CHEMISTRY COMMUNITY

VSU_3F

Hi Dr. Lavelle, Thank you so much for everything! You are the best professor any student could ask for. You care so much for all of your students and always go above and beyond to help us succeed. Your support and encouragement always meant the world to me! I will miss you and your class greatly. I'...

VSU_3F

Thank you so much for your help, Dr. Lavelle!

VSU_3F

I too prefer handwriting my notes, especially for my STEM classes.

VSU_3F

I don't think there will be a curve.

VSU_3F

I will be taking 14BL next quarter! :)

VSU_3F

If it is a zero-order reaction :)

VSU_3F

Yes, as others have mentioned -k is the slope of a first-order and zero-order reaction. You can gather this from the integrated rate law too. Both of the coefficients of t for first and zero-order reactions is -k.

VSU_3F

Because the system is our reference point, if it is doing work, we say that it is losing energy, which is why work done by the system has a negative sign.

VSU_3F

Hello! For part d, I wrote the oxidation half-reaction as Au+ (aq) --> Au3+ (aq)+ 2e-. The solution manuel had their oxidation reaction as Au(s) --> Au3+(aq) + 3e- (aq). I got the same overall reaction as the textbook solution, but it wasn't very intuitive for me to have Au(s) on the reactant side o...

VSU_3F

I figured out where I went wrong. The sulfur on the reactant side actually has a +2 oxidation state, not a +3.

VSU_3F

Hello, I'm having trouble with the oxidation half-reaction of this redox reaction: I got sulfur to have a +3 oxidation state on the reactant sides and a +6 oxidation state on the product side. And since we have 2 sulfurs on the left and right side, we have a +6 charge on the left and a +12 charge so...

VSU_3F

Balancing in basic solutions almost follows the same procedure as balancing in acidic solutions, but you must add OH- to both sides at the end to balance out the H+ you added.

VSU_3F

As others have mentioned, spontaneity really refers to the idea the energy states of the products and lower than those of the reactants. I believe in lecture that Dr. Lavelle mentioned C(diamond) --> C(graphite) occurs very, very, very slowly at room temperature (so slowly that it basically is nonex...

VSU_3F

For exothermic reactions, we're always going from reactants with higher free energy states to products with lower free energy states, which creates a net release of energy. For endothermic reactions, we're always going from reactants with lower free energy states to products with higher free energy ...

VSU_3F

I see, thank you!

VSU_3F

Hello! So this came up in Kate's workshop today. We were using the Nernst equation to find the concentration of H+ in the product side of the reaction. I've attached her work below. So for this scenario, we were allowed to input 1.0 mol/L as our Cl- concentration and 1 bar as our H2 concentration in...

VSU_3F

I emailed Dr. Casey, and she said that all students on the waitlist are automatically enrolled! :)

VSU_3F

Yes, only gases! Solids are not considered to have a concentration, and liquids are always assumed to be in extremely high volumes, so they don't tend to shift a reaction either way.

VSU_3F

As others have mentioned, it will likely be sometime next week. Last time, it was released on a Wednesday, so maybe we'll know close to the middle of next week?

VSU_3F

I''m taking Chem 14BL, LS 7C, Cluster 80CW, and an intro to biomedical research class (Biomedical Research 5HA).

VSU_3F

It is due the Sunday following Week 8 (2/28) at 11:59 pm

VSU_3F

Sorry, nevermind. I realize that K was not the quantity specified for part g. Please disregard my last post.

VSU_3F

Nope, that seems like the correct thing to do. Temperature can be thought of as adding or removing heat, so you would want to know whether your reaction is endothermic or exothermic, hence finding dH for the reaction Since we're assuming the delta H is the same across all temperatures, would we jus...

VSU_3F

Hello!

Enthalpy of vaporization is always per mole. So you know how much heat is needed to vaporize 0.579 mol of CH4(l). Use that to find how heat is needed to vaporize 1 mole of CH4(l). That is your enthalpy of vaporization of methane.

