CHEMISTRY COMMUNITY

Lakshmi Davuluri 1E

I'm not positive that this is correct. However, I think since it says per mole of gas and the number of moles aren't given, you can assume that n=1.

Lakshmi Davuluri 1E

I think that K is the rate constant when you take into account the forward and reverse reactions (multiple reactants) but k only takes into account the forward reaction when you only have one reactant. I'm not sure though. I think this is correct. K is equal to k/k', so it's taking into account the...

Lakshmi Davuluri 1E

Hi, I'm not entirely sure if I understand your question, bu I think that the problem is asking for the delta U in regards to moles of CO specifically. So instead of just leaving it as -2.058 kJ, you need to divide that by the number of moles, which you get by converting the grams of CO to moles of C...

Lakshmi Davuluri 1E

I also think the temperature would decrease because, like you said, of the relationship between volume and temperature. Using PV=nRT, you can see that if pressure, the number of moles, and the gas constant are all constant, then temperature would have to decrease. The temperature would decrease to c...

Lakshmi Davuluri 1E

- If a reaction is reversible, then \Delta S_{tot}=0 and \Delta S_{surr}=-\Delta S_{sys} -If a reaction is irreversible, then \Delta S_{surr}=0 so \Delta S_{tot}=\Delta S_{sys} Also would like to add that I think work doesn't equal zero in isobaric (constant pressure) conditions ! w=-P \Delta V so ...

Lakshmi Davuluri 1E

Hi! I think that the organic chemistry tutor on Youtube does a good job and has really helped me this quarter

Lakshmi Davuluri 1E

Hi! I'm sure that khan academy or the organic chemistry tutor on YouTube have some videos that work through examples on how to do problems with balancing reactions.

Lakshmi Davuluri 1E

Hi, I also got the same answer as you. I think the answer key must have rounded a number somewhere because I got lnQ=14.79, and they used lnQ=15. I think they just forgot to put the number before 10^6? Even if you use lnQ=15, you don't get just 10^6. Hope this helps

Lakshmi Davuluri 1E

You can determine the overall order of a reaction by taking the sum of the orders of the individual reactants

Lakshmi Davuluri 1E

Lavelle's website says Sunday, March 14th, from 9:30-11:00

Lakshmi Davuluri 1E

Just like Will said, when you’re balancing the equation, think of Fe2O3•3H2O as one unit. if you put a coefficient infront of the Fe2O3•3H2O, then that coefficient is distributed to the 3H2O.

Lakshmi Davuluri 1E

So under the skeletal reaction you have that Mn has a charge of 3+ and the O4 has an overall charge of 4-. O has a charge of 2-, so if there are four oxygen atoms, then the total charge would be 8-. In order to get the 1- charge for the whole molecule, Mn would have a charge of 7+ on the reactants s...

Lakshmi Davuluri 1E

Alvin Lai 3J wrote:I think it is the moles of electrons transferred, which you can find my looking at the redox reaction?

Yeah it the moles of electrons that were used in balancing the redox reaction

Lakshmi Davuluri 1E

The number of moles of electrons that were used in balancing the redox reaction is plugged into the delta G equation.

Lakshmi Davuluri 1E

My TA said that we should always put the aqueous solutions on either side of the salt bridge

Lakshmi Davuluri 1E

I don't think changing the amount of aqueous solutions applies here, just gases.

Lakshmi Davuluri 1E

At constant pressure: delta H = delta U + P(delta V)
Under conditions of constant pressure and volume: delta V would become zero, and delta H would equal delta U.
Under conditions of constant pressure and temperature: delta H = delta U + (delta n)RT

Lakshmi Davuluri 1E

q is the heat consumed or released by the system, while enthalpy is the heat at constant pressure. q only equals delta H when q=qp, which is under conditions of constant pressure.

Lakshmi Davuluri 1E

Degeneracy is the number of different states that energy can exist and entropy increases as there are potentially more ways for energy to exist, so as degeneracy increases, entropy does as well.

