CHEMISTRY COMMUNITY

Maia_Jackson_2C

Adsorption and Absorption differ in how they function and the state of the species involved. The above answers explain it well! But, why is it important to understand the difference between the two? In the last lecture he mentioned that reactants sit on the surface of a catalyst (adsorption) becaus...

Maia_Jackson_2C

Changing the mass of the cathode doesn't change the concentration of the cathode solution therefore changing the mass of the electrode would have no change / affect on the cell potential. I think you are referring to the second part of that question but he is asking about the first part where the v...

Maia_Jackson_2C

It is basically the process of speeding up a reaction by providing a new pathway and lowering the activation energy.

Maia_Jackson_2C

I think this means that you can assume that the reaction has the same intermediates (same pathway) in the forward and reverse directions.

Maia_Jackson_2C

Yeah I think the rate determining step is usually the slowest step, and it is also called the elementary step sometimes.

Maia_Jackson_2C

we can assume that a reaction has microscopic reversibility when the reaction has the same intermediates (same pathway) in a forward and reverse reaction. Also I don't think we need to know a lot about this topic but I may be wrong.

Maia_Jackson_2C

I think there would just be two reactants, so it would be bimolecular.

Maia_Jackson_2C

I think you can change one variable to delta if you change another variable on the other side of the equal sign to delta.

Maia_Jackson_2C

The nernst equation is used to find the max potential before any current flow. It can be used to find cell potential, equilibrium constant, reaction constant, and potentially any other variable in the equation.

Maia_Jackson_2C

Are galvanic/volteric cells normally closed? I though they were open but in that case O2 would be available in other redox reactions and that doesn't seem to be the case.

Maia_Jackson_2C

You add platinum or another inert electrode when both oxidized and reduced species are in the same solution. It is easy to tell when you look at the phases in the equation, since the electrode is always a solid, so whenever one side of the cell doesn't have a solid you need to add an inert electrode...

Maia_Jackson_2C

Yeah I think the book reversed it because the value was negative and you typically shouldn't use negative values for the cell. And I think it is also more accurate to use E(cathode) - E(anode)

Maia_Jackson_2C

The symbol means standard of reaction, so for example standard reduction potential (E knot) is the voltage difference between two standard electrodes and it is a constant. It is not related to the degrees symbol for temperature

Maia_Jackson_2C

You can use the Nernst equation to solve for Q, and then once you find Q use it to solve for the concentration of H+ or OH- depending on the reaction. And you only use log when the temperature is 25 degrees celcius and the equation replaces RT/nF with 0.0591/n

Maia_Jackson_2C

I think this is because one of the half reactions is Ag+ (aq) + e- --> Ag (s) and another is AgBr (s) + e- --> Ag (s) + Br- (aq) so when you reverse the half reaction and add them together the Ag(s) cancels out, but you would still include it in the cell diagram

Maia_Jackson_2C

You can add H+ or OH- when it is a basic solution because H+ +OH- --> H2O
but you only do this if there are extra H+ on one side of the reaction and H2O on the other to cancel out extra molecules
Sorry this isn't very helpful I don't fully understand it either

Maia_Jackson_2C

But generally I'm pretty sure that an element is oxidized when the oxidation number is increased, and is reduced when the oxidation number decreases

Maia_Jackson_2C

My TA said we should add OH- or H+ when it is a basic solution if you have a redox reaction with leftover H+ on one side of the reaction, so convert it into H2O since H+ + OH- ---> H2O (only in a basic solution)
Sorry that's not very helpful I didn't fully understand it in section either...

Maia_Jackson_2C

In the answer key it says that an inert electrode like Pt is necessary when the solution has both oxidized and reduced species and also for gas/ion electrode reactions

Maia_Jackson_2C

I think you can just try the problem both ways with each one being the cathode and anode and choose the identification that gives you a positive E cell value.

Maia_Jackson_2C

How do you determine an elements oxidation number ? I think its just the number of electrons that were lost or gained during a reaction, so you would basically just count them. If you look back at when we studied transition metals in 14A we also learned about this there. You have to count the numbe...

Maia_Jackson_2C

Personally, I've always just memorized it as reduction being a reduction in the charge. So if the charge is reduced, then that has to mean an electron is gained. But does this always work? for example in molecules in which the net charge stays the same, but the number of electrons in each atom is d...

