CHEMISTRY COMMUNITY

Nerissa_Low_2F

Dr. Lavelle, Thank you for the past 2 quarters of chemistry. Your lectures always conveyed your enthusiasm for chemistry and for the wellbeing and education of your students and I greatly appreciated it, as well as the extensive amount of time you put into chemistry community, organizing UA sessions...

Nerissa_Low_2F

I think the rate law is only governed by the slow step, so the overall rate law would be that of the first step.

Nerissa_Low_2F

I think dashes are typically used to separate numbers from words but I'm not completely sure.

Nerissa_Low_2F

The activation energy is the energy required by a reaction to actually proceed to completion. Free energy of activation refers to Gibbs free energy. When something is referencing E>Ea, it's talking about whether or not the specific reaction or collision etc. has more energy than the activation energ...

Nerissa_Low_2F

They've already been posted in another thread.

Nerissa_Low_2F

I'm not sure how to find the rate constant here, but I believe the units for activation energy is kJ/mol. Also, I think we aren't being tested on the Arrhenius equation for the quiz.

Nerissa_Low_2F

If a gas is allowed to occupy a larger volume, there are more states it can possibly be in (because of the larger volume of space) therefore entropy increases. Similarly, if the temp of a gas increases, then the equation for deltaS = nC ln (T2/T1) is positive, indicating an increase in entropy.

Nerissa_Low_2F

Yes it is possible. Entropy can be equal to q/T, and the units for q is Joules and Kelvin for T, so the units of your answer can be Joules/Kelvin.

Nerissa_Low_2F

The only time I can really think of this is happening is during a phase change. During this time, you can think of there being no temperature change because the heat is being fully used to break bonds to go from solid to liquid or liquid to gas.

Nerissa_Low_2F

I believe it is because you're going from a higher energy to a lower energy, which is more favorable than the opposite direction.

Nerissa_Low_2F

I think reaction enthalpies are usually written in kJ/mol.

Nerissa_Low_2F

I assume this means at the buffer point? If you use the Henderson Hasselbach equation, you see that pH = pKA + (log[base]initial/log[acid]initial). At the buffer point, the [base]initial = [acid]initial, and the log of 1 =0. Therefore, at this point, pH=pKA

Nerissa_Low_2F

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Nerissa_Low_2F

I think it's because they don't disassociate completely in water, so there are still a substantial number of disassociated molecules with the carboxyl group in the solution.

Nerissa_Low_2F

I think ligands are written alphabetically in the name of the compound but not necessarily in the formula.

Nerissa_Low_2F

Atoms that are paramagnetic have unpaired electrons, and atoms that are diamagnetic have no unpaired electrons. I think it would be possible to have an atom with an even number of electrons be paramagnetic when you have an atom with 4 electrons in the p orbital; 2 electrons would be unpaired, making...

Nerissa_Low_2F

I think you would know when to draw a free radical by counting the number of electrons. Since radicals have 1 unpaired electron, they will always have an odd number of electrons as opposed to the even number of electrons that other compounds would have. As for which atom the unpaired electron would ...

Nerissa_Low_2F

Yes, Phosphorus and Sulfur are exceptions because they are in the 3p area of the periodic table. If we look at quantum numbers, Phosphorus and Sulfur both have n=3. Therefore they can both have l for up to 2. l represents the possible orbitals that an element can possess, and if it is possible for l...

Nerissa_Low_2F

So, when we write electron configurations, we fill in electrons for elements based on their order in the periodic table. This is the only "general rule" to follow. When the d orbital becomes available, you would still fill the 4s first, and then proceed filling those in the d orbital. The ...

Nerissa_Low_2F

Basically, these orbitals are the wavefunctions of electrons within an atom, so they are mathematical expressions. When we take the square of the wavefunction, we obtain the probability of finding an electron within an area. The reason they are pictured as "spaces" is because people typica...

Nerissa_Low_2F

Maybe we'll get the quizzes back in our discussion section next week? I'm not completely sure, but I do know that even though the quizzes and tests are assigned a numerical score, the final grade is based on a curve.

Nerissa_Low_2F

I'm pretty sure the value was derived experimentally. My TA said that there was a lot of quantum mechanics behind the final derivation of the value that we don't need to know for this class.

Nerissa_Low_2F

I believe the question is saying that there is 510 g of the aqueous solution total, and then asking you to figure out the masses of each part of the solution.

CHEMISTRY COMMUNITY

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Re: Saying Thank You to Dr. Lavelle

Re: ch15 #51

Re: Naming [ENDORSED]

Re: Activation Energy vs Free energy of activation

Re: Quiz 2 Prep answers

Re: Quiz prep 2 #11

Re: Increase in entropy

Re: Units of entropy [ENDORSED]

Re: Clarification on heat transfer but no temp change

Re: Spontaneous

Re: Units of Reaction Enthalpies

Re: Half way stoichiometric point

Re: Post All Chemistry Jokes Here

Re: Bronsted Acids Clarification

Re: 17.31 d

Re: Fall 2014 Quiz 2 prep #9

Re: Free Radicals

Re: Octet Rule Exceptions? [ENDORSED]

Re: General Rules on Writing e- Configurations [ENDORSED]

Re: S, P, D, and F Orbitals [ENDORSED]

Re: Quiz Grades

Re: Why h/4π?

Re: Clarification for G.7