CHEMISTRY COMMUNITY

Armo_Derbarsegian_3K

If a question asks us to identify a mechanism (2013 5C), would nucleophilic substitution be as valid an answer as SN2 (given as answer)?

Is there SN1?

Armo_Derbarsegian_3K

Generally, you should use the log equation when also looking for pH. Otherwise, I think it is easier to use the natural log.

Armo_Derbarsegian_3K

Tina that is so interesting! Thank you so much for sharing. I really enjoy Dr. Olah's work! Great read!

Armo_Derbarsegian_3K

The lone pairs on oxygen!

Armo_Derbarsegian_3K

The rate law should be the indication.

Armo_Derbarsegian_3K

When you replace the intermediate in the rate law when doing pre equ problems, what do you do if you end up with another intermediate in your answer? ClO-+H2O<->HClO+OH- fast HClO+I-<->HIO+Cl- slow HIO+OH-<->IO-+H2O fast so the rate would be k2[HClO][I-] but HClO is an intermediate; K=[OH-][HClO]/[C...

Armo_Derbarsegian_3K

On page 73 of the CR, K was set equal to k1/k'1, why do we assume it is for the NO+NO<->N2O2 reaction? Why is K not equal to [NO2]^2/[N2O2][O2] reaction, k2/k'2? How do we determine this? Please elaborate this step. Also--is "k=2k2*K" going to be given? It seems we just set it equal at the...

Armo_Derbarsegian_3K

K=e^13.2 is what the CR does, but when I do it, I use the full number e^13.20417583... which varies my answer from the given, 540365, to 542626.

This is a fairly large difference and the final answer is worth 4 pts. Would we get points for doing it either way? CR or round at end way.

Armo_Derbarsegian_3K

How many orders can you have? Does it keep going?

Armo_Derbarsegian_3K

also LEO GER

Lose electrons oxidized

Gain electrons reduced

Armo_Derbarsegian_3K

It would be nice if they were provided right before the quizzes, so we could check our answers. This was done for 14B last year and I can see the quiz answers in my roommates old course reader from last year. That format worked much better than what we have now.

Armo_Derbarsegian_3K

No, they should be given.

Armo_Derbarsegian_3K

Some students benefit by working with peers, but others (like me) are more successful working independently. I did very well last quarter on the quizzes with the provided answers. My only incentive was my grade so I refrained from going straight to the answers. I think the method last quarter was ve...

Armo_Derbarsegian_3K

Would it be possible to get a PDF of the answers to the practice quizzes so we can study them at our own pace?

Armo_Derbarsegian_3K

When you need to cancel things from both sides. You switch signs to flip it.

for example:

A+B->C dH=x
2C->2A+D dH=y

you would multiply the top one by 2 to get A and C to cancel.

Armo_Derbarsegian_3K

So if the equivalence point is near pH 8, why wouldn't we want one that changes color near pH 8? How exactly did you get that range of pHs for the indicator?

Armo_Derbarsegian_3K

Why does pKin. (pKA) have to be close to pH at stoichiometric point to be an effective indicator?

Armo_Derbarsegian_3K

For the equation that says:

p=hv/c -> 1pt of the grade

will h=pc/v also earn the point?

because both multiplied across -> pc=hv=E

Armo_Derbarsegian_3K

More oxygen=more electronegative, which stabilizes the negative charge on I, making it MUCH easier for the H to break off and protonate water to create hydronium, thus more acidic.

Armo_Derbarsegian_3K

You can approximate if you remember to check for the 5% rule after...to make sure you are within range.

Armo_Derbarsegian_3K

Exactly, you get a BO of 0. This means there is no bond, hence no stability as a molecule.

Armo_Derbarsegian_3K

A decrease in partial pressure means a decrease in concentration, which would shift the rxn left. I don't think partial pressure vs pressure matters because because a decrease in pressure would cause the same effect.

Armo_Derbarsegian_3K

It is in an anionic complex.

K is a cation, so to make a neutral "salt" the [CoF6] must act as an anion and take a negative charge, hence cobaltate.

Armo_Derbarsegian_3K

Fe is [Ar]3d6 4s2
Fe2+ loses the 4s2
Fe3+ loses an electron from the d
so Fe3+ is [Ar]3d5

Not really an exception, just remember a full d/half full d will be much more stable(chromium/copper)

Armo_Derbarsegian_3K

Look at bonding.

For example, ethylene makes a C::C and each C also makes 2 H bonds, therefore each C makes a total of 3 bonds = sp2. When you look at sp3, which is the full octet, sp2 is missing a p, and that is your un hybridized p. This is where the pi bonding happens.

Armo_Derbarsegian_3K

For this problem: I solved for moles of each element by taking the given values and dividing by their atomic masses. i.e.: Carbon 3.27g x 1molC/12.011gC = 0.272molesC then I divided by the smallest value, got the ratio, multiplied by a number to get rid of fractions to get the answer, found grams of...

Armo_Derbarsegian_3K

In the ring, the carbons each make two bonds, either to another carbon or to the nitrogen. Since the problem states hydrogens have been omitted and we know carbon likes to make 4 bonds, we would add in two hydrogens to each carbon to fulfill their octet. Therefore, each carbon actually makes 4 bonds...

Armo_Derbarsegian_3K

Hello, For this problem I don't understand how you find the hybridizations. Do we assume the carbon will just make two hydrogen bonds? Or is it possible it may form a double bond? Would we have to calculate FC and figure that out, or can we assume it will just get 2 H bonds as problem only states th...

Armo_Derbarsegian_3K

For example: SF6's wedge and dash depiction (page 95 of CR) we show the wedges towards us and the dashes away, but we draw lines for on the top and bottom. How do we determine which get wedge and dash, and which get solid lines. Alternatively, for CH4 how do we determine which get the wedge and dash...

Armo_Derbarsegian_3K

You want a formal charge of -1 because it is BrO-. The molecule should have a total formal charge of -1. Therefore it will make a single bond and each atom will have 3 lone pairs.

Armo_Derbarsegian_3K

The other oxygen should form a double bond as well. So you get Xe with a lone pair, both Fs single bonded with 3 lone pairs, and the 2 Os double bonded with 2 lone pairs. :O: .. || .. :F - Xe - F: .. || .. :O: This Lewis structure will give each atom a formal charge of 0 and will be the biggest cont...

Armo_Derbarsegian_3K

I understand that since S is in the n=3 -> l=0,1,2 meaning it can use the d orbital when bonding. However, in the example in class, we get 2 S double bonds with O and 2 S single bonds with 0, giving S 12 shared electrons, while the d orbital can hold 10. So what happens with the other 2 electrons no...

Armo_Derbarsegian_3K

To start off,

You should visualize it as v/c=3600cm^-1
then solve for v
that will give you a.

for b,
use the equation E=hv
and solve for E.

for c,
multiply the E from (b) by Avogadro's number to determine the energy for a mole.

Armo_Derbarsegian_3K

Hello, I did #10 and I was looking at the answers in the back and noticed: "USE BOHR EQU. FOR H ATOM (ANY 1e- H-LIKE ATOM, Li^2+): En = -2.178x10^-18 J (Z^2/n^2) for n=1,2,3.... j=joules, z=atomic number, n=energy level" What is the difference between that equation and using -hR? When do w...

Armo_Derbarsegian_3K

I believe the formula finds the (minimum) energy needed to remove a single electron. Therefore, I think it is an increase in kinetic energy of a single electron.

CHEMISTRY COMMUNITY

Search found 35 matches

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