CHEMISTRY COMMUNITY

ColmConnolly3D

I believe we only need to know this for ideal gasses, and the values are given on the equation sheet, but I obviously can't say for sure

Hope this helps!

ColmConnolly3D

What part do you not understand? Iĺl try to give a somewhat broad explanation but lmk if you need help with a specific part. Basically, it is saying that for the first proposed mechanism, since water is not present in the bottleneck step, water's concentration will not affect the rate at which the r...

ColmConnolly3D

I think you should be good just knowing the equations on the sheet, the most important thing is remembering how to use each one! If you are struggling with this, I would recommend going through some homework problems just using the equations sheet so as to make sure that you can pick out the correct...

ColmConnolly3D

The only way I know at our level to figure this out is to count the number of bonds broken, and assume that the more ordered phase would have the least bonds broken. However, in general, the intermediate steps will always be higher than the reactants or the products, otherwise, the reaction would no...

ColmConnolly3D

Hi! You are going to need to use the Arrhenius equation ( k = A e^{\frac{-E_{a}}{RT}} ) to determine k for the original pathway, then for the catalyzed pathway. k will be exponentially higher for the catalyzed pathway due to the change in activation energy, so expect a large ratio as you will be fin...

ColmConnolly3D

https://www.youtube.com/watch?v=YWl4KHsUxuw

I found this video super helpful!!!

ColmConnolly3D

You can tell because you will see it in the steps of the reaction, but not in the overall chemical equation. I suppose that you could see that the catalyst was not being modified in the steps of the reaction? But I would be wary of using that method and I doubt you will be asked to do so.

ColmConnolly3D

Salt bridges connect anode and cathode and help keep reduction and oxidation solution from building up positive/negative charges, essentially keeps the cell working I really like what you said here, I just felt like it would be helpful to add that they move ions from one side of the cell to the oth...

ColmConnolly3D

The best way I have learned to do this is to use the mnemonic “Red Cat An Ox”, which tells me that oxidation occurs at the anode and reduction occurs at the cathode. Since the reaction at the anode is the source of electrons for the current, the anode is the negative terminal for the galvanic cell.

ColmConnolly3D

Hi Karen, the only equations I can find that we are responsible for are as follows Nernst Equation Ecell = Estd - ((RT)/(nF)) ln([P]/[R]) (This can be written many ways, but this is the most universal) Nernst Precursor Delta(G) = -nFE and then some more conceptual stuff like redox balancing, cathode...

ColmConnolly3D

Nope, you have to look up the table in this case (as far as I know), especially for transition metals, patterns are spotty in this area.

ColmConnolly3D

You should write the entire molecule, as we kind of treat it as a unit when performing these kinds of calculations

ColmConnolly3D

I would just look at the number of moles of electrons that are transferred in the reaction that you are looking at, that is what n refers to here

ColmConnolly3D

Yup! A good way to think about this for me is that any energy gained through heat must be equal to the amount lost through work (or vice versa) for the temperature to remain constant (or that it would have to go somewhere else like in a phase change). Thus, if q=-w, ∆U = q + (-q) = 0 (And vice versa...

ColmConnolly3D

Isobaric is just another word for constant pressure. This means that you are likely dealing with an irreversible expansion, so w = -P∆V, and that heat (q) will equal enthalpy (∆H).

Hope this helps!
Colm

ColmConnolly3D

I work part-time as an EMT and it can be really tough to manage with school. It is hard to come back from a shift and get work done, so I usually try to plan my workload around working, but I obviously can't do this for every night I work. Sometimes I will bring reading into work, as I usually have ...

ColmConnolly3D

As per this page on ChemLibreTexts , \Delta S = \frac{q}{T} for processes occuring at a constant temperature not a constant Internal energy. I think that this is just a condition of the equation given, as no rationale is listed but it would make sense that work could change in this type of scenario ...

ColmConnolly3D

From what I remember from last quarter, the final consisted of questions very similar to the midterms in both difficulty and style. I agree with the others that the best practice is to work through all of the given textbook problems and try to attend as many UA Sections as you can. Consistency is ke...

ColmConnolly3D

I think that doing textbook problems helps with making sure you know how to work the equations quickly and effectively, but for understanding the concepts nothing has been as good for me as talking it out with a peer. Find a study buddy and work through some problems together!!!

ColmConnolly3D

For me the best place to start is always the units. If I can look at the units in the question and the units required of the answer, usually I can determine what equations will manipulate those units in such a way as to take me to the right answer. The key to this method is memorizing the units of e...

