CHEMISTRY COMMUNITY

vuongnaomi1L

Screen shot of the question below. I am asking about part A. I am having trouble getting the half-reactions. I know that it has to do with h2 and hcl, but I don't understand why the solutions manuel has H+ going to H2 and vice versa for the anode and cathode. If anyone can help me out that will be g...

vuongnaomi1L

Screenshot of the question is inserted. I am specifically asking for part C. For part C, the standard potential of the anode half-reaction is 0.27 and the standard potential of the cathode half-reaction is 0.79. In the solution manual, we have to use the equation Enot= Enot cathode - Enot anode. Why...

vuongnaomi1L

This is the right equation. Disregard the second equation in my question

vuongnaomi1L

Balance each of the following skeletal equations by using oxidation and reduction half-reactions. All the reactions take place in basic solution. Identify the oxidizing agent and reducing agent in each reaction. PICTURE OF REACTION EQUATION BELOW. I am just confused about why the oxidization half re...

vuongnaomi1L

The presence of a catalyst provides a reaction pathway in which the activation energy of a certain reaction is reduced from 125 kJ⋅mol−1 to 75kJ⋅mol−1. (a) By what factor does the rate of the reaction increase at 298 K, all other factors being equal? (b) By what factor would the rate change if the r...

vuongnaomi1L

I think the best thing to do is review the textbook problems. It's almost impossible to redo every single one of them, so I recommend just doing one problem of each type from the topics we've already been tested on (unless there's a specific concept that you're struggling with) and all of the ones ...

vuongnaomi1L

I do handwritten and just copy the slides, then I make another set of notes that I personalize.

vuongnaomi1L

I feel this too :( I just get plenty of rest and just commit to a schedule of doing things that need to be done.

vuongnaomi1L

molecular shape and structure in Chem 14A and thermodynamics (yes i know) from 14B!

vuongnaomi1L

The Perks of Being a Wallflower :)

vuongnaomi1L

It was my birthday weekend, so I made so many french macarons so celebrate. Not very healthy, but my favorite snack/dessert!

vuongnaomi1L

Sunday morning on March 14, but check the time on the professor's website!

vuongnaomi1L

Running and stretching!

vuongnaomi1L

Reading (recreational reading) and excercise are my best self-care tips!

vuongnaomi1L

Im too busy at the moment haha, but during spring break im definitely going to binge watch some 90's classics.

vuongnaomi1L

I feel like reading and cardio really helps!

vuongnaomi1L

4J.13 Determine which of the following compounds are stable with respect to decomposition into their elements under standard conditions at 25 °C (see Appendix 2A): (a) PCl5(g); (b) HCN(g); (c) NO(g); (d) SO2(g). 4J.15 Which of the following compounds become less stable with respect to the elements a...

vuongnaomi1L

Suppose that 50.0 g of water at 20.0°C is mixed with 65.0 g of water at 50.0°C at constant atmospheric pressure in a thermally insulated vessel. Calculate ΔS and ΔStot for the process. I understand that since there is a temperature change we have to use ∆S=nCmln(Tf/Ti) in order to solve the equation...

vuongnaomi1L

wait, I reread this question again, and since C6H6 is a liquid does that mean we have to find the change in gas molecules?

vuongnaomi1L

4.7 4.7 (a) Calculate the work that must be done against the atmosphere for the expansion of the gaseous products in the combustion of 1.00molC6H6(l) at 25 °C and 1.00 bar. (b) Using data in Appendix 2A, calculate the standard enthalpy of the reaction. (c) Calculate the change in internal energy, ΔU...

vuongnaomi1L

A technician carries out the reaction 2SO2(g)+O2(g)→SO3(g)at 25 °C and 1.00 atm in a cylinder fitted with a piston and maintained at constant pressure. Initially, 0.030 mol SO2 and 0.030 mol O2 are present in the cylinder. The technician then adds a catalyst to initiate the reaction. (a) Calculate t...

vuongnaomi1L

Is there a way to tell whether or not a reaction can be favorable depending on ∆U?

vuongnaomi1L

(a) Calculate the work that must be done against the atmosphere for the expansion of the gaseous products in the combustion of 1.00 mol C6H6(l) at 25°C and 1.00 bar. (b) Using data in Appendix 2A, calculate the standard enthalpy of the reaction. (c) Calculate the change in internal energy, ΔU, of th...

vuongnaomi1L

When work is negative, the system is doing work and when the work is positive, then work is being done on the system?

