CHEMISTRY COMMUNITY

Jiyoon_Hwang_2I

Daniel Chen 2L wrote:But doesn't the concentrations of the reactants and products in a concentration cell have to be different in order to generate voltage?

I think in a concentration cell the two electrodes are composed of the same solution, but they differ in concentration so voltage will be generated.

Jiyoon_Hwang_2I

So adding h2o, h3o+, and oh- balances the charge and elements or do you then have to add certain elements? Are you limited to just those 3? If elements other than H and O need to be balanced in the equation you should use coefficients, but when you balance a redox reaction the only things being add...

Jiyoon_Hwang_2I

ursulavictorino1K wrote:So an aqueous solution isn't considered a liquid?

No, only pure liquids (l) and solids (s) are exempt from calculating k. Aqueous solutions describe a solution or mixture containing a solute dissolved in a solvent which is water.

Jiyoon_Hwang_2I

You should calculate Q and K and if Q < K then the reaction is product favored and will shift to the right

Jiyoon_Hwang_2I

I think if the amount of water that is added is given then you can use that value to recalculate the concentration of the solution the water is added to. If the concentration of the solution is changed then there will be a change in the voltage

Jiyoon_Hwang_2I

ursulavictorino1K wrote:If they are always given as recutions, How do you know which to flip to an oxidizer?

If they give you the cell potentials of each reaction, the reaction with the lower cell potential is the one that is oxidized so it will be flipped

Jiyoon_Hwang_2I

If they provide you with the cell potentials of each reaction, the reaction with the lower cell potential is the one that is oxidized.

Jiyoon_Hwang_2I

They both solve for the same delta G. You could also set these two equations equal to each other to derive the Van't Hoff Equation

Jiyoon_Hwang_2I

Ecell is the cell potential, whereas Eocell is the standard cell potential which is measured at standard conditions, such as 1 atm, 25 degrees celsius

Jiyoon_Hwang_2I

Usually in the problems they give the equations as gaining electrons (so the electrons are added to the left side of the equation), but anode's are oxidized which means electrons are lost, so the electrons should be on the right side of the equation which is why they flip the anode reaction.

Jiyoon_Hwang_2I

Changing the concentration of one solution in an electrode may affect the voltage of a cell

Jiyoon_Hwang_2I

Surface area does not affect the voltage of a cell because the voltage is affected by the electrodes, not the amount of material

Jiyoon_Hwang_2I

Does the inert metal always have to be platinum? How do you know when to use platinum or any other inert metal?

Jiyoon_Hwang_2I

I think he'll post final grades this Thursday because he said he would post Chemistry Community grades on Thursday.

Jiyoon_Hwang_2I

What if we live out of state and still haven't gotten our Test 2 back...?

Jiyoon_Hwang_2I

^^ I agree. The fact that he said he would upload chemistry community grades on thursday I'm assuming means he's going to upload all the other grades too. Especially since the final was multiple choice it shouldn't take long for him to grade and upload

Jiyoon_Hwang_2I

He said he's going to take the number of posts you have by wednesday and you should see your chemistry community grade up by thursday!!

Jiyoon_Hwang_2I

Because if pressure is not changing the only thing changing is volume

Jiyoon_Hwang_2I

Galvanic and voltaic cells are the same. The names are interchangeable

Jiyoon_Hwang_2I

E is an intensive property so it doesn’t depend on the amount of substances involved so multiplying the reaction equation does not multiply E

Jiyoon_Hwang_2I

The area under the curve usually represent work

Jiyoon_Hwang_2I

Yes. I think it should have been x10^4

Jiyoon_Hwang_2I

You should use the Arrhenius Equation when determining the effect of a change in temperature on the rate constant

Jiyoon_Hwang_2I

I think you just have to know that Q = 0 for adiabatic processes so when you use any thermodynamics equation set Q equal to zero and solve

Jiyoon_Hwang_2I

You should use the two equations for delta G and set them equal to each other. So set delta G = delta G(standard) + RTlnQ equal to delta G=-nFE

Jiyoon_Hwang_2I

Yes! The coefficients represent the power the concentrations are raised to

Jiyoon_Hwang_2I

As temperature increases the rate constant should increase

Jiyoon_Hwang_2I

The cathode is always the negative electrode

Jiyoon_Hwang_2I

Thank you for a wonderful quarter! I hope you are able to get some rest over break.

Jiyoon_Hwang_2I

For steady state, the first step of the reaction must be slower than the second step. On the other hand, for pre-equilibrium, the first step must be faster than the second step. how do you know which step is the fast step and which is the slow step Im not exactly sure either but sometimes they will...

