CHEMISTRY COMMUNITY

Darlene Lien 3E

For a visual example:

A + B --> C + D
D + E --> B + F

B would be the catalyst
D would be the intermediate

Darlene Lien 3E

Hi, I believe both are the same but represent different approaches. If we were to approach the problem using the forward and reverse rates, like the book does: rate (forward) = k[NO2]^2 and rate (reverse) = k'[N2O2] By setting the two rates equal to each other: k[NO2]^2=k'[N2O2] and, therefore, k/k'...

Darlene Lien 3E

Hi, from what I understood, adsorption occurs when the reactants stick/sit on the surface of the catalyst (heterogeneous). This is different from absorption which is when a substance is completely taken up by another (like water being absorbed by a sponge --> it goes inside the sponge). Because ther...

Darlene Lien 3E

Hi, I think the key part to look at here is the activation energy.

Step 1 is the fast step and step 2 is the slow step; therefore, the activation energy of step 2 is higher than that of step 1. This explains why the "hump" for the second step is higher than that of step 1. Hope this helps!

Darlene Lien 3E

Hi, for the rate, since the reaction order for C is 0, you don't have to include it in your rate law. Your rate law for the reaction should just be rate=k[A][B]^2 (since [C]^0=1). Hope this helps!

Darlene Lien 3E

I don't have specific videos, but the Organic Chemistry Tutor has been pretty helpful for me!

Darlene Lien 3E

5 points per week, and since this is Week 10, 5x10=50 points.

Darlene Lien 3E

k' represents the reverse reaction! For more info about it, you could look through 7C.2.

Darlene Lien 3E

Q represents the ratio of products and reactants at any time during the reaction (not at equilibrium) while K represents the ratio at equilibrium. At equilibrium, Q=K.

Darlene Lien 3E

I would assume the final would be the same this quarter as well!

Darlene Lien 3E

I have the same problem as well. I believe it's under maintenance!

Darlene Lien 3E

Temperature affects the rate constant, K!

Darlene Lien 3E

Dr. Lavelle derives it in Wednesday's lecture (lecture 23) too, so it might be worthwhile to take a look at it again. But, basically, ln2=0.693.

Darlene Lien 3E

The half-life of a zero-order reaction is t1/2=[A]0/2k. This is also in Wednesday's lecture (Lecture 23). Hope this helps!

Darlene Lien 3E

The info is on Dr. Lavelle's website too! https://lavelle.chem.ucla.edu/wp-conten ... hedule.pdf

Darlene Lien 3E

Hi, what I did was, I left out chlorine from the redox reaction because its oxidation state doesn't change. So, both the half reactions didn't have chlorine (half reactions focus solely on gold and nitrogen), but after writing the overall equation, I then added in the chlorine ions to the balanced e...

Darlene Lien 3E

Hi, for me, I think of balancing a basic solution as almost continuing off from how you would balance an acidic solution. I normally start the problem as if it were an acidic solution, and then adjust with OH- to replace H+. Hope this helps!

Darlene Lien 3E

Yes, you could also refer back to lecture 18 for further clarification! Dr. Lavelle goes through two methods that you could use to solve for the standard reduction potential of the cell (one of which being cathode minus anode).

Darlene Lien 3E

Yes, solids and liquids are typically excluded from the equilibrium constant because the concentrations remain constant throughout.

Darlene Lien 3E

I also heard that a lot of people dropped the class too, so I'd keep an eye out!

Darlene Lien 3E

I agree with the comment above. Although the exam would be multiple choice, it would still be good for us to know how to draw the diagrams that way we have the foundation to identify and solve problems relating to them.

Darlene Lien 3E

Kandyce Lance 3E wrote:How often does it update? If its been updated for Midterm 2 i don't have to worry about it for the final right?

Usually you don't have to update it too often I believe, but to be safe I would if there's an update.

Darlene Lien 3E

I also agree with the above comments! I just use the ones that would cancel out the units.

Darlene Lien 3E

The way I look at it is, I use the rate constant whose units would cancel out in the equation, since each rate constant is pretty much the same but representing different units. The rate constant 0.08206 L.atm/K.mol could be used if your pressure is in atm, volume in liters, temperature in Kelvin, a...

