CHEMISTRY COMMUNITY

Lucy_Balish_3G

I couldn't think of any way to solve it unless we are given more information. We either need values for the enthalpy and entropy or the values for Gibbs free energy of formation for the products and reactants. Depending on what we are given, we take one of those approaches to the problem.

Lucy_Balish_3G

The reaction order also tells us what units we will expect for k as well as how much the rate will change when the concentration of one of the products is changed.

Lucy_Balish_3G

Think about how the reaction with the lower Ea can proceed easier without an increase in temperature because it takes less energy than the reaction with the higher Ea that cannot. So when you increase the temperature, it makes much more of a difference to the reaction with the higher Ea because it w...

Lucy_Balish_3G

We did use this equation, but since Mg2+(aq) + 2e- -> Mg (s) has a lower reduction potential, it is going to be oxidized meaning it is the anode. So when we subtract the values: -1.66V - (-2.36V), the sign changes because we are subtracting a negative number. So it ends up being -1.66+2.36 which giv...

Lucy_Balish_3G

I'm not quite sure, but regardless of its state, the AgBr still has an Ag+ ion but the ion is solid. So in this case the ion is solid, but regardless is an ion.

Lucy_Balish_3G

Since in the example you mentioned the MnO4- and Mn2+ are both part of the cathode so they are written on the same side, and because they are both in the same phase (aq) they are separated by a comma. Because neither of them is a solid, we see the need to write Pt(s) in the reaction as well in order...

Lucy_Balish_3G

My TA had this helpful overview

• Salt bridge: Two vertical lines (||)
• Different phase: One vertical line (I)
• Same phase: One coma (,)

Diagram order:
Inert Metal – Anode – salt bridge - Cathode – Inert Metal

Lucy_Balish_3G

Yes, that is the definition of a first oder reaction because the n value has to be one or else it is not first order.

Lucy_Balish_3G

No, because the cell potential is the potential an electron experiences, the coefficients of the equation have no effect on this.

Lucy_Balish_3G

For example, if you have Au +(aq) --> Au(s) + Au 3+(aq), you can see how Au+ is both reduced to Au(s) because it gains an electron but also is oxidized to Au3+ because it loses 2 electrons. Both of these instances occur in the same reaction.

Lucy_Balish_3G

Since corrosion happens when a substance is oxidized or exposed to oxygen and it loses electrons and gets smaller, it helps to put a surface in between the substance and the surroundings so that the air will not oxidize the substance. A common example is paint, but for a more chemical approach you c...

Lucy_Balish_3G

Yes, even if the cell itself might have the two flipped, you always put the anode on the left and the cathode on the right to show the order the electrons are flowing.

Lucy_Balish_3G

This essentially means that you write in order of what touches what. In other words, in this case, the solid electrode is touching the crystals that are forming on the metal, and those are touching the aqueous solution ions. So for example you might put Pb(s)|PbCl2(s)|Cl-(aq) since from the diagram,...

Lucy_Balish_3G

The general outline is that the anode is always placed on the left side, and the cathode is placed on the right side. The salt bridge is represented by double vertical lines. So it will look like : Electrode material being oxidized (anode) | aqueous species || Aqueous species | Electrode material be...

Lucy_Balish_3G

Yes, you are allowed to have leftover OH- and H+. Which one you have is determined by if it is in acidic or basic solution, and to convert from acidic to basic you just add OH- which becomes H2O on one side and then OH- on the other indicating basicity.

Lucy_Balish_3G

Since delta S = nR*ln(V2/V1) and pressure and volume are inverses in the ideal gas law equation, you can substitute the inverse for pressure into the equation which gives delta S = nR*ln(P1/P2).

Lucy_Balish_3G

We use this equation (delta U = 3/2 nR delta T) when we are solving for the total internal energy of an ideal gas. But if we are trying to find the internal energy for any other system that is not an ideal gas, we use delta U = w + q.

Lucy_Balish_3G

Detla G naught implies standard conditions (25 C, 1atm) but if that is not the case then you would just refer to it as delta G.

Lucy_Balish_3G

Container A is the highest because it has the most particles because that one mol is monoatomic so there are 6.022x10^23 atoms in that container which makes it higher than B and C who are both diatomic meaning there are 3.011x10^23 atoms in each. But since C is vibrationally active, it is higher tha...