Lakshmi Davuluri 1E

Example 5J.5 in the textbook explains how use van't Hoff's equation really well. Hope this helps.

Lakshmi Davuluri 1E

They write it this way because regardless if q(reaction) is positive or negative, q(calorimeter) will always have the opposite sign. For this problem, q(reaction) is negative because it's an exothermic reaction. If the reaction is exothermic, then that means the calorimeter is gaining that heat, whi...

Lakshmi Davuluri 1E

I also don't think degeneracy has any units because it's just the number of states raised to the number of particles. Like Saumya said, I think it's just a quantitive term.

Lakshmi Davuluri 1E

In example 4.F.4 page 302, it shows they used the calculation \Delta S=C v ln(T2/T1) with C v =nC v,m So C v should be equal to (3/2)R, or 12.5. However, the textbook shows 20.79, which is actually equal to (5/2)R, or C p . Is this a mistake? Should Cv be used, or is the calculation correct in usin...

Lakshmi Davuluri 1E

Delta G naught and delta S naught are the changes in Gibbs free energy and entropy at equilibrium, while delta G and delta S are relevant under other conditions.

Lakshmi Davuluri 1E

To answer your question about part (a), the change in enthalpy is a positive number for the dissolution, so the change in entropy for the surroundings is negative. For the dissolution to be favored, the change in entropy must be positive because spontaneous/favored processes require an increase one ...

Lakshmi Davuluri 1E

To add on to what was said about steam causing worse burns than boiling water, steam has a lot more kinetic energy than boiling water. This means that when the steam touches skin, it releases a lot more energy upon the skin than boiling water does.

Lakshmi Davuluri 1E

Hi! So just by looking at the equation q=m*C*delta T, we know that m and C are always going to be positive. This means that the only term that can affect whether q is positive of negative is delta T. If the final temperature is lower than the initial temperature, then the overall change in temperatu...

Lakshmi Davuluri 1E

Joey_Okumura_1E wrote:The answer for 4C.3a is missing from the solution. Did you get the following answers:

4C.3a ΔH = 765 J, T_f = 343 K

Yeah, I also got that for part (a)

Lakshmi Davuluri 1E

One main difference between irreversible and reversible expansions has to do with pressure. Irreversible expansion happens under a constant pressure, which means the external and internal pressure are equivalent. For reversible expansion, the external and internal pressures are different.

Lakshmi Davuluri 1E

Just to add on to what Lillian said, q(calorimeter)+q(reaction)=0, so q(calorimeter) = -q(reaction).

Lakshmi Davuluri 1E

The C-H bonds are not taken into account because both structures have 6 of them, so there is no difference in regards to the C-H bonds between the structures. The question only asks for the difference between the C-C bonds.

Lakshmi Davuluri 1E

The addition of an inert gas, like argon, does not affect the reaction.

Lakshmi Davuluri 1E

sort of going off the original question, can someone briefly explain what deprotonation is again please? The percentage deprotonation is the ratio of the concentration of dissociated [H+] ions to the concentration of initial acid [HA]. You use it to find what percentage of the weak acid has dissoci...

Lakshmi Davuluri 1E

Last quarter it took about a week for exams with no partial credit, so I assume that the grades for this exam will take longer because there is partial credit.

Lakshmi Davuluri 1E

It might be helpful to think of heat as a regular product or reactant. Just as if we raise the concentration of reactants, the forward reaction will be favored. If the reaction is endothermic, which means heat is on the reactants side, then increasing the temperature ( kind of like "raising the...

Lakshmi Davuluri 1E

An increase in H20 is not affecting the equilibrium constant, it is just affecting the direction in which the reaction proceeds. Just like for any increase in product or reactant, the reaction will favor a direction in order to deal with the increase.

