CHEMISTRY COMMUNITY

Julianna_flores3E

I am pretty sure that they mean the same thing in the context of the work we are doing, however, as stated above EA could also mean electron affinity.

Julianna_flores3E

The anode is typically on the left, however, in one of the UA sessions, there was a diagram that had the anode on the right and the cathode on the left in order to make sure that the students knew the difference between oxidation and reduction reactions.

Julianna_flores3E

I am pretty sure that pseudo rate constants are just the rate constants that we find when we put multiple reactants in excess and measure the effect that another reactant has on the rate of rxn. It's is called a pseudo rate law because we know the rate law with respect to whatever reactant was not i...

Julianna_flores3E

I think that a catalyst is present in the beginning because the intermediates are part of the equation for a very short time, but the catalyst is in the reactants and the products of the rxn. I could be wrong though.

Julianna_flores3E

I am pretty sure that it varies by reaction. As long as it is produced and consumed by the final product then it will be an intermediate.

Julianna_flores3E

If the university is offering dorms I am definitely going to try to move in. Hopefully, they have double occupancy as well as single.

Julianna_flores3E

Anode=Oxidation and Cathode=Reduction. Most of the times the anode is on the left as well.

Julianna_flores3E

The salt bridge allows for the reaction to occur continuously because it neutralizes the charges of both sides.

Julianna_flores3E

E naught will be 0 at equilibrium because ln(1) =0 and this will make the equation equal 0.

Julianna_flores3E

The slowest step is the rate-determining state because you can only go as fast as your slowest part.

Julianna_flores3E

I try to make a part of my day not school-related. If I let school take the entire day it would, but If you do something that you like I find it easier to focus and turn out quality work.

Julianna_flores3E

Exactly, the circle above means standard conditions.

Julianna_flores3E

Midterm 2 was not as bad I thought it was going to be. With that being said I definitely did go study more than the first midterm. I think that the questions were fair, but there was definitely more conceptual stuff than I thought there would be.

Julianna_flores3E

n= moles of electrons so in the chemical equations however many electrons are transferred equals the number of moles of electrons transferred.

Julianna_flores3E

Standard Gibbs free energy does not change it is fixed under standard conditions. While Gibbs free energy does change and can be calculated using $\Delta G=\Delta G^{\circ}+RT ln(Q)$

Julianna_flores3E

I would recommend going through the outline and textbook slowly so that you can really see what you know and don't know. Also, I would do the problems provided in the syllabus.

Julianna_flores3E

State property is a property where it doesn't matter the path it takes the final and initial values will be the same. Essentially, for a state property in order to get to a final value, there are multiple paths that it can take but the value of the final will not be indicative of the path, but rathe...

Julianna_flores3E

A salt bridge keeps both solutions neutral by allowing an ion not involved in the redox reaction to be transferred. For example, a salt bridge could transfer Cl- to replace the negatively charged electrons that come into the solution in order to maintain neutrality.

Julianna_flores3E

I'm pretty sure it is derived through the delta g = -RTln K and delta g = delta H - T delta S

Julianna_flores3E

At a constant volume, work can be defined as force x distance since there is no change in volume (distance) then the equation if force x 0 = 0 therefore, work =0.

Julianna_flores3E

If Q<K then the reaction will favor the products and when Q<K the reaction will favor the reactants.

Julianna_flores3E

Residual entropy is just based on the possible microstate and shows the amount of disorder.

Julianna_flores3E

The delta S of the surroundings would just be -delta H/ T and make sure your units are correct and you should get it.

Julianna_flores3E

For this problem, we would use delta H for q (heat) and -P delta V for work. Then if we use PV=nRT in the form (-P delta V = -delta n RT) since the values equal to -P delta V are constant and assuming that 298K we can just add the values together to get delta U. I think. Hope this helps.

Julianna_flores3E

Try to go to step up sessions and workshops if you can they really help.

Julianna_flores3E

The Cp denotes that it is the specific molar heat at a constant pressure. There are also other specific molar heat such as Cv which is the specific heat capacity at a constant volume.

