CHEMISTRY COMMUNITY

Anusha 1H

It just depends on the information you're given
If they give you moles, for example, you would in most cases just use n and not convert to grams because the estimation would throw off your answer

Anusha 1H

R is the gas constant

Anusha 1H

Yes, because temp doesnt change

Anusha 1H

I think like the section on thermodynamics, knowing how to derive the equations makes it easier if you get stuck on a problem, but I dont think we have to memorize them as long as we know what to use in what situation

Anusha 1H

you use the half life equation for second order reactions first. With it, you can use the half life and [A]0 value they give you to solve for k. once you have k, you can solve for the t values by using the integrated rate law equation (you'll have to use (1/n)([A]o) where 1/n represents the concentr...

Anusha 1H

you basically just have to look at the ratios of the gases they're asking for.
For a) the ratio of N2 to H2 is 1:3, so the rate N2 is consumed at would be 1/3 the rate of consumption of H2

Anusha 1H

in basic solution, we first have to balance O by using H2O just like acidic solution. in order to balance the H, though, we have to add H2O to the side of the reaction that needs it and then balance the other side with OH- in the 7th edition, toolbox 6K.1 helps explain balancing redox in both acidic...

Anusha 1H

n refers to the amount of electrons being transferred
so for example if we're dealing with the equation
Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s)
the n value would be 2 mol because 2e- is being transferred in the reaction

Anusha 1H

they basically have the same purpose, so i dont think its important to know the distinction for this class

Anusha 1H

cell potential is basically the force (voltage) that exists between the two parts of the battery

Anusha 1H

Usually questions specify whether a salt bridge or porous disk is being used, so i dont think we have to worry about it.
They basically do the same thing

Anusha 1H

I think this weeks should be Gibbs Free energy or electrochem (so the new outline)

Anusha 1H

usually the more complex molecule is the one with a greater entropy because there's more bonds and therefore more vibrational energy states. for a, HBr will have the greater entropy because it is the larger molecule. for b, NH3 would have the greater entropy because it is more complex for c, the liq...

Anusha 1H

It's stops at Gibb's Free Energy! so until the end of all the entropy material

Anusha 1H

i think for this week we're supposed to do the second and third law homework or Outline 4.
Im assuming next week will be Outline 5

Anusha 1H

I think its going to depend on whatever he covers in the next two lectures because the website says "Chemical Equilibrium, Acid and Base Equilibria, Thermochemistry, Thermodynamics to the end of entropy" is all on the midterm.

Anusha 1H

This equation is calculating the change in the entropy of a system
Temperature is in Kelvin and Heat is in joules because the units for entropy is (J K^-1)

Anusha 1H

if work is positive, it means that work is done on the system by the surroundings an example would be compression if work is negative, it means that work is done by the system on the surroundings an example would be expansion in this case, work is positive because work is done on the air (compressio...

Anusha 1H

H is enthalpy which is the amount of heat released or absorbed, and it is NOT dependent on the path taken to obtain whatever current state it's in

q is the energy transferred to a system as heat and it depends on the path the reaction takes

Anusha 1H

Basically anything that is already in its standard form and doesnt have to change to form.
Elements are an example of this because theyre in their natural state

Anusha 1H

yes, i think they are always exothermic because when the acid and base react, bonds are formed to create the salt and water. This releases energy to the surroundings which would constitute an exothermic reaction

Anusha 1H

(Sorry the tables I'm referring to are labeled this way in the 7th edition!) Yes, for part a), you use the Kb value for NH3 in table 6C.2 to find the Ka value of its conjugate acid NH4+ (KaKb = Kw) For part b), you can use table 6D.1 to find the Ka value of common cations in water, which includes Al...

Anusha 1H

The pH value is based on the amount of hydronium ions in the solution. Because we assume that strong acids are fully deprotonated in solution, this means that there is a much higher concentration of H+ ions in the solution than in weak acids, which leads to a lower pH value. Basically, the higher th...

