CHEMISTRY COMMUNITY

Evelyn Silva 3J

Oxidation occurs at the anode and reduction occurs at the cathode

Evelyn Silva 3J

Yes, the oxidizing agent is the molecule that is being reduced. This is because it oxidizes other molecules since it itself is being reduced.

Evelyn Silva 3J

When you have aqueous solutions without a metal conductor you add Pt. It doesn't affect the reaction because it is an inert electrode, and it only helps conduct the electrons from anode to cathode.

Evelyn Silva 3J

There isn't a specific order you need to have them in as long as the products and reactants are on the correct side, but you do need to make sure all the states and coefficients are correct.

Evelyn Silva 3J

The reaction at the anode is oxidation and the reaction at the cathode is reduction. Once you identify which species is being oxidized and which is being reduced you can label them because the anode may not always be on the left.

Evelyn Silva 3J

I believe a UA said we won't need to use it, but Dr.Lavelle shows it to us to understand where it came from.

Evelyn Silva 3J

This is because the temperature is constant and there is no heat being transferred. Since q=-w they cancel out and the internal energy is zero.

Evelyn Silva 3J

Yes! You figure out the oxidation states of TM by looking at the oxidation states of the atoms its paired with.

Evelyn Silva 3J

Thank you this is so helpful!

Evelyn Silva 3J

It doesn't necessarily mean that w=0. We know for sure that q=0 but there can still be work being done.

Evelyn Silva 3J

Products to reactants (left-->right) is the forward reaction and reactants to products (right-->left) is the reverse reaction.

Evelyn Silva 3J

Free expansion occurs when a gas is expanding without any opposing force. This means that there is no external pressure and thus, there is no work being done inside the system because the gas is not pushing against anything.

Evelyn Silva 3J

I think either one is fine as long as you use the same units in your calculations. Unless the problem specifies which one to give your answer in both are correct.

Evelyn Silva 3J

I think the UA session are a great way to review the content. I like doing the practice problems they provide because most of them are different from the ones in the textbook and it helps me see if I can actually understand the concepts rather than just memorizing how to do the same textbook problems.

Evelyn Silva 3J

I use Kelvin all the time because if you need to use the gas constant somewhere in the problem it won't work with Celsius.

Evelyn Silva 3J

deltaH= q when there is a closed system and the pressure is constant

Evelyn Silva 3J

Yes you are correct. When you draw the Lewis structure you should determine what type of molecule it is and based off of that you have to choose the correct value of Cv,m.

Evelyn Silva 3J

I agree with the responses above. After drawing the Lewis structures you can compare the structures on the reactants and products side to see which bonds are broken and which bonds are formed. Sometimes it can be a little tricky with larger molecules, so I like to include all the bonds being formed ...

Evelyn Silva 3J

Last quarter I went to my TA's office hours and we went over my midterm one on one. You won't be able to see the questions by yourself, but they can help you see where you went wrong.

Evelyn Silva 3J

Since you know that Cp=7R/2 you can use the equation Cp=Cv+R to solve for Cv. To do this you would have:
Cv=Cp-R = (7R/2)-R = 5R/2. Then you would substitute in the ideal gas constant. If you look at the textbook example 4C.1 they work out a problem that shows this conversion.

Evelyn Silva 3J

If the question asks for the amount of energy in Joules you need to convert the L*atm using the conversion factor (1L*atm=101.325J). The question should always tell you the units you should have in your final answer but usually, you will always need to have it in J or KJ.

Evelyn Silva 3J

Last quarter he would usually take about a week to release scores, but I'm sure we'll get an email soon with an update.

Evelyn Silva 3J

I'm taking BL next quarter. I want to take it as soon as possible while the stuff I learned from 14A and 14B is still fresh in my brain.

Evelyn Silva 3J

If the equilibrium constant is less than 10^-4 you can assume the change is neglible. However, if you have a value that's really close to 10^-4 or exactly that value then I like to calculate the percent ionization just to make sure the assumption is correct. If you get something greater than 5% you ...

Evelyn Silva 3J

Yes, the diatomic molecules are already in their most stable state, so the standard enthalpy of formation is zero.

Evelyn Silva 3J

Kc is used when you are given the molar concentrations and Kp is used when you are given the partial pressures. But they are both equilibrium constants and they mean the same thing.

