CHEMISTRY COMMUNITY

DarrenKim_1H

Multiple factors indicate whether to change the sign. For example if work is done on the system it is positive and if it is done by then it is negative

DarrenKim_1H

Thank you all for all the help in Chemistry these past two quarters

DarrenKim_1H

The steady state method can only be used if the first step of a reaction is much slower than the second step, whereas the pre-equilibrium approximation requires the first step to be faster.

DarrenKim_1H

It simplifies the mathematics behind calculating rate law ex) E+S→ES(1) Rate of formation of ES = k1 [E][S] ES→E+S(2) Rate of decay of ES = k-1 [ES] ES→E+P(3) Rate of formation of P = k2 [ES] In this example because E + S -> ES occurs much faster than the second step. We can assume E, S and ES are a...

DarrenKim_1H

Against a constant pressure, work = -PdeltaV. So with expansion where volume is increasing and pressure remains constant. work also increases

DarrenKim_1H

Not sure exactly what you mean. Reaction intermediates are formed from one reaction, like any other part of the equation and then completely consumed in the following reaction

DarrenKim_1H

Dear Professor,

Thank you for all your wisdom the past two quarters. With Online Finals out of the way, I hope you can finally get some sleep and rest.

DarrenKim_1H

Lactic acid's chemical formula is:

CH3CH(OH)COOH (aq) + H2O -> H3O+ + CH3CH(OH)COO-

Because lactic acid produces hydronium ions, I think we can ensure that it will react with HbO2- to decrease it's concentration

DarrenKim_1H

From the graph, you can see that A has a initial P equivalent to 27.5. After the reaction occurs it's final P is around 17.5, meaning that the change in concentration is around 10 kPa. The same as C and double as B, which is why A and C have coefficients of 2

DarrenKim_1H

I believe it is still K

DarrenKim_1H

I believe that we will be given intermediates. The only step we need to know is how to identify them

DarrenKim_1H

I believe for endothermic reactions, like you said if the temperature is raised forward rxn is favored.

In contrast, I believe for exothermic rxn, if T is raised backward rxn is favored

DarrenKim_1H

I think it's because since q = mCdeltaT and you set qhot equal to qcold, you can divide both sides by the specific heat capacity of water to cancel them out, since you're finding the equilibrium temperature.

DarrenKim_1H

I got the same answer for e^14.8. Since, for the rest of the question you don't have to use that value. I left it as 2.6 x 10^6.

DarrenKim_1H

OILRIG

(Oxidized is Losing & Reduction is Gaining)

When something is oxidized it loses an electron because it gives that electron to something else, which makes it a reducing agent.

When something is reduced, it gains an electron by taking it from something else, making it an oxidizing agent

DarrenKim_1H

I don't think we need to know any specifics regarding it. I believe it is another way to show the electronegativity trend across the periodic table

DarrenKim_1H

I believe so

DarrenKim_1H

So basically, I think you split the Au+ as the half reactant for both parts. From there, writing the half reaction is pretty standard

DarrenKim_1H

Reduction (at cathode): Au+(aq) + e- ->Au(s)

Oxidation (at anode): Au+(aq) -> Au 3+(aq) + 2e-

DarrenKim_1H

I believe it's because hydrogen is the most standard element to use for measurements. I believe' hydrogen's potential is assumed to be zero at all temperatures

DarrenKim_1H

W max is the max amount of work a system can do. G is the max work one at a standard temperature and and pressure. So they can be related.

