CHEMISTRY COMMUNITY

Michael Lee 2I

Just make sure all of your units are consistent.

Michael Lee 2I

The units of k are always mol * L/s

Michael Lee 2I

The units of k are always mol * L/s

Michael Lee 2I

It technically can be used for all orders because order does not play a part in the equation.

Michael Lee 2I

Intermediates are not included in the overall rate law.

Michael Lee 2I

If a zero order is in the reaction, the rate law is k[A]^0 = k. Therefore it has is not based on the concentration, only the rate constant.

Michael Lee 2I

I agree

Michael Lee 2I

I think it's more of just the exponent of the [A]. For example k[A]^2 would be a second order reaction

Michael Lee 2I

Basically whatever the exponent of [A] is the order of reaction.

Michael Lee 2I

Ya, around number 43.

Michael Lee 2I

The order of reactions describes how the concentrations and the rate of the reaction are correlated/related.

Michael Lee 2I

Yes. I agree

Michael Lee 2I

n is just the amount of electrons transferred in the reaction. Usually you need to find the half reactions first.

Michael Lee 2I

This also brings up the fact that a positive E which is a negative G means the process if favorable

Michael Lee 2I

The E value will always remain the same because it is an intensive property.

Michael Lee 2I

Correct, only aqueous and not solids or liquids.

Michael Lee 2I

The salt bridge keeps the two solutions neutral so that the redox reaction can continue.

Michael Lee 2I

I agree. E is an intensive property which makes it independent of the mass.

Michael Lee 2I

No, integrals were just done to understand the derivation.

Michael Lee 2I

The heat is changing within the isolated system. Not changing from the surroundings.

Michael Lee 2I

I think you just have to balance the equations.

Michael Lee 2I

Both delta S and delta H remain constant from what Lavelle told us in class today.

Michael Lee 2I

Probably both chapter 8 and 9. Chapter 11 might be in there too

Michael Lee 2I

I think that most of the equations will deal with closed systems.

Michael Lee 2I

qsys + qsurr = 0 is a "perfect system" with no extra released/absorbed.

Michael Lee 2I

Most questions give the overall heat change but you should know how to solve for it. Doing a lot of problems in the textbook will give you a clearer idea on whether you need to know how to solve for the overall heat change.

Michael Lee 2I

For specific molar heat capacity you either need gas at constant volume: Cv or at constant pressure Cp.

Michael Lee 2I

It just works with the questions to assume the specific heat capacity is the same as the solvent.

Michael Lee 2I

The only thing you really need to know, since the path taken does not matter, is final state - initial state.

Michael Lee 2I

Usually the problem states whether it involves Reaction Enthalpy or Enthalpy of Formation. Sometimes both are needed if given the information.

Michael Lee 2I

Are we going to have to use more than one method of finding the enthalpy in any problem?

Michael Lee 2I

How is heat capacity a state function if it is an intensive property?

Michael Lee 2I

Can someone explain Hess’ law and how it pertains to the net reaction? I followed during class but was a bit confused about this.

Michael Lee 2I

Just to add on, sigma bonds also allow the molecules to rotate, whereas they are in a locked position with pi bonds.

Michael Lee 2I

Hydrogen only has one electron. Because of this it cannot satisfy the octet rule and won't be the central atom

Michael Lee 2I

I think they would rather have us say that the equilibrium sits/lies to the left or right. Using the word shift implies that there is a movement and change which is not true.

Michael Lee 2I

There are no difference between the formuals. However, Q refers to the reactants and products whereas K refers to the chemical equilibrium.

Michael Lee 2I

You only need to write it out if the question needs the full electron configuration.

Michael Lee 2I

Yes, bent and angular both mean that the atoms are not bonded to the central atom linearly.

Michael Lee 2I

If the difference is <2 then the bond should be covalent. If the difference is >2 then the bond is ionic.

Michael Lee 2I

It's true that the expanded octet applies to those with an energy level of n=3 and higher. I think the most common ones we are going to see will be Phosphorus and Sulfur.

Michael Lee 2I

Nope! the exam covers material from chapters 1-3, which does not include the VSEPR model.

Michael Lee 2I

Are there any exceptions to the electron affinity increasing across the period?

Michael Lee 2I

So when an electron is involved in resonance structures, it is considered delocalized.

Michael Lee 2I

A cation is an atom that has lost an electron, therefore it is positive. An anion has gained an electron, making it positive. Atoms can lose or gain more than one electron.

Michael Lee 2I

No the most we would need to know is 400 nm being violet light and 700 nm is red light. Everything in between would either be given or it would not be asked for.

Michael Lee 2I

You definitely should ask your TA because they are the ones grading, but I would say go with v because that is standard.

Michael Lee 2I

Using 2.998 x 10^8 is definitely the more accurate number. However rounding it to 3 won't cause too much of an error bound, so I don't think you would get the question wrong. But 2.998 is definitely the safe option because it is more accurate.

Michael Lee 2I

Do we still use Balmer's series if the electron is going excited from n=1 to n=2. I know that we use it when the electron goes from n=3 to n=2. Or do we use Lyman's series even though the electron ends up in the visible light spectrum?

Michael Lee 2I

The molar mass of the metal hydroxide M(OH)2 is 74.10 g.mol^-1. What is the molar mass of the sulfide of this metal?

I was confused on finding the molar mass of the sulfide. Is there another number I am supposed to know to use to solve this problem?

Michael Lee 2I

First you find the mass of KNO3, which is the percentage (5.45%) x the total mass of the aqueous solution (510g). Then you need to find the mass of the water before the KNO3 is added to it which is the total mass of the aqueous solution (510g) subtracted by the mass of KNO3. So to prep the solution ...

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