CHEMISTRY COMMUNITY

DBaquero

I believe Professor Lavelle told us to aim for the lowest first number. If the first number would be the same either way then aim for the lowest second number as well (and so on).

DBaquero

Professor Lavelle said we only have to know IUPAC naming because the quiz and final will only have naming questions asking specifically for IUPAC naming or questions not specifying whether you have to name the molecule in IUPAC or common naming. That being said, for some substituents the IUPAC name ...

DBaquero

Also, it should probably be noted that you have to add only H+ and H2O for balancing redox reactions in acidic solutions. If the solution is basic then once you are done adding the H+ and H2O like before, you have to add OH- to both sides to balance out the H+, then you combine the H+ and OH- on the...

DBaquero

Yeah if E°is negative then it would require outside energy in order for the reaction to occur.

DBaquero

So a reaction with a negative deltaG is spontaneous because energy is released into the surroundings. Essentially it takes less energy to break down the reactants than it does to form the products. Spontaneous reactions are more stable because they have a lower entropy

DBaquero

Yeah that's correct. Because deltaU is equal to deltaH - PdeltaV, if PdeltaV equals 0 then deltaU will equal deltaH. If there is no change in volume (deltaV=0), that would in turn make PdeltaV=0 so deltaU=deltaH. Hope that kinda cleared things up.

DBaquero

Yeah I'm inclined to agree. The practice exams and quizzes in the course reader were probably the most helpful resource for me, especially because they had some more difficult, or at least different, questions than the textbook. Of course, like everyone else is saying attending UA (or TA) office hou...

DBaquero

The equilibrium constant (in this case Kb) is larger if the concentration of product (in this case [OH-]) is larger, so it stands to reason that high concentrations of OH- means a stronger base. You take the -log of the Kb to get the pKb which is the same as the log(1/Kb) meaning the smaller the Kb ...

DBaquero

Bisethylenediamine refers to two ethylenediamine ligands, not two diethylenetriamine ligands. In the case of two (dien) ligands it would be bisdiethylentriamine. Bis (or tris, quadris,etc) are used when the typical prefixes (di,tri,tetra, etc) can lead to ambiguity or confusion because adding the re...

DBaquero

So basically you have the centralized carbon atom attached to the three methylene groups. Two of the methylene groups are attached to the central carbon with a single C---C bond and have a lone electron pair on the carbon. The third methylene group is attached with a double C==C bond and there are n...

DBaquero

So in general there are a few ways to tell if you might be dealing with an exception to the octet rule. 1.) When you add up all of the valence electrons for the molecule, you get an odd number. So something like nitric oxide (NO) will have 11 electrons, so there isn't any way you can satisfy the oct...

DBaquero

So the first thing you want to do is convert the 140.511 keV to Joules. There are 1.6022x10^-19 Joules in 1 eV so there are about 2.251x10^-14 J in 140.511 keV (140.511 keV * 1000 eV/keV* 1.6022x10^-19 J/eV = 2.251x10^-14 J) Now we know that we can calculate the frequency with Planck's constant. (2....

DBaquero

I would say a better description of the photoelectric effect would be:
Energy of incoming light- work function = Ek of ejected e-
That's basically what is happening in the photoelectric experiment

The second answer looks good.

CHEMISTRY COMMUNITY

Search found 13 matches

Re: Numbering

Re: Common vs IUPAC naming

Re: Adding H20 AND H+

Re: Question regarding: "E° + Value means reduction is spontaneous"

Re: Gibbs Free Energy and Entropy

Re: calculating delta U [ENDORSED]

Re: Most Helpful Resource

Re: pKb and pKa

Re: Bis

Re: Homework Problem 3.95

Re: Breaking the Octet Rule

Re: Homework Problem 1.23 [ENDORSED]

Re: Photoelectric effect AV Post assessment #8 and #14