CHEMISTRY COMMUNITY

ariana_apopei1K

michelle wrote:Will you post the answers here as usual?

I think so since the sheet says "Full solutions will be given only during the review session." so maybe regular answers will be posted

ariana_apopei1K

We probably don't have to graph any major things. I would just know the graphs we learned in lecture and what k means, what the y intercept means, etc.

ariana_apopei1K

Elle_Mendelson_2K wrote:Does anyone know what the midterm average was?

I don't remember the exact number but I think it was in the low to mid 80's percentage wise

ariana_apopei1K

Julia Lee wrote:Does anyone know if Karen posted worksheet 9 yet?

no it would have been posted here, week 8 is the most recent

ariana_apopei1K

the slowest step controls the rate since you cannot get any faster than the slowest part

ariana_apopei1K

I was also wondering this, when I look it up online I see a lot of people say it's non-spontaneous

ariana_apopei1K

delta G is zero when the reaction is at equilibrium. delta G standard is the same value, whether it's zero or not doesn't matter, because its the calculate of delta G at certain standard conditions.

ariana_apopei1K

Josephine Lu 4L wrote:could someone please explain #8 part a on the electrochemistry worksheet? How do we know that S2O8 2- is reduced and NO is oxidized?

I also have this question

ariana_apopei1K

I think those half reactions are only considering what is happening to the reactant ion in question. In reality, it would produce HCl, but since each half reaction only considers one ion, they don't include the other. But then again, since HCl is a strong acid, it would fully dissociate anyway. oh ...

ariana_apopei1K

For the rxn Cl2(g) + H2(g) --> HCl (aq) the half rxn's are shown as Cl2 (g) + 2e- -> 2Cl- (aq) and
2H+ (aq) + 2e- --> H2 (g).

How come HCl is dissociated? is it because of its acidity?

ariana_apopei1K

so for the cathode i did O2 --> OH- and treated it as a half reaction in a basic solution because of the OH-. once you balance both sides with H+'s, OH-'s, etc you should get 4e- + 2H2O + O2 -> 4OH- For the anode side you would solve normally, like in an acidic solution. so O2 -> H+ would balance ou...

ariana_apopei1K

it helps if you look at the equation deltaG=deltaH - T(deltaS). The more negative deltaG is, meaning exothermic, the more chances for delta G to be negative too, meaning spontaneous.

ariana_apopei1K

the oxidizing agent will be in the reduction half rxn, and the reducing agent will be in the oxidation half rxn

ariana_apopei1K

I went to a step up session and the UA mentioned that many people have been having issues with weird questions, so I think they'll give us normal ones on the test. At least I hope so haha

ariana_apopei1K

think of adding H+ the same as balancing the equation. the left side has 6 H's and the right only has 4, so you add 2 H's to even it out

ariana_apopei1K

I think they're all correct! I was stuck on the one with N2 since it isnt a noble gas, but some places online say that it can be used as an inert gas, so yes there wouldn't be any shift

ariana_apopei1K

i believe so yes. its the same as if you would convert the number of moles by multiplying by avo first

ariana_apopei1K

Celine Hoh 2L wrote:Quick question for Worksheet 1 ques8:

When I2 is added, neither products nor reactants is favored as I2 is a solid.
What happens when I2 is removed, does the reaction not shift too?

Yeah, since it is a solid it doesn't affect it either way

ariana_apopei1K

Megan_Ervin_1F wrote:when will you post the answers to hotdog?

He mentioned that he won't be posting answers this time

ariana_apopei1K

Search up hotdog!

ariana_apopei1K

Lyndon is probably going to make one that he'll go over in his review session this monday

ariana_apopei1K

For the first law worksheet, how do you do #2? First you have to do stoichiometry to find how many moles of Pb is made from 49.7 g PbO. You multiply the number of moles of product times the enthalpy value given, which gives you the q to plug into your regular q=mC s (Tf-Ti). Once you solve for mass...

ariana_apopei1K

q_p=n(5/2)R(Tf-Ti) with q_p= 765 J

ariana_apopei1K

Hi, I just checked my notes and i wrote "assume P=1 atm, use PV=nRT), meaning the pressure is constant at 1 atm. Maybe he only said this verbally or something and it wasn't on the slides and you might have missed it? I hope this helps!