Julianna_flores3E

It is essentially theoretical because it would take such a long time to actually occur and there is no perfect system that would convert all heat to work.

Julianna_flores3E

W is positive when work is being done on the system and negative when the system is doing work.

Julianna_flores3E

Hi, I think that in most problems the C will have the same units that you are given, or is calculated using the units that you have. The specific heat capacity or the molar heat capacity can be used depending on your units and the equation, I think. You should use Cp is the specific heat at a consta...

Julianna_flores3E

Work is the energy transfer from one thing to another in a system something could do work on it like compressing something with a piston or in the case of the piston work can be done by the system onto the piston which would cause expansion. Hope this helps!

Julianna_flores3E

Residual entropy is basically just the different positions the atoms in a molecule can take. Something that is symmetrical like O2 will have a residual entropy of 0.

Julianna_flores3E

The energy is not being used to raise the temperature, but rather to break or create bonds.

Julianna_flores3E

Endothermic reactions will always be positive and exothermic reactions will always be negative.

Julianna_flores3E

It would be 1/K

Julianna_flores3E

Since Kelvin and Celsius change in the same way you would not need to convert. They change in the same way just with different values.

Julianna_flores3E

Yes, we will. Earlier today Dr.Lavelle sent out an email regarding the midterm as well.

Julianna_flores3E

Once you have found the pOH you need to find [OH-] which would be 10^-pOH then you can see that [OH-] and [BH+] are the same. You then need to find the value of B and then find the % protonated. Hope this helps.

Julianna_flores3E

Yes taking the -log of [OH-] and [H3O+] will result in the pH and pOH. Likewise, if you have the pH or pOH and need to find the concentrations of either you would put the pH or pOH as the negative exponent of 10. In some of the Hw for week 2 this could have been useful. Hope this helps.

Julianna_flores3E

Yes, the heat energy is being used to break bonds and therefore, cannot raise the temperature.

Julianna_flores3E

I would recommend working through the outlines and doing the book problems that they suggest. Then if you don't understand something go to office hours or UA sessions.

Julianna_flores3E

I think that your equations work just make sure it is your final ionized value(found through the ice table) divided by the initial molarity and to multiply by 100.

Julianna_flores3E

Basically $10^-4$ is so close to 0 that it can be treated like 0, but you can still do the calculation with the value the result would just be insignificant.

Julianna_flores3E

Hi, I know that when Q<K then its a forward reaction and it shift right. It shifts left when it is the opposite. But, what exactly does this shift indicate? I know it's important to know whether it is a forward or reverse reaction but what is the importance of identifying the shift? Do we always ne...

Julianna_flores3E

I do believe that you need to know the weak/ strong acids and bases or at least know the structures and classes that contribute to making a strong/weak acid or base.

Julianna_flores3E

Yes, I believe they are the same format, maybe check the syllabus to see if it says anything. Hope this helps!

Julianna_flores3E

Yes, Q will equal K at equilibrium, while K will not change values (unless the temp is changed) Q will.

Julianna_flores3E

Since there are fewer points available there should be fewer points available on midterms and final, but I am not sure if that is where the points will come from. Maybe check the syllabus to see the breakdown of points for the midterms and final.

Julianna_flores3E

I think that unless specified by a change in temperature it is ok to assume that K does not change. If the temperature is mentioned and there is an indication that it changed then K would change.

Julianna_flores3E

Yeah based on what we have learned I think what you are saying is good. If you want to see if you are going to need to know more maybe check the syllabus and learning objectives.

Julianna_flores3E

Correct, when K is greater than 10^3 the products will be favored. Lower energy equates to more stable species because higher energy molecules have more repulsive forces than attractive therefore more stable species have more attractive forces than repulsive forces.

Julianna_flores3E

The addition of a reactant will cause the production of more products because it will make Q<K and in order for Q to equal K the reactants will need to form more products. It will be the same for products as well. Hope this helps!