DarrenKim_1H

I think its G* = -nFE is used at equilibrium. The equation you gave is if it's not at equilibrum

DarrenKim_1H

It's to maintain charge as the electrons are transferred from the anode to cathode

DarrenKim_1H

melting and Vaporization are endothermic processes, because they require energy to turn ice into water, and water into vapor

DarrenKim_1H

So basically, U = q + w refers to the fact that it is the total energy that the system is involved with, i.e the energy from the heat generated or loss plus the energy of the work done on or against the system

DarrenKim_1H

It refers to the energy contained within a system

DarrenKim_1H

It means Kw is equal to the concentration of H3O+ ions times the concentration of OH- ions, since Kw = 10^-14

DarrenKim_1H

I believe hydrogenation is given as a Hess's Law question, I don't think there is any specific way to calculate it within what we have learned so far

DarrenKim_1H

I think you did your Kelvin conversions wrong. It's suppose to be 55.85 + 273.15, so 329 K and 328K

DarrenKim_1H

So first, you set q(hot water) = -q(cold water) to find the temperature at equilibrium. Then you use the S = Cln(T2/T1) subbing in the Temperature at equilibrium to find the change in entropy for the hot and cold water individually, before you add them together to find the total entropy change

DarrenKim_1H

residual entropy is the remaining entropy left in molecules at 0 K

DarrenKim_1H

Because it does not matter HOW the heat was added or HOW much the temperature changes.

DarrenKim_1H

We don't know what the specific bond enthalpy for breaking a pi bond. So we break both and record the bond enthalpy for a, say, carbon-carbon double bond as different from a single bond

DarrenKim_1H

We use variation of Cm = q/n delta T

where Um is the molar entropy and given as:

Um = (3/2)RT

DarrenKim_1H

I think I just keep in mind q and w are the non-state functions.

DarrenKim_1H

With Hess's law, you can add or subtract different parts of the equation in order to get delta H

DarrenKim_1H

more gas implies a volume expansion, aka an increase of delta V, which according to PV = nRT, means there's an increase in the second part of the equations enthalpy = internal energy + nRT

DarrenKim_1H

I think part of it is understanding that burning fuel is an open system, meaning that both mass and heat are exchanged in the process. Other than that, it probably means to understand how to calculate heat, internal energy change, and other things we learned in class using the chemical equation for ...

DarrenKim_1H

A) Correct B) Closed, because coolant in a refrigerator coil is what "cools" down the refrigerator so heat is transferred C) Isolated, because nothing is exchanged D) Correct, both heat and volume of gas liquid is changing E) Yes closed, because the mercury changes in response to temperatu...

DarrenKim_1H

Hi Ayushi,

because since the reaction is endothermic and the temperature raises, we can assume that all the heat is absorbed by the calorimeter

DarrenKim_1H

The human body generates 100 W of heat, which we can set equal to q since the human body generates 100 joules per second of heat. So, q = 100 J

DarrenKim_1H

I believe its because they are fundamentally different constants. Cv is under constant volume and Cp is under constant pressure, so they are different

DarrenKim_1H

I think it was more to derive the equation at the end. However, I think he said that we may use the integral for reversible reactions and the derivation of it for irreversible reaction

DarrenKim_1H

I believe if you are calculating x using the information from Table GE.2, the values you would get for x are 1.47x10^-2 or 6.81x10^-3. But you would use 1.47x10^-2 since anything around 10^-3 is too small (I might be wrong about that, but thats my general rule of thumb lol)

DarrenKim_1H

I believe dividing by 100 is to convert the units from KPa to bar

DarrenKim_1H

It might just be a minor rounding error. What did you input for -logX? If you put -log(0.16), then you would probably get 0.8

DarrenKim_1H

They are just constants found probably through experiments

DarrenKim_1H

It is an ice box question, so you need to find x. From the table, the information you need to find is Kc to calculate X

DarrenKim_1H

Peer learning sessions refer to all the drop-in hours, step-up sessions, and workshops you can attend. Step up sessions are sessions designed to help students struggling with concepts. In comparison, workshops are designed for students that have already a deep understanding of concepts and need more...

DarrenKim_1H

I think Ka is usually 10^-x of something whereas pKa refers to the x

DarrenKim_1H

I don't think it is directly, but we turn in weekly homework at discussion

DarrenKim_1H

You can find x using the quadratic formula x = -b + or - square root (b^2-4ac) / 2a

DarrenKim_1H

I believe so, 10^-3 is the correct answer I think

DarrenKim_1H

I believe whichever K value is less is the thermodynamically more stable one

DarrenKim_1H

Part a is very similar to the example we did with the reaction vessel in lecture today. namely, you have to solve for x.

