CHEMISTRY COMMUNITY

Mona Lee 4L

If you search "Hot Dog" you can find the practice midterm that Lyndon, one of the UAs, made.

Mona Lee 4L

DeltaU = q + W. Therefore, the internal energy of a system does not change only when q and W are zero.

Mona Lee 4L

If reactants and products are at different states or reactions take place at different pressures, the enthalpy of the reaction can be different. Therefore, the standard reaction enthalpy refers to the enthalpy when all reactants and products are in their standard states at 1atm. Reaction enthalpy re...

Mona Lee 4L

When you move the piston up, the system is doing work on the piston so the system loses energy. Because the system loses energy, the W is negative. When you move the piston down, the opposite occurs.

Mona Lee 4L

Work is the integral of the product of force and displacement. In class, he used that definition of work to derive the formula for work done by a piston. However, you can just use the formula using delta V because that is more applicable to the scope of this class.

Mona Lee 4L

No, there is generally no limit. Essentially, the "surroundings" refer to the environment of the reaction.

Mona Lee 4L

Since we can only use scientific calculators for this class, this question is beyond the scope of our tests.

Mona Lee 4L

This question was previously answered here: https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=51&t=40454. Essentially, because water is a liquid it is not included in the equilibrium expression. However, the addition of water will decrease the concentration of C6H12O6, thereby causing the reac...

Mona Lee 4L

For this question, the pKa represents the relative strength of the acids. The lower the pKa, the higher the conjugate Ka, and therefore the weaker the base. Arylamines and Alkyamines refer to the class of organic bases that anilines, methylamine, and ethylamine fall under. I don't think we need to k...

Mona Lee 4L

In general, the net ionic equation will be water + ion <=> conjugate acid / base + oh-/h3o+

Mona Lee 4L

Not always. For example, you use Ka2 for H2SO4 because Ka1 is very strong. However, in most cases, you can disregard the second because it is much smaller and basically insignificant.

Mona Lee 4L

For reactions involving a strong acid and base, you can assume 100% protonation / deprotonation, and the single arrow represents the heavy skew towards products.

Mona Lee 4L

Both flask 3 and 4 are at equilibrium. The diagram displays the dissociation over time, so flask 3 is merely the first instance when the flask reaches equilibrium. The fact that flask 3 and 4 have the same ratios is a clear sign that the reaction has reached equilibrium in flask 3.

Mona Lee 4L

Hello! today we went over the basics of equilibrium including the definitions of equilibrium and equilibrium constants. You could skim through the textbook as outlined in the syllabus for a review. Also, Dr. Lavelle has video modules on his website that cover the same material as lecture.

Mona Lee 4L

No. In a Kc expression, you do not need to include the solute. The change in the concentration of any solute will be insignificant.

Mona Lee 4L

To calculate the pH of a weak acid solution, you need to use Ka and set up an equilibrium equation, which is beyond the scope of this quarter. For questions involving strong acids and baes, you can solve them stoichiometry as a neutralization reaction and then calculate the pH using the remaining H+...

Mona Lee 4L

Essentially, electrons that are more distorted are more likely to be attracted / repelled by neighboring dipoles, making them more polarizable.

Mona Lee 4L

To calculate the pH of a solution with a weak acid, you need to use the Ka equilibrium constant and set up an equilibrium equation, which is not covered in this course.

Mona Lee 4L

Pi bonds are formed with the unhybridized orbitals. Usually, they are the p-orbitals, but they can also be formed with d-orbitals.

Mona Lee 4L

Essentially, polarity is directly connected to electronegativity. You can also think of it in terms of the size of the atom. As you move down a group, the valence electrons are further away from the nucleus and experience more shielding. As a result, the atom will be less able to pull electrons in t...

Mona Lee 4L

Essentially, to determine if a molecule is polar, you must first look to see that the bonds are polar and that the molecule is asymmetrical. To tell if the bonds are polar, you look at the difference in electronegativity. On notable nonpolar bond is C-H. Then, look at the shape of the molecule and t...

Mona Lee 4L

Both structures are correct lewis structures, and the one you are suggesting would have a lower energy and would be the dominant lewis structure. Because this question was asking about shape, you don't necessarily need to draw the lewis structure with the lowest energy. Both lewis structures would r...

Mona Lee 4L

Both structures are correct lewis structures, but the one you are suggesting would have a lower energy and would be the dominant lewis structure. Because this question was asking about shape, you don't necessarily need the lewis structure with the lowest energy. Both lewis structures would result in...