Lucy_Balish_3G

To add on to what was already said, the BF3 molecule only has one appearance regardless of how you draw it, it will always look the same. But if you think about the possibilities for COF2, the orientation of the O and 2 Fs around the carbon has more options which means greater disorder even though t...

Lucy_Balish_3G

Every time you see the ° that means the system is standard (25 C or 298 K and 1 atm) but whenever the system has a different temperature or pressure than you just use G not G°.

Lucy_Balish_3G

Yes, to add on to what others have said entropy of the universe is the same as entropy of the system plus entropy of the surroundings which is the total entropy.

Lucy_Balish_3G

Yes, that was the same reasoning I used when solving this. Basically when it's lost its negative and the colder temp one (200K) gains it so it is positive.

Lucy_Balish_3G

From what I remember, the temperature of which he spoke was the temperature the surroundings approach after the reaction goes to completion.

Lucy_Balish_3G

The best way for me to make sure I pay attention is to watch the lecture and write down not just what is on the slides but more so the things he says because that is what really helps understand. Forcing yourself to copy the material while listening to it is a really good way to solidify information.

Lucy_Balish_3G

I agree with what was stated above, the V can never be 0 because all matter takes up space, but the delta V can be 0 if the system is isolated.

Lucy_Balish_3G

The work function is negative because the system is doing work, using energy, and losing it to the surroundings. So the work function for the system will be negative if the gas expands. But if the gas is compressed, work is done on the system (not by the system itself) so it will be a positive value.

Lucy_Balish_3G

Basically, the system responds to a change in pressure to match the pressure of the surroundings. So yes, the two will be equal but only if the system adjusts to the surroundings, even if infinitesimally.

Lucy_Balish_3G

I agree, I don't think it does.

Lucy_Balish_3G

In Lavelle's lecture from today (2/3), he talks about a 2 particle system with 4 microstates and says the degeneracy would be W=4=2^2. I'm a little confused by this, because in the discussion above, it says that W=X^n, where X is the number of microstates and n is the number of particles. So, would...

Lucy_Balish_3G

Adding on, to convert between them is super simple if you are given one but need the other. Kp=Kc(RT)^(delta_n) where delta_n is the change in the number of moles (moles products - moles reactants)

Lucy_Balish_3G

Adding on, to convert between them is super simple if you are given one but need the other. Kp=Kc(RT)^(delta_n) where delta_n is the change in the number of moles (moles products - moles reactants)

Lucy_Balish_3G

We can assume x is insignificant when the equilibrium constant is less that 10^-4.

Lucy_Balish_3G

To add on to what everyone else has said, when you look at a phase change graph, each new state is represented as a line with a new slope. So when the slope of the line changes, the phase is changing for that substance.

Lucy_Balish_3G

Just study lectures from weeks 1-3, the cutoff is Friday the 22nd.

Lucy_Balish_3G

I am taking it right now with 14B and it's really not too hard. The online format involves a lot of simulations but I feel as though I understand the ideas even though we don't get the "hands on" aspect.

Lucy_Balish_3G

So basically the temperature does not change during the phase change but after the liquid is completely switched to vapor and more heat is still added then the temperature changes? You can see this by looking at the slope of the heating curve graph. During phase changes, the slope of the graph is f...

Lucy_Balish_3G

Since we are focused on the donation or acceptance of protons (not electrons) at least with regard to pH and Ka calculations, I would assume that yes, we are more focused on Bronsted-Lowry which donate and accept protons instead of Lewis which donate and accept electrons.

Lucy_Balish_3G

The change for products is negative because they are going to form reactants. Because reactants are being formed, their C is positive.

Lucy_Balish_3G

Since K is the equilibrium constant at a certain temperature, changing the temperature changes K to a new value for that new temperature.

Lucy_Balish_3G

When a substance is heated, the heat has to go somewhere. So for phase changes, the heat goes into breaking the bond. We might think at first that heating raises its temperature, but that is actually not true because all the heat is going into breaking the bonds.

Lucy_Balish_3G

The most helpful for me was to think Kw constant allows us to find hydronium ion concentration or hydroxide concentration when only one is given.

Lucy_Balish_3G

It would be an ice table but instead of starting with 0 for your initial concentration of products, you start with .2M. Then you add to the left and subtract from the right and solve as you would normally! Don’t forget that you only ever include aqueous and gaseous in your equilibrium constant.

