam...

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