ariana_apopei1K

from what I understood we use Cp when pressure is constant and Cv when the volume is constant to find out how much heat has been gained or lost but I'm not sure if you were looking for something more specific about them

ariana_apopei1K

I noticed this too, sometimes they even give the answer in both J and KJ. I doubt you'd be marked wrong on a test for having one over the other though

ariana_apopei1K

Jim Brown 14ALec4 wrote:My handwriting is terrible but

Thank you so much!! life saver

ariana_apopei1K

Hi,
Does anyone have the notes from today's lecture (1/30, Wednesday) and is willing to share them? I was sick and couldn't make it today :(

ariana_apopei1K

For 7th edition some parts of 4C, 4D, and 4E include stuff that we went over in lecture already but you might still have to skip around

ariana_apopei1K

Something is a state function when the path to get from point A to point B does not matter. Since for enthalpy all we care about is what the values are from the start and from the end, that makes it a state function.

ariana_apopei1K

What are the initial values and the Kc value that we have to use for 4d worksheet 1? are the initial values the values from 4b?

ariana_apopei1K

Ray Guo 4C wrote:Don't we need the volume to solve question 5 on worksheet 1?

you can calculate the volume from the info given using PV=nRT!

ariana_apopei1K

For #3 on Worksheet 2, how can you find the Kb value? I'm not quite what is the first step or what the base is. So if percent protonation was 6.7%, then 6.7 = 100 x (x/.1) should give you x. For the ice table I don't think you even need to know the base, so i just wrote " oct + h20 --> OH- + o...

ariana_apopei1K

Hi all, In part a of this problem, understand how to set up the ice box and how to get to the set up (x^2)/(0.057-x), but I don't know where the solutions manual got 1.8x10^-5 to set it up as equal to. I know that's the Kb of NH3, but are we supposed to know that by heart? or is there a table somewh...

ariana_apopei1K

Basically the acid is so weak the overall solution can be considered neutral. Such a small concentration of H3O+ is pretty much not affecting the pH anymore in the grand scheme of things

ariana_apopei1K

Also whenever a question specifically gives you an initial value and a Kc value! you can kinda guess you'd have to end up with a quadratic equation to solve the question after that

ariana_apopei1K

Not 100% sure, but I think it might include acids/bases since we still have a couple lectures until the test. I'm sure Dr. Lavelle will let us know soon

ariana_apopei1K

We'll probably be given anything we need to solve the problem, I doubt we have to memorize any values like the ones from the tables some textbook questions have us use.

ariana_apopei1K

Q changes as a reaction occurs, and K is when said reaction reaches equilibrium. If the reaction does't reach equilibrium, you can't use K, only Q

ariana_apopei1K

I'm not sure if there's a specific place to get answers, but if you post any of the specific questions on here people will definitely help out!

ariana_apopei1K

Haowen_Liang_3E wrote:for 33a, how do we set up the equation with 1.5mol/L and 0.1155mol/210ml to find the mL require for the 1.5M solution?

1.5 x L1 = .55 x 210 then solve and you should get 77 mL

ariana_apopei1K

If you're asking about the ligands for coord compounds we have to memorize these ones: https://lavelle.chem.ucla.edu/wp-conten ... pounds.pdf

ariana_apopei1K

Carissa Young 1C wrote:will there be more worksheets for the recently covered material like acids and bases and coordination compounds ? Thanks

I think this last worksheet was the last one since our final is on Sunday when there would normally be a session

ariana_apopei1K

It's called ferrate rather than iron because of the overall negative charge on the complex. It's the same with the colbaltate on letter c of the same question!

ariana_apopei1K

I have access to the online sapling stuff, but whenever I go look for the practice problems Dr Lavelle was talking about in lecture some time ago, all I have listed are some training modules on how to draw shapes using the sapling system and some practice math problems. Does anyone know where the pr...

ariana_apopei1K

I don't know for sure but if we can just work out the lewis structures and use the formal charges from there for the ligands, then we technically don't have to memorize them. I do think knowing a few of the common ones by heart might be helpful though so you can have more time for other things on th...

ariana_apopei1K

You said that in order for hydrogen bonding to occur, the hydrogen in the individual molecule has to be bound to either O,N, or F to hydrogen bond with another molecule, but in CH3CH2OCH2CH3, none of the hydrogens are actually bound to the oxygen, but in number 3 d. of this worksheet, it says that ...

ariana_apopei1K

You said that in order for hydrogen bonding to occur, the hydrogen in the individual molecule has to be bound to either O,N, or F to hydrogen bond with another molecule, but in CH3CH2OCH2CH3, none of the hydrogens are actually bound to the oxygen, but in number 3 d. of this worksheet, it says that ...

ariana_apopei1K

On the Week 6 Step-Up, why is #1a. only dipole-dipole and dispersion forces? Isn't there hydrogen bonding also between the hydrogen of one CH3CF3 and the fluorine of another CH3CF3 molecule? No, the H in the individual molecule must have hydrogen bonded to N, O, or F first for the interaction youre...

