CHEMISTRY COMMUNITY

apurva-3E

It is cumulative!

apurva-3E

I think he basically means that the reactions that catalysts play a role in are zero order reactions, due to the fact that the reaction does not depend on the concentration of the reactant and how it varies. Therefore, for a zero order reaction, k is the driving factor in the reaction speed.

apurva-3E

Mann Mukti UCLA!

apurva-3E

I really like How I Met your Mother.

apurva-3E

We can use the reaction: t(1/2) = [A]0/2k

apurva-3E

We should do it.

apurva-3E

I found 7A less difficult work-wise, but more difficult in the sense that you actually had to challenge your brain to apply the information you are learning.

With enough practice though, I think the class will be fine to take.

apurva-3E

Try a ton of textbook problems and really try to understand why that's the answer. Then put it into a flashcard software and study it so you won't forget.

apurva-3E

I am sure everything will be fine. Dr. Casey is apparently a really good professor.

apurva-3E

Take some deep breaths and try reaching out to other people in the class. Without them, I wouldn't be functioning.

apurva-3E

In basic conditions, you would want to add OH- to the redox half reactions.

apurva-3E

It is due on sunday!

apurva-3E

I think it would be to identify whether the reaction takes place in acidic conditions

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I think the electron flow would be a great way to tell between the anode and the cathode

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Heat generated in a reversible process.

apurva-3E

I work 15 hours a week and it is rough, I find that doing a lot of time management and really microplanning your day makes it really smooth to get work done. Never save studying for after a shift!

apurva-3E

I think there isn't enough energy for the electrons to flow between the anode and the cathode anymore.

apurva-3E

Normally there is a temperature value that is given for the problem.

apurva-3E

For an isolated system, Q = 0 because DELTA(U) will equal zero since nothing can interact with the system. Since no work can be done to the system, Q must equal zero.

apurva-3E

The salt bridge allows for the neutralization of the beakers. The redox reaction's transfer of electrons will result in highly charged beakers. The salt bridge allows to balance these out.

apurva-3E

145 KJ

apurva-3E

m is mass and n is mols.

apurva-3E

Isolated system, it shouldn't lose heat!

apurva-3E

Since it is an isothermal reaction, the internal energy change must be 0.

apurva-3E

Look at the units you are given and then select the gas constant with the same units.

apurva-3E

Grind the textbook problems and write a guide on how to solve those problems.

apurva-3E

8.314J/K*mol, but a good way is to look at the units of R values and what you are given

apurva-3E

State functions depend on the starting and end states

apurva-3E

I am sure there will be more larger point questions with partial credit within the MCQ answer choices.

apurva-3E

Clean working environment!

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CP – CV = R, i don't know why that was on sapling though because we didn't learn it.

apurva-3E

Should be half of the outline we covered so far and another one,

apurva-3E

apurva-3E

Should be sometime before week 7

apurva-3E

Should be formatted in MCQ on CCLE

apurva-3E

Keep your work organized so you can catch yourself on the mistakes!

apurva-3E

In general the formation of bonds = exo
the breaking of bonds = endo

apurva-3E

Std. reaction enthalpies are the values of enthalpy for a substance / reaction at standard conditions (pressure, etc.)

apurva-3E

Around a week later.

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Heat doesn't depend on the final and initial conditions of a reaction. There are many things that can affect it, making it not a state function.

apurva-3E

Enthalpy measures heat. Heat is the actual energy that flows.

apurva-3E

Enthalpy is a state function, which means they are additive and can be manipulated.

apurva-3E

Yes, endothermic reactions will always have a positive deltaH values.

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I completely understand but there's no point in thinking about it now. We should just peacefully wait and work to improve ourselves for next time!

apurva-3E

If it is a gas, we would include it in the ICE table.

apurva-3E

Not sure, maybe ask some upperclassmen?

apurva-3E

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A good bit should be covering this topic.