Maia_Jackson_2C

When writing half reactons why is the number of electrons written in the equation being added? For example in this oxidation: 5Fe = Fe3+ + 5e- Why is the 5e- added and not subtracted? I think this is because the net number of electrons cannot change, since the reaction is balanced and no new electr...

Maia_Jackson_2C

But do you have to write the electrons in a chemical equation to show reduction or oxidation?

Maia_Jackson_2C

Also you're not alone!! I did not feel good about that midterm either and basically everyone I have talked to about it so far has said the same thing :(
But I agree that practicing is definetly one of the best things you can do to prepare.

Maia_Jackson_2C

We have to know the first page of the thermodynamics outline, so I think that has parts of 4J.1 and 4J.2 but I'm not entirely sure.

Maia_Jackson_2C

Usually you can tell from the context of the question whether or not there is a change in pressure or volume, but if not then I think you usually use Cv (that's what they said in the review session today but they didn't say why)

Maia_Jackson_2C

The rule is that the reaction will favor the side with less moles of gas for a volume decrease, you didn't mention gases so maybe the moles you counted weren't for gases?

Maia_Jackson_2C

You would square the constant when the stoichiometric coefficients of the reaction are doubled.

Maia_Jackson_2C

I would think that the question would have to tell you whether or not there was a volume change, but I'm not sure about gasses.

Maia_Jackson_2C

K is Boltzmann's constant and it is approximately 1.380649×10−23 J/K

Maia_Jackson_2C

I think you have to think about the shape and the definitions of cis and trans to figure it out. Since trans means the two atoms are opposite each other, then there are only three options in the picture they show in the question. Since cis means the molecules are next to each other there are 12 poss...

Maia_Jackson_2C

If you can figure out which bonds are broken and formed without drawing the lewis structures then you shouldn't need to draw them

Maia_Jackson_2C

I think you would just do the C-C bond enthalpies for the bonds that break and form and subtract them, so it would be 3(837) kJ/mol - 6(518)kJ/mol = -597 kJ/mol (assuming you're talking about part a)

Maia_Jackson_2C

I think it comes from the equation q=delta H=n*C*delta T but they didn't give you moles or mass so there is no n, and you just solve for C= delta H/delta T

Maia_Jackson_2C

But why is Hess's method more accurate than the third method we learned?

Maia_Jackson_2C

The distance is 3.87nm... but did we learn how to calculate this? I thought we just need to know that delta Hf degrees of oxygen gas is zero

Maia_Jackson_2C

Are you talking about what we covered in lecture on Friday or Wednesday?

Maia_Jackson_2C

H2SO4 is a strong acid, so the equilibrium concentration of products is the same number as the initial concentration of H2SO4. Then you can calculate pH using -log of the concentration of hydronium

Maia_Jackson_2C

You would make an ice table and use Ka to find x (the change in concentration) and the concentration of hydronium, and then find the pH using -log of this concentration. Then you would find pOH using pH+ pOH=14

Maia_Jackson_2C

If they weren't assigned in the homework do we have to know about them for the test?

Maia_Jackson_2C

Yeah I'm pretty sure you would need to be given Ka and Kb, because using the pH you can find the concentrations of H3O+ and OH-, but not the constants. I think they would give them to us on the test.

Maia_Jackson_2C

I'm not entirely sure but I think for sig figs you just use the least given amount, and it doesn't really matter where it comes from unless otherwise stated.

Maia_Jackson_2C

Yeah, this is right as long as the temperature stays constant

Maia_Jackson_2C

In the answer key it says that its an endothermic reaction because it requires energy to break an X-X bond, but the values of b can be calculated.

Maia_Jackson_2C

When you calculate ph you normally do -log(10^a negative number), and the negatives cancel out. If you get a negative ph, that means you would have
-log(10^a positive number), and the negative would not cancel out. So the concentration would be larger than 1.

Maia_Jackson_2C

Can you post the question that you need help with?

Maia_Jackson_2C

I think you should get the same number for both, but I'm not entirely sure.