Darlene Lien 3E

I wouldn't worry too! I believe they usually get updated completely towards the end of the quarter before grades are submitted.

Darlene Lien 3E

Hi, just from looking at the outlines and the lectures, I don't believe those topics are covered, so I don't think they should appear on the midterm.

Darlene Lien 3E

The final temperature is 343K, and the change in enthalpy is 765 J.

Darlene Lien 3E

The answer should be 1.45x10^5 J or 1.45x10^2 kJ. Hope this helps!

Darlene Lien 3E

I think it also depends on the UA, but quite a handful of them do general reviews at their workshops before midterms!

Darlene Lien 3E

Also, here's the link to the syllabus if you want to get more info.
https://lavelle.chem.ucla.edu/wp-conten ... 4BSYLL.pdf

Darlene Lien 3E

Hi, yes, I believe the P in the formula is the external pressure because by definition, w=-P(external)(deltaV).

Darlene Lien 3E

Hi, after obtaining your balance chemical equation, you would add up the moles of gas for reactants and moles of gas for products. The delta N/change in moles would be moles of product-moles of reactants. Since there are 6 moles on the products and 7.5 moles on the reactants, the change in moles is ...

Darlene Lien 3E

To add on, you can also check out the Chempendix on Sapling! They have a list of strong acids and bases there as well.

Darlene Lien 3E

To add on, you can check out this comment/thread!
viewtopic.php?f=129&t=73924&p=311928&sid=3d86a902716aa023b0f3d18e2c4dba7e#p311928

Darlene Lien 3E

I recommend doing the textbook problems and attending UA sessions!

Darlene Lien 3E

Like everyone else said, you would most likely be able to view them with your TA!

Darlene Lien 3E

Dr. Lavelle usually sends out an email once they're out too, so we'll know once he sends out the email!

Darlene Lien 3E

Yep! I don't believe the delta H limitations apply to Hess's Law.

Darlene Lien 3E

We should be given them, so I wouldn't worry!

Darlene Lien 3E

You would ignore the second ionization when Ka1>>Ka2. I think there are a few questions in Topic 6E that you could refer to in the outline that cover this. Hope this helps!

Darlene Lien 3E

I wouldn't worry about having to know or memorize Ka or Kb values. They should be given!

Darlene Lien 3E

Hi, personally I recommend doing the textbook problems and redoing the ones that you had trouble with. Also, attend UA sessions if possible!

Darlene Lien 3E

Hi, I believe regular molar concentrations would just refer to concentrations not at equilibrium, so basically at any time during the reaction. For instance, you would use these concentrations to solve for Q if needed. Equilibrium concentrations are explicitly the concentrations at equilibrium. Hope...

Darlene Lien 3E

To add on, Dr. Lavelle mentions this approximation method a few times, such as in Lecture 6, so definitely take a look back if you want explanations on the 5% rule too. Hope this helps!

Darlene Lien 3E

You would use Kelvin. If you look at the constants and equations sheet, the units for the rate are in Kelvin. Hope this helps!

Darlene Lien 3E

I agree with the above comments. Solids and liquids are not included in the ice charts.

Darlene Lien 3E

In 14A, there was ample time to complete the midterms. I had enough time to double check and work through the questions at a good pace.

Darlene Lien 3E

If there are also any other topics that you are unsure about, I also recommend watching videos or searching up other examples to further your understanding.

Darlene Lien 3E

Hi, to add on, for that question if we were to calculate the ionization, we see that (1.3E-3/0.10) x 100% gives 1.3%, and since this is less than 5%, we know that this approximation is valid. Hope this helps!

Darlene Lien 3E

Hi, I believe Dr. Lavelle mentioned in the lectures that if K<10^-4 then the change in x is very small where it essentially becomes negligible, but I believe it could sometimes work when K<10^-3. Like the comment above said, a good rule of thumb is to use the 5% rule to check if the estimation works!

Darlene Lien 3E

Hi, I agree with the above comment. You use Kc=160 from Table 5G.2 and set up your equilibrium ratio accordingly. :)

Darlene Lien 3E

Hi, for 2.00 mL of 0.175M KOH(aq) after dilution to 0.500 L: you could use the dilution formula M1V1=M2V2 to solve for the concentration of OH- after dilution. The concentration would be 7.0x10^-4 M. Then you would use this value to find the pOH and then pH. For 13.6 mg of NaOH dissolved in 0.350 L ...