Jiyoon_Hwang_2I

Weak acids have higher pH because they’re essentially closer to 7 (neutral pH) and strong acids are on the farther side of the pH spectrum which is why they have a lower pH

Jiyoon_Hwang_2I

I think when you use the equation E cell = E cathode - E anode you don't use the flipped cell potentials. Instead, you use the cell potentials that were given in the table so it would be E = 1.23 - 0.80 = 0.43. Otherwise you can add 1.23 + (-0.80) which is also 0.43

Jiyoon_Hwang_2I

But you should be careful of whether they're asking you to balance the equation in acidic or basic conditions because the way you balance H is different for both conditions

Jiyoon_Hwang_2I

When solving for the solubility product, it is similar to finding the equilibrium constant, k, but only the concentration of the ions are included in the solubility product

Jiyoon_Hwang_2I

Thank you so much Dr. Lavelle!! I am so so soooo glad I was able to take both Chem 14A and Chem 14B with you! I really appreciated your quick email updates and the effort you put in to allow your students to succeed in the course. I'm sure this situation with finals becoming online was stressful for...

Jiyoon_Hwang_2I

The internal energy is 0 because isothermal means the temperature is constant

Jiyoon_Hwang_2I

Its because some of the concentrations are squared or cubed so the units are also squared/cubed

Jiyoon_Hwang_2I

In a pseudo rate law you assume that the other reactants are in large excess and you measure the isolated reactant

Jiyoon_Hwang_2I

He actually just sent an email out saying that it will be on CCLE. Theres going to be 6 questions with multiple parts and each question will be about 5 points and the final is a total of 180 points

Jiyoon_Hwang_2I

I think n is 1 electron instead of 2 because the question asked for the Ka of HF and not 2HF

Jiyoon_Hwang_2I

I don't think we'll get partial credit for our work. I think the final might be short answer so you fill in the answer instead of it being multiple choice, but there may be some multiple choice questions like there were on the midterm.

Jiyoon_Hwang_2I

How are we showing our work if the final is on CCLE?

Jiyoon_Hwang_2I

Can someone explain why E=0 for question 5?

Jiyoon_Hwang_2I

Everything is worth 50 points except the midterm which is worth 120 points and the final which is worth 180 points so the entire course is 500 points

Jiyoon_Hwang_2I

The TAs dont even get to see the answer key until about one hour before we take the test.

Jiyoon_Hwang_2I

For steady state, the first step of the reaction must be slower than the second step. On the other hand, for pre-equilibrium, the first step must be faster than the second step.

Jiyoon_Hwang_2I

Catalysts don't affect the equilibrium because they speed up the forward and reverse reaction rates equally.

Jiyoon_Hwang_2I

I am not sure if there will be a lot of questions on it, but I would still study it sufficiently because 14A's final did not have a lot on acids and bases so it might be possible that 14B has more.

Jiyoon_Hwang_2I

A positive standard reduction potential represents a spontaneous reduction reaction.

Jiyoon_Hwang_2I

Changing the concentration of a solution will increase the concentration gradient between the two solutions, causing the voltage potential to increase. Since the two solutions are further away from equilibrium, Le Chatelier's principle will be implemented to bring the two solutions back to equilibri...

Jiyoon_Hwang_2I

The purpose of the salt bridge is to maintain the neutral charge in the solutions as electrons flow between the two solutions.

Jiyoon_Hwang_2I

Gibbs Free Energy tells you the maximum amount of work that can be done under a system with constant temperature and pressure.

Jiyoon_Hwang_2I

Solids and liquids are pure and they don't affect the reactant amount so they are omitted from the equilibrium expression. Also, the concentrations of these pure solids and liquids are calculated by dividing their density by molar mass and their density remains constant regardless of the amount of p...

Jiyoon_Hwang_2I

If the questions provides you with initial concentrations and asks you to solve for the equilibrium compositions you should use the ICE table.

Jiyoon_Hwang_2I

Yes! It is the same textbook

Jiyoon_Hwang_2I

I think you use the concentrations instead of the partial pressures because the given value of K is Kc and they are asking you for Qc so you should keep it consistent by using concentrations instead of partial pressures.

Jiyoon_Hwang_2I

I think it is because Br- is a conjugate base of HBr which is a strong acid therefore Br- is a weak base and weak bases do not affect the pH and K+ is a conjugate acid of strong base KOH so it will also not affect the pH

Jiyoon_Hwang_2I

I don't think so. We technically only need to know up to 6C and one section in 6D.3 and one in 6E.1

Jiyoon_Hwang_2I

I think the only one he talked about was cisplatin

Jiyoon_Hwang_2I

I think it's because they're using the log rule that when you do log x - log y you can also write it as log (x/y) and in this problem they were finding the change in pH so they were doing log(final) - log(initial). So, the final would be 0.12 x [HCl]o and the initial is just [HCl]o

Jiyoon_Hwang_2I

It is platinate because the coordination compound [Pt(Cl3)(NH3)] has an overall -1 charge and when there is a negative charge on the coordination compound, the metal name changes to -ate at the end.