Lucy_Balish_3G

Be sure you are using 2x and also remembering to square the concentration of NO!! I forgot that the first time around.

Lucy_Balish_3G

The acids and bases we learned in 14A are those typically suggested to memorize and have a solid understanding of. I would say that knowing the more common ones can only help you!

Lucy_Balish_3G

Instead of equating the two (pKa and pH), the main idea is that they have the same relationship as each other with regard to pKb and pOH respectively.

Lucy_Balish_3G

If you want a full explanation for this question, the chemical equilibrium part 4 video on Lavelle's website was helpful and describes the relationship between temperature increase, product and reactant increase, and the value of Q. It starts at 37:50 and the link is https://lavelle.chem.ucla.edu/wp...

Lucy_Balish_3G

To set up this problem, you need to make an ICE table so for initial conditions BrCl=1.84 x 10^-4 M Br2=0M and CL2=0 M. Then, you know that 18.3% of BrCl gas remains at equilibrium, so you can find the equilibrium concentration of BrCl by multiplying its initial concentration by 0.183. Doing so you ...

Lucy_Balish_3G

Only include aqueous and gaseous substances in an ICE table, never a liquid or a solid. So just make sure you double check which phase it is in.

Lucy_Balish_3G

These are some useful tables to keep track of R values and units given.

Lucy_Balish_3G

When you get two values for X, ignore a value that is negative as well as a value that is greater than the initial reactant concentration or pressure. In other words, the value of X should be positive and smaller than the initial concentrations on the left side of the reaction.

Lucy_Balish_3G

I think you are correct that it is sp3. As has 5 valence electrons and F has 7. So 5+7x3= 26 electrons. In general, you can think of having a central atom and know that you only need to pair three atoms around it for this problem because there are only 3 Fs. So knowing that the valence for each F wi...

Lucy_Balish_3G

Angular and Bent refer to the same molecular geometry and can be used interchangeably.

Lucy_Balish_3G

Rachel Kho Disc 1I wrote:Wait so does this also mean that tetrahedral molecules are usually polar? Are there any examples of nonpolar tetrahedral molecules?

CCl4 is an example of a molecule that is tetrahedral and nonpolar.

Lucy_Balish_3G

It helps to identify when the molecule is not symmetrical, because this usually means it has a dipole moment. To see where the arrow points (towards the more negative) you have to think about which element is more electronegative. Hope this helps! https://www.chemtopper.com/myblog/wp-content/uploads...

Lucy_Balish_3G

There's typically at least 2 or 3 textbook problems so it helps to make sure you understand them. But also many of the problems are written by the TA's too so it's a balance.

Lucy_Balish_3G

Something I find helpful for conceptual questions is going back through those learning outcomes sheets on the chem 14A Lavelle website and making sure I not only understand but can explain the list of concepts on the bottom and actually have a grasp of what they mean and how they can be applied.

Lucy_Balish_3G

The order of the ligands does not really matter because they will be alphabetized when naming the compound anyways. As long as they come after the transition metal cation (central atom) then you should be okay.

Lucy_Balish_3G

en stands for ethylenediamine, is a bidentate ligand (involves 2 electron pairs), is a neutral molecule, and has the molecular formula NH2CH2CH2NH2.

Lucy_Balish_3G

I heard from someone in 14B and BL currently that with classes online, it is not as challenging as it might normally be to balance both, so if you are up for the workload and have a solid foundation in chemistry she suggested to go for it. But honestly, most people take BL with 14C so you would be t...

Lucy_Balish_3G

I think memorizing where metalloids are located is the most important because if you know where metalloids are, you also know that metals are to the left of it and nonmetals are to the right. The way I memorized the location of metalloids on the periodic table is that I imagine a mental staircase t...

Lucy_Balish_3G

I have found that the textbook simply gives too much information and the lectures are more to the point of what we need to know. It is also too time consuming to try to learn from the book and from the lectures, so I typically watch lectures and either check khan academy or consult the book if I am ...

Lucy_Balish_3G

I agree, pretty sure we are expected to memorize these shapes and know their names, how to draw them, and the bond angles.

Lucy_Balish_3G

Since each carbon is bonded to three other atoms (double to carbon, single to hydrogen, and a single to hydrogen) it would be sp2 hybridization.

