CHEMISTRY COMMUNITY

Anna Yang 1A

Also I would recommend setting up an appointment if you want to take a longer time to go through problems because sometimes they go through it quickly with you in order to help the other people in office hours.

Anna Yang 1A

Some are ok with 50 by the end of the quarter, but I think others might grade it based on 5 per week.

Anna Yang 1A

I planned out my week to get through one or two of the topics per day and to attend UA sessions.

Anna Yang 1A

It starts at 9:30 but I'm pretty sure the zoom proctoring opens at 9:15 if you want to log in early.

Anna Yang 1A

In 14A, he didn't reuse questions from midterms on the final. However, there were a lot of questions from the textbook.

Anna Yang 1A

For second-order half-life, you can't use 0.693/k. Instead you have to use 1/k[A](initial). When plugging in the half-life time you'll see that they yield different values due to the different equations for the different orders.

Anna Yang 1A

I think it is possible but just like we're not focusing on higher orders than zero, first, and second, fundamentally we're learning the basics and keeping it simpler.

Anna Yang 1A

For this question, I assumed it was a typo and when I used the .030 mol/L instead of the .30 mol/L I got the right answer.

Anna Yang 1A

However, for overall order of the reaction, you would add up the n, m, l values so a reaction with two first orders would have an overall order of 2.

Anna Yang 1A

If it is first order, its n value will be 1. If zero or second, n will be 0 or 2 respectively.

Anna Yang 1A

In the Solutions Manual, at the end it comes out to lnQ=15 and then somehow computes Q=10^6 when an answer between 2.7*10^6 and 3.3*10^6 would be more accurate.

Anna Yang 1A

It only changes in sign if the reaction is flipped. In magnitude it doesn't change as it isn't dependent on how much of the substance there is (or how much of a factor its amount was increased by).

Anna Yang 1A

Textbook problems! I recognized a lot of them on the 14A Final, but Dr. Lavelle has also mentioned that he pulls some questions from the textbook to encourage us to practice the assigned textbook problems. Also the UA workshops have been very helpful especially with thermodynamics and electrochemist...

Anna Yang 1A

I find it easier for me to identify the reduction half-reaction because I always think of reduction as reducing the charge. Also a helpful way to remember the galvanic cell is AN OX and RED CAT, where anode is oxidation and cathode is reduction.

Anna Yang 1A

I got that answer as well and it marked it as correct so your answer should be right?

Anna Yang 1A

Can someone please help me with the balanced equation for: N2H4(g)+ClO3-(aq)⟶NO(g)+Cl-(aq)? I know that Cl is reduced and that N is oxidized, but I'm not sure how to balance it. Ox: 8OH- + N2H4 --> 2NO + 6H2O + 8e- Red: 6e- + 3H2O + ClO3^- --> Cl- + 6OH- LCM for electrons to cancel is 24 so multipl...

Anna Yang 1A

They served as a pretty good example of one for an everyday object if they are closed and insulated.

Anna Yang 1A

Both are equilibrium constants and calculated the same way.

Anna Yang 1A

For the first midterm, lecture and discussions were cancelled on that Friday, so I believe the same will be the case for this Friday.

Anna Yang 1A

Midterm 2 will be proctored via the Zoom links on CCLE under its tab. The final was proctored for 14A.

Anna Yang 1A

I work part-time, but my job has been flexible for me and I've been able to cut down to only weekends so I can attend the extra workshops alongside my lectures. I definitely rely on a planner to keep up with assignments and studying though.

Anna Yang 1A

Does anyone know if Midterm 2 will cover all the material from this quarter we have learned so far? Or only cover what we have learned since Midterm 1?

Anna Yang 1A

Thank you so much!!

Anna Yang 1A

Arielle Sass 2A wrote:Hi! You’re right, the answer at the end of the textbook seems to be incorrect (it says 8 joules) but I believe the solutions manual booklet has the correct answer, which is 28J (for both w and delta U)

Hi where do you find the solutions manual booklet?

Anna Yang 1A

Using the correct gas constant is dependent on the units available and the numbers change as a result of such conversions between different units. Ideally, you want to choose the one with units that match those that are given in your problem so that cancellations can occur. The difference between 8....