VSU_3F

Yes, there should be! We had a UA session last quarter on Veteran's Day. So UA sessions still occur on holidays, I believe.

VSU_3F

As others have mentioned K could refer to Kc or Kp. You have to look at the units to determine whether we're dealing with Kc (moles/L) or Kp (atm). Typically with gases, K usually refers to Kp. And for solids, liquids, and aqueous solutions, K refers to Kc.

VSU_3F

Yes, it does! Thank you so much!

VSU_3F

Hello! So for this problem, the balanced equation is 2HCl (aq) + Zn (s) --> H2(g) + ZnCl2 (aq). It seems like when we have to calculate the enthalpy of the reaction from the standard formation of enthalpies, we consider HCl to be on both sides of the equation. Is that because the reaction is taking ...

VSU_3F

Got it, thank you so much!

VSU_3F

Hello everyone! I've posted the problem below. So after we balanced the equation, the problem became 1 C6H6 (l) + 7.5O2 (g) --> 6CO2 (g) + 3H2O (g). I thought we liked having our balanced equation with whole-number coefficients, so I found the work done by 2 moles of C6H6. Do we always have to find ...

VSU_3F

Just to clarify, C(subscript) is for representing something remaining constant? thank you Yes, I believe so. Cp is the amount of heat required to raise one mole of gas one degree Celsius at constant pressure and then Cv is the amount of heat required to raise one mole of gas one degree Celsius at c...

VSU_3F

How do we know the volume is constant in this problem?

VSU_3F

Yes, it does! Thank you both very much!

VSU_3F

It's also important to note that delta U will equal q in this problem since the gas is at constant volume, meaning no work is being performed.

VSU_3F

Hello!

There was this formula in sapling Q18. I've attached it down below. I don't remember us ever going through this formula in lecture. Does anyone know where this came from? Thank you!

VSU_3F

Hello! I believe that you need to find the enthalpy of vaporization of one mole of enthane. So you would take 4.76 kJ/0.579 mol to get 8.2 KJ/mol.

VSU_3F

Also, Sapling Q18 mentioned that we could use this equation to calculate a change in internal energy if it corresponded to a temperature change. I don't remember us ever going over this. Could someone also please explain this equation to me? Thank you.

VSU_3F

Hello!

I saw that the Week 3/4 Sapling assignment had some questions regarding constant‑pressure molar specific heat and constant‑volume molar specific heat. Did we go over these in detail during lecture yet? If not, will we?

VSU_3F

Yes, as everyone mentioned above, you must first check the units! That's the most important thing. But now that we're doing a lot of thermodynamics and work problems (which has the unit of Joules), we'll be using R= 8.314 J/Kmol a lot more often.

VSU_3F

I also felt like it was harder than I expected. I didn't feel as good after this midterm as I did with last quarter's midterms.

VSU_3F

If the K value is less than 10^-4, X is negligible.

If less than 5% of the acid is deprotonated when you ignore X, then X is also negligible.

VSU_3F

reva_bajjuri wrote:wait to bomb calorimeters not exchange heat/energy with the surroundings?

Yes! Because it is an isolated system.

VSU_3F

Yes, as others have mentioned, they will most likely be out by the end of next week! It usually takes a week.

VSU_3F

Textbooks do vary on the classification of this acid, for you can consider it as a strong acid for this class!

VSU_3F

Yes, as Tanya mentioned, in chemical reactions, bonds are always being formed (energy is released) and broken (energy is released), but exothermic reactions have a net release of energy and endothermic reactions have a net absorption of energy. We're really just looking at the final energy of the pr...

VSU_3F

Yes, exactly! As Ashlen mentioned, the heat energy is solely going in to break the intermolecular forces of the system. It is not increasing the kinetic energy of the molecules during a phase change, which is why temperature remains constant during phase changes.

VSU_3F

Temperature is still constant because the heat that is being absorbed by the system is solely breaking the bonds of the system. That's why during phase changes, temperature remains constant.