Lakshmi Davuluri 1E

Yeah I agree with the people above. I also just assumed that K meant Kp because the other column was labeled as Kc

Lakshmi Davuluri 1E

To answer the first part of your question, the 5% rule applies to both percent protonation and ionization

Lakshmi Davuluri 1E

Yeah "p" is just shorthand for -log, so you're right to say that pKa is just the negative log of Ka

Lakshmi Davuluri 1E

Doing the textbook problems listed in the outlines are extremely helpful. Also, going through the bullet points of learning objectives on each outline is a good way to check your understanding of that unit.

Lakshmi Davuluri 1E

While pH and pKa both measure acidity, they do so in different ways and are not exactly equal. pH is calculated by -log[H3O+], and pKa is calculated by -log [H3O+][conj base]/[acid]. Same goes for pOH and pKb. Hope this helps!

Lakshmi Davuluri 1E

Your approach to this problem is correct. Make sure you are using 1/Kp in this case because you are starting with the products.

Lakshmi Davuluri 1E

Ellison Gonzales 1H wrote:Could someone list what terms are used when talking about Q? I get a bit confused on which terms are used for K or Q

K is the equilibrium constant, while Q is the reaction quotient.

Lakshmi Davuluri 1E

Aqueous solutions are included. Only solids and liquids are omitted because their concentrations do not change significantly.

Lakshmi Davuluri 1E

He mentioned in one of the lectures that he was going to cover thermodynamics later

Lakshmi Davuluri 1E

If the reaction requires heat (endothermic) while forming product, then increasing the temperature will result in product formation being favored. On the other hand, if the reaction gives off heat (exothermic) while forming product, then increasing the temperature will realist in reactant formation ...

Lakshmi Davuluri 1E

I agree with you on (i), but I came up with something different for the other two. For (iii), I agree that the change in pressure wouldn't affect the NO. However, I thought that it would be because it was inert gas being added, not because of the number of moles. As for (ii), I'm a little confused m...

Lakshmi Davuluri 1E

Yeah, I agree with Claire. I think you could get a reasonable answer by approximating, but you're probably better off just doing the math for a more accurate answer. Also, since Dr. Lavelle never mentioned using approximation in a problem with a quadratic, I think it's safe to say to just do the math.

Lakshmi Davuluri 1E

I think Dr. Lavelle said that we will mainly just see quadratics. If we are presented with a cubic expression, then it will most likely be a problem in which approximation can used to solve it.

Lakshmi Davuluri 1E

Do the Q and the K values follow the same trend/pattern when describing on whether a system will favor reactants/products? For example in the lecture, it states if K is small, then it would favor the products (shift right) since there's more reactants at equilibrium. And if Q is smaller than K, the...

Lakshmi Davuluri 1E

You can determine an electron donating group by look at the central atom of the group and noting if it has a lone pair or not. If it has a lone pair on the central atom, then I think it is safe to assume that it is an electron donating group. Feel free to correct me if I'm wrong. Hope this helps!

Lakshmi Davuluri 1E

https://www.google.com/url?sa=i&url=https%3A%2F%2Fslideplayer.com%2Fslide%2F11923338%2F&psig=AOvVaw2ZxNlAAo1AHy7aHRwbiWoJ&ust=1607842862051000&source=images&cd=vfe&ved=0CAIQjRxqFwoTCNDG9sjvx-0CFQAAAAAdAAAAABAD Here is picture depicting the amphoteric line against the metalloi...

Lakshmi Davuluri 1E

Did Lavelle already send out an email regarding the instructions for the final?

Lakshmi Davuluri 1E

I found the [OH-] after finding the pOH. I just did 10^-(pOH). I found that this is the easier way for me since you already needed to find the pOH. Hope this helps!

Lakshmi Davuluri 1E

H20 is considered monodentate because only one lone pair bonds a central atom, while the other lone pair points away from the central atom. Hope this helps!

Lakshmi Davuluri 1E

Hi! I don't know exactly how much of acids and bases is going to be on the final, but I'd say everything Dr. Lavelle has gone over is probably fair game. I would assume only a small portion of the test will be on acids and bases because he'll want to get in all the other units too.

Lakshmi Davuluri 1E

Yeah, it's basically calculated the same way. Like it was stated above, if you have the pH, then you can just subtract the pH from 14 to get the pOH.