Darlene Lien 3E

Hi, generally you would use ICE charts to help when you have an initial concentration and an equilibrium concentration. When you solve for the change in concentration, you are able to get the equilibrium concentrations and use Kc accordingly. Typically, when the question gives or states that the con...

Darlene Lien 3E

Hi, to add on, I believe pka is more specific than pH. While pH measures the concentration of H+ ions, pKa typically is an indicator of the strength of an acid. Hope this helps!

Darlene Lien 3E

You're correct. For this class we won't have to worry about solving cubic equations, but as Dr. Lavelle mentioned in the lecture, we can use that same approximation even for quadratic equations when K is very small as well. Hope this helps!

Darlene Lien 3E

I agree with the above comment, and I believe it would follow Le Chatelier's Principle.

Darlene Lien 3E

Yep, and I believe Dr. Lavelle mentioned that for this class you can omit the "-x" since the equilibrium constant will be very small, so we won't have to solve the cubic equations.

Darlene Lien 3E

Hi, the reaction quotient has the same form as K, but instead, it uses the activities evaluated at an arbitrary stage of the reaction (not at equilibrium). Hope this helps!

Darlene Lien 3E

Hi, there is a link on his website! https://lavelle.chem.ucla.edu/wp-conten ... PL_ALL.pdf

Darlene Lien 3E

Hi, I had the same issue as you! I think there might be an error in the textbook. The K value I got from the chart was 6.1E23 too.

Darlene Lien 3E

Yes, you would divide the moles of each by 3.00 to get the concentration. Then you would plug these values in to solve for Q. Hope this helps!

Darlene Lien 3E

Hi, so when the homework tells you to look at Table 5G.2 for the K value, you look at the temperature they give you in the problem and then locate the K value that has that respective temperature. Hope this helps!

Darlene Lien 3E

I don't believe the points carry over. I think you must complete another 50 posts this quarter.

Darlene Lien 3E

In the textbook, they also list the strong acids and bases, so definitely check that out!

Darlene Lien 3E

I think sometimes it takes a while for the grade to update, so if you wait a bit it should update!

Darlene Lien 3E

Hi, I think you're supposed to look at Table 6C.2 to get the kB values!

Darlene Lien 3E

Hi, I think that in this case HI is actually a stronger acid than HCl because you would compare the bond length first before looking at electronegativity. The HCl bond length is longer/weaker which allows it to lose H+ more easily; therefore, HI is a stronger acid than than HCl.

Darlene Lien 3E

I think they should be an hour each since they're held during what would have been the TA's office hours, which is only an hour long.

Darlene Lien 3E

I believe it has a -4 charge.

Darlene Lien 3E

I've had the same issue too! I think I let it it set there for a while, and it randomly came back.

Darlene Lien 3E

Also, most of the time, acids begin with a Hydrogen. So, HCN is a weak acid. Hope this helps!

Darlene Lien 3E

Hi, there are two ways you could do this.

The first way is:
1. Find the pH from the given pOH. pH=14-pOH.
2. Solve for [H+] from the formula pH=-log[H+]

The second way is:
1. Solve for [OH-] from the formula pOH=-log[OH-]
2. Solve for [H+] from the formula [H+]=(1x10^-14)/[OH-]

Hope this helps!

Darlene Lien 3E

Hi, the same thing is happening to me too!

Darlene Lien 3E

Yes, since pH measures the concentration of hydrogen ions in a solution, strong acids fully dissociate; thus, strong acids typically have a lower pH. Hope this helps!

Darlene Lien 3E

Hi, I personally don't have any strategies, but I found it a little easier the more I did the practice problems. Also, there might be YouTube videos that you could search up that could give tips on naming the coordination compounds. Hope this helps!

Darlene Lien 3E

Definitely check Toolbox 9C.1 since it goes more into depth for naming coordination compounds!

Darlene Lien 3E

I believe ferrate will always be used because for some metals, Latin names are used, and iron is one of them. Hope this helps!

Darlene Lien 3E

Hi, also, I think Dr. Lavelle posted a worksheet on his website that would help with naming coordination compounds, so definitely check that out! Hope this helps!

