CHEMISTRY COMMUNITY

Mukil_Pari_2I

Actually, correction: know how catalyst rules apply to enzyme rules and enzyme kinetics. After all, an enzyme is just an organic catalyst.

Mukil_Pari_2I

It does not matter. And also, if A is a reactant, you would put a negative sign in front of the d[A]/d[t] because the reactant is being used up.

Mukil_Pari_2I

Well, I would assume you do not really need to know enzyme kinetics given that it does not show up on the outline. I would just know that an enzyme is a catalyst.

Mukil_Pari_2I

I would assume it would depend on the equation. We would always be given enough information to figure out the answer.

Mukil_Pari_2I

Because you would have to calculate the entropy due to the temperature change, the reversible equation can be used for the first part. You would have to consider each portion separately. So, first you would assume the the temperature as constant (isothermal), for the compression. Then, you would ass...

Mukil_Pari_2I

The inert electrodes are written in cell potential diagrams when either electrode does not have a metal.

Mukil_Pari_2I

It is also important to note that you have to identify when you have to flip the Ecell standard based off the question. Even though the reaction is given one way, the problem may be asking for the reverse of the reaction (like the oxidation when only the reduction is given). In those cases, you will...

Mukil_Pari_2I

Tony Ong 3K wrote:Will the final have pseudo equilibrium constants on it where we have to calculate it?

I would assume there would always be a way to calculate it even if it is not given to us. But if it is given to us, that would also work.

Mukil_Pari_2I

As for water, water or any other molecule that is not the solvent, if it is part of a reaction, it can be in the rate law. However if the solvent is water, you do not include water in the rate law since you have excess amount of water so that the concentration of water does not contribute to the rea...

Mukil_Pari_2I

The units of the rate are always M/s. Therefore, you always want to find a way to make the units equal to M/s. So the units of K would always be in relation to the units of the concentration. For example, for a zero order reaction, the units of K would be M/s.

Mukil_Pari_2I

so if I'm understanding this correctly... zero orders are just equal to the constant (k). how would we write them in the k[A][B]^2 thing or would be leave them out? Well, for a zero order reaction, since the rate law is fully independent of the concentration of the reactant, the rate law equation w...

Mukil_Pari_2I

For your second question, the [NO]^2 has to do with the concentration being squared not the formula so it would not be equivalent to N2O2.

Mukil_Pari_2I

It just means that with respect to the triiode ions, the reaction rate is just the reaction constant.

Mukil_Pari_2I

The catalyst is introduced in the step with less activation energy since that is the purpose of the catalyst.

Mukil_Pari_2I

Anything that is not a final product is an intermediate. So if the product in the elementary reaction is not the same as the product of the overall reaction, then it is an elementary reaction.

Mukil_Pari_2I

It would be used in a situation where you are trying to find a K value at a different temperature. For instance, on test 2, we were given the K value at one temp and we were asked to find the K value at a different temperature since the K value changes with temperature. In that situation, we would h...

Mukil_Pari_2I

Go over past final exam questions. additionally, previous midterm questions are quite helpful for standard potential questions.

Mukil_Pari_2I

The approximation can be utilized when solving ice problems if that condition is true.

Mukil_Pari_2I

In the overall reaction law, the equation would just be the reaction constant.

Mukil_Pari_2I

The intermediate is just that--the intermediate. It shows up only in the elementary reactions and is not in the final product.

Mukil_Pari_2I

Residual entropy is the difference in entropy between a non-equilibrium state and crystal state of a substance close to absolute zero.

Mukil_Pari_2I

I would add that Gibbs free energy is the maximum of reversible work that may be performed by a thermodynamic system at a constant temperature and pressure. Entropy is a thermodynamic function that describes the randomness and disorder of molecules based on the number of different arrangements avail...

Mukil_Pari_2I

I believe Gibbs free energy is 4J. Additionally, electrochemistry is 6K to 6M.

Mukil_Pari_2I

^Dr. Lavelle said that we should study based off the homework fro Gibbs Free energy as well as the electrochemistry material.

