CHEMISTRY COMMUNITY

SimranSangha4I

catalysis is basically the process of increasing the rate of a chemical reaction by adding a substance known as a catalyst, it accelerates the reaction.

SimranSangha4I

the difference is that the slow step is considered the determining step, and that is what we use when we are trying to find the rate of a mechanism.

SimranSangha4I

correct because that would mean its formed and then used up.

SimranSangha4I

Well it really depends on where it is actually taking place but if it's in the side with a lower concentration would increase the concentration. In a cathode solution, it would decrease the cell potential.

SimranSangha4I

He said he will let us know when that information is available! So until then, I would just anticipate that it will be the same date/time.

SimranSangha4I

Since it was formed by the actual reaction, it isn't part of the rate law.

SimranSangha4I

When halved, the concentration would not be changed so the Ecell wouldn't be changed.

SimranSangha4I

A is essentially the frequency of collisions in the proper orientation and it is unique.

SimranSangha4I

It would be moles per liter per time (in seconds)!

SimranSangha4I

The slow step would be the rate determining step, so that would be used in finding the rate-law.

SimranSangha4I

Gibbs free energy! If the delta g is negative then it is going to be spontaneous.

SimranSangha4I

An oxidizing agent, or oxidant, gains electrons and is reduced in a chemical reaction and vice versa.

SimranSangha4I

It can be confusing at time but the best way to look at it would be associating it with the opposites of the term. So the element being oxidized is the reducing agent.

SimranSangha4I

When the cell is positive, the reaction is spontaneous, which also means there's a negative Gibbs free energy.

SimranSangha4I

From what I remember from 14A, it should be more like Test 1 in terms of difficulty.

SimranSangha4I

Normally his tests are a pickup on what is the half way before the next major test. The midterm and final are cumulative and will require you to know everything up until that test but the tests between (so test 1 and 2) will only cover a particular outline. So, test 2 covers everything taught AFTER ...

SimranSangha4I

You should try to be as specific as possible and show the redox half reactions to show and indicate you understand what is actually happening. I normally think of the reactions as OILRIG (oxidation is loss [of electrons]) (reduction is gain [of electrons]) so in this particular case, Fe is the one t...

SimranSangha4I

Lavelle doesn't offer extra credit opportunities.

SimranSangha4I

In relation to Le Chatelier's principle, I think it would just be how the reaction would react to a change in the shift in products and reactants based on concentrations.

SimranSangha4I

A salt bridge is a device used in experiments that allows ion transfer between two half reactions in the galvanic cell, neutralizing them.

SimranSangha4I

Positive would occur when the work is actually able to be done. When there is work being done on the system itself, the work would actually be negative.

SimranSangha4I

I think the best things you can do for 14b are office hours, review sessions, and practicing problems like the ones found in our assigned homework problems for the different outlines we cover.

SimranSangha4I

In an oxidation reaction, an electron is lost. In a reduction reaction, an electron is gained.

SimranSangha4I

LBacker_2E wrote:Is this related to the relationship between ∆G and the direction of a spontaneous reaction? Does a spontaneous reaction have a -∆G and proceed in the forward direction?

Yes, when a reaction is spontaneous, it has a negative delta G

SimranSangha4I

Yes! That sounds right since the heat of the universe doesn't change.

SimranSangha4I

Yes, exclude the two when calculating the Q or K.

SimranSangha4I

Cm would be the molar specific heat for the equation.

SimranSangha4I

Within an adiabetic system, delta U would equal work because there is no heat transfer that can take place within the equation Delta U=q=w.

SimranSangha4I

The concept of an extensive property would be things that alter depending on the amount of material.

SimranSangha4I

All material from outlines 1-4!

SimranSangha4I

It's primarily focused on gas molecules and their coefficients.

SimranSangha4I

It's just the constant 1.380649x10^-23 J/K, hope that helped!

SimranSangha4I

A pretty common example you can probably relate to would be a really good thermos or even a hydroflask!

SimranSangha4I

The first law of Thermodynamics states that heat is energy and in relation, cannot be created or destroyed.

SimranSangha4I

Work can be defined as the product of the force used to move an object times the distance the object is moved.

SimranSangha4I

It depends on the UA normally, some provide worksheets while others don't!

SimranSangha4I

Yes, if the answer comes back more than 5% you should go back and use the quadratic formula.

SimranSangha4I

It would be proportional so the change would essentially be the same.

SimranSangha4I

A state function does not depend on the path taken to reach that specific value, so heat would not be considered a state function.

SimranSangha4I

The phase would require a certain amount of energy to change. Within enthalpy, the phase change needs to be combined with the bond enthalpy.

SimranSangha4I

It actually wouldn't affect the equilibrium because it does not affect the concentrations.

SimranSangha4I

Yep! Leave out H2O, solids, and other liquids.

SimranSangha4I

The reaction would end up shifting towards the side in which there are fewer mole because that would effect the equilibrium!

SimranSangha4I

The lower the pKa, the stronger the acid

SimranSangha4I

H2O would have the concentration of 1 as well as it is normally in excess.

SimranSangha4I

Q is the constant through the entire equilibrium while K is the constant at equilibrium!

SimranSangha4I

You shouldn't use units when calculating K as a constant but for molarity, you should use moles per liter (moles/L).

SimranSangha4I

Kc is often the equilibrium constant found in concentrations while Kp deals with the partial pressures.

SimranSangha4I

Kp is used within partial pressures while Kc is the constant within concentrations.