Jiyoon_Hwang_2I

For strong bases you should know that group 1 and group 2 oxides and hydroxides from strong bases. For strong acids HI, HCl, HBr, HClO4, HClO3, H2SO4, HNO3

Jiyoon_Hwang_2I

In order for a ligand to be polydentate it must have on or more lone pairs because it has to have a lone pair to donate

Jiyoon_Hwang_2I

You can identify if a molecule is basic if it accepts a proton (H+) and if it is acidic it will lose/donate one of its protons. On the other hand, it can be amphoteric if it can both accept and donate a proton. So, H2O is amphoteric because it can accept a proton to become H3O+ or it can lose a prot...

Jiyoon_Hwang_2I

You should know the main strong acids, such as HI, HCl, HBr, HNO3, etc

Jiyoon_Hwang_2I

EDTA is ethylenediaminetetraaceto and it is a chelate that removes metals from solutions. For example, edta may be prescribed to remove toxic metals, such as lead, from the blood.

Jiyoon_Hwang_2I

I think it just helps you to picture the shape of the molecule, including the number of lone pair electrons to determine the shape. I am not 100% sure if it serves any other purpose, but typically I think its just used to determine the shape.

Jiyoon_Hwang_2I

If the coordination compound has 6 ligands then it will be an octahedral complex, but not all of them have 6 ligands. Some can have 4 ligands which means it can be a tetrahedral complex or square-planar complex.

Jiyoon_Hwang_2I

A coordination compound is formed when a ligand (a molecule or ion) forms coordinate covalent bonds with the central atom which is usually a transition metal. The coordination number of the coordination compound is determined by the number of bonds the central atom forms with the ligands. For exampl...

Jiyoon_Hwang_2I

Ligands are ions or molecules that are attached to the central metal atom. They usually have one or more lone pair electrons that bind to the central atom to form a coordinate covalent bond.

Jiyoon_Hwang_2I

I'm not 100% sure if you'll have to memorize the table, but I think it'd be good to know the general rules. For example, if a complex has a negative charge add -ate to the end of the metal name. Also, add -o to the end of the anion name (i.e. F- is fluoride, but add an -o cuz its an anion and it bec...

Jiyoon_Hwang_2I

I actually got C4H8O. First, I used the grams of CO2 given to determine the grams of carbon. Next, I used the grams of H2O to determine the grams of hydrogen. Then, add these values together and subtract from 2.00 g of the sample compound to determine how many grams of oxygen there are. Afterwards, ...

Jiyoon_Hwang_2I

You can think of sigma bonds as basically a single bond. So, if two atoms have a single bond between them that means it has one sigma bond. On the other hand, every other bond after the single bond is considered a pi bond. For example, a double bond is a sigma bond and 1 pi bond and a triple bond is...

Jiyoon_Hwang_2I

Usually, lone pairs cause more repulsion, so the bond angle decreases. For example, NH3 has 1 lone pair on nitrogen and its bond angle is 107 degrees, but H2O has 2 lone pairs and the bond angle is 104.5 degrees. So, the more lone pair electrons on the central atom, the more repulsion, causing the b...

Jiyoon_Hwang_2I

Dr. Lavelle briefly introduced induced dipole-induced dipole during one of his lectures last week so I think it'd be good to go over it. However, london dispersion forces, dipole-dipole, and hydrogen bonds were emphasized more so you should def know these more in depth.

Jiyoon_Hwang_2I

Polarizable means an atom's electron cloud can be easily distorted by the strong pull of another atom. On the other hand, polarizing power means an atom has enough charge to cause distortion to another atom's electron cloud. I think it doesn't necessarily mean polarizability is only related to anion...

Jiyoon_Hwang_2I

Yes. The larger the difference in electronegativity, the larger the dipole moment.

Jiyoon_Hwang_2I

Polarizability tends to decrease from left to right on the periodic table and increases going down the periodic table.

Jiyoon_Hwang_2I

The arrows are usually pointed towards the more electronegative atom (or the atom with the partially negative charge)

Jiyoon_Hwang_2I

There is a coordinate covalent bond between NH3 and BF3. The bond between the nitrogen and boron is a coordinate covalent bond because both electrons come from the nitrogen atom.

Jiyoon_Hwang_2I

It depends on the shape of the molecule (i.e. trigonal planar, tetrahedral, etc.). You determine the shape based on the number of atoms and the number of lone pairs on the central atom.

Jiyoon_Hwang_2I

My TA said you don't have to write out the px, py, or pz so I think you can just combine them and write 1s^2 2s^2 2p^3

Jiyoon_Hwang_2I

For the first part of the question you would use the formula Δp ≥ h/4πΔx. Plug in Planck's constant for h and 350*10^-12 m for Δx. After plugging in all of the values you should get an answer close to 1.51*10^-25 kg*m*s^-1. Then, use the formula Δv = Δp/m (this formula is derived from the equation p...

