CHEMISTRY COMMUNITY

Rishika Yadav 3D

No because reactions with strong acids and bases only go in one direction, while reactions with weak acids and bases reach equilibrium and then occur at a constant rate.

Rishika Yadav 3D

It means they dissociate completely in solution. This correlates to being strong because it means that there are more H+ or OH- ions in the solution, making it more acidic or basic respectively.

Rishika Yadav 3D

Lone pairs on O, F, or N can form hydrogen bonds with a H that is connected to an O, F, or N.

Rishika Yadav 3D

You can also know if a compound is amphoteric by the periodic table (refer to the diagram he showed in lecture of the diagonal stretch that tend to form amphoteric compounds).

Rishika Yadav 3D

HClO2, because the oxygens are more electronegative, pulling electrons away from the hydrogen. This makes the hydrogen easier to break off.

Rishika Yadav 3D

I heard the finals would be graded by Friday, but I'm not sure when they'll be put up on myucla.

Rishika Yadav 3D

The first bond between two atoms is a sigma bond. Each following bond is a pi bond.

Rishika Yadav 3D

I believe it is also the only hexadentate molecule we are supposed to know.

Rishika Yadav 3D

De Broglie can be used when mass or velocity is given - or both, to find the wavelength.

Rishika Yadav 3D

For number 21, why is Fe 2+? I remember him saying in the review that it was because of the nitrogen bonds, but I don't remember why.

Rishika Yadav 3D

A ligand is a neutral molecule or anion that donates an e- pair to form a coordinate covalent bond.

Rishika Yadav 3D

Marty Hockey wrote:For number 20 how do we know whether the expected geometry of the compound is square planar or tetrahedral?

I think that because there are no lone pairs on the Fe, it will be tetrahedral

Rishika Yadav 3D

Yes, they are the same. They are always present because electrons in different atoms can always affect one another and create temporary dipoles.

Rishika Yadav 3D

You can use the de Broglie equation for other things, if they have mass. keep in mind that this means the de Broglie equation cannot be used for photons because they do not have mass.

Rishika Yadav 3D

I believe the TAs are giving them back during discussions.

Rishika Yadav 3D

For our purposes, I don't think we need to worry too much about double or triple bonds. They are just one region of electron density.

Rishika Yadav 3D

Yes, because it can donate an H+ (making OH-) or accept an H+ (making H3O+).

Rishika Yadav 3D

Yes, the VSEPR formula is the AXE formula.

Rishika Yadav 3D

I like to think of it in terms of VSEPR - 2 bonds/electron pairs = linear, so sp hybridization; 3 bonds/EP = trigonal planar, so sp2 hybridization, and so forth. The hybridization level is equal to the number of bonds/EP.

Rishika Yadav 3D

Yes, I believe we have to memorize these and know from the names.

Rishika Yadav 3D

Dr. Lavelle hasn't said anything about a practice final, so I would assume not. However, there are many review sessions and some problems available here on chem community.

Rishika Yadav 3D

Dr. Lavelle also said that he would be finishing the syllabus material on Wednesday, and then Friday will be review I believe.

Rishika Yadav 3D

I think we need to know how to do it for the final because it was covered in class and is on the outlines/homework.

Rishika Yadav 3D

Yes, they can each form a hydrogen bond. This is because they are both negatively charged, and will push away from each other but towards a partially positive hydrogen.

Rishika Yadav 3D

Inorganic compounds are able to lose H+ ions much more easily than organic compounds, making them stronger.

Rishika Yadav 3D

My TA told me that if there are lone pairs, always write less than the respective number of degrees. Thus I think you are right, see saw should be less than 90 an 120 degrees.

Rishika Yadav 3D

No, only on the central atom. Lone pairs on the outer atoms d not affect bond positioning with the central atom.

Rishika Yadav 3D

the 4s orbital is lower in energy, so the electrons would rather be there because it is more stable.

Rishika Yadav 3D

c = wavelength x frequency ==> c/wavelength = frequency
E = h x frequency ==> E = h x c/wavelength ==> E x wavelength = hc ==> wavelength = hc/E

Rishika Yadav 3D

I think a general understanding will suffice, as long as you understand how they work.