Mukil_Pari_2I

Most formulas were provided on the midterm. I feel that even those that were not provided could have been figured out based off the ones given too us.

Mukil_Pari_2I

If the equilibrium constant is larger than 1, that means product formation is favored. If the formation of something is favored, it usually means its the more stable molecule.

Mukil_Pari_2I

Usually you will be given some information as to the conditions of the situation. Based of that, you should be able to figure out if it is constant pressure or constant volume.

Mukil_Pari_2I

Calculus is probably not required but its a good idea to understand the conceptual backing behind the formulas. That would include the calculus derivations.

Mukil_Pari_2I

Anjali_Kumar1F wrote:When is Lyndon going to post his midterm review ws??

He will probably post today or tomorrow before his midterm review session on Monday. I would recommend searching up the word "hotdog" to find his worksheet because Dr. Lavelle recommended that to be the name of his worksheet.

Mukil_Pari_2I

When a gas expands against external pressure, it has to to transfer some of its energy over to the surroundings so it reduces its overall energy because of its negative work.

Mukil_Pari_2I

No. Isolated systems don't allow either energy transfers or matter transfers while closed systems still allow energy transfer. For example, closed systems could still absorb heat.

Mukil_Pari_2I · Re: Friday 02/01/2019 lecture What is your email?

What is your email?

Mukil_Pari_2I

Looking up and doing extra practice problems online also helps reinforce key concepts.

Mukil_Pari_2I

For the reaction, q would be negative if that helps.

Mukil_Pari_2I

Yes, if the force being applied is in the opposite direction of the displacement, the work will be negative.

Mukil_Pari_2I

Dr. Lavelle said in his email today that he will be announcing when the sessions will be most likely on Tuesday or Wednesday.

Mukil_Pari_2I

I believe we do need to know how to set up the equation by ourself, however I believe the Ka or Kb value will be given to us or enough information will be given so we can find either value. To set up the equation, you first need to determine whether the molecule dissolved in water is a weak acid or ...

Mukil_Pari_2I

Since the Al is a cation, it is surrounded by water molecules. So the equation would be Al(H₂O)₆3+ + H₂O --> H₃O+ + Al (H₂O)₅OH2+

Mukil_Pari_2I

I believe the Chem_Mod said that this week's homework should be from the Acids and Base Chapter and the material we just learned last week.

Mukil_Pari_2I

For part a, you're given that the pH of 0.10 M HCLO2 is 1.2. pH is -log[H3O+], so to solve for [H3O+] you have to do 10^-pH. In this case, it would be 10^-1.2=0.06 M H3O+. Then, you solve for the Ka (acidity constant), which is (0.06^2)/(0.1-0.06). The pKa is the -log of your Ka value. Part b is th...

Mukil_Pari_2I

For pH, the sig figs in the final answer are calculated based off the least amount of sig figs in the answer. Then that amount of sig figs is used after the decimal point.

Mukil_Pari_2I

Only Q should be affected because q is in terms of what is happening during the reaction. K is just the overall equilibrium constant and it should not change.

Mukil_Pari_2I

As Dr. Lavelle said in lecture, less than 5% is the rule. You can check with the quadratic formula to confirm though.

Mukil_Pari_2I

If the approximation is not within 5%, then the answer would not be valid. You would have to use the quadratic equation to figure out the value of x.

Mukil_Pari_2I

Dr. Lavelle's tests in 14A were a good balance of computational and conceptual questions. 14A also did not cover deriving things like schrodinger's equation: rather it focused on the concept. Other than that, there really should not be any real difference.

Mukil_Pari_2I

Well, the reaction quotient Q is supposed to come into play when we try to figure out which direction a reaction is likely to proceed, based of pressure or the concentrations of products and reactants.

Mukil_Pari_2I

I would assume so because in the reaction, the reactants become the products. Therefore it would make sense for there to be a negative x on the reactant side and a positive x on the products side.

Mukil_Pari_2I

If it has more than h+ proton that can dissasociate.