Mona Lee 4L

When determining hybridization, it is important to know how many areas of electron density are around an atom. Like we discussed in class, CH4 and NH3 have the same hybridization, even though one is tetrahedral and one is trigonal pyramidal. As a result, you need to know what type of bonds are aroun...

Mona Lee 4L

Yes. Although both the tetrahedral and see-saw shape have four atoms bonded to the central molecule, the lone pair in the see-saw shape causes the difference in name. Every VSEPR structure (AX4 vs. AX4E1) has a different shape name.

Mona Lee 4L

a. Xenon has more electrons, so it is more polarizable and has stronger LDF than Ar. Because of the stronger LDF, the melting point will be much higher because it requires more energy to break the stronger bonds. b. H2O has hydrogen bonding while C2H5OC2H5 is nonpolar and only has LDF. Because LDF a...

Mona Lee 4L

If there is a lone pair, there will likely be a net dipole pointing in the opposite direction. In general, if a molecule has lone pairs, it is likely to be polar.

Mona Lee 4L

For every molecule, there are a lot of negatively-charged electrons held around a positively-charged nucleus. If a small shift in this electron cloud occurs, it will create a small dipole whereby one part is more positive and one more negative. This small dipole will then induce a change in nearby m...

Mona Lee 4L

Hydrogen bonds are incredibly polar because of the significant difference in electronegativity between hydrogen and O, N, F. They're essentially just very strong dipole-dipole bonds.

Mona Lee 4L

You can think of "rod" vs. "sphere" as straight-chain vs. branched molecules. Essentially, straight chain molecules are going to have fewer atoms "sticking out." so they can pack closer together, increasing the LDF.

Mona Lee 4L

Essentially, if the formal charge does not add up to the overall charge of the molecule, that lewis structure is not valid. Then, you can use process of elimination to figure out where to expand the octet.

Mona Lee 4L

Yes, because the most electronegative atom will draw the most electrons, so should have the most negative formal charge.

Mona Lee 4L

Essentially, the elements in rows 1&2 will never have more than 8 valence electrons. In rows 3+, n=3 so there are p shells, which allow atoms to expand their octet.

Mona Lee 4L

When it comes to these types of questions, always write out the electron configuration then look where the electrons are and if they're any different than the ones that follow the expected behavior.

Mona Lee 4L

In general, we don't really need to worry about heavier transition-metal elements because they don't usually follow normal patterns. But in general, the prime reason for their behavior will be their abnormally large size because they have so many more protons and electrons than elements in the first...

Mona Lee 4L

In general, for these questions: 1. list out the electron configuration 2. start removing e- from the end of your list, because these are the electrons furthest from the nucleus and the least stable

Mona Lee 4L

Remember that Ephoton = WorkFunction + KEejectede-. If Ephoton is less than the work function, the kinetic energy of the electron would be negative, and the electron would not be ejected.

Mona Lee 4L

In general, the work function will always be the difference between the energy of the photon & the kinetic energy of the electron. Thus, if they give you ways to find these quantities, you can be responsible for finding the work function.

Mona Lee 4L

My TA specified that the following would be on the test: photoelectric effect, deBroglie and wavelength, Heisenberg uncertainty, atomic spectra, and quantum numbers. We won't need to know electron configurations or multi-electron systems.

Mona Lee 4L

Different metals have different threshold energies depending on the nucleic pull on electrons. There are many factors that affect the nucleic pull. ie: electrons far away from the protons will be "shielded" from the nucleic force and as a result, easier to remove.

Mona Lee 4L

Here is my solution for that question: Mhz = 10^8 Hz

Mona Lee 4L

Light as a wave has two key characteristics: amplitude and frequency. According to this model, the energy of a stream of light should be directly proportional to the amplitude. Essentially, "high-intensity" light should have high energy. However, with the discovery of the photoelectric eff...

Mona Lee 4L

Another way to think about reactants is to think about which one will produce the least product. Ie: calculate the product stoichiometrically assuming each reactant is limiting. The correct answer will always be the reactant that produces the least amount of product.

Mona Lee 4L

Thank you so much for your advice! It's really reassuring to hear from a successful past student that what might seem to be a challenging class will be rewarding in the long run!

Mona Lee 4L

In general, I try to start with elements that appear in the least number of compounds. Then, I make sure there are no odd numbers of elements if there is an even number on the other side (ie: double H2O if there is an O2 on the other side). Lastly, try to balance single-element compounds (ie: O2) la...

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