CHEMISTRY COMMUNITY

Payton Schwesinger 1J

This problem requires you to go from the molecular structures to the formula... would we be expected to be able to do this on the test? I was having trouble naming the compounds to put in the formula because they seemed more complicated.

Payton Schwesinger 1J

From my understanding that is correct!

Payton Schwesinger 1J

I believe you will be given the observed rate law (experimentally determined) and the elementary reactions. Usually I believe you are also told which reaction is the slow one, unless they want you to determine that (as in problem 15. 89).

Payton Schwesinger 1J

I believe he said in lecture today that we would be given the steps! I don't think we are expected to have the knowledge to come up with the equations; we just need to know how to analyze them.

Payton Schwesinger 1J

You can determine which type of reaction it is by looking at how the rate changes in response to concentrations. Based on this, you can determine which order reaction it is and use the correlating equation.

Payton Schwesinger 1J

If it is second order, then I believe that means that as concentration of reactants doubled, then the rate would quadruple (concentration =2, n=2---> 2^2=4)

Payton Schwesinger 1J

Does [A] always mean [A] initial? Or does it have to be explicitly shown? ( [A]0)

Payton Schwesinger 1J

I was wondering this too. Maybe you have to compare the ones with the same oxidation state/number of electrons if there is more than one option?

Payton Schwesinger 1J

I think you assume that it is standard states unless it is otherwise specified- for anything that is not in standard state they usually give you the concentrations/partial pressures.

Payton Schwesinger 1J

What is the difference between how you use information from a Galvanic cell and information from a Concentration cell? Do they allow you to calculate the same thing with different formulas? Or do they give different information?

Payton Schwesinger 1J

How does it affect it differently if you are balancing a redox reaction in a basic solution, acidic solution, or neutral solution? What changes in the way you go about balancing the equation?

Payton Schwesinger 1J

When you are given a redox reaction to balance, is there a general set of steps you can use to do it? What do you balance first, second, etc?

Payton Schwesinger 1J

Yes I believe they will be- the midterm covers ch 8, 9, and part of 11-- if you look at the syllabus above the homework problems are listed exactly which sections from each chapter were covered in class, and therefore will probably be tested. (There were a couple of sections not covered from those c...

Payton Schwesinger 1J

Your body is generating the 100 J/s, which means that the surroundings is absorbing 100 J/s (positive) which is why the entropy of the surroundings is a positive value.

Payton Schwesinger 1J

K usually refers to the equilibrium constant at equilibrium pressures, while Kc is the equilibrium calculated using the equilibrium concentrations. I believe that gases are usually given in partial pressures though.

Payton Schwesinger 1J

For the equation deltaG = deltaH - delta(TS), when calculating the standard enthalpy of the reaction, can you also use Hess's law or mean bond enthalpies? Professor Lavelle mentioned using standard enthalpies of formation in class but not the other two methods so I wasn't sure.

Payton Schwesinger 1J

In problem 9.43, when calculating the entropy changes for the different volume and temperature changes, they used the number 75.3 KJ/ K mol as the specific molar heat capacity. How did they calculate this value?

Payton Schwesinger 1J

You need to have a value in moles so that the units will cancel-- multiplying "n (mol)" times "R (J/ mol K) allows the moles to cancel so you get the correct units. If it is given, you use that, but sometimes you must solve for "n" using the ideal gas law.

Payton Schwesinger 1J

In part b, you are asked to calculate the enthalpy change, which they find using standard enthalpies of formation. Why is water excluded from this calculation?

Payton Schwesinger 1J

I was confused on this too.. Is there another way to do this without adding them together and ending up with fractions?

Payton Schwesinger 1J

I believe that one way energy could be transferred is by compressing the gas inside, which would increase the energy in the system.

Payton Schwesinger 1J

On the left side of your equation you have to also account for the amount of heat that will be used to melt the ice into water before the temperature equalizes.

Payton Schwesinger 1J

I think for this equation it is a liquid because it is a combustion reaction, which always produces carbon dioxide and water (liquid) as products. I think for other equations the states should be given unless it is obvious which one it is.

Payton Schwesinger 1J

I also think that if the system is doing work on the surroundings, then energy is being released and therefore the work has a negative value... am I thinking about this right?

Payton Schwesinger 1J

I believe that it is because during the phase change, all the heat added is being used to break the bonds between the molecules to convert them to a different state, and so the temperature of the substance will not begin to rise again until the phase change is complete. At the freezing or boiling po...