Is that correct?

vuongnaomi1L

Does ∆U always depend on w, q, and T? I am asking since there are two equations for ∆U, which are ∆U=3/2nR∆T and ∆U=q+w.

vuongnaomi1L

When q is zero in ∆U=q+w, does this mean that there is no heat transferred to the surroundings? When q is negative, it means that heat is released and when q is positive this means that there is heat absorption?

Can someone let me know if my statements are correct, or if I'm wrong.

vuongnaomi1L

505598869 wrote:I was wrong to bet against the bucks.

LOL

vuongnaomi1L

I think consistently practicing is the best place to start. Going to UA sessions, practice textbook problems, sapling, all of those will help with the midterms and finals. Remember not to cram!

vuongnaomi1L

Attack on Titan

vuongnaomi1L

I feel the same way, but don't feel too bad over it :) I find what has been helpful so far is going to UA sessions!

vuongnaomi1L

Thanks for sharing this! It made me feel better about my mistakes on the test too!

vuongnaomi1L

I never related to a chem community post more. Last quarter's midterms and this quarter's midterms are vastly different in terms of how challenging they were. I think we just need to try our best and see how it goes. We did the best we could and that's what matters the most.

vuongnaomi1L

Hopefully by the end of the week!

vuongnaomi1L

I am waiting next year as I want to get some hands on experience!

vuongnaomi1L

I completely relate to this! I think as hard as it is, try to incorporate some healthy habits to keep our mental wellness in check, like listening to music, going in walks, or reading books. Hope this gets better!

vuongnaomi1L

Unfortunately, I do not think there is a reset button. What has been helpful to me is to take notes on the sapling questions so I can come back to it later.

vuongnaomi1L

Professor Lavelle said that to be safe, whenever Ka < 10-4 then we are good to use the approximation. I don't think that will change in an exam setting either.

vuongnaomi1L

Professor Lavelle has stated to be safe, whenever Ka < 10-4

vuongnaomi1L

I actuallly dont read the textbook and whenever I have a question I either ask the UA's or go to the professor's office hours.

vuongnaomi1L

PLF sessions are also very helpful!!

vuongnaomi1L

thank you so much for your update! this was so inspiring!!

vuongnaomi1L

If you get the e-text, then it comes with the solutions manuel! not sure if that is included in the looseleaf version.

vuongnaomi1L

if both are positive, choose the one that will not result in a negative number when you plug it in the equations for equilibrium constants. For example, if an equation to find the equilibrium concentration of a reactant is 0.5-x and you have values 0.2 and 0.8 from your quadratic equation, 0.8 will ...

vuongnaomi1L

You should use the units given in the question, but mol/L is standard.

vuongnaomi1L

Yes midterms are during lecture. I believe this was in the syllabus.

vuongnaomi1L

P is for Pressure, V is for Volume in Liters, n is for number of moles, R is for gas constant, and T is for Temperature in Kelvin

vuongnaomi1L

How To Be Single!! Dakota Johnson is such a great actress imo and this movie really shows how everyone's successes and timings are different! Really good to wind down after a stressful event (class).

vuongnaomi1L

I think Dr. Lavelle assigns different due dates. It might not be the end of every week.

vuongnaomi1L

Yes, our midterms will be during lecture for this quarter and not during discussion like it was last quarter.

vuongnaomi1L

I find that going to workshops and step-up sessions before exams helped a lot. For this quarter, I'm going to try to go to at least one step-up and two-workshops per week so I can keep up with those worksheets!

vuongnaomi1L

The best thing that I can do differently for myself this quarter is to make a schedule to realistically space out my studying, and force myself to stick to it. Cramming does not work, especially in a class like this. I find that it is better to study ahead of time so I can retain the information bet...

vuongnaomi1L

Yeah, I felt rushed too. Many people did. I dont think the alloted time was enough. Good luck to you though! Just remember that theres always chances to do better and grad schools like to see upward trends :)

vuongnaomi1L

So H2SO3 is an acid, but the lewis structure (which I inserted below) has lone pairs to attract H+ ions from H2O, which will make OH- ions. So why is it an acid and not a base? What makes H2SO3 lose H+ ions but not attract them?

vuongnaomi1L

This question asks to classify whether certain molecules are acids or bases. For part C, since KOH has an H, I thought this means that it could give off a proton and be an acid. Why is KOH a base and not an acid?

vuongnaomi1L

This is the pic, forgot to insert.

vuongnaomi1L

In part d, the question asks the ligand can be polydentate, and give the maximum number of places on the ligand that can bind simultaneously to a single metal center. For oxalate ion, the formula is C2O4 2-. I attached a picture. Since there are lone pairs on 4 of the atoms, what makes it bidentate ...

vuongnaomi1L

FrancescaHawkins3J wrote:In my understanding, CO is a monodentate ligand because when one of the lone pairs becomes a bonding pair, it repels the other lone pair, which becomes too far away to create a second bond.

