CHEMISTRY COMMUNITY

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!

Darlene Lien 3E

Hi, to add on, I don't think a double bond would work as well because then the formal charge for the double bonded Fluorine would be +1 as opposed to 0. Hope this helps!

Darlene Lien 3E

I would recommend going to the UA sessions if you are unsure about certain topics. The UA sessions really help especially if you're having trouble on a topic. Hope this helps!

Darlene Lien 3E

I love this thread! Don't worry, we all got this!!

Darlene Lien 3E

Hi, Dr. Lavelle mentioned during today's discussion section that it should be up on Thursday!

Darlene Lien 3E

Hi, for this problem E=(1/2)mv^2 and de Broglie's equation won't work because this problem concerns photons. E=(1/2)mv^2 involves the kinetic energy of an electron while de Broglie's equation is used for electrons/things with mass (not photons). Therefore, you should use E=hv and c=(lambda)v or lamb...

Darlene Lien 3E

Hi, I was confused with this problem as well! I think it is essentially asking you to first solve for the mystery metal, M. Then you would find the molar mass of the metal sulfide, so Mystery metal + Sulfur. First, you would solve for the molar mass of (OH)2, which would be 34.02 g/mol. Then you wou...

Darlene Lien 3E

Hi, for this question the work function is given and since kinetic energy equals to 0, the energy per photon is equal to the work function. Therefore, you could solve for the wavelength by using the formula lambda=hc/E, where E=3.607E-19 J. Hope this helps!

Darlene Lien 3E

Hi, for this problem to find the energy of the electron, you would first use de Broglie's wave equation to solve for the velocity of the electron, but make sure to convert the micrometers to meters first! Then you would use the velocity you get, and plug it into the kinetic energy formula, E=(1/2)mv...

Darlene Lien 3E

Hi, although we probably don't need to use temperature conversions now, I think either way it would be good to know the conversions since there are only a few.

Darlene Lien 3E

Hi, I think some of the UA worksheets might have answer keys available, but for the most part, in order to get the answers you have to attend the UA session. Hope this helps!

Darlene Lien 3E

Hi, 3.37 x 10^-19 J was the energy you get from part a. Hope this helps.

Darlene Lien 3E

Hi, I believe he will also send out an updated one on Wednesday!

Darlene Lien 3E

Another good way to get practice problems might be to google them. There are a lot of good practice problems on the web, so I'm sure just searching them up would be a good way to get practice problems. Hope this helps!

Darlene Lien 3E

Hi, the webcams are highly recommended. There are a lot of good webcams on Amazon, and I think anything with 4+ stars should be sufficient. Hope this helps!

Darlene Lien 3E

Hi, the equation used to find the energy per photon is E=hc/lambda which is a combination of E=hv and c=v(lambda). Hope this helps!

Darlene Lien 3E

Hi, to add on, when using the Rydberg equation v=R[(1/n1^2)-1/n2^2)], the n2 value should be the value you're solving for, and n1=2. I wanted to just clarify since putting in the wrong values for n would result in different answers. Hope this helps, too!

Darlene Lien 3E

c= 3.00x10^8 m/s is the speed of light. In the lecture today, it was used to derive the de Broglie equation.

Darlene Lien 3E

As with what the others said, the units of the work function should be Joules (or Joules per electron), but if you are asked in terms of moles then you would most likely have to use Avogradro's number accordingly. Hope this helps!

Darlene Lien 3E

They have different masses!
The mass of an electron is 9.109 x 10^-31 kg. The mass of a proton is 1.672 x 10^-27 kg. The mass of a neutron is 1.674 x 10^-27 kg.

Darlene Lien 3E

No, E(photon)>E(remove e-) is correct as is. I think the phrasing might be the part that is a little confusing to comprehend because of "unless," but it could help if you read the statement as "e- is not emitted even for high intensity light unless E(photon)>E(remove e-)." Hope t...

Darlene Lien 3E

In simple terms, the energy of a photon + the energy required to remove an electron = the energy of the removed electron (the excess/what remains). Hope this helps!

Darlene Lien 3E

The key idea to note from this is that each photon must have enough energy to remove an electron. In addition, we must consider the idea that one photon interacts with one electron. Increasing the intensity/amplitude of the light only increases the number of photons, not the energy of each individua...

Darlene Lien 3E

So does the fact that the work function is in kJ instead of J.mol^1 not apply to this question? For part b when it asks for the energy required to remove an electron from one sodium atom, then the unit conversion would matter. You would have to convert the work function from kJ to J. In part a, the...

Darlene Lien 3E

Hi, there is no need for a unit change for that part of the question. To solve for the kinetic energy of the ejected electron you would use the formula: E=(1/2)mv^2. m=mass of an electron which is known to be 9.109x10^-31 kg v=6.61x10^5 m/s You would then plug these values into the formula according...

Darlene Lien 3E

The correct answer I believe is "wave, particle." Hope this helps!

Darlene Lien 3E

Yes, I personally use what the above comment says to check how many/what my posts were. "Quick links" --> "Your posts"
It should show all of the posts that you made on Chemistry Community. Hope this helps!

Darlene Lien 3E

Hi, I believe that you wouldn't round in between steps until your final answer. This would make your answer as accurate as possible. Hope this helps!

Darlene Lien 3E

Hi, personally, what helped was to first look at the given information and try to apply it to what you know about the topic. Given: volume of CuSO4 (250 mL), molarity of CuSO4 (0.20 M), molar mass of CuSO4 (calculate to get 159.62 g/mol) Want: mass of CuSO4 (g) By looking at this, we know that we ne...

Darlene Lien 3E

Hi, in case you wanted to refer back to tips for sig figs, you could visit Dr. Lavelle's website! There's a pdf titled "Everything you want to know about sig fig" explaining sig figs in depth. Hope this helps!

Darlene Lien 3E

Personally, I've been using two decimal places as well for the molar masses, and my answer hasn't been off by too much. I think either should work. Hope this helps!

Darlene Lien 3E

Hi, just in case you wanted to refer back, I believe in the third Audio-Visual Focus topic (Limiting Reactant Calculations) Dr. Lavelle mentioned that limiting reactants and limiting reagents could be used interchangeably. Hope this helps!

Darlene Lien 3E

Yes! If you check Dr. Lavelle's website, there is a link to textbook errors titled "Solution Manual Errors 7th Edition."

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