CHEMISTRY COMMUNITY

rkusampudi

The more the molecular complexity, the more the disorder, and thus the more the entropy.

rkusampudi

I further defined a reversible
process as one that can be reversed by an infinitely small change in a variable

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I believe that the 4 points for the question indicated 4 answer choices that were correct.

rkusampudi

An adiabatic process indicates that q=0.

rkusampudi

The name's Bond. Ionic Bond. Taken, not shared.

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A photon checks into a hotel and is asked if he needs any help with his luggage. He says, "No, I'm traveling light."

rkusampudi

Making bad chemistry jokes because all the good ones Argon

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Did you know that you can cool yourself to -273.15˚C and still be 0k?

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Don't trust atoms, they make up everything.

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On the tests they are usually given.

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This is a useful equation to figure out the units:

K = M^-n+1 x s^-1
M = mol/liter
n = total reaction order

rkusampudi

They seem to be interchangeable, but I'm not sure

rkusampudi

Perhaps because it is looking at the concentration and not the number of moles.

rkusampudi

Im assuming that you would treat each element individually in that case.

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A large equilibrium constant means that the products are favored.

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The units for voltage is joules per coulomb. The unit for Farraday's constant is coulombs per volt. Multiplying these and the number of moles together leaves joules per mol which is the unit of Gibbs free energy.

rkusampudi

The test should not have this on there.

rkusampudi

Of the equations you listed, only the first two equations allow for you to calculate K, the last equation is Q which is the reaction quotient not the equilibrium constant.

rkusampudi

This is so helpful!
I think one good thing to add would be that reversible work occurs under ideal conditions and is very rare in real life.

rkusampudi

Will we be tested on the derivation of this equation?

rkusampudi

For part A is it also possible to use the work equation?
As in: q/T = -w/T

How would that work?

rkusampudi

I would add that if OH- and H+ ions are on the same side, then they are written as water.

rkusampudi

Because the transition metals have variable oxidation states, which ones are we expected to memorize or will they be given?

rkusampudi

In today's (Wednesday week 5) lecture, Dr. Lavelle put water in the half-reaction for reduction. Is there a reason to do so?

rkusampudi

Because it doesn't say otherwise and because all the given information assumes a standard state, it is a safe assumption to make.

rkusampudi

I think the simplest approach is to look at the shape of the molecule as well as the number of different elements involved (e.g. switching the positions of two oxygens doesn't count as a different arrangement). Here there are four atoms around the central sulfur atom and they are of two types

rkusampudi

I think using nR allows us to cancel out the moles in the denominator but I am not sure.

rkusampudi

Will our midterm basically cover chapter 8,9, and 11?

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There are no calculations on heat and internal energy but you do have to understand the concepts.

rkusampudi

If you're not part of the solution, you're part of the precipitate.

rkusampudi

Q: Why does hamburger yield lower energy than steak?
A: Because it's in the ground state.

rkusampudi

Did you know that you can cool yourself to -273.15˚C and still be 0k?

rkusampudi

I think there are some homework problems that use those equations indirectly in order to derive other values. But I don't think those equations will be tested on directly because they were not touched on in a detailed manner in class.

rkusampudi

Will the formula and equation sheet be posted on the website like it is for Chem 14A?

rkusampudi

The solution indicates the use of the equation q=mC delta T. However the mass is not shown in the solution or indicated in the question. How do you solve this question without knowing the mass?

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From the textbook: The state function that allows us to keep track of energy changes at constant pressure is called the enthalpy, H: H = U + PV U, P, and V are the internal energy, pressure, and volume of the system With this being the case, I am assuming that the great the volume, the greater the e...

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Does anyone know what topics will be covered in Test 1? Is it only Chapter 8?

rkusampudi

In a conceptual paragraph in the textbook, it said that when a solid melts, the molecules can move more freely,
but the interactions between them remain as strong. Does anyone understand why this is true?

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Just to make sure, salts will not be covered on the final exam due to the fires.

rkusampudi

A change in temperature will always favor the exothermic reaction whether it is an increase or decrease.

rkusampudi

Drawing the Lewis structures definitely helps because you are essentially looking at the symmetry of dipole moments in the molecule. If there is no one direction of electron density it is non-polar and if there is it's polar.

rkusampudi

The solution manual sometimes uses cyano but the table in the textbook that has the steps for naming uses cyanido. Are they interchangeable?

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Do we just have to memorize ligands and their charges, or is there a manner in which we can find them?

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Could someone clarify the relationship between lone pairs and bond angles? I understand that the bond angles are "strained" by the lone pairs but am not quite sure how to show that in our answers.

rkusampudi

Just to clarify, Dr.Lavelle uses the terminology "bent" when describing molecular shape while the textbook seems to use "angular" to mean the same thing. Are those terms interchangeable?

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Does resonance have an impact on the molecular shape? How do we decide which Lewis structure to base our VSEPR theory on?

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As I was reading the textbook, I noticed that many molecular shapes have bond angles that are different from what is expected of that shape. Is there any way to predict these exceptions or do they have to be memorized?
EX: the HOH bond angle

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In one of the answers on this forum, the moderator answered that an atom that donates two electrons is a Lewis base. I'm a little confused about the relationship between coordinate covalent bonds and Lewis Acids and Bases. Thank you for the help!

rkusampudi

The trend in electron affinity is best applied to elements that do not have a full octet and thus benefit from additional electrons that allow them to be more stable. Thus the halogens have extremely high electron affinity because the only require one more electron to fulfill their octet, but the tr...

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Could someone explain the layout of the f-block on the periodic table?

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I just have a pretty simple question about the manner in which we write electron configuration. In lectures in was written in order of the periodic table (putting 3d after 4s) but in my discussion, we were told to order it in terms of principal quantum number. Which way is correct?

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The correct answer is C. The textbook says if one property is known the other cannot be known simultaneously.

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During Wednesday's lectur of Week 3, can someone explain what Dr.Lavelle meant by the term e-p distribution?

rkusampudi

I would like to add that there is also the Paschen series which has a lower energy level of n=3.

rkusampudi

In one of the questions on our test, it asked for the net moles of gas produced. However, in the calculation, it added the total number of moles produced on the product side (including that of H2O which I assumed was in liquid form). When questions ask for net moles produced do we just include the m...

rkusampudi

In one of the questions on our test, it asked for the net moles of gas produced. However, in the calculation, it added the total number of moles produced on the product side (including that of H2O which I assumed was in liquid form). When questions ask for net moles produced do we just include the m...

rkusampudi

All combustion reactions follow the basic setup of hydrocarbon + O2 (g) -> CO2 (g) + H2O (l).
You put the hydrocarbon in this set up and balance the equation as necessary.

rkusampudi

I guess the quickest way is to estimate the molar mass of the empirical formula in your head without doing any calculation. and compare that to the given mass. If it is not close, you know that the molecular formula is a multiple of the empirical formula.

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