This makes perfect sense! Thank you!

vuongnaomi1L

So the lewis structure of CO is C and O having triple bonds and there are 1 lone pair on each C and O. Since there are two lone pairs on the CO molecule, why isnt it a bidentate ligand?

vuongnaomi1L

Since Fe is a transition metal and has unfilled d orbitals, it can have up to 12 valence electrons when bonding; for example in the molecule Ba(FeBr4)2. Usually, when it is not bonded, it just has 2 valence electrons.

Can someone confirm or deny whether the statement above is correct?

vuongnaomi1L

To my understanding, Bronsted acids and bases use the accepting or donating of H+ ions in their definition, while lewis acids and bases go by accepting or donating electrons in their definition. Is this correct?

vuongnaomi1L

The definition of a bidentate ligand is one that is able to donate 2 electron pairs. Since each N has 5 valence electrons and it bonds to Co, H, and another H, then it only has a lone pair left. Wouldnt this make each NH2 a monodentate ligand? I have attached an image of the coordination compound be...

vuongnaomi1L

The definition of a bidentate ligand is one that is able to donate 2 electron pairs. Since N has 5 valence electrons and it bonds to Co, H, and another H, then it only has a lone pair left. Wouldnt this make NH2 a monodentate ligand?

vuongnaomi1L

Can someone explain why nickel has an expanded octet?

To my understanding, there are 8 electrons in nickel's 3d state, and there are 2 electrons in its 4s state. So how does it have an expanded octet?

vuongnaomi1L

Since nickel has n=4, which means it contains the f-orbitals, is this why it has an expanded octet?

vuongnaomi1L

Why is SO2 acidic when it accepts a proton?

vuongnaomi1L

In cisplatin, why do the chlorines unbind from the ligand, and not the amine groups? Shouldn't the bond between platinum and ammonia be more sensitive to breaking off as opposed to the chlorine?

vuongnaomi1L

My TA said it was about 30 questions, and partial credit might be given to questions with multiple parts. I'd check with your TA just to make sure :)

vuongnaomi1L

I don't think extra credit is offered, but I would ask just to make sure :)

vuongnaomi1L

I totally understand your point of view, but he is allowing many people to pass with a C- by just earning half the points in the course. I know several of us really want to pass with an A, including myself, but the pandemic sadly puts everyone in a difficult position. Do not let earning a grade low...

vuongnaomi1L

I'm doing well! I'm just glad its the holidays. the holiday season is so overrated but is my favorite season of the year!

vuongnaomi1L

Ever since CCLE came back up this late morning, my lectures wouldn't load. I tried clearing out cookies and logging back in, but does anyone know what else I can do?

vuongnaomi1L

I think that questions that are harder and have more room for error have partial credit, but just to make sure I recommend emailing the professor or your TA :)

vuongnaomi1L

Procrastination is my worse enemy, and I think I'm getting a little bit better, but I definitely still have room for improvement :)

vuongnaomi1L

I find that going over the lecture notes and re-explaining the concepts to myself in my head are beneficial, as well as doing the outline homework. I only do the questions, I don't read the textbook though because I am very much a visual learner.

vuongnaomi1L

I find that going over the outline questions and attending as many workshops as I can are very helpful! For older material I was planning to review the modules on the professor's website.

vuongnaomi1L

I honestly struggle with this so much so I don't even know if I'm qualified to answer, but my best tips would be to listen to instrumental music (dont listen to music with voices and lyrics, very distracting), have snacks and water on hand, take breaks, and be in a not too cold or not too hot enviro...

vuongnaomi1L

I really like to paint! or read a new book! :)

vuongnaomi1L

I think we will have class since we only do really have Thursday and Friday off, but I would check with the TA or ask the professor during OH to make sure :)

vuongnaomi1L

I live near the woods, so I like to run around there during daylight :)

vuongnaomi1L

So I understand that BaBr2 is more ionic than BeBr2 because since electronegativity decreases down a group, then the electronegative difference between Ba and Br is greater than the electronegative difference between Be and Br2. However, I am confused because then why is NaCl a stronger ionic bond t...