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Probably 50 minutes

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reactant / product

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If the pressure doubles (only by halving the volume) the concentration would go up.

apurva-3E

I believe Dr. Lavelle might add some stuff from outline three as that will be covered in Week 3 Material.

apurva-3E

You can rearrange the equation pv=nrt and turn n/v into concentration.

apurva-3E

The equilibrium constant has to be smaller than 10^-3

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An acid that can only donate one proton.

apurva-3E

You would be given the units most of the time.

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([Conjugate Acid]/[Base]) x 100% = Percent Protonated
([Conjugate Base]/[Acid]) x 100% = Percent Ionized

apurva-3E

I think it would be structured in a similar fashion as chemistry 14A/.

apurva-3E

Online, there are lists that help us ascertain which acids are strong and which bases are strong. Generally we have to memorize that list. But we can also think on what makes an acid strong.

apurva-3E

Yes, post them on that page.

apurva-3E

Exactly, when k > 10^3 it favors the products

apurva-3E

You can upload the image online through like imgur or another site and then insert the link using the image function on chem community.

apurva-3E

Exclude it when it's a solid or liquid.

apurva-3E

Reactants are usually negative while the products are positively charged.

apurva-3E

All of the lectures are asynch

apurva-3E

Sapling has the solution manual linked on their website.

apurva-3E

Did you purchase the multiterm access? If not, all I had to do was check "use $72 credit"

apurva-3E

Deep breaths

apurva-3E

H3PO4 more oxygens will stabalize the charge better on the resulting anion

apurva-3E

https://qph.fs.quoracdn.net/main-qimg-d2d1d21c7c6b4263a81890dd72f4cc6b As seen on the drawing above, the geometry of the molecule shows that the polar bonds do not cancel each other out. Always draw the shape out because simple lewis structures won't be able to tell you if something is polar or not.

apurva-3E

Email your TA for help! I think my TA has talked to some people about midterm mistakes.

apurva-3E

Same thing! Just a different way of wording it!

apurva-3E

You must take both of them into consideration. I would prioritize IMFs

apurva-3E

I don't think it matters.

apurva-3E

You use it when converting between mass and mols of a substance. This substance can be an element or molecule.

apurva-3E

use the one on the formula sheet

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So you want to determine the molecular shape using VSEPR geometry and identify how electronegative elements are arranged. This can give you a glimpse into whether a compound is polar or not.

apurva-3E

Most likely the molecular shape but I think it will be specified on the questions in the test.

apurva-3E

Contact the TA's for the worksheets! They all have their contact information posted on the CCLE site.

apurva-3E

They cannot rotate because there are p orbitals overlapping side by side. Try to visualize this in your head after looking at a picture of the pi bond.

apurva-3E

Electronegativity!

apurva-3E

Complete all of the LA worksheets, textbook problems, and skim through the textbook.

apurva-3E

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Consult a chart for the angle size based off of shape and the presence of lone pairs.

apurva-3E

Pretty behind on lectures but good overall!

apurva-3E

I think the format will be very similar to chemistry 14A

apurva-3E

Calculate the polarity vectors and see if they cancel out. Must draw lewis structure.

apurva-3E

Lone pairs would be present

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They are super useful, definitely go to them for help!

apurva-3E

Consider the difference in electronegativity as well as the shape of the molecule.

apurva-3E

They will be covered in time.

apurva-3E

Sapling HW 50 points total
Chemistry Community 50 points total
Midterms 110 each, Final 180, total 500 points.

apurva-3E

Single Bond = Sigma
Double Bond = Sigma and Pi Bond
Triple Bond = Sigma and 2 Pi Bonds

Determining bond strength between sigma and pi bonds won't be tested. However, single bonds are longer and weaker than double bonds.

apurva-3E

Remember to look specifically at the phosphorus atom. Don't worry about the entire structure.

apurva-3E

Shape is the region of electron density. Polarity kind of depends on shape but also difference in electronegativity.

apurva-3E

1 Sigma 2 Pi Bonds

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They are highly electronegative

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Acids accept electrons while bases donate them!

apurva-3E

These are all intermolecular forces. INTER = holding multiple molecules together. While INTRA = what holds one molecule together.

apurva-3E

They are induced dipole induced dipole interactions. Basically as the electron density fluctuates, molecules form temporary dipoles with each other.

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