Mukil_Pari_2I

Proton transfers are shown by the transfer of a hydrogen proton. Here, there was no proton transfer. Instead, an OH bond was transferred from acetic acid and replaced with NH2. And the OH bonded with last H atom remaining.

Mukil_Pari_2I

No, it is not the only amphoteric compound. Other examples are HCO3 and HSO4.

Mukil_Pari_2I

Based off past final exams, the final exam will be a cumulative. Think of it as the midterm but on steroids with the stuff that we just learned.

Mukil_Pari_2I

BenJohnson1C wrote:Is this also true for the upcoming final? (fall 2018)

Calculating Ka values, based of what Dr. Lavelle has said, will not be on the final.

Mukil_Pari_2I

The only equations given are the ones on the sheet given on every test. The Formal charge equation is just FC = Valence - [Lone Pairs + (1/2 Bonding Pairs)]. It would behoove you to memorize it.

Mukil_Pari_2I

Based off previous finals, its cumulative. Even if Dr. Lavelle does not isolate a concept such as balancing equations, he may give us something where we have to balance the equation as a key step in another related problem. Basically blending concepts.

Mukil_Pari_2I

Or you could be safe and write bent/angular on the test.

Mukil_Pari_2I

If you want to figure out the which has more ionic character for a specific molecule, just use the general trend to generally find the difference. We will not be expected to calculate the electronegativity difference unless the electronegativity values have been given to us. In the case of the probl...

Mukil_Pari_2I

Does this mean that the material covered in this past Monday's lecture won't be included? No it as stated on Dr. Lavelle's website, the test will only cover 3.12-3.16, 6.4-6.5, and 4.1-4.7 which do not include Monday's lecture. Hi guys. What do we need to know about Acids and Bases for Test 3? Need...

Mukil_Pari_2I

When determining hybridizations, you have to take into account molecular shape. When looking at molecular shape, you have to look at both bonded pairs and lone pairs. Furthermore, you have to take into account all of the electron densities on the atom that you are looking at, usually the central ato...

Mukil_Pari_2I

A good way to study conceptually is just explaining the concepts to yourself without looking at your notes. For instance, writing the concepts on a whiteboard and explaining it to yourself as you go along writing.

Mukil_Pari_2I

Dr. Lavelle also posted on his website what the test would cover - Test 3 covers: End of Bonding from 3.12 (6 Ed.) and from 2D (7 Ed.); and all of Molecular Shape and Structure (see Syllabus and Outlines for details).

Mukil_Pari_2I

Soyoung Park 1H wrote:Is the test over both outlines 3 and 4?

The test will be up to whatever he covers during lecture until a couple days before the test. He will probably specify as we get closer to the test but presumable Outline 3 and 4, since they were not on the last test, will be covered on the next test.

Mukil_Pari_2I

Are you talking about the 6th edition 2.7 homework problem? The problem does not talk about resonance structures.

Mukil_Pari_2I

The most important lewis structure would be the most stable, lowest energy lewis structure with the right bond angles.

Mukil_Pari_2I

In addition to the aforementioned response, this post from 2016 may help: viewtopic.php?t=16153

Mukil_Pari_2I

Every chapter builds on previous concepts we have learned. Whilst it may not be an explicitly cumulative test, there probably still will be concepts in the third test from prior tests (ie dimensional analysis, energy calculations).

Mukil_Pari_2I

At the very least, its a good idea to know the general concept behind the Ep equation even if we are not using it for calculation questions. Also, general rule of thumb, whatever Dr. Lavelle goes over in class is fair game for the third test or the final exam.

Mukil_Pari_2I

Degenerate orbital are just orbitals that have the same energy level and thus (for hydrogen atoms), the same energy.

Mukil_Pari_2I

For this, you actually have to calc the mass of the nitrogen gas molecule. The way to do that is first find the molar mass of nitrogen (in this case of n2 bc nitrogen is diatomic). Once you have found the molar mass, divide the molar mass by avagadro's number to find how many gs each molecule weighs...