VSU_3F

The (l) means liquid. So whatever compound has that subscript, it is in a liquid state.

VSU_3F

Hannah_Butler_2E wrote:How do you find Ka for the part b of this problem, AlCl3? I don't see its conjugate base in the provided table.

Hey Hannah! You'll find the Ka value in table 6D.1 under the category small, highly charged metal cations.

VSU_3F

As the others have mentioned, if you consider the inverse relationship between Ka and Kb, the higher Ka is, the lower Kb is. When you multiply Ka and Kb, they equal the constant Kw which is 10^-14 at 25 degrees Celcius. Kw is a constant, so as you increase one thing in the equation, you must proport...

VSU_3F

Yes, you're exactly right and so are all the users that posted before me. A great way to test this is by using numbers. It's important to note that n (the number of moles) stays constant!

VSU_3F

Another helpful way to remember Q and K is to plot Q and K on the number line. If Q is ever to the right of K (greater than K), draw an arrow pointing from Q to K. Remember that Q will always want to go to K. An arrow pointing from right to left would mean that the reactants are favored. Likewise, i...

VSU_3F

Along with what all the others have mentioned, you can also think about this problem in terms of LeChatlier's principle. The external environment has more heat than it did previously, so to minimize the effect of that change, the reaction will use up the heat and try its best to restore the external...

VSU_3F

Yes, you're exactly right. To determine whether the products or reactants are increasing, you have to first look at how the reaction will shift. Because Q is 0 and less than K initially, we know that the reaction will shift right, meaning product concentration will increase. Therefore, the change in...

VSU_3F

Hello Devin! The three types of learning sessions are step-up, workshops, and drop-ins. Step-up sessions and workshops are extremely great ways to get extra practice on problems and practice before exams as the UA's go over past exam questions, textbook problems, and more. All of the problems they g...

VSU_3F

Hello! Water is one of the reactants of this chemical reaction, so the reaction will use the extra water to produce more product. You can think about this problem mathematically too. If the concentration of water increases (number in denominator increases), the numerator (concentration of products) ...

VSU_3F

Yes, you're exactly right. As others have mentioned, exothermic reactions have a negative delta G as the energy of the products is lower than the energy of the reactants. This is why exothermic reactions are spontaneous as things are always workings toward stability. Conversely, endothermic reaction...

VSU_3F

Even if you add more reactant to a reaction that already has excess reactant, the reaction would still work to achieve equilibrium. In this case, the forward reaction would occur at an even faster rate until equilibrium is reached but note that K is still unchanging.

VSU_3F

Hello! I think when the textbook said to disregard significant figures, they meant to not round to two significant figures until the very end. I think a good rule of thumb is to not round until you reach your final answer. I hope I interpreted their statement correctly, haha. Since you essentially g...

VSU_3F

Thank you so much Dr. Lavelle for being the best professor any student could have asked for! You are always so kind, patient, and more than willing to help us. Your hard work and dedication is so apparent and very much appreciated by all of us. I'm very much looking forward to next quarter! Hope you...

VSU_3F

Hello!

Just curious, what would the conjugate acid and bases be in this chemical reaction? Since this reaction creates a neutral solution, would we still have conjugate acid and base pairs? Thank you so much!

VSU_3F

The arrow is always pointing towards the atom with the stronger electronegativity because that is the atom the electrons are slightly more attracted to. To determine if a bond is polar or not, you would have to look at the electronegativity difference (0-0.4 is nonpolar and 0.5 or more is polar, I b...

VSU_3F

If there is a lone pair on the central atom, you know that the bond angle will be less than 109.5. This is because the lone pair repels the bonding electrons and pushes the rest of the atoms bonded to the central atom down. You don't have to know that it's 107 degrees. You just have to know that it ...

VSU_3F

Yes, whenever you are dealing with something that has a charge, that usually is a good indicator that it is an ion.