Jiyoon_Hwang_2I

I think you use the equation E = hc/λ. You would plug in Planck's constant, the speed of light, and the wavelength which would be 633 nm (633 * 10^-9 m). Then, you would get your answer in J/photon. Then, multiply this answer to (2.4 * 10^21 photons). Your answer would be in joules, but since they i...

Jiyoon_Hwang_2I

I think it's two shared electrons because in his example SO4 2-, for the oxygen, the equation was FC = 6 - (6 + 2/2) so the numerator in the fraction stands for the number of shared electrons and for SO4 2- the oxygen has a single bond with sulfur so the value of S was 2 since two electrons were sha...

Jiyoon_Hwang_2I

They're both correct. Both ways indicate a negative 2 charge.

Jiyoon_Hwang_2I

The first exception was that a half full d^5 and full d^10 subshell have lower energy. So, for chromium the electron configuration would be [Ar] 3d^5 4s^1 and the electron configuration of copper would be [Ar] 3d^10 4s^1. The other exception was that the 4d state has lower energy than the 5s state, ...

Jiyoon_Hwang_2I

Black body radiation is emitted by an object (the "black body") which absorbs all frequencies of light and emits radiation solely based on the wavelength rather than the composition of the object. However, I believe Dr. Lavelle said we don't need to know what black body radiation is.

Jiyoon_Hwang_2I

Yes, I think constructive and destructive wave properties would apply to sound. Constructive waves of sound would cause the sound to be louder and destructive waves would cause the sound to be quieter. This is because in constructive waves, the amplitude of the waves increase, whereas the amplitude ...

Jiyoon_Hwang_2I

I believe Wien's law is used to determine either the temperature or wavelength based on which variables are provided in the problem. However, I don't think we would need to know Wien's law as of now because on Outline 2 Dr. Lavelle did not put 1B.11 as an option which was the question about Wien's L...

Jiyoon_Hwang_2I

The ground-state electron configuration for sodium would be 1s^2 2s^2 2p^6 3s^1. However, you can also use a shortcut by determining the closest noble gas to the given element (in this case neon is the closest noble gas to sodium) and then writing the electron configuration starting at this noble ga...

Jiyoon_Hwang_2I

I don't think it would be related to wavelength because the wavelengths of visible light all differ depending on what color light it is. So, the problem groups the series based on their absorption level because all the lines in the Lyman Series will always transition to/from n=1 and all the lines in...

Jiyoon_Hwang_2I

This question is basically saying that all the lines in the Lyman Series end at n=1 and all the lines in the Balmer Series ends at n=2. So, UV lines which are part of the Lyman Series will always transition to/from the energy level n=1 and visible light which is part of the Balmer Series will transi...

Jiyoon_Hwang_2I

Use the equation T=constant/λ, where the constant is given as 2.9 mm*K (2.9*10^-3 m*K) and λ is 700 nm (700*10^-9 m).
If you plug these numbers into the equation you would get T=(2.9*10^-3 m*K)/(700*10^-9 m)=4.1*10^3 K

Jiyoon_Hwang_2I

First, I would do 32W * 2.0s to find the amount of energy, in Joules, that is emitted in 2.0s as stated in the first question. Then, use the equation E=hc/λ to determine the energy per photon. For h use Planck's constant (6.626*10^-34 Js), c use the speed of light (2.99792*10^8 m/s), and λ use wavel...

Jiyoon_Hwang_2I

He said that as the energy level increased, E (the energy) decreases. Basically as n approaches infinite, E approaches zero. This is because n increasing means the electrons are moving further away from the nucleus, so the interaction between the electrons and the nucleus, E, decreases and approache...

Jiyoon_Hwang_2I

I would recommend writing the state of the molecules when writing the equation and usually the problems provide you with the state. For example, they'll say "in an aqueous solution of..." However, if the question doesn't tell you what state the molecules are at I don't think you will be pe...

Jiyoon_Hwang_2I

You should always round your final answer to the correct number of sig figs. If you're doing addition/subtraction, round your answer to the least number of sig figs from any of the numbers in the given problem. If you're doing multiplication/division, round your answer to the least number of decimal...

Jiyoon_Hwang_2I

You may think that there are multiple correct answers for stoichiometric coefficients, but usually the stoichiometric coefficients should all be simplified to their lowest possible values. So, if you think there are multiple correct answers try dividing each coefficient by one number and if all the ...

Jiyoon_Hwang_2I

Yes, the test is only on the high school review unit that we learned until today. Although we started the quantum unit and will continue until we take the tests in our discussion sections, he said it won't be covered on next week's test.

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