Part b is essentially the same question, except you are given F2 and are calculating its dissociation instead of Cl.

DarrenKim_1H

I believe for this question we use Kc

DarrenKim_1H

I do not think we have yet learned a case where we would multiply K by a factor of n. As for why we raise it to the factor of n. Remember, whenever we have a coefficient in front of a reactant/product, we raise its concentration by that coefficient as well. i.e. the K for N2 + 3H2 -> 2 NH3 is [NH3]^...

DarrenKim_1H

I believe c is another variable for a constant

DarrenKim_1H

So, I think overall KClO is a salt comprised of K+ a weak acid and ClO- is a strong base. So the anion for ClO- will react with water to form:

ClO- + H2O -> HClO + OH-

Therefore it forms a basic solution

DarrenKim_1H

I would go about answering this question by first asking myself: Does this molecule form an acidic or basic solution when it reacts with water? For example, in part a where you're given NH4Br, you know that NH4+ is the conjugate acid for weak base NH3 such that: NH4^+ + H2O -> NH3 + H3O+. Hence, sin...

DarrenKim_1H

If we are looking at the same question, they give us the molecule as KOH. Since it is a base it will react with water to from hydroxide (OH-) ions. Therefore, we know to use pOH = -log[OH-]

DarrenKim_1H

By oxidation state, are you referring to the charge of an individual atom (ie Fe^3+)? If so, let me try to explain with an example. Suppose you have the molecule Fe(OH)3. Here , we know OH has a negative charge of 1 but the overall compound is neutral. Therefore, Fe has to have a positive three char...

DarrenKim_1H

So I would start the problem by drawing out B(OH)3's Lewis Structure. From there, you notice that the structure has no delocalization of electrons so therefore no relevant resonance. If you draw out the structure for the reaction B(OH)3 + 2 H2O -> H3o+ + B(OH)4^-. You can see that since Boric acid d...

DarrenKim_1H

I believe pH refers to the concentration of H3O+ ions in a solution. So a conjugate base wouldn't affect this number.

HCl + H2O -> H3O+ + Cl-

In the equation, Cl- is the conjugate base of strong acid HCl, but its not an H3O+ ion so i don't think it affects pH

DarrenKim_1H

Because in Inorganic acids, the hydrogen is bonded to any highly electronegative element, whereas for organic acids the hydrogen is restricted to being bonded to an O atom, which is bonded to a C atom

DarrenKim_1H

I believe they are different atomic models. So a question would specify whether to place them in one

DarrenKim_1H

Because there are three ammonia molecules. Therefore, TriAmmine

DarrenKim_1H

So that coordination compound is shown to be neutral. So you would find the net negative charge of 3 So4^2-, which is -6 (3 times -2). In the cation, NH3 and H2O are both neutral so Cobalt is the only component that has a charge which is +3. Overall, its a net positive charge of +6 since (2 times 3 ...

DarrenKim_1H

It represents the positive charge on a cation. For example a copper (ii) sulfate CuSO4, represents a Cu^2+ ion which attaches itself to the sulfate anion (SO4^2-)

DarrenKim_1H

Because hydrogen atoms from other molecules are attracted to these electronegative regions

DarrenKim_1H

Probably questions from the Outline on Acids and Bases

DarrenKim_1H

Specific shapes...rather, specific areas of electron density correspond to different hybridization levels. For example, tetrahedral geometry corresponds to sp^3, octahedral corresponds to sp^3d^2 etc.

DarrenKim_1H

I believe he will. There is going to be a slight adjustment he makes to all our grades at the end

DarrenKim_1H

I believe it is only connected to the oxygen atom.

DarrenKim_1H

Yes, Dr. Lavelle will adjust grades accordingly he may decide the cutoff for grades to be 250/500 points, but I do not believe there is any "set" percent ranges for the grades. Therefore, I don't think I can answer your second question.