Julianna_flores3E

Lakshmi Davuluri 3K wrote:Did Lavelle already send out an email regarding the instructions for the final?

I do not think he has.

Julianna_flores3E

To determine an electron-donating group you would see if the central atom is stable and what charge it has. This may lead to an understanding if it would donate electrons. Hope this helps.

Julianna_flores3E

I think that there are many metals (transition metals) but most consistently metalloids that form amphoteric compounds.

Julianna_flores3E

Yes, bent and angular are the same things. In the molecular geometry review, the diagram says bent or angular.

Julianna_flores3E

Electron geometry is the positions of electrons within an atom, while molecular geometry is the positions of atoms within a molecule.

Julianna_flores3E

Just to expand denticity is the amount of donors that can bind to central atom. This, I’m pretty sure, would be the binding sites allowed per different ligand.

Julianna_flores3E

I think that with a double bond the bond is stronger and also the formal charge would be 0. If there is a single bond and lone pairs the formal charge would be negative one and the the strength of the bond would be weaker.

Julianna_flores3E

I don’t recall them all during the lecture, but on the chempendix it says that the 6 strongest acids are Perchloric acid HClO4, Hydrochloric acid HCL, Phosphoric acid H3PO4, Nitrous acid HNO2, Hydrofluoric acid HF, and Carbonic acid H2CO3. Hope this helps!

Julianna_flores3E

Yes, as stated before it has four because it is the s and p orbitals. Remember that the p orbital is p(y) p(x) and p(z) this added to the s orbital equals four.

Julianna_flores3E

I agree you just put a space between the inside of the bracket and outside, but also the ion outside the bracket is named normally not with prefixes.

Julianna_flores3E

Why is it that if in H2C(C*)H2 that an even numbered carbons will cause the hydrogen to be coplanar?

Julianna_flores3E

None of the structures specifically give bond strength. VSEPR gives the geometry of the molecules by showing the regions of electron density in the form of atoms and lone pairs.

Julianna_flores3E

There should be no resonance structures, the last one you are missing is the carbons forming a triangle shape with their bonds. hi when do we know not to include a resonance structure? or should we just assume that for all questions? thank you so much for your response btw I think resonance structu...

Julianna_flores3E

By looking at the structure you can see the atoms that are more highly electronegative. If the highly electronegative atoms are on one side of the central atom I think it will be polar.

Julianna_flores3E

When talking about hybridization you need to consider regions of electron density. Therefore when determining hybridization you consider atoms and lone pairs. For a double bond, the hybridization of each atom becomes sp2 instead of sp3. Hope this helps.

Julianna_flores3E

I think that it has to do with the size and electronegativity of the atoms. Since the electron clouds repulse one another they want to be as far away as possible. Hope this helps!

Julianna_flores3E

Also, the molecular geometry for this is tetrahedral!

Julianna_flores3E

Why are there brackets separating the parts of potassium phosphide and ammonium chloride? Just a little confused.

Julianna_flores3E

I am studying the questions in the textbook given by the outlines. As well as the learning objectives.

Julianna_flores3E

Is there a specific reason why some elements are metals, nonmetals, or metalloids in the same group?

Julianna_flores3E

I believe that you have to do both filling the octet and simultaneously figure out the formal charge in order to see the characteristic of the bonds.

Julianna_flores3E

Lewis acids accept electrons and lewis bases give them away. I think that there would be a correlation between cations and anions since they have + a d - charges.

Julianna_flores3E

A delocalized electron is shared between atoms it is not bound to one atomic shell.

Julianna_flores3E

I'm pretty sure that the differences in electronegativity is experimentally determined. The type of bond between 1.5 and 2 depends and could be either but have characteristics of both ionic and covalent.

Julianna_flores3E

I think that the p orbital that you are talking about has to do more with quantum numbers rather than valence electrons? Correct me if I am mistaken.

Julianna_flores3E

It is crazy how fast each week is going. This time is particularly stressful, but I am optimistic that things will settle down and I will be able to get everything done.