ColmConnolly3D

285.83 kJ/mol is the standard enthalpy of formation you want, not -241.82. Your equation looks right, I think you just accidentally had the wrong value.

Hope this helps!
Colm

ColmConnolly3D

The way that I think about it is that in a reversible process, the energy lost to heat is minimized and the energy that goes into work is maximized. Since the two systems are in constant quasiequilibrium. Since this is impossible in the physical world (as it would take infinite time), no process is ...

ColmConnolly3D

I cannot recommend the organic chemistry tutor enough. https://www.youtube.com/watch?v=NyOYW07-L5g&list=PL0o_zxa4K1BWziAvOKdqsMFSB_MyyLAqS&index=110 https://www.youtube.com/watch?v=GL3gCBUua4c&list=PL0o_zxa4K1BWziAvOKdqsMFSB_MyyLAqS&index=111 https://www.youtube.com/watch?v=E7s-hIoy...

ColmConnolly3D

I cannot recommend the organic chemistry tutor enough. https://www.youtube.com/watch?v=NyOYW07-L5g&list=PL0o_zxa4K1BWziAvOKdqsMFSB_MyyLAqS&index=110 https://www.youtube.com/watch?v=GL3gCBUua4c&list=PL0o_zxa4K1BWziAvOKdqsMFSB_MyyLAqS&index=111 https://www.youtube.com/watch?v=E7s-hIoyN...

ColmConnolly3D

First calculate the calorimetry constant of the calorimeter: (1.05kJ/2.19C)/0.700L = ? kJ/C*L This lets you know how much energy it takes to raise a liter of solution in the calorimeter by 1 C. Then you can do essentially the same thing in reverse to see how much heat the reaction released. Hope thi...

ColmConnolly3D

Depends on what you are trying to solve! The best practice is probably to track the units for the problem you are trying to solve, but in general, if you want volume use 0.08, and if you want energy use 8.13.

Hope this helps!
Colm

ColmConnolly3D

The thing is, they are changing but ever so slightly. At the level of precision we are working with, these changes will not be a factor in our calculations.

Hope this helps!
Colm

ColmConnolly3D

Heat energy = energy that can cause a change in temperature
Work energy = energy that can do work to change a system's physical properties, such as pressure or volume. While we have only calculated this for gasses, it is present for solids and liquids as well, just imperceptibly.

ColmConnolly3D

Enthalpy is the amount of heat-energy present in a system, as well as the work-energy available in the system. Thus, H = U + PV. At constant pressure, since the work-energy will not change, the change in enthalpy will be equal to the heat energy change of the system.

ColmConnolly3D

It is kind of representative of our perspective. From our perspective, the "system" is the gas being manipulated, and the "surroundings" are the container that it is being manipulated in. Since energy is being transferred from the system to the surroundings, we say it is being lo...

ColmConnolly3D

Path A is isothermal reversible expansion, meaning that In the reversible process, so the pressure of the surroundings is always going to be less than the pressure of the system, by an infinitesimally small quantity. Thus, W = nRT ln(V1/V2). As long as you use the correct constants, this process sho...

ColmConnolly3D

Heat energy causes burns, not just temperature Steam has to contain enough heat energy to raise the speed of the H 2 O molecules to reach the temperature of the gas, as well as to break the intermolecular interactions holding the liquid H 2 O molecules together. Thus, 100 o C liquid H 2 O molecules ...

ColmConnolly3D

Just a quick note as to why:

For at the level of precision we will be dealing with solids and liquids are incompressible. Thus, a change in PRESSure cannot affect the concentration of such substances.

Hope this helps!

Colm

ColmConnolly3D

Say I have to subtract then divide: 0.7552g - 0.7241g = 0.0311 g <- Here we keep 4 decimal places because we are subtracting. For my next step, I would like to divide by 768.95 g/mol Will my rounded answer be 4.044*10^-5 mol or 4.04*10^-5 mol? The question boils down to can subtraction reduce the ac...

ColmConnolly3D

Strong acids are those found here, they will dissociate completely in water, all other acids and bases will require a Ka/Kb calculation to find their h30+ concentration

ColmConnolly3D

Use the Ka to solve the dissociation equation (Ka=[CB][H30+]/[A]) to find the [H30+] concentration, then take the -log of that to find the pH!

Hope this helps!
Colm

ColmConnolly3D

A Monoprotic acid is one that will dissociate into a single proton and a conjugate base when it dissociates. It would change a reaction by increasing the pH.

Hope this helps!
Colm

ColmConnolly3D

I bet we will have to consider phase changes, especially since there were homework problems that included them.