Lakshmi Davuluri 1E

CIs configuration refers to two groups being on the same side of a molecule, while trans refers to the two groups being on opposite sides of each other.

Lakshmi Davuluri 1E

Hi, if you're talking about the sheet on sapling, then there's a column that tells you on about the denticity.

Lakshmi Davuluri 1E

Sahaj Patel- Lec 2 and Dis 2L wrote:That seems correct, but remember that the overall compound will have a +1 charge that is cancelled out by Bromine!

Hey I'm just a little confused because there are two Bromines. So wouldn't the compound have a 2+ charge that the two bromines cancel out? Feel free to correct me if I'm wrong

Lakshmi Davuluri 1E

Tanya Nguyen 1E wrote:What is (en)? Is it just a symbol for a bidentate ligand or is it a specific ligand?

it's ethylene diamine

Lakshmi Davuluri 1E

George_Yin_3I wrote:Delocalized pi bonds occur when pi bonds extend to more than 2 atoms. There is an exercise on Sapling about that.

Yeah I think it is question 16

Lakshmi Davuluri 1E

When talking about hybridization you need to consider regions of electron density. Therefore when determining hybridization you consider atoms and lone pairs. For a double bond, the hybridization of each atom becomes sp2 instead of sp3. Hope this helps. Like she said, you just need to focus on the ...

Lakshmi Davuluri 1E

America Alvarado wrote:Where are the dates for the test?

I think you can find it on myucla under your study list

Lakshmi Davuluri 1E

There should be no resonance structures, the last one you are missing is the carbons forming a triangle shape with their bonds. hi when do we know not to include a resonance structure? or should we just assume that for all questions? thank you so much for your response btw I think resonance structu...

Lakshmi Davuluri 1E

The third structure should utilize a triangular ring, with one hydrogen bonded to each of the two carbons and two hydrogens bonded to the remaining third carbon.

Lakshmi Davuluri 1E

Aliya 2J wrote:If we stacked up with 40 from before, do we need to do more to have them be from this week?

I think so. I think we need to have at least five posts per week

Lakshmi Davuluri 1E

I think the sp3 hybrid orbital looks like a somewhat distorted p-orbital with unequal lobes

Lakshmi Davuluri 1E

If the chlorines are on opposite sides of the molecule, then their dipole moments cancel out

Lakshmi Davuluri 1E

I'm just going through the textbook problems. I found that they were the most helpful in preparing for the last midterm.

Lakshmi Davuluri 1E

Is there a difference between valence shell configuration and ground state configuration?

Lakshmi Davuluri 1E

Madison Muggeo 1G wrote:
darchen3G wrote:Single bond: 1 sigma
Double bond: 1 sigma 1 pi
Triple bond: 1 sigma 2 pi
Quadruple bond: 1 sigma 2 pi 1 delta

What exactly is a delta bond? Have we gone over this?

I think a delta bond occurs when all four lobes of a d orbitals interact with one another

Lakshmi Davuluri 1E

Veronica Macias 3C wrote:Also, what type of bonds are atoms with an electronegativity difference between 1.5 and 2?

I think it could either be ionic or covalent

Lakshmi Davuluri 1E

The 4s subshell is filled before the 3d subshell because the 4s orbital has a lower energy level than the 3d orbital. In terms of writing out the electron configuration, you write the subshell with the lower quantum number first. That is why 3d comes before 4s when written.

Lakshmi Davuluri 1E

Julianna_flores3L wrote:I think that the p orbital that you are talking about has to do more with quantum numbers rather than valence electrons? Correct me if I am mistaken.

Yeah, I agree. I think this has to do more with the shape and the wave function, but I'm not positive

Lakshmi Davuluri 1E

Is it safe to assume that most of the time on exams the molecular formula will be given along with the name of the molecule?

Lakshmi Davuluri 1E

My guess is we will have to draw some structures with expanded octets because Dr. Lavelle took the time to go through an example in one of the lectures. As far as being able to tell whether an element has an expanded octet, I don't really know. The example that Dr. Lavelle used was PCl5, which had f...

