CHEMISTRY COMMUNITY

Daniela_Martinez_3B

I going to binge watch some Netflix, but I'm mostly looking forward to reading more books!

Daniela_Martinez_3B

same here. I regret not attending UA sessions! Next quarter I definitely want to improve my time management so that I can attend those regularly.

Daniela_Martinez_3B

Finals week is tough, but a cup of tea always helps me calm my nerves

Daniela_Martinez_3B

I'm not sure if there is another way to determine ionic character besides difference in electronegativity, but if there is I would also like to know

Daniela_Martinez_3B

A photon checks into a hotel and is asked if he needs any help with his luggage. He says, “No, I’m traveling light.”

Daniela_Martinez_3B

electron geometry takes into account both bond electron pairs and lone electron pairs whilst molecular geometry only takes into account bond electron pairs

Daniela_Martinez_3B

en is ethylenediamine (H2NCH2CH2NH2)

Daniela_Martinez_3B

I think their saliva triggers a histamine release which is what causes the itchiness.

Daniela_Martinez_3B

I'm also stressed about finals, but I'm determined to end on a good note! Also, really excited for winter break :)

Daniela_Martinez_3B

The organic chemistry tutor on youtube has a great video on sigma and pi bonds that really helped me visualize them!

Daniela_Martinez_3B

My friend who previously took this class said she didn't get her final back when she took it, so I think your best bet is to study the textbook problems.

Daniela_Martinez_3B

yeah I think redoing the homework problems since week 1 will be best and it's a good idea to start doing those this week

Daniela_Martinez_3B

A delocalized pi bond is a pi bond in which the electrons are free to move over more than two nuclei. A regular pi bond is a localized pi bond in which the bonded electrons are confined to a concentrated region between two separate atoms.

Daniela_Martinez_3B

I think any scientific calculator should be okay. As for the periodic table, I go back and forth between the one professor Lavelle gave us and one that I found online that has more info.

Daniela_Martinez_3B

This is super helpful, thank you!!!

Daniela_Martinez_3B

if the dipole moments cancel out it's non-polar!

Daniela_Martinez_3B

1 millimole = 10^-3 moles

Daniela_Martinez_3B

Hydrogen will never be a central atom because it never forms more than one bond!

Daniela_Martinez_3B

Taking my dogs for a walk really helps me relax, but I also like to binge watch some netflix!

Daniela_Martinez_3B

I feel the same way, but I'm not letting it discourage me! All we can do is move forward and study harder for the final :) you got this!!

Daniela_Martinez_3B

This molecule has 6 electron dense areas (5 outer atoms and one lone pair) so you can expect a square pyramidal geometry. The lone pair on the central atom is pushing the other atoms closer and so the bond angles will be slightly less than 90 degrees and 180 degrees.

Daniela_Martinez_3B

For Sb the normal configuration would be [Kr] 4d^10 5s^2 5p^3, but the questions asks for Sb^(3+) which means it has lost 3 electrons and so it becomes [Kr] 4d^10 5s^2 For Sn the normal configuration would be [Kr]4d^10 5s^2 5p^2, but the questions asks for Sn^(4+) which means it has lost 4 electrons...

Daniela_Martinez_3B

Yes, you're correct for part a! For part b, it's invalid because l=0 so ml must also equal 0. Part c is also invalid because the maximum value l can have is n-1 and n=4 so lmax=3.

Daniela_Martinez_3B

LDF are the only intermolecular force that exists between non polar molecules or atoms. You can tell if a molecule will be non polar if 1) the molecular shape around the central atom has no lone pairs, or if it does it's either square planar or linear 2) all atoms around the central atom are the sam...

Daniela_Martinez_3B

I believe Dr. Lavelle addressed this during his office hours, but yes hydrogen bonding is its own thing!

Daniela_Martinez_3B

I'm definitely going to focus more on the textbook problems! The sapling problems and my lecture notes aren't really a priority, but I'll still take some time to review those too.

Daniela_Martinez_3B

Hey! I also took regular high school chem and most of the material we've covered in this class is completely new to me. I didn't learn much in high school chem besides some of the topics on the fundamental section.

Daniela_Martinez_3B

I feel you! It's really easy to get distracted at home, so it's really important to have a designated place where you can study with no distractions. I definitely recommend using the pomodoro technique. I also like listening to lofi beats on youtube to help me focus lol.

Daniela_Martinez_3B

Resonance structures with low formal charges are favored over high formal charge!

Daniela_Martinez_3B

There's a sapling hw question about whether the bond has ample or overwhelming bond character so I'm really confused on this too! I know that it's essentially asking if the bond is more of a single bond or more of a double bond based on the bond lengths, but the ample/overwhelming thing is really th...