Hope this helps!
Colm

ColmConnolly3D

It is my understanding that standard reaction enthalpies like those found in the table here show how much heat energy would be required to form a given reactant from its constituent parts in their standard states.

Hope this helps!
Colm

ColmConnolly3D

You should be able to use the [OH-] you found in (i) times the dilution factor (think 5 -> 500) to find the [OH-] of the original solution.

Hope this helps!
Colm

ColmConnolly3D

Hi there,

What is your question on this topic? Just so you know, you should ask a specific question about a problem, I'm not sure if you get credit for just generally asking for help.
I hope this doesn't come off as rude, just letting you know so you can get your points.

All the best,
Colm

ColmConnolly3D

I think that this question may be beyond the scope of this class, but I would imagine that if you were in the situation you described, and you were dealing with gaseous reactants, for optimal accuracy you would want to proceed with a correction for the new pressures after the addition of an incompre...

ColmConnolly3D

Ok, thanks for the help Sara! I'll make a quizlet :(

ColmConnolly3D

The percent ionization will be the amount of charged particles at equilibrium (for a monoprotic acid, this is just the H30+ concentration) divided by the initial concentration (in this case 0.00214M) of the acid. This will tell you what percentage of the acid has dissociated to form charged species....

ColmConnolly3D

Hi all,

On exams, will we be specifically told when we are dealing with a "weak" acid or base? I have a hard time with memorization, so I dont want to memorize the strong acids table unless I need to.

Thanks in advance!
Colm

ColmConnolly3D

Reaction quotient tells us the relative amounts of reactants and products at a given time in the progress of the reaction.
Hope this helps!
Colm

ColmConnolly3D

Q describes a reaction in progress, K is Q when the reaction is at equilibrium. So for a given reaction, we can compare Q to K to see how that reaction will proceed (Q>K: More reactants will be created than products, Q<K: More products will be created than reactants) unless Q=K, in which case the tw...

ColmConnolly3D

Can you clarify your question?[code][/code]

ColmConnolly3D

We will have to be given the concentrations of the P and the R. Then the problem will focus on manipulating the given concentrations to find an unknown concentration.
Does that make sense?
All the best,
Colm

ColmConnolly3D

Just to clarify: It is perfectly fine to use the actual molar amounts in the ICE table, as the volume of the container will not change between steps of the reaction. When you then plug the values from the ICE table into the K equation, you need to make the conversion to concentration. This is becaus...

ColmConnolly3D

They are just looking at different parts of the acidic process. Acids and bases are just substances that become charged in water. The Bronstead-Lowry definition simply calls substances acids or bases based on how their protons behave while the Lewis definition is about how their electrons behave. Th...

ColmConnolly3D

Amphoteric and Amphiprotic essentially mean the same thing, Amphiprotic is just specific to the Bronstead Acid Definition. While Amphiphoteric means a compound that can act as either an acid or a base depending on its milieu, a definition that applies to all ways to describe acids, Apmhiprotic speci...

ColmConnolly3D

Nope, these are different chemical processes. As I understand it, Ocean Acidification is a phenomenon driven by the increased atmospheric carbon dioxide we are producing while acid rain comes from sulfur and nitrogen oxides released into the air. Ocean acidification comes from the high atmospheric C...

ColmConnolly3D

Where are our participation points? I checked both ccle and ucla and I didn't see them To find them on MyUCLA: ⋅ Click "Classes" in the navigation bar, and follow the link "Exam and Homework Grades" under the category "Grades and Transcripts". ⋅ On ...

ColmConnolly3D

I really focused on textbook problems. I know that they can get a bit tedious, but I really think that practicing is the best way to learn this stuff. Especially with the heavy emphasis on VSEPR and Lewis Structures, you have to build muscle memory so that the answers will come quickly and you will ...

ColmConnolly3D

You should still get the class but get switched to another discussion section, especially since Lavelle is known for offering all of the students on his waitlist into the class. Honestly, as long as you can get to the ccle page, you should be fine.

Hope this helps,
Colm

ColmConnolly3D

All of the resonance structures are not equivalent because one of them will be most favored by the system. One way to think about this is with lewis structure formal charge. The lowest formal charge system will be favored over the others, even though they will all be a part of the explanation of tha...

ColmConnolly3D

https://opentextbc.ca/chemistry/chapter/19-2-coordination-chemistry-of-transition-metals/ This Canadian textbook uses H 2 O, but everyone else above also makes a good point as Oxygen would be the atom actually bonded to the Cobalt (In the [Co(H 2 O) 6 ] 2+ example shown here: https://opentextbc.ca/...