Evelyn Silva 3J

If Ka is close to 10^-3 then I always check the percent ionization because sometimes it is greater than 5% and in that case, we can't assume the change is neglible. But if Ka is extremely small like 10^-12 then you can automatically ignore the x.

Evelyn Silva 3J

A general rule of thumb my teacher told us in high school is that all the halogens, Nitrogen, and Oxygen are written as diatomic molecules when they are by themselves.

Evelyn Silva 3J

When we are given a strong acid we know that it is completely dissociated because almost all the HA molecules are deprotonated. This allows us to assume that the molar concentration of H3O+ ions is the same as the initial concentration of the strong acid, so we can just calculate pH without using th...

Evelyn Silva 3J

I like to check the percent ionization if the Ka value is really close to 10^-3. Sometimes you can't assume x is insignificant, and if you get a value greater than 5% you would have to include x because the change is not neglible.

Evelyn Silva 3J

As mentioned above, the concentrations that they gave us were already at equilibrium so there wouldn't be any "change" that we have to take into account. Therefore, we can solve this problem without having to use an ICE table.

Evelyn Silva 3J

As mentioned above, you use what is given in the problem and if you need to convert your values you would just use PV=nRT to convert in between concentration and pressure. To find concentration you use conc.= P/RT and to find pressure you use P=nRT/V.

Evelyn Silva 3J

Since you know that A has a coefficient of 2, you can look at the graph and see that its partial pressure decreases from about 35 to 25. Likewise, the partial pressure of C increases from 0 to 10. The difference between both is the same, and because they both increase/decrease by the same amount the...

Evelyn Silva 3J

After you plug in both values of x, you should see that one of them gives you a negative value for the equilibrium concentrations. Since this is not possible you can automatically rule it out, and therefore the other value of x is the correct one. If you have positive values for all equilibrium conc...

Evelyn Silva 3J

For this question you want to start off by writing down the equilibrium expression of the original reaction which would be K= [HI]^6[N2]/ [NH3]^2[I2]^3 In order to solve this problem you know that you have to manipulate the given equations in order to have the same K expression as the above. You can...

Evelyn Silva 3J

When there is an endothermic reaction, you can think of heat as a reactant. Increasing the temperature makes K increase because the equilibrium favors more products and less reactants. We know that K=[P]/[R], so if your denominator is smaller then the value of K will be bigger. In an exothermic reac...

Evelyn Silva 3J

You should also memorize the units you have to use when using the equation. For example, temperature is in Kelvin and volume is in Liters.

Evelyn Silva 3J

In problems where you have k=10^-4 or anything smaller, you can just ignore the "x" because the change in concentrations will be very small. Therefore, the concentrations will essentially be the same and we just write down the approximate value. If k is less than or equal to 10^-4 do you ...

Evelyn Silva 3J

Basically 1000mL= 1L, so when you do the conversion you change 500mL to 0.5L

Evelyn Silva 3J

Since we were given the concentrations in this problem, we automatically assume we have to use Kc even though the compounds are given in gaseous states. You usually just go by whatever is given in the problem.

Evelyn Silva 3J

Yes, you are correct. If the reaction is at equilibrium then the reverse and forward rates must be the same.

Evelyn Silva 3J

Also if you go on Sapling and click on "Resources" you should be able to see the "Atkins 7e SSM." It's the solution manual for all the odd numbered textbook problems and it shows you how to work them out.

Evelyn Silva 3J

In problems where you have k=10^-4 or anything smaller, you can just ignore the "x" because the change in concentrations will be very small. Therefore, the concentrations will essentially be the same and we just write down the approximate value.

Evelyn Silva 3J

I usually keep 5 decimal places just to be safe, but like mentioned above since the test is multiple choice I think you should be fine.

Evelyn Silva 3J

I agree with the post above, you basically just break separate them into ions and determine whether they were an acid or a base. Also remember to balance out the equations and take into account the oxidation numbers. For example, in part b Zinc would become Zn(OH)2 because it has a +2 charge.

Evelyn Silva 3J

In this case you would have to consider the electronegativities of each atom because they are not being bonded to H. You only consider atomic radius when you have a binary acid.

Evelyn Silva 3J

An example you should know is cisplatin, which is a chemotherapy medication that has two Cl atoms on the same side. It bonds to two neighboring guanines and it stops DNA replication.

Evelyn Silva 3J

Yes, first you write the electron configuration for a neutral atom and then you add or remove the electrons from it

Evelyn Silva 3J

Also Table J.1 in the textbook shows a list of common aqueous strong acids and bases. I recommend memorizing them and it will make it a lot easier.