VSU_3F

It's important to note that intermolecular forces occur between different molecules. If we're solely looking at HSe and HS on their own, we only can really look at intramolecular forces, or more specifically, the strength of their covalent bonds. Hope this helped!

VSU_3F

Hello! The whole idea about the least electronegative atom being the central atom is just a guideline, so there will be some exceptions like in these instances. I believe what takes precedence here is the fact that because we're donating a proton, we know that the H will be bonded to the oxygen, and...

VSU_3F

As others have mentioned, it is extremely hard for HF to dissociated into its constitutive ions as it is the smallest atom out of all the others. Since how easily a bond dissociates is a measure of how strong an acid is, with easier dissociations equating to stronger acids, HF is not considered a st...

VSU_3F

Hello!!

So NH4 gets produced when the lone pair that was originally on the NH3 forms a bond with the H+. As a result, nitrogen will now have a +1 formal change, which explains why NH4 has a +1 charge overall.

VSU_3F

As mentioned above, I believe he meant that what we designate as the pi bonds and sigma bond doesn't matter. You could have labeled the top bond as the sigma bond and the other two below it as pi bonds, for example.

VSU_3F

Hi! So both the electron densities of water and CH2 2- have a tetrahedral arrangement and because the 2 lone pairs are pushing down on bonds, we know that the bond angle will be less than 109.5 degrees not more than 109.5 degrees. Therefore, a number like 104.5 degrees makes sense for both of these ...

VSU_3F

For this problem, I recommend you draw out the Lewis structures and determine how many bonds the central metal cation ion will form. Also be sure to determine if the ligands are monodentate, bidentate, etc. The number of bonds the metal cation can form is the coordination number.

VSU_3F

Another way to do this is finding how many grams of carbon, hydrogen, and oxygen you have by taking their respective percentages and multiplying them by the total molar mass. From there, you'll find how many grams of carbon, hydrogen, and oxygen you have. To go from grams to moles, you'll have to ta...

VSU_3F

You're welcome!! :D

VSU_3F

I wanted to point out that when there's a double bond, one of them will be a sigma bond and the other will be a pi bond. Likewise, when there's a triple bond, one of them will be a sigma bond and the other two will be pi bonds. We don't have to know which is which. Just know that with any type of bo...

VSU_3F

As others have said, I believe they're the same thing! Free radicals are involved in oxidizing elements, and I feel like an element with an odd number of electrons would want electrons from other substances, so I believe they are interchangeable terms.

VSU_3F

It's important to note that most organic molecules have C-H bonds, so even though CO2 has carbon in it, it's an inorganic compound because there are no C-H bonds.

VSU_3F

Hello! For me, it was helpful to think about the ammonia example that Dr. Lavelle used in lecture. When we were dissolving a transition mental in water, we saw that when we added ammonia to the solution, the ammnoia could break the bond that was originally formed between the central metal cation and...

VSU_3F

Last time, I believe they were posted on Thursday and a little after noon, around 1-ish.

VSU_3F

From my understanding, if you have a double bond, one will be a sigma bond and the other will be a pi bond. Similarly, if you have a triple bond, one will be a sigma bond and the other two will be pi bonds. As for now at least, these are the only two types of bonds we need to know. I believe Dr. Lav...

VSU_3F

705512695 3K wrote:So the longer the bonds are the less dissociation energy the structure will have?

Yes, you're exactly right! Bond length and bond strength have an inverse relationship. :)

VSU_3F

To count the number of electrons in an expanded valence, you would count all of the bonding electrons and lone pair electrons. So all of the bonding pair electrons would count as 2 and each lone electron would count as 1. For example, if the central atom S has 4 bonds and one lone pair of electrons,...

VSU_3F

I know that Dr. Lavelle mentioned in lecture that hydrogen bonding is -20 kJ/mol. Hydrogen bonding is a special type of dipole-dipole bonding, and this is an example where dipole-dipole bonding is greater than dipole-induced dipole bonding. If we were to compare, I think we would have to look at it ...