DarrenKim_1H

Induced dipole occurs when a nonpolar moelcule is turned dipole by an ionic molecule. Whereas dipole moments occur when the molecule already has a different charge

DarrenKim_1H

Heavier halogens have more electrons in their shell which increases Van Der Waal forces and boiling point

DarrenKim_1H

The Lecture we had the Friday before the midterm up until the lecture we had last friday.

DarrenKim_1H

I believe in lecture, he explained that the angle less than 120 degrees results from a stronger pull from the central atom.

DarrenKim_1H

sp hybridization refers to the mixing of the s and p orbitals. What the answer implies is that decreasing the amount of p orbitals from 3 to 2 increases the bond angle and the answer asserts that increasing the amount of s orbitals for sp hybridization will in contrast increase the bond angle.

DarrenKim_1H

I believe 1/r^6 denotes a inverse relationship between energy and distance. Whereas, the other one you described is the actual equation. It's kind of like how wavelength = c/frequency is the actual equation where as wavelength = 1/frequency denotes that wavelength and frequency are inversely related.

DarrenKim_1H

Yes, most dipole moments we have discussed are temporary. Since bonds are just areas where electrons are most likely to be found, there is no "permanent" dipole moment.

DarrenKim_1H

Like you said, the SI unit is Coulombs times Meter and it denotes the magnitude of charge times the distance between them

DarrenKim_1H

I believe cancelling out has to do with the symmetry of a molecule. For example C2Cl4 has no net dipole moment because the molecule is symmetric so the moment cancels out

DarrenKim_1H

I would assume so. Although depending on when your section is, Wednesday lecture may be the first lecture where we cover material on outline 4

DarrenKim_1H

Group 1 Elements Be Like:

What If I...gave you my one valence electron...to form a complete octet...aha ha, just kidding...unless?

DarrenKim_1H

Generally, the first step requires understanding the course material in-depth. Most Learning Assistants I've had have told me that they generally received an A or higher in the course. Afterwards, I believe it's only a matter of time commitment and submitting the application by the due date. I've li...

DarrenKim_1H

Hi Anna,

I believe what you are referring to is Slater's rule and it allow you to estimate the effective nuclear charge from the real number of protons in the nucleus and the effective shielding of electrons in each orbital "shell."

DarrenKim_1H

I think we would just have to know that electronegativity increases bottom to top in groups and left to right across periods

DarrenKim_1H

To my understanding, don't s orbitals get filled first before d orbitals

DarrenKim_1H

Bismuth is undergoing the inert-pair effect

DarrenKim_1H

On my.ucla.edu sign in and under the Academics tab, click Exam and Homework Grades

DarrenKim_1H

In this question it is asking for ions with a positive three charge. So for part a for example: Ar on the periodic table has 18 electrons. Add 18 with the 6 from 3d^6 and you get 24 which would normally be Cr on the periodic table. However, since you are dealing with ions with a +3 charge, the answe...

DarrenKim_1H

Exceptions to the octet rule generally have to deal with molecules that have an odd number of electrons, which is different from normal orbitals being able to accept an even level

DarrenKim_1H

Because completely filled orbitals are generally more stable than partially filled orbitals

DarrenKim_1H

From what I remember in lecture, electrons are typically first paired in the s orbital and are placed parallel first in every other orbital

DarrenKim_1H

I was reading up on this...and from what I can garner the answer is a bit complicated. You're welcome to also search for answers but overall it just seems that it is more favorable for an electron to occupy an empty 4d orbital rather than deal with pairing with an electron in the 5s orbital.

DarrenKim_1H

Hund's Rule directly states that other sub levels besides s, have electrons placed in individual orbitals before being paired. In contrast, Aufbau's principle can be explained in three points 1. The number of electrons in an atom is equal to the atomic # 2. Each added electron will enter the orbital...

DarrenKim_1H

According to Hund's rule, electrons repel each other in the same subshell which is why they must occupy different orbitals with parallel spin

DarrenKim_1H

Wein's Law is used to calculate wavelength using a constant divided by the temperature in Kelvin. Since we have not gone over Black body radiation in lecture yet, I believe it will be better explained by Professor Lavelle in a future lecture.

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