Julianna_flores3E

In an ionic bond formed between atoms with large differences of electronegativity. Covalent bonds are formed by sharing electrons. Ionic bonds are usually between a metal and nonmetal, while covalent bonds are usually between 2 nonmetals.

Julianna_flores3E

I took Honors Chemistry but not AP Chem at my school. We learned a little bit of everything, but a lot of the sections that are supposed to be familiar are not. It might also be because I took chem as a sophomore. Anyways, I am in your boat I am grateful that Dr.Lavelle covers the fundamentals of ea...

Julianna_flores3E

Honestly, I just try and do my best to feel prepared, and don't really worry about the rest.

Julianna_flores3E

I tend to just change where I study. I like to designate different areas to different things because I tend to get used to an environment (my bedroom) an dI tend to lose focus. If you can change locations even within your house it may help you focus.

Julianna_flores3E

HAppy Halloween! I worked and then watched Halloweentown while doing homework. :)

Julianna_flores3E

Hi, the way that I think of it is that since they are the same element the radius changes based on how many electron there are. The Rb- has more electrons then Rb and Rb+ and then Rb has more electrons than Rb+ so its radius will be larger. Hope this helps.

Julianna_flores3E

The Lyman series is the UV so n1=1 while the baller series is in visible light so n1=2.

Julianna_flores3E

According to Pauli Exclusion principal there can only be 2 electrons in each orbital and they must have different spins.Hope this helps

Julianna_flores3E

I think that since 1 meter=1,650763.03 you would have to convert m to nm and divide by the number of wavelengths to get a single wavelength. Then having the wavelength you could identify what part of the spectrum it is on. With the wavelength that you find before you can find the frequency and then ...

Julianna_flores3E

I think frequency and wavelength are independent of amplitude. However, Increasing amplitude does increase the intensity of the wave.

Julianna_flores3E

A baseball is so large that it follows classical mechanics. However, if you wanted to find the wavelength you would convert mph to meters/s and ounces to kg. Then you would use wavelength=Planck's constant/ momentum (mass x velocity). Then you should get the wavelength. Hope this helps.

Julianna_flores3E

You would follow basically the same procedure as the top part of the problem, except you would need to use Avogadro's # to convert to atoms and then convert to KG because it is the SI unit needed for the calculations. Then you would be able to follow the uncertainty principle with the mass of the he...

Julianna_flores3E

On the website is says to round only your final answer, but to keep track of sig figs throughout your calculations. There is a couple pdfs on the website if you are still confused.

Julianna_flores3E

In my work I followed the same procedure as you. Maybe when calculating you are not separating the terms properly. That is the only thing I can think of that sometimes messes up my answer.

Julianna_flores3E

It was unexpected because electrons have a measurable mass and are particles, but when they passed through the the crystal displayed diffraction patterns they showed that they also have wave-like properties.

Julianna_flores3E

My TA also said that equations will be provided for us for this class. For future use it would probably be helpful to start remembering them.

Julianna_flores3E

In his email he said that he would send another email of when to start working on them.

Julianna_flores3E

I think it is just kelvin. No degrees.

Julianna_flores3E

Hi, a higher frequency will cause shorter wavelengths and vice versa. They are inversely proportional therefore when the wave frequency is high the wavelength will be short and if the frequency is low then the wave will be long. Hope this answers your question.

Julianna_flores3E

You would need to convert the amount of KMnO4 to moles and the ml into liters. Once you find that, you know the initial molarity and the initial volume then you would use M1V1=M2V2 since you also know the final volume.

Julianna_flores3E

The question gives you the theoretical yield in moles the only thing you need to do is convert to grams. As for question 10 you just have to work from ml that they give you to the product since it said it had excess of the other reactant. Sorry I don’t remember the names. Hope this helps!

Julianna_flores3E

The lectures will be asynchronous and the office hours are live.

Julianna_flores3E

I just assumed a 1:1 ratio as well since the Br and Mg were not mentioned in the products. Hope this helps!

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