Anna Yang 1A

For this, you will actually use heat capacity of the calorimeter, or C(cal). Use the formula: C=q/Change in T. In this scenario, -q(rxn)= q(cal) so q(cal)= 2.03kJ. Divide this q by the change in temp of 3.18C. This will give you the heat capacity of the calorimeter that you need to solve the rest of...

Anna Yang 1A

Putting energy or work into the system is positive as the system is gaining energy. Meanwhile, if the system does work, it loses energy and thus the work is negative.

Anna Yang 1A

Consider the two formulas: q=mC(T final - T initial) and change in H fusion + q ice = -q water. With these, I set up an equation where change in H fusion + mC(T final - T initial) (for the ice)= -(mC(T final- T initial) (for the water). You can calculate change in H fusion by multiplying the mass of...

Anna Yang 1A

So the cut off for Week 5 posts wasn't Wednesday?

Anna Yang 1A

For example, an open system would be a beaker of water without a lid. In this scenario, water can evaporate. In a closed system, the beaker would be sealed and volume would not be lost through that process.

Anna Yang 1A

Try to match it to the units given in the problem as these will cancel to leave you with the correct units for the answer the problem asks for.

Anna Yang 1A

For this question, for Path A, I considered the equation for work=-nRT.ln(Vfinal/Vinitial). Based on this, you need the amount of moles which you can calculate using PV=nRT, and solve for n as n=(PV)/(RT). From here, plug in the values of moles, R(8.314 J/mol.K), K, and the final and initial volumes...

Anna Yang 1A

^Yea once you get your value for w=-P(change in V) just multiply it by 101.325J/L.atm, the conversion factor to cancel out the L and atm and end up with J

Anna Yang 1A

Is it because the standard enthalpy of formation for N2 is 0 because it is in its most stable form?

Anna Yang 1A

This question is asking you to compare the stability of Benzene when using a resonance structure versus using 3 single bonds and 3 double bonds. If you look at the table, there is a C-C bond with also a dotted line and its bond enthalpy is 518 kJ/mol. This bond is representing the bond obtained thro...

Anna Yang 1A

Anna Yakura 2F wrote:
Anna Yang 1A wrote:Does anyone know how to download lockdown browser or if there is a specific link we need for it?

The Respondus lockdown browser download link is on the CCLE website under Site Info! Should be near the very bottom of the page.

Thank you!

Anna Yang 1A

Does anyone know how to download lockdown browser or if there is a specific link we need for it?

Anna Yang 1A

Every single textbook problem he assigns lol. Also UA sessions. I find Matthew's on Wednesdays very helpful for review of concepts before midterms!

Anna Yang 1A

There's also a Week 3 assignment, but its combined with Week 4 and due at the end of Week 4 instead of Week 3.

Anna Yang 1A

Hi! When I solved this problem I referred to the list of strong acids and bases from the textbook. When you look at that, you should find similar parts of the strong acids and bases within the salts listed. If the salt has a part of a strong acid in it (as it was derived from the strong acid), then ...

Anna Yang 1A

I think it was mentioned that if the x value you find is less than 5% of the initial concentration, then the approximation can be used. But since its greater than 5%, you have to use the quadratic as K is not small enough to approximate it as not having an affect on the initial molarity (i.e. cannot...

Anna Yang 1A

On the syllabus it says that the midterm is on a Friday during class time. Is this true because for 14A last quarter, it was during our respective discussion sections?

Anna Yang 1A

Liquids and solids are not included because liquids tend to be in excess and remain virtually unchanged (and therefore cancel out in the numerator and denominator when calculating K) and solids don't have a concentration. Gases are included, and therefore H2O as a gas counts.

Anna Yang 1A

It's just a means of organization to help you better see your known and unknown values and how values find relation to each other (such as x-something is one substance's concentration at equilibrium while x or coefficient*x is another substance's concentration).

Anna Yang 1A

I could be wrong, but I thought Dr. Lavelle had said in lecture that because usually there is so much more H20(l) or solvent than the rest of the reactants that the change is so small it is often not measurable; and therefore, we don't include solids and liquids into the K expression, as they do not...

