CHEMISTRY COMMUNITY

405268063

The reduction reaction should be Br2 + 2e- --> 2Br-. You would multiply this whole thing by 5 to balance the electrons when trying to write the overall reaction.

405268063

I think that's unclear right now, but Dr. Lavelle should be sending out an email with all of that information probably later today.

405268063

I believe that an elementary rate law corresponds to the rate law for one step of an equation for a multi-step reaction. The overall rate law refers to all the steps of a reaction, and that is determined by the slowest step.

405268063

Yes, like the previous responders have said, k prime generally refers to the reverse reaction rate and k refers to the forward reaction rate.

405268063

Zero order reaction rates are independent of [R]. This means that the rate law is just k which makes its units mol/(L x s).

405268063

The standard cell potentials are written so that each compound is gaining electrons (aka cathode reaction) so if we want the equation where electrons are actually being lost, you have to reverse that standard cell potential value.

405268063

Yes, you basically just need to know that however many electrons you have in your overall balanced equation will be plugged into n.

405268063

Like everybody else is saying, a concentration cell just happens when both the anode and cathode have the same electrodes in it but with different concentrations. The most important thing to remember for concentration cells though is that E standard cell is equal to 0.

405268063

I believe this means that the reaction's delta G is negative so it is technically spontaneous, but the reaction won't normally proceed because it has such a high activation energy.

405268063

I believe that the slowest step of a multi-step reaction is the rate-determining step.

405268063

Yes there is a half-life for zero order reactions and it is: [A initial] / 2k

405268063

I think we will be getting a review packet that will be gone over in Lyndon's session... at least that's how it worked for Chem 14A. Dr. Lavelle would make an announcement tomorrow in class or sometime this week about that.

405268063

I think HI in this case is a solid, and we don't include solids or liquids in the rate law expressions. As far as I know, I believe that if a product is gaseous or aqueous, we would include those.

405268063

For zero order, the rate law makes a linear graph, which means that the rate just goes at constant k (slope). For first order reactions, the graph is also linear if you plot ln[A] versus time. For second order reactions, the graph is linear is you plot 1/[A] versus time.

405268063

I was wondering when we add an inert electrode to a cell reaction. Is it when the states are aqueous? The solution manual just says "An inert electrode such as Pt is necessary when both oxidized and reduced species are in the same solution."

405268063

Yes, there is a difference. The reducing agent is the one being oxidized and the reduced species is reduced in the reaction.

405268063

Flip whichever one will best fit the equation given. I like to think of it as an addition problem.

405268063

I think Kc can signify K for all intents and purposes at least on the test. Kc is generally most used for the equilibrium constant for a rxn that is made up of aqueous solutions though, and Kp for for gases since Kc stands for concentration and Kp stands for pressure.

405268063

I believe that in class, we have been mostly using Gibbs free energy in kJ/mol but as Esha said, you can technically put it into whatever units you want it to be in.

405268063

Also, I believe Dr. Lavelle said today that in an electrolytic cell, electrical energy is converted into chemical energy,

405268063

In cases like this, I usually look at whether the other molecule involved the reaction is being oxidized or reduced, and then use process of elimination.

405268063

I think practice problems are the way to go for thermo. Personally, it is helpful to go through different types of problems and have set techniques for solving them.

405268063

Yes, I believe electrons always move from the anode to the cathode.

405268063

There's not really a huge difference. You can apply what we've learned about delta H and delta S to help you find delta G now.

405268063

Yup, H2O as a gas is still included in K. Basic rule of thumb is just to exclude anything designated (s) or (l).

405268063

I believe you only use hydroxide and H+ to balance a solution only if you are told it is acidic or basic. I would agree with everyone else that you should probably use water to balance things out in a neutral solution.

405268063

I memorized all the basic ones and I find that it's really helped me in the homework. On tests, I think he usually gives you the charge but I have found that just knowing the charges and the atoms is way easier for his class.