Lucy_Balish_3G

Since electron affinity is based on how much each electron wants to achieve an a stable octet, noble gasses with a stable octet therefore do not desire to gain or lose electrons.

Lucy_Balish_3G

I found this table to be really useful for studying shapes and angles. https://8dd969e7-a-ff053077-s-sites.googlegroups.com/a/coe.edu/principles-of-structural-chemistry/chemistry-and-climate-change/molecular-shapes/CD%20Spectrum.png?attachauth=ANoY7crF6LZuLn5XS4eaA7LRqj1A37qe5pqikDjjrIUlsarDWzmRkpK...

Lucy_Balish_3G

Because CO2 has two double bonds, there is space for the molecule to accept electrons and shift them to the O, breaking the double bond and replacing it with a single bond. In this manner, CO2 can act as an electron acceptor and bond to an electron rich atom or molecule.

Lucy_Balish_3G

If two H2SeO4 molecules were to interact, hydrogen bonds would take place between the hydrogens in one molecule (which are bound to an O) and the oxygen in the other which gives rise to hydrogen bonds.

Lucy_Balish_3G

It's fundamentally the same idea (how the charge on one atom affects the charge on another) just different terms because the atoms have opposite charges. For cations with a positive charge, their ability to distort an anion and skew the electron cloud is known as their polarization power. The tenden...

Lucy_Balish_3G

The oxygen anion has a charge of 2-, not 1-, so you would add two electrons and get 2p^6. Maybe the 1- charge was a typo, because when oxygen forms an anion it carries a 2- charge.

Lucy_Balish_3G

I think the main point of Schrodinger's wave function equation is how it helps describe atomic orbitals. I remember it being a double derivative but other than that I'm pretty sure there was nothing else calculation-based he went over.

Lucy_Balish_3G

Being more polarizable is tied to having more electrons, so the bigger the atom (the further down the periodic table) the more electrons it has. Consequently, it has greater interaction potential energy because the charges are greater so with more interaction attraction energy, the larger molecules ...

Lucy_Balish_3G

To determine the number of unpaired electrons, you have to draw the electron configuration for each ion (adding or subtracting electrons depending on the charge) and then filling in each subshell, spin up then spin down. If ever you are left with an subshell that is not full where there are not enou...

Lucy_Balish_3G

Negative energies imply that energy is given off by what is taking place. In this case, the more negative energy forces for ion-ion attractions and hydrogen bonds are more favorable and give off more energy than dipole-dipole and dipole induced-dipole. The reverse is true that a greater energy must ...

Lucy_Balish_3G

No, we do not have to memorize electronegativity values. They will be given to us in a table when we need to perform certain calculations. However, you should familiarize yourself with the electronegativity trends and know that F, O, and Cl are highest.

Lucy_Balish_3G

From my understanding, the calculation for a dipole moment involves multiplying the charge by the distance between atoms. So the longer the bond length (ie. single bond), the greater the dipole value will be. A shorted bond length (double or single bond) will have a smaller value for distance betwee...

Lucy_Balish_3G

You should always put the lone pairs on the central atom when you are drawing a lewis structure. That way you can ensure that the outer electrons have filled valence shells with a low formal charge and consequently know you are finding the most stable configuration.

Lucy_Balish_3G

What helped me most understand shielding was the example of people standing around a fire pit. Those closest to it block you from some of the heat, but you can still feel the warmth of it even if there are people standing between you and the fire. So because s-orbital electrons are closer to the nuc...

Lucy_Balish_3G

To add on to Nika's explanation, by adding two double bonds, the formal charges lowered for 4 of the 6 atoms which we learned today is much more stable. When drawing Lewis structures, the one with the lower formal charges is a better representation of the molecule.

Lucy_Balish_3G

Delocalized electrons means that the electrons in a molecule are not paired to just one atom. They are able to move freely because they are shared equally between different atoms.

Lucy_Balish_3G

The extra electron added at the end is due to the overall -1 charge of the nitrate ion (NO3-). Nitrate is one of the polyatomic ions and is pretty common, so I would say you should be familiar with the formula and the overall charge.

Lucy_Balish_3G

You can find the molar mass of a Helium atom on the periodic table: 4.002 g/mol. Since we are using Joules with Planck's constant, convert grams to kilograms, so 4.002 x 10^-3 kg. And make sure to multiply by Avogadro's number so 6.002 x 10^23 He atoms/mol in this question so that you get the value...