Rishika Yadav 3D

Because one is positive and the other is negative, they are attracted to each other.

Rishika Yadav 3D

I found that actually sitting down and reading the textbook and going through the worked out problems really helped me. I also plan on doing as many of the homework problems as possible - practice is the best way to understand it.

Rishika Yadav 3D

By forming a bond atoms become more stable. This is why it takes energy to break a bond - they want to remain in a more stable state and will not willingly go back to being less stable.

Rishika Yadav 3D

You multiply 5.00 m/s by 2 because the error is 5.00+/-5.00 m/s, so it is in either direction. This the error is the absolute value (5) times 2.

Rishika Yadav 3D

Accommodating more electrons often works for the d-orbital. An element with a principle quantum number of 2 cannot be expanded because there is no 2d-orbital.

Rishika Yadav 3D

l correlates to the orbital. As you increase l, you get to another orbital. So, l=0 is s, l=1 is p and so on. I found the quantum numbers to be quite confusing but I think if you just practice them more they become easier to understand.

Rishika Yadav 3D

Does anyone know if the midterm will be up to the quantum world or will it be what we cover next too?

Rishika Yadav 3D

You are right, and just as a tip to help you remember, I like to think of it as the energy required to create an ion, thus ionization energy

Rishika Yadav 3D

We can do the quantum homework for this week.

Rishika Yadav 3D

Knowing Lavelle, there may be, but if not there are hours and hours of sessions on his website per week that you can go to for help.

Rishika Yadav 3D

I think so, because that is all we'll have learned by then, but I'm sure he/our TAs will tell us in class.

Rishika Yadav 3D

Can someone please explain why two electrons have a different spin if they are paired versus if they are parallel?

Rishika Yadav 3D

Electrons, protons, and neutrons all have a set mass that does not change from element to element. What changes the mass of elements is the number of protons and neutrons it has (electrons do not have a noticeable effect because their weight is so small that it is essentially negligible).

Rishika Yadav 3D

I'm not 100% sure, but I figured it was because carbon is essential to life - it is a main focus point in all forms of science and understanding it is necessary in the study of life and how the world works.

Rishika Yadav 3D

Yes, I believe it does because of the structure and strength of the material.

Rishika Yadav 3D

Can someone please explain why it was unexpected that electrons were not emitted even for high intensity light in the experiment?

Rishika Yadav 3D

If an electron is closer to the nucleus, it is more stable and thus releases energy (in the form of light) when going to a lower energy level. Conversely, when they are in higher energy levels, they are more unstable and have more potential energy within them.

Rishika Yadav 3D

I'm not sure when we'll get our tests back, but I'm pretty positive we will be getting them back. Also, it is true that we will not have a final grade until the end, and that there will be no letter assigned to individual exams - we will only be given a letter grade at the end of the quarter.

Rishika Yadav 3D

Hey! So, this isn't necessarily a limiting reactant question because there is only one reactant. It focuses more on percent yield. For this question, the chemical equation is CaCO3 (s) --> CaO (s) + CO2 (g) . We know that we used 42.73 g CaCO3, but we need to know how many moles that is. To figure t...

Rishika Yadav 3D

My T.A. said that we should not focus too much on them, but if we want should go over them in our free time because it isn't covered in depth in lecture. Also, he said that sig figs are not usually looked for too much on tests, as long as you round reasonably. I would say that it is good to review t...

Rishika Yadav 3D

When doing the problems in section, the T.A. was showing each step in a lot of detail. On the test, do we have to show every step or is it okay to skip some that you know already or from practice?

Rishika Yadav 3D

I think the small amount 'lost' is due to rounding - if we went out further in the decimal places then we would likely have gotten a value slightly closer to 8.000. Additionally, in real life we will nearly always be slightly off of the theoretical value, due to things like solution sticking to the ...

Rishika Yadav 3D

My T.A. said that we don't need to worry about sig figs too much, and that we should focus more on always having units.

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