405268063

Yes, if you have more possible states (aka higher W), there is more entropy in the system. Also, as the previous responder said, I would use the equation when we are given the number of particles/microstates in the question.

405268063

I don't think that this equation shows that entropy always increases in a system. I think during the review session today, we were told to use this only for phase changes.

405268063

In a reversible reaction, delta S of the system is equal to delta S of the surroundings because delta S of the universe is 0. But I'm pretty sure that the general equation you described in your question applies to all reactions.

405268063

If you look at the graph Dr. Lavelle drew of q versus T early on in thermochemistry, you can see that at the phase changes, the graph plateaus.

405268063

I like to think of it as a balancing act. If you decrease the volume on one side which increases the pressure, the reaction will favor the side with less molecules because the pressure needs to go somewhere else, if that makes sense.

405268063

I believe that if the delta U value is negative, it just means that the temperature was decreased which caused the gas molecules to slow down.

405268063

I just wanted to add that you might be able to avoid doing the quadratic equation is the Ka value is less than 10^-3 or percent ionization is less than 5%.

405268063

For more clarification on your question, are you asking about how Le Chatelier's works?

405268063

Yup, I just wanted to add that bomb calorimeters are used to measure heat of combustion at constant volume and that a calorimeter is used at constant pressure.

405268063

In an irreversible reaction, reactants will create a product that cannot be then turned back into reactants.

405268063

I would think that since heat is required to make those phase changes, the reaction would be endothermic. Therefore, delta H would have to be positive.

405268063

Yes, as the previous response said, you must add the enthalpy of the phase change.

405268063

I think the method you are referring to is using bond enthalpies to calculate delta H. In class, Dr. Lavelle said that this was a less accurate way of finding delta H than using Hess' Law. I'm not quite sure if that means we would be tested on it, but it is possible he could give us bond enthalpies ...

405268063

From a mathematical standpoint, Delta H is the change in enthalpy and this is equal to change in U +P*delta V.

405268063

Not entirely... combustion does include these processes (for example, carbon and H2O molecules are oxidized in a combustion reaction, but all of these actions are not the same.

405268063

I believe we are supposed to use the ideal gas equation only when we need to convert from Kp to Kc (from bars to moles).

405268063

I believe that my TA told me the first test would cover both equilibria as well as what we learned up until Wednesday about Acids/Bases.

405268063

I would look at proton/electron transfers to identify acids and bases (depending on whether you are talking about Lewis/Bronsted). If something is donating a proton or accepting electrons, it is an acid. If something is accepting a proton or donating an electron, it is a base. Salts, as a previous p...

405268063

I believe that buffers are mixtures of weak acids and their conjugate bases, so whenever an acid or base is added, it can use either its products or reactants to neutralize the solution.

405268063

I personally like to think of the system trying to alleviate stress within itself. So if pressure is increased on one side, it wants the reaction to go the other way to relieve it.

405268063

I like to think of the "quick way" Dr. Lavelle referred to in class as just a balancing act. If there's more pressure on one side, the reaction will want to go to the other, if that makes sense.

405268063

n/V is also a way of writing molarity since its units are mols/L. This is nice for understanding how you can use the ideal gas equation to determine Kc as well I believe.

405268063

If Q=K, as everybody has been saying, you know the equation has reached its equilibrium. However, K really is the right way to label the ratio of reactants to products at equilibrium.

405268063

Yes, this is the same case for solids. Their concentrations do not change.

405268063

K never changes since the equilibrium will always have the same ratio of reactants to products. However, Q values can change depending on how far along into a reaction you are since they are not calculated at equilibrium. Depending on if the equation is shifted to the left or the right, there can be...

405268063

Yes, the definition of a ligand is just that it is a Lewis base and can donate at least one pair of electrons.

405268063

Yes, I was also told by my TA that it would be easiest just to memorize the above list for strong acids.

405268063

I believe that all of the questions asked about Heinsenberg's have included values for delta x and/or delta p (or some portion of delta p like mass).

