CHEMISTRY COMMUNITY

Jenna Kovsky 1I

In this particular problem, you're asked to draw the mechanism, so you do have to come up with the steps. In the question, it tells you that the reaction is "acid-catalyzed", and suggests that you use H+ or H30+. (remember that if something is in basic conditions, it means that there's OH-...

Jenna Kovsky 1I

You don't have enough information to calculate the delta H at 398K, the way you did for 298K in the first part of 1B. So you have to break it up into steps. Then you can combine the steps, since enthalpy is a state function and doesn't care about how you got from point a to point b, as long as you s...

Jenna Kovsky 1I

For your first question, when a part of the main chain is put in parentheses because it's being repeated, then that's just to make things simpler, and it can still be written out if you want to write it without the parentheses. Side chains, however (for example, the methyl group), do need to be in p...

Jenna Kovsky 1I

If you mean the fact that you take the oxidation half reaction that happens in the anode and reverse it, then you can do that so that it's written as a reduction half reaction. If it's written that way, then you can use the standard reduction potential for that half reaction to calculate the overall...

Jenna Kovsky 1I

That seems right to me! Where did you get this problem, maybe you can look up the solution online or something?

Jenna Kovsky 1I

That seems right to me! Where did you get this problem, maybe you can look up the solution online or something?

Jenna Kovsky 1I

Actually, now that I think about it, doesn't q=deltaH at constant pressure?

Jenna Kovsky 1I

The first part (calculating the final temperature) is fairly straightforward. I used the equation q=n*C*deltaT. For the heat capacity (C), you were told to assume that it is an ideal, monatomic gas at constant pressure, so you would use 5/2 R as your heat capacity (this is given on the formula sheet...

Jenna Kovsky 1I

If you look at page 27 in the course reader, you see that we're given the equation deltaU=3/2*nRdeltaT for an ideal gas. It follows that if the expansion is happening isothermally, then since deltaT is 0, deltaU is as well. So deltaU is zero for the isothermal expansion of any ideal gas. If you're c...

Jenna Kovsky 1I

The phases are what's important here. In the 1st reaction (2Al(s)+6HCl(aq)--->2AlCl 3 (aq)+3H 2 (g)), the AlCl 3 is aqueous , but if you look at the reaction that you're trying to end up with, it has solid AlCl 3 , so you need to use the 4th reaction (AlCl 3 (s)--->AlCl 3 (aq)) to cancel out the aqu...

Jenna Kovsky 1I

The fact that he called it "symmetrical" confused me too. In this case, it's less like the antibonding being "symmetric" to the bonding, and more like a reflection. Imagine a horizontal line between the bonding and the antibonding; if you fold the paper along that line, the sigma...

Jenna Kovsky 1I

I heard Potassium and Oxygen went on a date....It went OK

Q: What do you do with a sick chemist?
A: If you can't helium, and you can't curium, then you might as well barium

Q: What is the name of 007's Eskimo cousin?
A: Polar Bond

Jenna Kovsky 1I

The only one that matters as far as naming is pz in the molecular orbital. Sigma has to be pz, so then px and py are both pi. Other than that it doesn't really matter, but in the atomic orbitals, most people go in alphabetical order (px, py, pz).

CHEMISTRY COMMUNITY

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