ariana_apopei1K

Generally polarizability refers to the ease with which an e- cloud of a particle can be distorted. Polarizing power though, is the ability something has to distort said e- cloud. A smaller atom will have less polarizability, I think because the actual clouds are smaller and the e- are held more tigh...

ariana_apopei1K

Yes, try to know the shapes along with their formula (like AX2E, etc) and what the bond angles are. I made a chart ranging from linear to octahedral and it helped me visualize it better. There are also illustrations in the book and online!

ariana_apopei1K

It has London dispersion forces becasuse they are interactions that happen for all examples that you'll be given. They're slightly negative parts of molecules that occur by chance as e- are constantly moving and may group up for shorts periods of times, causing dispersion. This is weak and temporary...

ariana_apopei1K

Hi! So it important to know a major difference bw them is that σ bonds allow for rotation because of cylindrical symmetry, whereas pi bonds are side-by-side interactions that do not allow rotation, or else the bond would break. When there are pi bonds, that restriction in movement means there are no...

ariana_apopei1K

I was also a little confused about this today. I think it might have something to do with empty orbitals? Like the empty 2p orbital for his C2H4 example.

ariana_apopei1K

What is the difference between molecular geometry vs electronicgeometry? Electronic geometry describes the shape of a molecule based on the number and arrangement of electron pairs around a central atom. Molecular geometry is describing the shape of the molecule in terms of bonds. So if a molecule ...

ariana_apopei1K

Can someone explain how to draw KrF4 You put Kr as the central atom and arrange F around it, then count that there are 36 total e- to use. Subtract 4 single bonds, giving you 28 e. Once you give all the F their octet, you're left with 4 extra e-. Fluorine does not allow for expanded octets, so the ...

ariana_apopei1K

I think you can generally find the answers to those kinds of questions by using the trend we learned, but there's a diagram listing some of the specific numbers on pg 97 (fig. 2d.2 at the top of the page). I doubt they'd give us any questionable comparisons to make where the specific numbers would b...

ariana_apopei1K

Overall dispersion/van der waal forces occur among all molecules, but they're weak and only happen by chance. Hydrogen bonds are much stronger although more specific in nature. I don't think we'll need to memorize any particular example (although H2O is an easy one to remember for hydrogen bonds) bu...

ariana_apopei1K

Ethan Breaux 1J wrote:When was this discussed in lecture?

I don't think it was discussed yet, but I found this from a different quarter if it's any help! viewtopic.php?t=16454

ariana_apopei1K

Hi all,
Does anyone know if Dr. Lavelle structures chem 14b similarly to 14a? Meaning if there are still lots of available resources like the workshops and all that? I know chem community is still available to us but I was just curious!

ariana_apopei1K

Also I found this for why Mg and Be are equal. Their 2s orbitals are full, making them very stable. Adding an e- would create an awkward arrangement that it does not prefer over having a full subshell. Nitrogen also has a very low affinity because its 2p orbital is half full, meaning more stable.

ariana_apopei1K

Hi I also had trouble with this and looked around online for some answers. I think it's because Oxygen is smaller than Sulfur, and has higher electron repulsion. This repulsion makes less energy be released when an e- is added, since it is being made even less stable. I guess it't just an overall ex...

ariana_apopei1K

They have the same trend on the table too, increase as you go from left to right and decrease as you go from top to bottom.

ariana_apopei1K

Anjali 4A wrote:Some questions give the radius instead of the diameter, in that case would we multiply the radius by 2?

I think that's correct yeah!

ariana_apopei1K

You would multiply by 2 when you're given something like +/-0.5 as your value. This uses the same reasoning as the first answer on this post, the two outermost values we can get are .5 and -.5, so when you subtract them (.5--.5) you get 1, the same thing as multiplying .5 times 2.

ariana_apopei1K

Thank you! For 6.b. isn't it asking for the work energy/ the threshold energy? In the solution, E=hv is used to find energy, but I thought that was energy of the incoming photon and it's asking how much energy is required to remove the electron. So wouldn't you find the kinetic energy of the photon...

ariana_apopei1K

Alexa_Henrie_3B wrote:Yes the position in question 1 is just the diameter, 2x10^-4 m.