405268063

If we look at each ion in KBr, we get K+ and Br-. K+ is a Group 1 metal, which means it has too small of a charge to affect the pH of the solution. If Br- accepts a hydrogen like it wants to (because it's an anion), it will form HBr, which is a strong acid and immediately completely dissociate into ...

405268063

Yup, if there are no cations or anions attached to it, you also add "ion" at the end of the name.

405268063

As far as I know, coordination compounds involving polydentate ligands are considered chelates.

405268063

Just like for Ka, Kb is calculated by this equation: [B+][OH-]/[BOH]. A lower Kb value indicates a stronger base.

405268063

I believe that the molecular name should stay the same. Cis and trans are used only to specify which isomer you are referring to.

405268063

I have personally not seen an instance where the transition metal has a negative anion charge if the charge is designated by a parentheses. If I see a roman numeral, I assume it's "plus whatever that charge is" and it works for me.

405268063

As everyone above has said, the pKa constant can tell you the strength of an acid, but this number is generally used for weak acids.

405268063

Yes, considering that the final will entail everything we have covered in lectures this quarter, I would assume naming will also be on the final.

405268063

Cisplatin, unlike transplatin, can bond to the same piece of DNA with two bonds to block DNA replication. This lack of replication can kill cancerous cells. The structure of transplatin makes it so that it can only bond at one site on DNA, which is not strong enough to stop DNA replication.

405268063

I'm pretty sure that in one of the UA sessions we were told to think of dipole-dipole in the gas phase as induced dipole-induced dipole, which is a very weak IMF.

405268063

Bronsted and Lewis acids essentially mean the same thing but there is a difference in what part of the atom you look at to determine if it's an acid or base. Bronsted acids are proton donors whereas Lewis acids are electron acceptors.

405268063

As far as I understand, if the ligand binds at more than one site (meaning it donates more than 1 electron pair), it can be considered polydentate.

405268063

Stronger IMF's lead to higher boiling points because they require more energy to break.

405268063

When you pick a lone pair position, you want to look for the least amount of electron repulsions. If you pick one of the three equatorial atoms for the positioning of a lone pair, we only have two electron repulsions happening. If you pick one of the two axial atoms for this positioning though, we w...

405268063

The dissociation energy is a measure of the strength of a bond. Stronger IMF's such as hydrogen bonding have a higher dissociation energy because it takes more energy to break a strong bond. Weaker IMF's like London forces have a lower dissociation energy.

405268063

Yes, it is possible for a molecule to have dipole-dipole bonds and also be nonpolar. If the polarities occur on opposite sides of a molecule between the same atoms, the molecule becomes nonpolar.

405268063

Yes, regions of electron density include both bonds and lone pairs.

405268063

I just wanted to add on that the book uses the term LDF and not Van der Waals, so I personally would stick with that.

405268063

I believe the example Professor Lavelle used in class was the two rod shaped molecules vs. two spherical shaped molecules. The bonds in the two rod shaped molecules were stronger than the ones in the spherically shaped molecules because they were physically closer together.

405268063

Boiling point is when a liquid goes to the gas phase, and this requires a release of energy, and this comes from the breaking of these intermolecular bonds. But, stronger bonds mean that they are harder to break. In this way, boiling point could be a useful tool for telling us how strong the IMFs are.

405268063

There is unequal sharing of electrons in coordinate covalent bonds. In a regular covalent bond, two atoms would be contributing one electron each, but in a coordinate covalent bond, one atom is contributing both electrons.

405268063

Hydrogen bonds form between oxygen, nitrogen, and fluorine atoms and hydrogen atoms. Oxygen, nitrogen, and fluorine are the only ones that are electronegative enough for this type of bond. But, I would also remember that hydrogen bonds are fairly weak.

405268063

I think you remove electrons with the highest electron affinity first, which in this case would be from the 4s orbital.

405268063

The acid would be the electron acceptor and the base is the electron donor.

405268063

Yes, you would want to get the FC as close to zero as possible to get the most stable Lewis structure.