Daniela_Martinez_3B

The number of electrons is equal to the number of protons of each element! An atom with a neutral charge is one where the number of electrons is equal to the atomic number. Ions, however, are atoms with extra electrons or missing electrons.

Daniela_Martinez_3B

The gain of an electron adds more electrons to the outermost shell which increases the radius because there are now more electrons further away from the nucleus and there are more electrons to pull towards the nucleus so the pull becomes slightly weaker than of the neutral atom and causes an increas...

Daniela_Martinez_3B

I believe the main group elements are any of the chemical elements belonging to the s and p blocks of the periodic table. The main group elements are important because they, along with a few light transition metals, are the most abundant elements in the universe, solar system, and on Earth.

Daniela_Martinez_3B

Yes! Remember that anything in the UV region involves n=1, anything in the visible region involves n=2, and anything in the infrared region involves n=3.

Daniela_Martinez_3B

Midterm season can get really stressful, but it's important to take time for yourself and de-stress! I usually just binge watch tv or read a book!

Daniela_Martinez_3B

I had the same problem, just input the full electron configuration without the comma and it should work!

Daniela_Martinez_3B

I think the questions are still being randomized unfortunately!

Daniela_Martinez_3B

The questions will probably specify which unit to use! I believe the exam is also multiple choice, so you can tell that way too.

Daniela_Martinez_3B

Any electromagnetic wave's frequency multiplied by its wavelength equals the speed of light. The speed of light is always 3x10^8!

Daniela_Martinez_3B

The same thing happened to me! In case you couldn't find it, here's the link to the solution's manual with other errors from the textbook https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/Solution_Manual_Errors_7Ed.pdf

Daniela_Martinez_3B

Hey! I definitely recommend going over the textbook homework problems and any notes you may have written during lectures. If you're still a bit confused on a certain topic, you can find tons of great youtube videos that explain it.

Daniela_Martinez_3B

Even though I've already done the homework problems, I plan on picking a few of them and re-doing them to study. I will also go over my lecture/module notes and do some of the problems from the worksheets we work on during the UA workshops!

Daniela_Martinez_3B

I was stuck on this problem too! You can solve this problem by using the Rydberg equation or you can solve it using En=-hR/n^2. However, using the Rydberg equation is faster. Since the line is observed at 102.6nm, we know that it's in the UV region and it must involve n=1. Since the electron is fall...

Daniela_Martinez_3B

If you're having trouble with the textbook problems, I'd definitely recommend going to the step up sessions/workshops. However, If you're doing well with the material and want extra practice the workshops are good reinforcements too!

Daniela_Martinez_3B

To solve this problem, you first need to convert 1850nm to m. Then, using the equation E=hc/λ solve for the energy per photon of light. To answer the second question, you can divide the total energy 11J by the energy per photon value you got from the previous question and that will give you the numb...

Daniela_Martinez_3B

Calculate the frequency of a photon emitted by a hydrogen atom when an electron makes a transition from the fourth to the second principal quantum level.

I'm having some trouble figuring out how to first approach this question. Any help is appreciated thank you :)

Daniela_Martinez_3B

I'm confused on this too, the sapling answer for my Q10 was off by 1% (78% instead of 77%), but I still got credit for answering 77%. I hope we can get some clarification on how much error sapling allows for as well.

Daniela_Martinez_3B

The reason why there are ions is because those compounds are being held together by ionic bonding. Usually if a compound is made up of a metal and a non-metal, its bonding will be ionic. For example, you can check the periodic table and see that Cu is a metal and Br is a nonmetal, so the compound Cu...

Daniela_Martinez_3B

Agreed! This is very important to understand. Almost every time I made a mistake in these kinds of questions, I found myself forgetting to use the molar ratios. Also, it's good to double check that the equation is balanced correctly or else the rest of the problem gets messed up. Your explanation is...

Daniela_Martinez_3B

Practitioners of the branch of alternative medicine known as homeopathy claim that very dilute solutions of substances can have an effect. Is the claim plausible? To explore this question, suppose that you prepare a solution of a supposedly active sub- stance, X, with a molar concentration of 0.10 m...

Daniela_Martinez_3B

To solve this problem, the first thing you want to do is find the mass of 2-butane by multiplying 0.35mL *0.81g/mL to get the amount in g of 2-butanone. 0.35mL *0.81g/mL= 0.28g Next you need to find the number of moles in 0.28g of 2-butanone by dividing by its molar mass ( 2-butanone's molecular for...

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