Perfect, thank you!

ariana_apopei1K

Hi everyone! I would love some quick clarification on question 1 from test 2: anytime we are given the diameter in this type of question we substitute it directly in for delta x? As in no manipulations or using formulas, the diameter covers the potential position for the molecule? Thanks and good lu...

ariana_apopei1K

Heyy I was wondering where I can find the answers for worksheet 5 (newest one)? Also for question 9 aren't there more than 2 ions that aren't isoelectronic with F-. I got 4 ions (Li+, Be2+,O-, N2-) that aren't isoelectronic. Thanks! I also got this as my answer! I think the answer sheet will be pos...

ariana_apopei1K

I'm sure someone would buy, if not the entire set, one or two parts of it! You wouldn't get all your money back, but people next quarter could definitely use it and would prefer to purchase it if it is a lower price. Sorry if that isn't much help! Best of luck to you :) Thank you! I'll try it out a...

ariana_apopei1K

Leela_Mohan3L wrote:O has an even amount of valence electrons (6) while Cl has an odd amount of valence electrons (7). Thus, Cl is the radical since it has an odd amount of valence electrons.

That's so much easier than I thought it would be haha, thanks!!

ariana_apopei1K

Hi all, part a asks for the lewis structure of ClO, and i understand it's a radical, but how did they determine that the single electron goes on Cl and not O? Does it have to do with electronegativity or something from the trends? Thanks

ariana_apopei1K

Hi everyone, I hope midterm season isn't too bad so far! I accidentally bought the e-text, loose leaf text, and solutions manual twice toward the beginning of the year. The return period just expired a couple days ago, and I was wondering if anyone had any specific advice on how I should go about tr...

ariana_apopei1K

The dots can go on any four sides, but keep in mind that two dots next to each other means that those two electrons are paired. If the 4 electrons that you're drawing aren't paired, the structure would have one dot on each of the 4 sides. I have a similar question from this: A line drawn between tw...

ariana_apopei1K

Chan Mee Lee wrote:Hi, is the review session in Young Hall where our lectures are or is it CS24?

Hey, the session is in CS24, which is the same place where the 2 pm lecture is held. If yours is in CS50 it will be right next door!

ariana_apopei1K

I'm a little confused on #7 in the Quantum World Part 1 worksheet. I just can't seem to come up with (1/ni)=0.0278 and I'm not sure why. Could someone explain it to me possibly? I follow her formulas and rationale, I'm just not getting the right numbers... I only get 0.0278 if I have 7.55x10^-20 be...

ariana_apopei1K

The same paper that's given to us each test is on Lavelle's website for the class if you want to see the specific things we are given to use without memorizing. I think it's called "constants and equations" or something along those lines

ariana_apopei1K

Imelda Mena 1J wrote:Hi! Where do these workshops take place? I would like to attend.

Covel 210

ariana_apopei1K

Does anyone know what information will be given to us...in terms of formulas and what not. Do we need to know the mass of an electron? Hi, everything that is given to us is in the "constants and equations" file on the 14a website! Mass of an electron is included in the list of things give...

ariana_apopei1K

For Quantum World Part 2 Worksheet, why is 11 D False? I saw it as a trick question with the "always" because higher frequency doesn't matter until the threshold wavelength is reached for the photoelectric experiment. After that, it does matter, but since the wavelength doesn't always rea...

ariana_apopei1K

Also, I am a little confused on what it means when the electron is ejected from the metal. Like is it light? Like what is the purpose of the electron being ejected? It has to do with the experiment we discussed in class where light (photons) hit a block of metal, leading to the e- being ejected. Th...

ariana_apopei1K

I think it's overall stating that we can't be sure of the speed or position of an object at the same time. If we stop an object to get its position, then we cannot know its speed and vice versa. This video really helped me understand the concept more in terms of wavelengths and what we're learning a...

ariana_apopei1K

I would just know that Ballmer includes visible light, whereas Lymann includes UV, and that Ballmer starts at n=2 and Lymann starts at n=1. Otherwise the questions will probably include anything more specific that we need to know.

ariana_apopei1K

I think it depends on what the question already gives you data-wise. If one of the equations isn't possible to use because a specific data point isn't given, then switch to the other and see from there.

ariana_apopei1K

Hi, I thought of the 150ml solution as a sort of starting point for the person doing the dilution. They would use this to get their first "sample" of 20mL. So try using 20mL as your V1!

ariana_apopei1K

I think Dr. Lavelle mentioned in my lecture that since we are still practicing the fundamentals to prepare for the first exam we can pull HW2 problems from the fundamental section of questions! He said HW in general is pretty flexible.

ariana_apopei1K

Probably, since it's part of the review from HS. Only a couple extra steps anyways, I'd know it just in case!!

ariana_apopei1K

I see extensive as something that can "change", such as the volume of a cardboard box. A bigger cardboard box could hold more water than a smaller one, and thus has a larger volume--this difference between the two means volume is an extensive trait. Intensive is when it's an intrinsic part...

ariana_apopei1K

Essentially, I don't think it changes how you get to the solution btw! Same as the previous parts of the problem.

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