Lucy_Balish_3G

The photons are essentially packets of light energy that help the electrons move between energy levels. To move up in energy level, an electron must absorb a specific amount of energy in the form of a photon. In the same way, to drop to a lower energy level, it must lose a specific amount of energy,...

Lucy_Balish_3G

To add on to Alex's response, it is most stable to have a full electron shell and second most stable to have a half filled shell. It is least stable to have almost filled levels. Thus having a full d shell (10d) and a half full s shell (1s) is a more stable configuration for Ag (as opposed to an alm...

Lucy_Balish_3G

E=pv comes from a simple combination of units. Knowing that E is in J which is kg*m^2/s^2, you can multiply momentum (kg*m/s) by velocity (m/s) to get kg*m^2/s^2. E=pc comes from knowing the velocity for a photon is the speed of light. So you essentially input C as the velocity for a photon because ...

Lucy_Balish_3G

Jacquelyn Challis 1H wrote:Regarding the midterm, can I use a graphing calculator or does it need to be scientific? I can't remember if Lavelle told us this info already or not.

I believe you can use a graphing calculator. If you have a TI-84 or TI-89 you should be fine.

Lucy_Balish_3G

It is necessary to have 2 detectors to ensure you have both final and initial values so that you can calculate the total distance travelled as well as how long it took. Distance/time = velocity.

Lucy_Balish_3G

They are all still in order and do a good job of paralleling his lectures. As Dr. Lavelle mentioned today in his lecture, we only need to know how to solve up to question 19 for the midterm. We still have to complete the rest of the assignment but it is not due until after the midterm, so complete t...

Lucy_Balish_3G

From what I recall, that is more of a physics concept so we really don't need to know much more besides that. I think you are all set with the knowledge that black body radiation absorbs light frequencies, so don't worry about it too much!

Lucy_Balish_3G

For me, I start by watching the lectures and making sure to attend them around the time when we normally would if they were live. This helps me stay on top of watching the videos and also feel like I have to still "go to class" and be accountable for my participation. During lecture I take...

Lucy_Balish_3G

Once atoms form molecules, they have a specific energy transitions and wavelengths that it absorbs which lets it have a unique spectral fingerprint as well. Hope this helps! Just like what Sophie said, both atoms and molecules have a unique "fingerprint" which can be depicted through spec...

Lucy_Balish_3G

The Paschen series is what you are asking about. This series includes transitions from n >= 4 to n = 3 state.

Lucy_Balish_3G

The wave model predicts that the amplitude of the wave (the intensity) determines how much energy the wave carries. But this classical model does not apply to light. It can explain its wavelike appearance properties but does not accurately explain its energy properties. This is because the photoelec...

Lucy_Balish_3G

A good way to think about this is that the energy levels have to match up. If there is not enough energy, it will not match the energy needed to jump the gap, so it will pass through because it is not useful to the atom. But when the energy is enough and matches what is required, the atom will absor...

Lucy_Balish_3G

We learned today that increasing the intensity/amplitude of a light (making it brighter) does not have any effect on the electrons emitted. This is because the energy required to emit an electron has not been met. To meet this required energy, a specific photon with a high frequency is necessary. To...

Lucy_Balish_3G

I once heard someone describe quanta as pennies of the quantum world. If you want to transfer energy, quanta are the smallest units that you can do so in. This analogy is super helpful! Thank you for sharing because it makes something so abstract seem as tangible as everyday coins. Definitely incre...

Lucy_Balish_3G

When doing a dilution, you usually pour more water in or pour a more concentrated solution in. You don't usually pour more solute in. For instance, if I had a bowl with 10 blueberries and 10 strawberries, I could reduce the ratio of blueberries by putting more strawberries. I didn't change the amou...

Lucy_Balish_3G

Yes, I agree. That was super helpful!! For me the way I try to remember is "precision means consistent" and "accuracy means correct".

Lucy_Balish_3G

The empirical formula is always the most basic, simplified ratio. Thus, the molecular formula will either be the same or a multiple of that base ratio depending on the molar mass.

Lucy_Balish_3G

Thank you so much for your encouraging advice. I appreciate you taking your time to post this, and now I am even more motivated to take notes during lectures and discussions. Best of luck in Med School!!

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