Maia_Jackson_2C

Also if you can't figure it out you can also go to the textbook store and they will help you

Maia_Jackson_2C

In the video modules he explained that inert gasses do not affect the reaction because they do not change the concentration of products or reactants. When pressure increases by decreasing volume, the concentration increases, so the reaction is affected, but adding inert gasses does not increase or d...

Maia_Jackson_2C

The principle basically says that chemical reactions will adjust to minimize the effect of any changes. When pressure increases by a decrease in volume, and there are more moles of gas on the reactant side, then the reaction will produce more products. When pressure increases by a decrease in volume...

Maia_Jackson_2C

Do we have to know this for the final?

Maia_Jackson_2C

I haven't heard this mentioned anywhere else, did he mention it in lecture? and should we do it this way on the final?

Maia_Jackson_2C

In row 1 the pH of H30+ with 1.5M is -0.176, so when you do ([H30+]= 10^-pH, it looks like, ([H30+]= 10^-(-0.176)=1.5M. The negative signs cancel out so the equation does work, even though the pH is negative.

Maia_Jackson_2C

Hybridization lowers the overall energy level by moving electrons around and creating hybrid orbitals to free up all of the electrons. The energy levels of the electrons will be between that of the two orbitals that are hybridized.

Maia_Jackson_2C

The bond angles for a and c are 120 degrees because of the trigonal planar shape, and 109.5 degrees for b because it is tetrahedral.

Maia_Jackson_2C

The lewis structures are all pretty straightforward, and none of these are radicals, but the bond angles depend on the shape and VSEPR theory. This is what I got for increasing bond angle: CH2 (2-) < CH3 (-) < CH4 < CH2 < CH3 (+) < CH2 (2+) The structures with a positive charge have a larger bond an...

Maia_Jackson_2C

PCl4+ is sp3 because there are four bonds, PCl6- is sp3d2 because there are six bonds (phosphorus has an expanded octet), PCl5 is sp3d because there are five bonds, and PCl3 is sp3 because there are three bonds and one lone pair

Maia_Jackson_2C

The first two isomers are polar because the C-Cl bond dipoles are not pointing in opposite directions and do not cancel out. The first form has the largest dipole moment because the bond angle between the two Cl's is the smallest.

Maia_Jackson_2C

Pyridine is polar because the N is not bonded to a hydrogen so the dipoles don't cancel out. Ethane is nonpolar because the dipole moments cancel out and there is no electronegativity difference between the two carbons. Trichloromethane is polar because the dipole moments for chlorine point are stro...

Maia_Jackson_2C

I believe it also has a negative 4 charge because of 4 singly bonded oxygens making it attractive to single electron atoms easily

Maia_Jackson_2C

Yes, I think that generally when the s character in a hybrid bond increases, the angle also increases

Maia_Jackson_2C

The only problem with this is molecules with double bonds can have the same formula as those with no double bonding, so they would still have the same shape and angle, but the single bonded molecule would be able to rotate while the double bonded molecule could not

Maia_Jackson_2C

The overlaps are referring to their electron in the orbitals overlapping making them shared

Maia_Jackson_2C

You can also write it like sp^3d or 2sp^3d

Maia_Jackson_2C

Basically if the electrons were closer together they would want to repel, making them unstable, but if they are as far apart as possible they have nowhere to go, so they are stable.

Maia_Jackson_2C

Bent or angular refers to when there are three or four groups of electrons around an atom, but only two of them are bonding groups, meaning the other(s) are lone pairs. One example is H2O

Maia_Jackson_2C

So for example in iodine trichloride the VSEPR formula would be AX3E2 because there are three bonds to the central atom and two lone pairs on the central atom.

Maia_Jackson_2C

Yes but for most of them you can just memorize the angles that go with each shape.

Maia_Jackson_2C

You should probably say that it is less than 109.5 degrees because that is technically more correct.

Maia_Jackson_2C

You can also use effective nuclear charge to describe the electronegativity trend, because it also increases in the top right corner of the periodic table. This means that atoms hold their electrons more tightly because the elements with more electrons and smaller radius have more protons, so there ...

Maia_Jackson_2C

The sigma bond is just a covalent bond, and every bond has a sigma, and the pi bonds represent the overlapping of another bond, which is why triple bonds have 2 pi's.