ColmConnolly3D

I managed to find the chart and a solid quizlet similar to the one detailed above! I hope these help whoever comes across this next. Just click on the Q logo to get linked to the quizlet (https://quizlet.com/209997502/vespr-flash-cards/). https://www.sigmaaldrich.com/content/dam/sigma-aldrich/articl...

ColmConnolly3D

I have also found this image quite useful, it was what let me eventually understand this topic. https://i.imgur.com/ThuMN1j.jpg It is definitely a model representation, however, you can see that while the sigma bond occurs on the line between the two double-bonded nuclei of the molecule, the pi bond...

ColmConnolly3D

Those are the only factors I consider and I did OK on the sapling problems. I would make sure you had all the shape angles memorized as they are somewhat counterintuitive for me.

ColmConnolly3D

Yes I would say that that makes sense

ColmConnolly3D

A delocalized pi bond is a resonant pi bond, basically, if you could write a resonant structure where a double or triple bond would take the place of a single bond, that would represent a delocalized pi bond.
Hope this helps!
Colm

ColmConnolly3D

Hi Augustina, hope your first year of college is going well! If atoms have a negative charge, they should repel any lone pair/bonded electrons around them while if they had a positive charge they would attract these regions of high electron density. This should also hold true for dipoles and other p...

ColmConnolly3D

The key things for me are when you count up, you start with the s orbitals and when you count down, like with ions, those are the first to go. You also need to remember the exceptions to this general rule with chromium (3d5 4s1) and copper (3d10 4s1) both moving an s orbital electron to the d orbita...

ColmConnolly3D

I’m not trying to rush him, just speculating on when we might have our scores back, I know a lot of people stress about not knowing how they did and not knowing when they will find out

ColmConnolly3D

I just want to add that for most molecules it would be good to make use of the octet rule first, iodine is kind of an exception to the standard operating procedure

ColmConnolly3D

I think it can also cause conformational changes in large molecules, like I know for a fact that some proteins shapes depend on hydrogen bonding, and it is the breakage of these hydrogen bonds that leads to protein denaturation at high temperature.

ColmConnolly3D

I imagine we will still have a lecture but I would wait for word from up top before jumping to conclusions.

ColmConnolly3D

There was also a problem with the last midterm that extended the grading time iirc, so I think it might be faster this time around.

ColmConnolly3D

Yes, this compound can serve as the lone pair side of hydrogen bonding, but not as the hydrogen-attached-to-electronegative-atom side of it.

ColmConnolly3D

I think this is a case of set subset interaction, where all coordinate covalent bonding is Lewis Acid-Base bonding but not vice versa, or I could be confised. This is a bit of a tricky topic for me but I thought I would throw in my two cents. Please correct me if I am mistaken!

ColmConnolly3D

Google is a good shortcut for the real world, but unfortunately, it won't help you on the test. If you are given the bond lengths though, we can see what resonance structures will be favored based on experimental evidence.

ColmConnolly3D

Just to add on a bit I believe it has to be a double bond due to the magnitude of the dipole involved. A polar double bond will have more of a charge disparity than a polar single bond, and therefore more of an ability to accept electrons.

ColmConnolly3D

The boiling points of these molecules will be different based on the strength of the intermolecular interactions that they experience. Their relative molecular weight also plays a part as compounds with heavier molar masses will take more energy to vaporize than lighter ones. CF 2 H 2 does not have ...

ColmConnolly3D

I think so, especially since Noble gasses are the exception to so many rules. But there is also some argument that He should be next to H I think? Like this is one of the arguments for why it belongs on the left side of the table iirc.

ColmConnolly3D

I'm pretty sure it is because the form with the lowest formal charge would also be the most stable! So it is naturally going to be favored due to the 2nd law of thermodynamics. Hope that helps!

ColmConnolly3D

Yeah, I'm pretty sure CO2 would have a resonant structure with one C=O bond and one C-O bond.
Just for a bit more clarity.

ColmConnolly3D

Hi All, As the scores are now in CCLE, I was wondering if there will be any way for us to review our tests and see what we did wrong. I'd like to learn from my mistakes, but I understand that the online environment does not always permit this. Perhaps we could schedule a meeting with our TAs 1on1 to...

ColmConnolly3D

They are in! Go and check if you dare.

ColmConnolly3D

My TA told us that the test will be graded on a straight scale with partial credit awarded for some wrong answers on certain questions to make up for the fact that they can't see our thought process. In another thread, someone mentioned that we should expect to have our scores in hand at the end of ...

ColmConnolly3D

Hi Lily, could you clarify your question a little bit? I'll do my best with what you have but I'm a touch unsure what I am answering. Electron affinity won't itself determine whether energy is released or absorbed, that has to do with what is happening to the electron. We would need to add energy to...