Evelyn Silva 3J

One of the UAs told us that his final for this class included all the material we covered and everything was equally distributed, so I think it will be pretty much the same thing for us.

Evelyn Silva 3J

You should disregard the prefixes when putting the ligands in alphabetical order.

Evelyn Silva 3J

If the overall negative charge of the complex is negative, you add -ate to the metal and then write the Roman numeral after it.

Evelyn Silva 3J

Yes, a nonpolar molecule can have dipoles, but if the arrows cancel each other when you draw them then the molecule has no net dipole.

Evelyn Silva 3J

The axial atoms lie above and below the plane of a triangle and they form 180 degrees. The rest of the atoms are the equatorial ones.

Evelyn Silva 3J

I like going on runs because it helps me destress. I usually go on trails, but when the weather is bad I just run on the treadmill.

Evelyn Silva 3J

I just remembered that in the lectures we have seen multiple carbons form a ring structure, so I tested it out and got the correct structure. Honestly I wasn't so sure either and I had a bit of trouble.

Evelyn Silva 3J

I think doing textbook problems are the most helpful for me. I would also try to review past worksheets from step-up sessions and workshops.

Evelyn Silva 3J

A delocalized pi bond is a bond when electrons can move to more than two nuclei. Usually they are found in resonance structures where there is a double bond.

Evelyn Silva 3J

The lone pair in the SO2 atom repels the O atoms down, giving the molecule a bent shape. Looking at the vectors, you will see that the molecule is polar. In CO2, however, Carbon does not have a lone pair, so the molecule is linear and nonpolar.

Evelyn Silva 3J

Cl is a halogen so all elements in group 17 only need one bond to get an octet. Also if you calculate the formal charge you'll see that Oxygen has a formal charge of zero with a double bond, which means that adding the double bond to one of the oxygen's would give you the most stable structure.

Evelyn Silva 3J

We started on Week 1 so since we have to have a minimum of 5 posts per week, you should have 35 posts by the end of this week.

Evelyn Silva 3J

Hydrogen cannot be the central atom because they only like to form one bond (its an exception to the octet rule). When drawing Lewis structures they are usually added at the end.

Evelyn Silva 3J

Yes it is possible to have an unpaired electron. This will usually happen when you have an odd number of electrons and they are known as radicals. They are very reactive.

Evelyn Silva 3J

Yes I had the same issue and it should say SO2, so it's an error.

Evelyn Silva 3J

If you get a list of successive ionization energies, whenever there is a significant jump that requires a greater amount of energy to take away an electron you should know that you are starting to remove the core electrons. By looking for the large jump in ionization energy, you can determine how ma...

Evelyn Silva 3J

The dashed lines in the picture tell us that the actual structure is a resonance hybrid. Because the electrons are delocalized, the double bond character is spread across both of the O bonds. Therefore, the length of both bonds is in between a single and double bond (it's not exactly either one) and...

Evelyn Silva 3J

As mentioned above, they would be drawn separately because they form an ionic bond. You would only draw them together if they formed a covalent bond and shared the electrons.

Evelyn Silva 3J

The format will be the same and we will be proctored through respondus. My TA told us to join zoom through a separate device and I think there will be more questions this time since there aren't many calculations involved and its mostly conceptual.

Evelyn Silva 3J

The octet rule is a principle that demonstrates how atoms share electrons until they reach a noble-gas configuration. An atom almost always wants to end up with 8 electrons in their outermost shell, but there are some exceptions to this rule with atoms such as P, S, and Cl. These atoms can have more...

Evelyn Silva 3J

A lone pair is a pair of valence electrons that don't participate in bonding. They are represented as dots around an atom.

Evelyn Silva 3J

There should only be 1 double bond from Carbon to Oxygen or else the C atom ends up getting more than 8e-. When you take out one of the double bonds in an oxygen add a lone pair. Also put a -1 charge on each of the Oxygens with a single bond. Hope this helps!

Evelyn Silva 3J

From what I've been told in the step-up sessions, 4s has a lower energy than 3d before the orbitals are occupied and this is why 4s is filled before 3d on the periodic table. However, after they are filled they switch and 4s has a higher energy than 3d. This is why when writing the electron configur...