Anna Yang 1A

In approaching this problem, I would first look at the smallest value of change so that you won't be dealing with fractions in your balanced equation. This smallest value of change would be B, while C is double that. Meanwhile, looking at A, note that it starts halfway between 25 and 30 and ends hal...

Anna Yang 1A

I could be wrong, but I feel like this might be a typo? Because I think it would make more sense if 2HD was by itself on one side and then the H2 and D2 were the products; in this case, it would be balanced and there would be no change because the moles of gas would be equal (both 2) on both sides. ...

Anna Yang 1A

For this one, I utilized two of the K equations given (Ka and Kc), but I think it is more open-ended and there are multiple ways to get to the solution. I also thought of it as substitution and modified Ka and Kc so that when they were multiplied together, they would produce K. From the equation, K=...

Anna Yang 1A

I was having the same problem, but for some reason it ended up working for me when I tried it in Google Chrome instead so maybe try a different browser?

Anna Yang 1A

Since Kc is [products]/[reactants], the value will change depending on which are the products and which are the reactants for the reaction. The reverse reaction will have these values swapped, so therefore, the reciprocal must be used.

Anna Yang 1A

K means equilibrium constant. Kc denotes its specification for concentrations while Kp denotes its specification for gases/pressure.

Anna Yang 1A

Because the equilibrium constant, K is found by dividing concentrations of products by that of reactants, if the larger number is in the denominator (the concentration of reactant) then K will be a smaller value. But I thought this only leads to favoring the reactants when it is so small that it is ...

Anna Yang 1A

Withdraw electron density comes from electronegativity, which can be described as a pulling power of electrons. When one atom has higher electronegativity than another in the same molecule, the one with higher electronegativity will better pull electrons towards it; it would also then be harder to r...

Anna Yang 1A

When a charge is delocalized, it basically means that it is spread out throughout the species rather than restricted to a certain area. This allows for a more stable structure because if a negative charge is restricted to one certain area that area will be negative, whereas if it is spread throughou...

Anna Yang 1A

I'm not sure what's been up with CCLE but the same happened to me. Halfway through Monday's lecture, the video was not able to load. And then today's lecture won't load either. I also tried the videos for slower connections as well but still ran into the same problem.

Anna Yang 1A

Hi, I could be wrong but my TA had mentioned that we might be sent a different Zoom link that the Zoom discussion link and that there would be another grad student proctoring. I'm not sure if they're splitting up into two different Zoom videos or both on the same one though.

Anna Yang 1A

Yea I'm pretty sure Dr. Lavelle said that today's lecture (Wed) is when he'll finish up the last syllabus, and Friday will be a review for the final during lecture.

Anna Yang 1A

I was having trouble submitting course evaluations, and I couldn't figure out why. In case anyone was in the same situation as myself, I was finally able to submit it by refreshing multiple times until a contrast mode box popped up and then clicking on it.

Anna Yang 1A

One way to think of this is that the N in NH2 can form further bonds outside of those shown in the image. In lecture, we had covered a similar example to this where the N in NH2 were both bonded to Co; however, they formed a ring (pentagonal shape) in that they also bonded with C atoms that bonded w...

Anna Yang 1A

At that point, the O has a negative charge after losing a proton (H+). With Cl on the opposite side of the molecule, the high electronegativity of Cl greater pulls the electron density towards it (greater than the C would) and thus giving its side more negativity of charge. Therefore, the negative c...

Anna Yang 1A

Why is H2O a monodentate ligand? I thought it would be at least bidentate.

Anna Yang 1A

For s-block and p-block elements, there are specific charge values for each column. For the transition metals, their oxidation state isn't really specified the same way because their charges vary. In order to determine these, you would need to look at the charges of the other atoms or molecules invo...

Anna Yang 1A

I think Dr. Lavelle had mentioned in his Monday lecture that both are correct, but that we'd be mostly using the -o in this class.

Anna Yang 1A

So all T-shaped molecules have a notation of AX3E2? And ICl3 (from the textbook 2E.11) would be an example of such?