SimranSangha4I

I think as long we continue to cover equilibrium during lecture this week it should be fine, but I also believe you should contact your TA!

SimranSangha4I

Yes! It assists with naming while showing the the biological significance, like cisplatin.

SimranSangha4I

I think chlorine can be an exception to the octet rule as long as the charge is 0

SimranSangha4I

I think the main thing you need to recognize is when they occur so sigma would be a single bond, a double would be both a sigma and pi bond, and a triple bond would be two pi bonds and one sigma bond.

SimranSangha4I

I know it's fairly late for this advice now but I think that one of the best ways to study would have been by doing as many homework problems as possible from the many sections, reading the text book, and looking over the past tests that we have been given!

SimranSangha4I

Bent and angular are the same thing!

SimranSangha4I

That's just about it! The homework problems really help with practicing those skills.

SimranSangha4I

Yes! The lone pairs are still considered sites for bonding to occur

SimranSangha4I

It can actually act as both, it really depends on the situation.

SimranSangha4I

Definitely cumulative! Best way to study in my opinion is to go over the assigned problems and previous test questions!

SimranSangha4I

The 0-14 primarily acts as a scale for most solutions.

SimranSangha4I

Ligands are ions that bond with central metal atoms.

SimranSangha4I

Sigma is a single bond (primarily the 'inital' bond) while a double bond would be the 'next' bond, being one sigma and one pi bond.

SimranSangha4I

linear wouldn't contain a lone pair while bent would.

SimranSangha4I

you need to make sure that you just label the bond sigma or pi.

SimranSangha4I

Yes they do! They get involved during the hybridization of period 3!

SimranSangha4I

Yes, every molecule experiences these forces, the strongest to weakest would be as follows: ion-dipole, hydrogen bonds, dipole-dipole, and lastly, Van Der Waals.

SimranSangha4I

Since there is still repulsion present between the bonds, the double bond isn't double.

SimranSangha4I

the electron arrangement is primarily the placement, the molecular shape would be a 3D representation of where the electron density is actually placed in relation to the electron-electron repulsion.

SimranSangha4I

There wound be four bonding pairs and one lone pair!

SimranSangha4I

Yes, it's a type of dipole-dipole but stronger than the normal dipole-dipole.

SimranSangha4I

for the most part, it depends on what type of element it is!

SimranSangha4I

I believe we'd need to know what they are and under what circumstances they occur since we will probably have to identify IM forces in assigned problems, hope that helped!

SimranSangha4I

I'd imagine that it only cover topics after the midterm! :)

SimranSangha4I

The trend would demonstrate that from right to left and top to bottom, the atomic radius would increase.

SimranSangha4I

Most likely, it'll be this week since there are so many of us!

SimranSangha4I

Acids accept while the bases donate electron pairs!

SimranSangha4I

it would be between -5 m/s to 5 m/s, so it'd be about 10m/s.

SimranSangha4I

A good place to read about it is 1F.3 but it essentially means when atoms have the same amount of electrons! Hope that helped!

SimranSangha4I

Yes, the trend would continue!

SimranSangha4I

It'll include everything from outlines 1 and 2, while including most of 3

SimranSangha4I

covalent bonds are only formed between two non-metals!

SimranSangha4I

I haven't been to every single one but I believe most of the sessions is the UA/TA going over the basics and then progressing into what we're covering in the actual class. I have heard that of some of the other students on occasion have been given worksheets.

SimranSangha4I

When picking a central atom, it's most common to pick the one with the lowest ionization energy.

SimranSangha4I

SimranSangha4I

For the blocks, the amount of valence electrons is based off of the group, or much rather the last shell's electron amount.

SimranSangha4I

You can either do Chemical Bonding or Quantum problems for week 4!

SimranSangha4I

What happens when electrons lose their energy?
A: They get Bohr'ed

SimranSangha4I

Honestly, I'm not sure because I think everything he's posted has been uploaded at once. Hopefully he will be posting some more modules.

SimranSangha4I

Personally, I go to the tutoring sessions and the step up sessions up in Covel! They really do help since they hit just about everything we cover in lecture, just in a slower and more explained way. Another thing I would recommend is getting some extra help from your TA at office hours. Good luck!

SimranSangha4I

The wavelength measures the distance between each wave! So the unit for wavelength would be in meters (m) but we also frequently use nanometers (nm). The conversion would be 1nm= 10^-9 m!

SimranSangha4I

it'll probably be in discussion this week!

SimranSangha4I

I would use M (initial) X Volume (initial) = M (final) X Volume (final)

SimranSangha4I

e=hv references the energy per photon and it was derived from the photoelectric effect.

SimranSangha4I

It's the disturbance.

SimranSangha4I

Yes, it can! Since molecular formula tells you the amount of molecules in a compound and the empirical formula tells you the simplest ratio in a compound, if a compound's molecular formula cannot be reduced anymore, the empirical formula would be able to be the same as the molecular.

SimranSangha4I

A scientific calculator should work well!

SimranSangha4I

Not necessarily, but the reactant with the smallest number of moles can be the limiting reactant if all reactants within the chemical equation react in a mole ratio of one to one.

SimranSangha4I

Well, actual yield is the mass of the product that was actually obtained while theoretical is just going to be what is expected, without any complications or opportunities for the mass of the product to decrease. So normally, the actual yield will always be less than the calculated theoretical due t...

SimranSangha4I

The way I had memorized them in high school was organizing them by the suffixes of the compounds like -ite, -ate, and -ide! (i.e. AgNO3 - Silver nitrate) Hope this helped!

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