CHEMISTRY COMMUNITY

Anna_Kim_2E

viewtopic.php?f=140&t=20850&p=59021&hilit=cell+diagram&sid=62cad8228751744d5735abc415246b2e#p59021

I was stuck on this too, but its due to counter ions that's coming back into half cell with Ag+.

Anna_Kim_2E

viewtopic.php?f=140&t=20850&p=59021&hilit=cell+diagram&sid=62cad8228751744d5735abc415246b2e#p59021

nevermind I found it.

Hope this helps you too Huongly!

Anna_Kim_2E

First the conventional way of writing the cell diagram is Reactant, Product, Reactant, Product etc. The reason AgBr(s) is in the diagram is because it is solid and Ag+ has been oxidized and that is why it is to included in the diagram. It is actually supposed to be the whole solid rather than the d...

Anna_Kim_2E

180 degrees - anti staggered would be the lowest energy conformation.

Anna_Kim_2E

iso on a substituent is used when it is not attached in a linear structure.

Anna_Kim_2E

Do we then get extra points for posting more than one each week?

I thought I heard somewhere we get like 2% extra credit??
Can anybody confirm this?

Thank you

Anna_Kim_2E

Yes you can figure out the order of reaction by looking at the units of k. Because the unit of rate is M/s, if you plug the unit of rate in 0 order, 1st order, and 2nd order rate law, you will be able to get the unit of k for different orders. For example rate=k[A]^0 for 0 order. Since you know that...

Anna_Kim_2E

The units of rate constant is varies for the different depending on the order of reaction.

Anna_Kim_2E

For second order, -d[A]/dt = k[A]^2 . Integration of both sides results in an equation 1/[A] = kt + 1/[A]0. Hence, in order to find out for the half life, we use this equation when it is in second order.

Anna_Kim_2E

For second order, half life is t = 1/k[A]initial . Therefore, t (half life) and concentration have 1/[A] relationship.
Whereas, for zero order, half life t = [A]initial/2k , which shows that t(half life) and initial concentration direct linear relationship.

Anna_Kim_2E

I am still having a bit of confusion.

So on the problem, on the attached picture, how is the sig fig of the answer determined? Do we just disregard the value that ln becomes? ln(12.90/7.60)

Anna_Kim_2E

Ka = [H+][F-]/[HF] and K = [H+]^2[F-]^2]/[HF]^2

therefore, when comparing Ka to K, K is exactly Ka square. Therefore you square root the value K in order to achieve Ka.

Anna_Kim_2E

I am having a bit of confusion for sig figs when ln, e, or log is involved. I am aware that for ln(X), if X has 3 signifiant digits then the value for ln(X) should have 3 decimal places. But in midterm 2014 Q2D, e^4.63 = 102, which also means that ln(102) = 4.63. 4.63 has 2 decimal places. Then, Sho...

Anna_Kim_2E

So last week in the review session, the TA said that entropy should always be written in Joules and when in kJ we should convert it to J. Is there a specific reason why it has to be in J/Kmol and not kJ/Kmol?

Anna_Kim_2E

You use subtraction Ecathode-Eanode when they are both in their reduction state. Hence, since the two half reactions are already in oxidation and reduction states, you can add their E° of the both equations. 0.8V+0.4V=1.2V

Anna_Kim_2E

Thank you!

Anna_Kim_2E

Anna_Kim_2E

Also to add on, anode, reaction losing electron, goes on the left side of the double bar, and cathode, the reaction gaining electrons, goes on the right side of the double bar.

Anna_Kim_2E

Can someone explain why in #11 does q go from -23.8 to 23.8? Because -23.8kJ is negative because it was released from converting PbO to Pb. Then, that amount of heat is transferred into the system, which is water. When heat is put into an system, it is an endothermic process and the sign should be ...

Anna_Kim_2E

Where do you find the prep quiz?
I only see the midterms from past years.

Do I have to buy a separate booklet or is it in the course reader? Then what page is it in?

Anna_Kim_2E

Just for a clear confirmation, then there is nothing due at our discussion sections before our quiz because we are not doing the turn-in quizzes. Is that right?

Anna_Kim_2E

I asked the guy sitting next to me if he had any Sodium Hypobromite…

He said NaBrO

Anna_Kim_2E

The previous thread says that we should,
"When working with pH, use the number of sig figs as the number of decimal places in the pH. For example, if the pH=-log(1.8 x 10^-5) then pH = 4.74 (two sig figs in 1.8 x 10^-5, and two after the decimal in 4.74)."

Anna_Kim_2E

Q:Did you hear about the man who got cooled to absolute zero?

A: He's 0K now

Anna_Kim_2E

I think so because anything we learned last week is fair game.

Anna_Kim_2E

Anna_Kim_2E

Having a half full d5 or full d10 subshell has lower energy, therefore, where there is d4 or d9 subshell, it is better to move one electron from the 4s and fill the d subshell.

Anna_Kim_2E

The different energy levels of the orbitals affect the overall size of the atomic radii. When the elements have the same energy level and orbital, the electrons have affect in the different electrostatic attractions that pertain to each element.

Anna_Kim_2E

I think it is nanometer that has been converted to meters.
When you want to convert, for example, 900nm into meters, because there are 1x10^9nm in 1 meter, or 1x10^-9m in 1nm , it would 900nm would be multiplied by 10^-9 to convert into meters.

Anna_Kim_2E

Anna_Kim_2E

I am definitely interested. I can do any day of the week and preferably during the day than the evening.

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