Anna Yang 1A

Dipole moments occur in polar molecules, so because in 3, the dipoles in the C-Cl bonds are equal (as they are the same) and directly opposite to each other, they completely cancel out leaving the molecule non polar. Therefore, 3 is out. In 2, while the dipoles are pointing in different directions, ...

Anna Yang 1A

So because of resonance, a pi bond can be delocalized as a result of the electron that makes up half of it moving around, or being delocalized itself?

Anna Yang 1A

I think another factor that affects whether the orbital is hybridized or not is the structure of the molecule. For example, C in CH4 has 4 hybrid orbitals while C in C2H4 has only 3 hybrid orbitals. This is because while the C in CH4 forms four single bonds and has four regions of electron density, ...

Anna Yang 1A

Electronegativity can be thought of as the pulling power of electrons by an atom. Therefore, those with higher electronegativity have negative poles as these atoms have the stronger pull on electrons and since electrons are then closer to them, they have an overall slightly more negative charge, or ...

Anna Yang 1A

From my understanding, electronegativity basically means the electron pulling power of an atom. If you think about trends of atomic radius on the periodic table, you can remember that as you go across a period, atoms become smaller as they pull electrons in tighter. Thus, Group 17 elements tend to h...

Anna Yang 1A

Dipole moments cancel when they are equal and opposite to each other. When polar bonds are in opposite directions though the bonds are polar themselves, they make the molecule as a whole non polar. CCl4 is an example as such as C-Cl bonds are polar themselves but because of the symmetry of its tetra...

Anna Yang 1A

I'm doing this as well, and I've been worrying about it too. If anything, I'm down for forming a study group for this so all of us can work with each other since we're more familiar with the 14 series than the 20 lol.

Anna Yang 1A

I have a Mac laptop and I just went to Finder then Applications. If you downloaded Respondus it should be there. Once you open it up, there's should be a little globe icon in the top left; if you click on it you'll find a little box that you can click on to check for recent updates. Mine was also up...

Anna Yang 1A

Just an additional comment because I was a bit confused about this, but regarding symmetry, dipole moments are in a sense reliant on shape. CO2 is non polar because of its symmetry as CO2 takes on a linear shape. Meanwhile, a molecule like SO2 would be polar as it takes on a bent shape becoming asym...

Anna Yang 1A

Is H (from the far left of the periodic table) being in a molecule that has an atom from the far right of the periodic table (such as O, F, Cl) a giveaway that there is a dipole-dipole interaction?

Anna Yang 1A

For this problem, first draw out the Lewis structures. What you should end up with is NO2- and CH3 are radicals as for NO2-, N has 5 valence electrons but after double bonding with 2 O atoms, it has one lone electron left and unpaired; therefore, it is a radical, making it highly reactive as it seek...

Anna Yang 1A

I just rewatched Lecture #9 from 10/21 as it had some more information about orbitals, their shapes, and their nodal planes, so I don't think that the information we will be tested on will go beyond the content covered in these two.

Anna Yang 1A

Oh that makes sense. Thanks!

Anna Yang 1A

In the Lewis structure for ONF (part c) I was wondering why N and F don't form a triple bond because if they did, wouldn't N still have 5 valence electrons (10/2 + 0) and F still have 7 valence electrons (6/2 + 4)?

Anna Yang 1A

I could be wrong, but I think the oxidation number is found from the number of electrons gained or lost in order for a certain atom to form a bond with another atom (I think of it as how many electrons it needs to gain or lose to have a complete shell). For example, oxygen has an oxidation number of...

Anna Yang 1A

For this problem how do you figure out the formulas for the nitrate and nitrite, etc ions? Or is it something we are supposed to memorize?

Anna Yang 1A

How do we know how many valence electrons the transition metals have?

Anna Yang 1A

I believe Dr. Lavelle had mentioned in lecture that this is based upon the general principle that unpaired electrons lead to higher reactivity (kind of like in ionization energy where group 17 are the most reactive because they only need one more electron to become stable) as they seek to complete t...

Anna Yang 1A

I believe closer proximity to the nucleus is linked to better shielding as the orbitals closer to the nucleus are able to shield the ones further away; thus the s-orbital shields the p, d, and f while the p shields the d and f and so on. Therefore, the d orbital would not be better at shielding than...