Evelyn Silva 3J

You can use either one, but it depends on what information is given in the problem. If you know the frequency or wavelength use E=hv. The E= work function + KE equation is usually used when the electrons are being ejected and it asks you to calculate the energy required to eject them.

Evelyn Silva 3J

Wavelength and frequency cannot be negative. Energy can sometimes turn out to be negative, such as when an electron transitions from a highter to a lower energy level. This is because energy is being released, but the wavelength and frequency are always positive.

Evelyn Silva 3J

Ground state is the state of lowest energy in an atom. This is when an atom is more stable. When it becomes excited it means it goes up to a higher energy level.

Evelyn Silva 3J

yes sorry for the confusion, but you can still post Sunday and get your points as long as its before 11:59

Evelyn Silva 3J

Energy is proportional to the amount of photons which is why they multiply E times 100. However, frequency and wavelength stay the same because each individual photon has the same wavelength and frequency regardless of the total amount of photons there are.

Evelyn Silva 3J

As mentioned above, there can only be 2 pairs of electrons in each orbital because of the Pauli Exclusion Principle and their spins must be paired (meaning they go in opposite directions).

Evelyn Silva 3J

Your TA checks you have 5 posts per week, so as long as you complete them before midnight every Sunday you should be fine!

Evelyn Silva 3J

The wavelength depends on the type of surface metal that is being used for the experiment. Some atoms have a greater binding energy, so they hold on to electrons more tightly than others and they require a higher frequency to be ejected. But UV light is usually the incoming light used for this exper...

Evelyn Silva 3J

I recommend using the same amount as in the worksheet given to us and try to keep as many decimal places you can in between calculations. Since the midterm is multiple choice then you can just pick whatever answer is the closest to yours.

Evelyn Silva 3J

When you're doing your calculations you should use meters for wavelength and Hz (s^-1) for frequency. However, if a problem specifies to give your answer in a specific unit then you would change it at the end.

Evelyn Silva 3J

Heisenberg's indeterminacy equation helps determine very precise measurements with subatomic particles that include protons and neutrons as well.

Evelyn Silva 3J

The question will usually give the actual yield in grams, which means that the theoretical yield should also be in grams of product. If they give another unit then you should convert it to that in order to be able to calculate the percent yield.

Evelyn Silva 3J

I had that same question! I think if we know the region of light the Balmer and Lyman series belong to and their energy levels we should be fine, since those were the only ones mentioned in the lecture. But I recommend knowing the name of the other one too just in case. (The Paschen series is the se...

Evelyn Silva 3J

Start off by converting 102.6nm to frequency using v=c/lambda. Since they said they use UV light, the Lyman series tells us that n1=1. Then use the Rydberg equation and plug in the frequency you calculated (v), Rydberg constant, and n1=1 to find the value of n2. (You should get 3) We looked at this ...

Evelyn Silva 3J

Yes, the amplitude doesn't have an effect because it describes the intensity of the light. As the photoelectric experiment shows, the intensity of the light doesn't increase the energy of the individual photon because light has more than just wave properties. This is why we have to adjust the freque...

Evelyn Silva 3J

Increasing the frequency would result in the photon having more energy (E=hv). Increasing the intensity would result in an increase in the number of photons, but this has no change in their individual energy. No matter how bright the light is (or how many photons there are) none of the photons will ...

Evelyn Silva 3J

He multiplied all the ratios because 1.33 isn't close enough to 1 for you to be able to round it down. This means you would have to multiply by all of them by 3, and end up with the ratio 3:4:3.

Evelyn Silva 3J

If you end up with a fraction multiply all the stoichiometric coefficients by the denominator. Just to be safe you should use whole numbers when you're balancing an equation.

Evelyn Silva 3J

You want to start by converting the mass of KCI to moles
(55.1g KCI)/(74.55g/mol)= .739 mol KCI

Then calculate the molarity (using M=n/V)
Make sure to change the units of volume from mL to L (125mL= .125L )

M= (.739 mol KCI)/ (.125L)= 5.91M KCI

Evelyn Silva 3J

It's best to keep the exact numbers for in between calculations and round until the end. If you round before then you risk having a final answer that is off by a little and it won't be as accurate.

Evelyn Silva 3J

I believe we do have until Sunday to post because it restarts on Monday, but I'm not sure!

Evelyn Silva 3J

Yes I agree! When he showed us the target pictures in the lecture it helped me visualize what he was trying to explain and understand the difference between the two.

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