405268063

Resonance structures I believe only apply to when there are multiple ways of drawing the exact same structure. For example, moving the location of a double bond would make another resonance structure.

405268063

As we discussed in class on Friday, increasing difference in electronegativity means increasing ionic character of a covalent bond, and that charge difference (electric dipole moment) can be measured.

405268063

Personally, I would just know generally what categories the different wavelengths fall into. X-rays/gamma rays are less than 3 nm, UV rays are up to 350 nm, visible light is from 400-700 nm, infrared light is 1000 nm, and microwaves and radio waves are greater than 3x10^6 nm.

405268063

Yup, I just wanted to elaborate on what the person before me said. The angular momentum (l) is based on the orbital you are talking about. So, it's 0 for the s-orbital, 1 for the p-orbital, 2 for the d-orbital, and 3 for the f-orbital.

405268063

The brackets denote a noble gas, and it's just used for efficiency purposes so you don't have to use a huge list to describe the ground state of an atom.

405268063

I agree with the person before me who suggested using double bonds first when you are trying to satisfy the octet rule after you have figured out how many total electrons belong in that molecule. After that, you can use double bonds to make the FC equal to 0 for resonance structures.

405268063

Yes, Prof Lavelle did mention that the bond lengths of these atoms are not actually shorter or longer in real life. This is contrary to what a lot of textbooks will tell you (at least in my experience), and I think it's important in understanding conceptually what resonance structures really are.

405268063

On tests, I personally would just use all of the figures in the constants the sheet gives you to get the most accurate number. Even in doing informal practice problems, I would use 6.626 because changing the number to 6.63 will give you a slightly different answer and sometimes you don't know if it'...

405268063

According to the equations Professor Lavelle gave us in the presentation in class, E of the photon = E of energy required to remove the electron + kinetic energy According to the Photoelectric Experiment, it is the incoming photons that release the electrons from the metal surface. You can't really ...

405268063

Yes, you are correct. The electron would be released but it really wouldn't travel since KE = 0.

405268063

The first equation you've listed is the De Broglie equation and that's used for electrons or objects that have a mass. You cannot use light with the De Broglie equation. The second equation you CAN use for photons (light).

405268063

My routine so far has been to go to lecture and then to come back to my room and then watch the video modules again so that I can try to recall the same information that he just went over. I try to spread out my homework problems over the week so that I'm forced to remember how to do certain things ...

405268063

Dr. Lavelle made a point in class on Friday that kind of cleared things up for me. He said that if you imagine two holes, particles would just go through one but waves would hit both holes at the same time (or at least that's how I interpreted his analogy). Diffraction of light tells us that it has ...

405268063

Wavelengths are grouped together in certain series so that the principal quantum number for their lower energy level is the same. For example, we know the Lyman series' lower energy level is n=1 and the Balmer series' is n=2.

405268063

I think you may be referencing when Dr. Lavelle emphasized at the end of his lecture that the De Broglie equation cannot be applied to light because you need a measurement for the momentum of the object, for which you need also need a mass. Light does not have a mass. Hopefully, this made things a l...

405268063

Could someone also explain if there is any correlation between amplitude and frequency?

405268063

I'm sorry this is such a late response to your clarification question (I just checked back into this post right now), but I thought that I might answer anyway in case you want to come back to this problem for a midterm/final. I think what you are asking is why the 37.5 g of HCl is on the bottom of t...

405268063

G27) Yes, when you see a dilution problem the equation you want to use is: (M initial)(V initial) = (M final)(V final) The problem gives us the final values: (M final)(V final) = (0.7436 M)(10 L) = 7.436 mol HCl (formula mass = 36.46 g) We want the same amount of mols of HCl on the other side of the...

405268063

Yes, that's what I would do. Keep more figures than you know you need for the sig figs at the end of the problem (you can figure this out by looking at the numbers given in the problem) and then round at the very end for your solution.

405268063

That symbol designates micrometers, which is a common unit of measurement for wavelengths of infrared light and biological cells.

Search found 102 matches