vuongnaomi1L

So what is the relationship of cations and melting points? So if cations that are smaller in size and have higher positive charges are stronger, then why does NaCl have a stronger ionic bond and a higher melting point than MgCl2. Mg is smaller than Na and has a 2+ charge, while Na has a 1+ charge. C...

vuongnaomi1L

To my understading:

When it comes to ion - dipole, hydrogen bonding, dipole-dipole interactions, and london dispersion forces, these are all intermolecular interatcions, not intramolecular interactions.

Can someone confirm or deny what I stated is correct?

vuongnaomi1L

To my understanding: For covalent bonds, increasing the atomic radius would result in a larger distance and a weaker bond, while for London dispersion forces, increasing the atomic radii would cause higher polarizability and thus a greater intermolecular attraction. Can someone confirm or deny what ...

vuongnaomi1L

For one of the questions on the week 5 and 6 sapling homework, there was an explanation of one of the homework questions that stated "dispersion forces do not involve electrostatic interactions of polar groups and are present between all molecules. I'm a bit confused by this statement since it ...

vuongnaomi1L

I'm confused about how sulfur can have 12 valence electrons in SO42- and SF6. I thought that since sulfur is in the third period, it could only go up to 10 valence electrons because sulfur's electrons can be excited to the 3d state. Any clarification would be appreciated.

vuongnaomi1L

So to my understanding, the 4s state is lower than the 3d state only when the 3d state does not get filled, or when the 4s state gets filled before the 3d state, correct? For example, in the electron configuration of Ca:[Ar] 4s^2, the 4s state would be lower than the 3d state because Ca does not hav...

vuongnaomi1L

Not going to lie, the online format is honestly very depressing. With that being said, this class, in particular, is so much better than all my other classes. I feel like this class is very accomodating given all the changes from in-person classes to online classes. Hope everyone is drinking water a...

vuongnaomi1L

For my TA, he said that we have to email him to see which questions we missed. To clarify, I think it would be best to email your TA :)

vuongnaomi1L

I'm pretty sure it will be about the same number of questions. For the difficulty level, I'm not sure if midterm 2 is going to be harder than midterm 1.

vuongnaomi1L

Since the average for the midterm was high, I doubt that there will be a curve. But please check with your TA to clarify :)

vuongnaomi1L

Hi! I'm pretty sure it will be the same format, with multiple-choice questions using Respondus.

vuongnaomi1L

So your objective is to find the difference between the incident photons and the energy of the ejected electrons, in eV/atom. So since they give you the wavelength, you would use that to find the energy per photon in joules, convert that to energy per photon in eV. You would subtract the 0.137 eV by...

vuongnaomi1L

For K (potassium) on the periodic table, after Argon, you would just jump to period 4 for K, which would result in the electron configuration of potassium, K: [Ar]4s^1. Therefore the energy of the 4s state will be lower than the 3d state unless I fill the 3d state for another element (ie, scandium),...

vuongnaomi1L

So to my understanding, for hydrogen, we don't specify orbitals with the principal quantum number (n=1,2,3 etc) with the angular momentum (l) or the orientation of the angular momentum because the energy for the electron in the orbital is the same? If I am correct above, then the interaction of mult...

vuongnaomi1L

This is a discussion question from week two and I'm confused for part c. Part C asks what is the wavelength of the radiation that caused photo ejection of the electron. So for this question, I thought we can use the equation E photon = h times c / wavelength because we are trying to find the wavelen...

vuongnaomi1L

So this question asks what region of the electromagnetic spectrum does this wavelength correspond. Can someone clarify how to determine which wavelengths go in which region of the electromagnetic spectrum? This is the module question. The meter was defined in 1963 as 1,650,763.73 wavelengths of radi...

vuongnaomi1L

10 multiple choice questions with varying difficulty!

vuongnaomi1L

Based on what we've learned so far, C would be the speed of light unless the question otherwise specifies

vuongnaomi1L

video modules, retake the module tests and try to get the questions rights without referencing your notes too much, and then text book problems!

vuongnaomi1L

We are supposed to use a scientific calculator for the exam. I'm not sure if we are allowed to use a scientific graphing calculator so I'd recommend to email the professor or your TA :)

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