Lakshmi Davuluri 1E

Hi! From what I understood from the lectures I think formal charge is used to mainly keep track of the charge of atoms and molecules while drawing the lewis structures. From there I think you can use the charge to predict things about the atoms and molecules, such as reactivity.

Lakshmi Davuluri 1E

The most energy is absorbed by gaining an electron for Kr because it is a noble gas and already has a eight valence e-

Lakshmi Davuluri 1E

Some elements, like sulfur and chlorine, can be expanded because they utilize d-orbitals from the third energy level and on

Lakshmi Davuluri 1E

I think they release energy because they become more stable once they have attained the electron(s) needed, and stable states have less energy that unstable states. Feel free to correct me if I'm wrong :)

Lakshmi Davuluri 1E

When talking about the wavelength and frequency, it is referring to one photon because all of the photons in the pulse have the same wavelength and frequency. The energy is multiplied by 100 because there are 100 photons with the same wavelength and frequency. If you look at the equation, E=hv, freq...

Lakshmi Davuluri 1E

Javier Perez M 2J wrote:By what day do we have to post every week?

Sunday by midnight

Lakshmi Davuluri 1E

I'm pretty sure none of the problems assigned from section 1.D are apply to this midterm.

Lakshmi Davuluri 1E

It depends on the problem, like what you're given and what you're converting to.

Lakshmi Davuluri 1E

We include the delta sign because it represents the heat that is supplied to carry out the reaction.

Lakshmi Davuluri 1E

I'm pretty sure my TA said between 20-30 question, and they will be multiple choice.

Lakshmi Davuluri 1E

Yeah you can use any type of calculator. It doesn't have to be a scientific one.

Lakshmi Davuluri 1E

So first off you want to use the equation c=wavelength*frequency and solve for the frequency of the 102.6 nm wave. Then, you would plug this into Ryberg's equation and use that to solve for n2. Hope this helps :)

Lakshmi Davuluri 1E

Yeah, you are right. Delta v is 10 m/s. This is one of the errors in the solution manual.

Lakshmi Davuluri 1E

Is the energy released from a photon always positive then? I'm pretty sure because the only reason the electron has a "negative" energy is due to to it coming back down from the reference state where the energy is zero. The photon would have a positive energy because it either has energy ...

Lakshmi Davuluri 1E

I'm not entirely sure about this but I believe that the energy of the photon given off is the negative of the change in energy of the electron when it drops to a lower energy level. Since the change in energy of the electron is negative, the energy of the photon would have the opposite sign, hence ...

Lakshmi Davuluri 1E

Are the questions he's adding still apart of the quantum unit? Does this mean we will be in this unit until the end of week 4?

Lakshmi Davuluri 1E

Yeah, it's an error. On Professor Lavelle's website there is a link that has all the solution manual errors and the corrections.

Lakshmi Davuluri 1E

Hi, I'm not sure I understand your question completely, but I'm pretty sure an electron will always start at the energy level of n=1. For the electron to jump to a greater energy level, it must be excited. For the example you are talking about, I think that they just implied that the electron starte...

Lakshmi Davuluri 1E

Are the three videos before the one about the photoelectric effect supposed to be review for the quantum unit?

Lakshmi Davuluri 1E

Hey, I was also confused, but I found this post where a lot people left explanations. viewtopic.php?f=13&t=63002

Lakshmi Davuluri 1E

I agree with Justin. I think if the question wanted the answer in units different from SI units it would state that.

Lakshmi Davuluri 1E

I don't know if anyone from Discussion 3K still needs this, but here is the link for the GroupMe. https://groupme.com/join_group/63092382/qDDFwhPF

Lakshmi Davuluri 1E

The textbook problems aren't being collected. You can find them on Professor Lavelle's website under "Learning Outcomes" in the outlines, and this week's problems would be in "Outline 1: Review of Chemical & Physical Principles."

Search found 100 matches