Anna Yang 1A

I think it is due to the overlap as the energy levels increase and become closer and closer to each other. When this happens, if you look at a visual you can see that the 4s shell is slightly lower in energy than the 3d shell, and thus, the 4s shell is filled before the 3d. I believe in lecture, it ...

Anna Yang 1A

Why is it that the d shell starts at the 3d when it is in the fourth row of the Periodic Table? Looking at the electron configuration for silver, it lists [Kr] which is at the end of the fourth row but then continues on to list 4d so I was thinking that might be affected by the d shell starting at 3...

Anna Yang 1A

As atoms increase in size (have more shells) going down a group, the electrons in outer shells are further and further away from the nucleus and are more shielded by the inner electrons. Therefore, they are not held as tightly in the atom and are more susceptible to being removed.

Anna Yang 1A

I had been a bit confused about this as well. I'm pretty sure that they both follow the same trends of increasing across a period and decreasing down a group because for ionization energy if the atom has only one electron in its outer shell, it has very low ionization energy and, therefore, it is ea...

Anna Yang 1A

Oh that makes sense. Thank you!

Anna Yang 1A

Just to clarify about this topic, I believe it was mentioned in today's lecture that ionization energy and electron affinity are alike to opposites of each other, yet both increase across a period and decrease down a group?

Anna Yang 1A

The empirical formula is the lowest ratio of the molecule's composition using whole numbers, so just make sure that the numbers of each atom in the molecular formula you end up with can't be divided any further without having a fraction. For these kinds of problems where all of the masses for the pr...

Anna Yang 1A

I've been a bit confused on how the point distribution works throughout the test... I think he said that some questions are worth more points than others, so what would be the base value of points for each question, if it even works like that?

Anna Yang 1A

Within the second shell (n=2), the shape (l) can take on either a p- or s-orbital. Since l=1 in this instance, it corresponds with the p-orbital, which has the possibilities for the value of ml being -1, 0- or 1. With these, px is interchangeable with -1, py with 0, and pz with +1. So in this case, ...

Anna Yang 1A

And just as a little side note, part of this problem is knowing which wavelengths correspond with their respective electromagnetic radiation. For the midterm, I'm pretty sure that we have to memorize at least the ranges for UV, visible light, and infrared instead of having the spectrum provided, and...

Anna Yang 1A

For the spectral lines, you would just do n(final)-n(initial) so that would be 5 (since it was excited up to this shell) - 1 (the ground state). And then as for the Rydberg equation portion for finding the wavelength range because the problem does not provide which energy level the electron ended up...

Anna Yang 1A

As mentioned above, light travels at 3.0x10^8 m/s, which is symbolized through c. Because wavelength times frequency is supposed to provide this value, c, they are inversely related to each other. Thus, a longer wavelength and lower frequency (as with infrared) would provide the same value as a shor...

Anna Yang 1A

A helpful way to think about this is to remember the experiment where incident light was passed through a barrier with two slits and the diffraction pattern was revealed on the other side. Here light acted like a wave because it was able to pass through the barrier and revealed diffraction patterns ...

Anna Yang 1A

I'm pretty sure that they used the formula: speed of light constant/wavelength = frequency, with the speed of light constant being 3.0x10^8

Anna Yang 1A

Continuing on with the above responses, if electrons acted like particles within the experiment where they were passed through a crystal, they would not have necessarily passed through it but would have bounced off of it, much like how Dr. Lavelle compared particles bouncing off the barrier to waves...

Anna Yang 1A

And as just some more discussion on the topic of atomic spectroscopy: as the person mentioned above, the ultraviolet wavelengths belong to the Lyman series. Keep in mind that the ultraviolet wavelengths are going to have a wavelength that is less than 400 nm (in this case 102.6 nm is less than 400 n...

Anna Yang 1A

Because within the photoelectric effect photons act like packets of energy, the particle model must be utilized as opposed to classical mechanic's wave model. In the context of the wave model, increasing the intensity of the light has to do with increasing the amplitude. However, when experimenting,...

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