ColmConnolly3D

Effective nuclear charge refers to the net positive charge an electron experiences in an atom. Since the only positive component of an atom of matter is a proton, and a fluorine atom would have 1 more proton than an oxygen atom, a fluorine atom would have more total positive charge to distribute amo...

ColmConnolly3D

^^for the person who asked for my solution, sapling said this was correct

ColmConnolly3D

Constants: λ=0.0197nm=(1.97*10^-11)m, h=(6.63*10^-34) J*s, m(electron)=(9.11*10^-31) kg λ=h/mv -> v=h/mλ -> v=(6.63*10^-34)/((1.97*10^-11)(9.11*10^-31))=(3.69427248464*10^7)m/s Ek = 1/2(m)(v)^2 -> Ek=(0.5)(9.11*10^-31)((3.69427248464*10^7)^2)=6.2165042e-16 J Then you just convert J to eV (6.2165042*...

ColmConnolly3D

I think a good tip is to use the function buttons. It is just a lot easier for me to keep track of like fractions and stuff when I use f1 to make them more clear.
Hope this helps!

ColmConnolly3D

So for this problem, we are given the wavelength of the matter-wave and asked to calculate the energy it outputs. To find the total kinetic energy of an electron from its wave-like properties, we need to use the Kinetic Energy equation, however, this equation requires the velocity of the particle wh...

ColmConnolly3D

According to his lecture on the topic, this would mean that the wave function for the electrons in a given atom would have experimentally verifiable solutions that fall into 5 distinct patterns or "lobes". As I understand it, this would be one function that could have 5 general areas of so...

ColmConnolly3D

Are the Hamiltonian functions something that would be unique to each atom, or will all atoms share a set of wave functions?

ColmConnolly3D

I do not believe that there will be any more Audio-Visual Modules, as from Outline 3 onward they are no longer present in the outlines. However, I don't think that this is conclusive it's just my best guess with the information available. Hope this helps!

ColmConnolly3D

I don't know specifically what kind of questions we might be asked on intensity, but it should factor into the wave equations as amplitude. From what I remember from calc, this should be a constant multiplying the rest of the wave equation, but it would be great if someone could give a more mathemat...

ColmConnolly3D

I honestly think one of the best ways to study is to teach others. I have a friend in this class, and we take turns explaining how to do difficult problems to each other in theory, then the other person follows the steps laid out and sees if they got the right answer according to the SSM on sapling....

ColmConnolly3D

The first line of the question tells us that 1 meter will contain 1,650,763.73 waves of the radiation being emitted. We can flip this on its head to say that 1 wave will have a length of 1 meter / 1,650,763.73 or 6.05780211×10^-7m (605.8 nm). Since visible light ranges from about 400-700nm, we know ...

ColmConnolly3D

Just as a fun aside, the closest we have come to making a black body material is a set of vertically aligned carbon nanotubes. These allow EM radiation in which then can bounce around in the tubes and get absorbed by the carbon at near-perfect levels. Their unique structure allowed a Japanese team i...

ColmConnolly3D

My TA told us that using our phone would not be acceptable but that seems to contradict what others are saying in this thread. In any case, I got this webcam https://www.amazon.com/gp/product/B089NWSGDR/ref=ppx_yo_dt_b_asin_title_o01_s00?ie=UTF8&psc=1 used off of amazon for dirt cheap (<$10). I'...

ColmConnolly3D

I found that the Audio-Visual Focus topics really helped me to shake off the rust I felt after not taking chem for a year. Even though the first couple seem a little basic, they are a great way to get your mind thinking about chemistry and get those neural networks firing! Besides that, everything w...

ColmConnolly3D

I would just like to add that stoichiometric coefficients do not necessarily represent the number of moles being used in a problem, but the ratios of the number of moles of various compounds in relation to each other. For example, if I was using electrolysis to separate water into hydrogen and oxyge...

ColmConnolly3D

If you think about it, 3000 only has one numerical value and 3 placeholder digits. For example, you would be just as accurate saying 3 x 10^3 or 3000. 3200 on the other hand would necessitate 3.2 x 10^3 to maintain the same level of accuracy. To some extent, this is just a convention of science as w...

ColmConnolly3D

Good to know, I watched the new lecture I was just thrown off at first because the music at the beginning was the same. Thanks for all the help guys!
Colm

ColmConnolly3D

I think there is some problem with today's lecture then? The only things that show up in my media resources tab are two bruincasted lectures. Is anybody able to watch live?

Colm

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