Payton Schwesinger 1J

If the reaction is reversed, then for the reversed equation you just switch the sign on the enthalpy from the original equation. To go from fusion to freezing, it would go from a positive to negative.

Payton Schwesinger 1J

The reactants have positive enthalpy because the bonds are being broken, which requires an input of energy, giving them the positive value. However, in the products, bonds are being formed, which releases energy, and therefore the enthalpy values are negative because energy is released.

Payton Schwesinger 1J

I was able to come up with the same values as the solution manual for Ka1; however, it seems like the order they listed the acids in is wrong. I got:
H3PO4 < H2SeO3< HSeO4- < H3PO3
because the smaller the Ka the weaker the acid. Am I missing something?

Payton Schwesinger 1J

When drawing the product of the reaction between PF5 and F-, how do you know that the resulting compound, PF6, has a negative charge?

Payton Schwesinger 1J

When looking at a group of acids and trying to determine which one is stronger, which is more important? Bond length or electron pulling power (electronegativity)? Do you determine only move on to looking at electron pulling power if the bond lengths are the same?

Payton Schwesinger 1J

I think you just have to figure out how many bonds the central atom is going to have, either with individual molecules or with a compound (such as CN) , which you can usually do just by looking at the formula.

Payton Schwesinger 1J

Does the number of lone pairs correspond to the number of ligands it can attach to? For example, if a molecule has an oxygen with 2 lone pairs then does that mean the molecule is bidentate?

Payton Schwesinger 1J

When you are drawing a Lewis structure, how do you know if you are going to be dealing with an expanded octet or not? I'm afraid I'm going to give a compound an expanded octet because if fits the number of electrons that should be present but end up with the wrong configuration.

Payton Schwesinger 1J

In order to determine if something is a radical, you just have to find out if there is an odd number of electrons; if there is, then it means you will be left with one unpaired electron and therefore the molecule will be a radical.

Payton Schwesinger 1J

I was confused after reading the book about the purpose of wave functions... the book seemed to go more in depth than Professor Lavelle did in the lecture and the homework barely addresses the concept. What specific ideas do we need to understand for the test? Just that wave functions are used to pr...

Payton Schwesinger 1J

The 4s orbital is at a lower energy than the 3d, so the 4s fills up before the 3d. Once the 3d fills up then it switches back and 3d is at a lower energy level than 4s. It is the same as this for the rest of the periods, so usually the d orbital is one number behind whichever period you are dealing ...

Payton Schwesinger 1J

Going along with this, is it correct that whichever Lewis structure results in the most formal charges of zero is the most stable? In other words, how do you use the calculated Lewis structures to determine stability?

Payton Schwesinger 1J

The first element in the P block has one electron in its P orbital, which is written as Px^1. As you move right, you will add one electron to the Py orbital and then the Pz orbital, and then begin to fill up the orbitals from the beginning until you reach the element at the end of that row, where th...

Payton Schwesinger 1J

I think that orbitals are made up of shells, which are made up of subshells. The subscripts used today represent how in the p orbital, there are three sub orbitals, each of which can hold two electrons.

Payton Schwesinger 1J

I think that orbitals are made up of shells, which are made up of subshells. The subscripts used today represent how in the p orbital, there are three sub orbitals, each of which can hold two electrons.

Payton Schwesinger 1J

So you can deduce if an electron is in the Balmer or Lyman series by looking at the wavelength of light it absorbs? and then in terms of energy levels if it is a Lyman series that means the electron is starting from or returning to the n=1 energy level and if it is in the Balmer series that means it...

Payton Schwesinger 1J

I believe that the reference sheet will be given to us in class.

Payton Schwesinger 1J

Whenever I click on the Chemistry 14A site at CCLE it goes directly to Professor Lavelle's website, and I cant find the link for media resources on his website anywhere. Where should I go?

Payton Schwesinger 1J

Okay thank you Miranda- so then you only have to look at the number of significant figures and just do the one with the least amount? You don't even have to really worry about decimal places?

Payton Schwesinger 1J

I was confused on this question as well. I understand that doubling it 90 times dramatically reduces the amount of solute per L of solution so that it almost becomes obsolete but I don't understand the solution manual's method of solving the problem. Could someone please help explain why they are tr...

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