Mukil_Pari_2I

Its a pretty general term actually. As far as I know, its just not associated with a single atom or a covalent bond.

Mukil_Pari_2I

I don't think so. Just know everything Lavelle goes over in lecture.

Mukil_Pari_2I

I am not 100% sure but I believe its because the Aufbau principle indicates that the 4s orbital fills before the 3d orbital and thus the energy is lower in the 4s.

Mukil_Pari_2I

it should be 5d. I think "6d" was prolly a typo.

Lina Petrossian 3D wrote:Isn't 2.27 part a) 5d instead of 6d?

Mukil_Pari_2I

Dr. Lavelle mentioned that E = hv can only be used for light. However, De Broglie's equation of lambda=h/p can only be used for objects with "rest" mass and velocity. "Rest" mass basically means anything that has mass when stationary. why is rest in quotes.. is there a significa...

Mukil_Pari_2I

So what you need to do is account for the per mol portion of the threshold energy. The energy given was in terms of how much energy is needed to eject 1 mole of electrons of that metal. You need to find how much energy is needed to eject one electron. This can be easily done by dividing by the conve...

Mukil_Pari_2I

all you need to do is multiply the frequency by plancks constant. The correct answer if you just do hv is 1.99*10^-18 J.

Mukil_Pari_2I

The way to figure it out from my understanding is that 4 is the energy level it is in (the principle quantam number n equals 4 in this case) and f is determined by what l is equal to. In this case, the value of l, which is 3, corresponds with the the l value for the f orbital. Thus the answer, if pu...

Mukil_Pari_2I

Yes. De broglie's equation (lamba = h / p) is the equation that cannot be used for photons. However, c=lambda/v, E=hv, E = (hc)/lambda, and E=(1/2)mv^2 can all be used with light.

Mukil_Pari_2I

The minimum uncertainty is change in velocity or delta (the triangle) v. In this case, delta v would be 1.65*10^5 m/s.

Mukil_Pari_2I

Dr. Lavelle mentioned that E = hv can only be used for light. However, De Broglie's equation of lambda=h/p can only be used for objects with "rest" mass and velocity. "Rest" mass basically means anything that has mass when stationary.

Mukil_Pari_2I

As energy levels increase, the separating space between them becomes smaller and smaller. At "very high" levels, this gives way to a continuous range of all allowed energies. To corroborate this, during office hours, Dr. Lavelle mentioned that the energy levels eventually become a continuu...

Mukil_Pari_2I

I thought it was everything until the end of this week no? Like including quantum numbers which we should go over in final lecture of this week.

Mukil_Pari_2I

Don't get frustrated about it! It is really just a guess and check most of time and it does take some time. One tip that I have personally incorporated is starting with the least amount of total atoms for the elements in the equation. For instance, if there are 3 atoms of C vs 5 atoms of F I would s...

Mukil_Pari_2I

It matters for things like sig figs. For instance, if your answer was 1330. but you only had 3 sig figs to work with, you would put it into scientific notation ie 1.33 *10^3 because this way you have the same answer and you have the right amount of sig figs.

Mukil_Pari_2I

This is just a general rule of sig figs that we must use I believe. A lot of the documents on the website do indicate that we should round to the nearest even number. This is because we will be less precise if we just round up (we would overestimate) and if we just round down (we would underestimate...

Mukil_Pari_2I

You need to first convert tristearin and o2 (the reactants) to moles. Then using the molar ration of 2 moles of tristearin/163 moles O2, find how much of one reactant is necessary to full react with the other reactant. Then, you can find how much O2 you need to react full with tristearin. Once that ...

Mukil_Pari_2I

Use scientific notation when dealing with very big numbers. Otherwise, it will be a lot more difficult to do multiplication and division without it. In addition, always round the number to the least amount of sig figs in the problem. This youtube vid on scientific notation really help me understand ...

Mukil_Pari_2I

Also, for sig fig rounding, if the last digit is above 5 round up (ie 2 sigfigs then 6.56 is 6.7) and if the last digit is below 5 round down (ie 2 sigfigs then 3.44 is 3.4).

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