Maia_Jackson_2C

I'm not entirely sure but this seems correct because if one photon needs a certain amount of energy to emit one electron, then the next photon would need that same amount of energy to emit the next electron, making the total energy double in the end.

Maia_Jackson_2C

Also acids are usually positively charged and bases are more likely negatively charged

Maia_Jackson_2C

I think it is when there is a polarizing cation and a polarizable anion in close contact

Maia_Jackson_2C

There is also an electronegativity difference, because coordinate covalent bonds are more polar

Maia_Jackson_2C

It's basically when you can place a bond in multiple places in the Lewis structure without drastically changing the structure or charge.

Maia_Jackson_2C

I don't think there is a specific number of electrons that each expanded octet can hold that we need to know, just know that it is more than 8

Maia_Jackson_2C

There would only be a radical if there is an unpaired electron, so an odd number of valence electrons. There is a good example in 2C.13 part b, where ClO2 has one unpaired electron.

Maia_Jackson_2C

I thought that all elements in or past the third period can have expanded octets because they have an empty d orbital where the extra electron can go.

Maia_Jackson_2C

Then if the bond length is always the same why is it important?

Maia_Jackson_2C

When you say 18 valence electrons is the most, does that mean 18 electrons fill the shell? and if the shell is full wouldn't that mean that there are no valence electrons?

Maia_Jackson_2C

I thought I heard that the octet rule is more of a guideline, but the exceptions are H, He, Li, and Be

Maia_Jackson_2C

Also electron affinity is the energy released when an electron is added to a gas phase atom

Maia_Jackson_2C

Also the valence of an atom depends on the number of valence electrons, for example if there were 3 valence electron then the atom should typically be able to form 3 bonds.

Maia_Jackson_2C

Just to clarify you get the total energy using the equation E = W(j/s) * t (s) to get 64 J

Maia_Jackson_2C

But since the question gives you the work function there is another equation that uses both which is
kinetic energy = energy of photon - work function (energy required to remove photon

Maia_Jackson_2C

Another equation that might be relevant here is E = W(j/s)*t (s)

Maia_Jackson_2C

Also I think there is a derivation for the equation that would explain why h/4pi is there, but Dr. Lavelle said it was too complicated for us to work through the whole derivation because it uses things we haven't learned yet.

Maia_Jackson_2C

Momentum is also important because some small objects may not have a measurable mass, but they have momentum, so we can still use the equation for them because of momentum.

Maia_Jackson_2C

Also wavelength is inversely proportional to frequency, increasing one decreases the other, so with a short wavelength of light, there would be a higher frequency and the electron would eject.

Maia_Jackson_2C

The main property of waves that I understood was diffraction patterns displayed when electrons passed through a crystal. It might help to look up a picture or diagram of this also.

Maia_Jackson_2C

Destructive interference and out of phase waves basically mean the same thing. Also if two waves are completely out of phase and are the same size, they can cancel each other out.

Maia_Jackson_2C

The only time I really noticed amplitude making a difference was in constructive and destructive interference, because it seemed to increase in constructive and decrease in destructive.

Maia_Jackson_2C

Also note that in constructive interference the amplitude gets larger and in destructive interference it gets smaller.

Maia_Jackson_2C

Another way of saying it is that it is used to find the energy needed to move electron, or threshold energy.

Maia_Jackson_2C

I started the problem by finding how many grams of gallium there were so I knew how much astatine I needed to balance the scale, which was 3.794 grams. Then I divided that by the molar mass of astatine, and got 1.8066 x 10^24 moles, and then I divided that by Avogadro's constant to get 3 atoms of as...

Maia_Jackson_2C

I think that either way of writing the answer works since 33.03 x10^2 is written with a number higher than 10 (assuming this problem came from the textbook?), but both answers are technically the same number.

Maia_Jackson_2C

Also while the symbol for molar mass is M, when given a molar mass in homework or test problems it will have units of g/mol, which may help distinguish it from molarity.

Maia_Jackson_2C

The problem is:
Calculate the amount in moles of Cu^2+ ions in 3.00g of CuBr2.

I thought the ratio would be 1Cu: 2Br because the 2+ attached to the Cu is referring to the charge, not amount, while the Br2 is referring to the amount, but on the solutions manual the ratio was 1:1

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