VSU_3F

I am in Dr. Lavelle's 3 pm discussion right now, and he said that there will still be 40 hours of UA sessions a week next quarter, but the exact dates and times of each UA session are not known yet as they are dependent on the UA's availability.

I'm not sure about the TA's though, sorry.

VSU_3F

My gut tells me that the shielding effect of sulfur takes precedence here just because there are more orbitals involved in sulfur and a lot more electrons. Even though N gained one more electron than sulfur did, it still has a lot fewer orbitals, so it should still be smaller!

VSU_3F

A compound that is a radical has an odd number of electrons. It is defined as anything that has only one electron when there should be a lone pair.

So to approach this problem, I would count the number of valence electrons of each species and draw out the lewis structures.

VSU_3F

I believe the textbook solution stated molecules with polar double bonds tend to accept electrons, which would make them a Lewis acid. I'm guessing the positive dipole on sulfur induced by the differences in electronegativity between sulfur and oxygen would attract some electrons.

VSU_3F

I'm in Dr. Lavelle's discussion section right now, and like many have said before, a hydrogen bond is not considered a coordinate covalent bond because it is not a covalent bond. Hydrogen bonds are a lot weaker than covalent bonds.

VSU_3F

First, determine the electron configuration of the neutral atom. Then, determine how many electrons from the ion are being removed or added. Remember that electrons are always removed from the highest energy level first because they are held the least tightly. Electrons are also added to the highest...

VSU_3F

Does an exception like this occur in every period down the periodic table when a second electron occupies the 2px orbital? Like are the ionization energies of Sulfur, Selenium, Tellurium, and Polonium all slighter lower than those of Phosphorous, Arsenic, Antimony, and Bismuth respectively? If no, ...

VSU_3F

As the others have mentioned, if you draw out the electron configuration of oxygen, you'll find that in one of the 2p orbitals, there are 2 electrons. Oxygen would be a lot more stable if its 2p subshell was half-filled (one electron in each orbital). Furthermore, electron-electron repulsion in that...

VSU_3F

Yes! A bond has ionic character if there is a significant difference between the electronegativities of the elements involved. Since a polar covalent bond fits this description, it would have ionic character. However, it wouldn't have as much ionic character as an actual ionic bond where the electro...

VSU_3F

Do you mean that you're leaving the numbers plugged into the formal charge formula? I'm not sure what you mean by your formal charges aren't simplified enough.

VSU_3F

You're welcome!! :)

VSU_3F

The reason why there's only one orbital is because of the third quantum number. The third quantum number will specify which orbital you're in. We don't know what orbital +1 is referring to (since there's no set standard for ml numbers...for example +1 could be 3dxy or 3dxz), but we know it's referri...

VSU_3F

Essentially for chromium and copper, it is more stable for their 3d to either be half-filled (1 electron in each orbital) or completely filled (2 electrons in each orbital). It is a more stable arrangement, or in other words, a lower energy configuration this way. This is why chromium and copper hal...

VSU_3F

Hi Charlene, You would only need to multiply by Avogadro's number if there asking you to find the electron affinity of a mole of atoms. In the first part, you only need to find the electron affinity of one atom. The electron affinity that you are finding is already in terms of eV per atom, so there ...

VSU_3F

Another friendly reminder is to make sure your wavelength is in meters! :)

VSU_3F

In this case, the grams of solute don't contribute to the overall volume, so it's negligible. Therefore, 500 mL is the total volume of this solution.

VSU_3F

Yes, your speed is correct!

The speed is actually reasonable because it is slower than the speed of light, which is the fastest recorded speed of anything we know. As long as the speed is slower than the speed of light, it is reasonable.

VSU_3F

Hi Samantha, I believe the textbook defined Z as the atomic number (# of protons) of any element, so when the book mentioned "+Ze" and "-Ze", it was really telling how to determine the positive and negative charge of any element respectfully, where e is the fundamental charge (on...

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