Darlene Lien 3E

Helena Hu 3E wrote:Will week 10 be review then?

I'm not entirely sure if Week 10 will be review, but I would assume that if anything we'll be wrapping up all the topics by then.

Darlene Lien 3E

I normally go out on afternoon jogs or evening walks! It really helps to clear out your mind :)

Darlene Lien 3E

Hi, so for sulfite ion, when you draw the Lewis structure, you would have a VSEPR formula of AX3E, which is trigonal pyramidal (sulfur as the central atom with one double bonded oxygen, 2 single bonded oxygens, and one lone pair on sulfur). A tetrahedral shape would have a formula of AX4 (no lone pa...

Darlene Lien 3E

Also, I believe he'll send out an email once the results are out!

Darlene Lien 3E

Hi, I also had that problem, and even if it doesn't show up after you refresh the page, I think if you wait a bit it'll appear normally. Hope this helps!

Darlene Lien 3E

Hi, I agree with the comment above. Since oxygen is more electronegative, it should have the more negative formal charge. Therefore, it would be better if arsenic's formal charge is 0, so there should only be one double bond in total. Hope this helps!

Darlene Lien 3E

Hi, to add on, I think it's also pretty common for people to take Chem 14B and 14BL separately! I wouldn't worry about it if you can't take them together :)

Darlene Lien 3E

Hi, since you could get a max of 5 points from 5 posts each week, I think by the end of Week 7, you should have at least 35 posts. Hope this helps!

Darlene Lien 3E

Hi, I believe that since we haven't covered nomenclature, we wouldn't have to know the names of them. However, for more common ones especially, it would be good to know. Hope this helps!

Darlene Lien 3E

Hi, to add on, I think for this unit with IMF, the main idea that we should get from it is that boiling points and melting points follow the same trend, where a stronger IMF=higher boiling/melting point. Hope this helps!

Darlene Lien 3E

Hi, also, I believe it was also mentioned that there is a possibility of having midterm exams during the lecture times and not discussions as well. To be safe, it is best to avoid class conflicts. Hope this helps!

Darlene Lien 3E

Hi, if I believe the greater the ionic character (less covalent character), the more soluble the compound is. In other words, more polarizable = more covalent = less soluble, less polarizable = more ionic = more soluble. Also, since larger anions are more polarizable, this would mean that smaller an...

Darlene Lien 3E

Hi, from what I've heard, I don't think there will be anymore Audio-Visual modules since I think Dr. Lavelle mentioned that they take a lot of time and work to make.

Darlene Lien 3E

I believe they are still being held!

Darlene Lien 3E

I'm not exactly sure if it is at STP, but I think -2 kJ/mol for LDF is just the typical energy released.

Darlene Lien 3E

I don't think we have to know how to write the electron configuration for an element in the f block, but let's say the element is in the 6p block, for example, then we would have to include 4f in the electron configuration, if that makes sense.

Darlene Lien 3E

By definition, a radical has an unpaired electron, so typically when you see odd numbers of valence electrons this could kind of tell you that it is a radical. For NO2-, there are a total of 18 valence electrons and the Lewis structure is drawn with N as the central atom with a double-bonded O and s...

Darlene Lien 3E

To add on, this has to do with the law of conservation of energy. So, when energy is used to break a bond, energy is also released when the bonds are broken. Hope this helps!

Darlene Lien 3E

I believe this has to do with the idea of polarity. Hydrocarbons typically are nonpolar, and therefore, would have induced dipole-induced dipole forces. However, in HF, the structure is polar and has partial charges: a partial positive on H and a partial negative on F; as a result, HF has dipole dip...

Darlene Lien 3E

I believe that in order for hydrogen bonding to occur, two things have to happen:

1. In the first molecule, Hydrogen is attached to N, O, or F.
2. In the second molecule, there has to be a lone pair on either N, O, or F.

Hope this helps!

Darlene Lien 3E

Hi, based on the examples in the lecture, I think it would be best to use dots to represent the lone pair electrons and lines to represent the bonds.

Darlene Lien 3E

Hi, I agree with the comment above about how resonance structures are essentially just changing positions of the bonds. A formal charge of 0 would be most stable. Hope this helps!

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