Anusha 1H

For this question, we actually have to refer to table 6C.1 which the acidity constants. In the case of a), you know that you have to work with CH3CO2-, so we can use the Ka value of acetic acid (CH3COOH) to find the Kb value of its conjugate base. Since Ka for acetic acid is 1.8 x 10^-5, the Kb valu...

Anusha 1H

Because the molar concentrations of solids and liquids don't change during a reaction, they are not included in calculating the K value

Anusha 1H

I believe all chemical reactions need to have some sort of energy change, so they are all classified under the categories of either endothermic or exothermic.

Anusha 1H

When Q < K, that means that the ratio of products to reactants is less than it is at equilibrium, so the forward reaction is favored because by producing more products, the ratio of products:reactants will increase.

Anusha 1H

Molarity and Partial Pressure are directly related.
We can tell from PV=nRT when the equation is written
- (n/V) = P/(RT) or
- concentration = P/(RT)
that when the pressure increases the concentration will also increase.

Anusha 1H

it helps if you write out the K values for all of the equations 2BrCl >>> Br2 + Cl2 H2 +Cl2 >>> 2HCl ----------------------- 2BrCl + H2 >>> Br2 + 2HCl K1 = [Br2][Cl2]/[BrCl]^2 K2 = [HCl]^2/[H2][Cl2] K3 = [Br2][HCl]^2/[H2][BrCl]^2 = K1 x K2 Hope this helps!

Anusha 1H

Use the same approach you use to find K (products/reactants).

a) for example would be -- Q = 1/(P BCl3)^2

Anusha 1H

Table 9C.1 in the seventh edition is a good reference point. I think the most important ones to know, though, are aqua, ammine, nitrosyl, and carbonyl

Anusha 1H

Yes, ligands have to have at least one lone pair

Anusha 1H

Yea I think for the sake of this class we can say that.
All the molecules are deprotonated

Anusha 1H

I think a good example for this idea is ethane, C2H4, because both of the carbon atoms in the molecule are sp2 hybridized and they will each have one unpaired electron in the p orbital. So this parallel overlap of the p-orbitals will be unhybridized.

Anusha 1H

Hydrogen bonding is unique because it relies on a partially positively charged hydrogen atom that is attached to an electronegative atom bonding to another electronegative atom. The more electronegative the atom is, the stronger the Hydrogen bond will be. It's because of these qualities that water h...

Anusha 1H

Drawing out all the molecules helps.
Single bonds are the longest and weakest, while triple bonds are the shortest and strongest.

So, NO+, for example, has one triple bond, so you know it will have a shorter N-O bond length than NO2- which has one double bond and one single bond.

Anusha 1H

p-p orbital overlaps can be sigma or pi bonds.
it depends on the type of overlap.
i thought these pictures were helpful

Anusha 1H

I think all the book is referring to here is the shape the orbitals form because of their regions of electron density

Anusha 1H

We only use up to sp3 because there are only three p orbitals. Anything beyond that, will go into the d orbital
ex. sp3d - five dsp3 hybridized orbitals which would be trigonal bipyramidal

Anusha 1H

You're right about the partial negative charge, which is determined by the electronegativity of the atoms in the molecule. I think the reason Lavelle made that distinction in class is because a partially negative atom with at least one lone pair has a high density of negative charge, so it is ideal ...

Anusha 1H

Because Be has the lowest electronegativity, it would be the central atom. It forms two bonds, so it will bond to each of the carbons, and each of the carbons will have three hydrogen bonds.
Because of the two bonded pairs, this structure would be linear in respect to the Be.

Anusha 1H

We can determine bond angles using the VSEPR theory. Because of electron repulsion, molecules form shapes that will give them the lowest repulsion energy. The basic idea, though, is to look at how many bonds a central atom forms and how many lone pairs surround it. (In this case triple bonds and dou...

Anusha 1H

I dont think we need to know about them for this midterm.

Anusha 1H

I don't think we have to know how to write out the compounds. They will probably give us the formulas on the test. Because the compounds are ionic or polyatomic, the Lewis structures are separated into cations and anions. This is because the electrons are not being shared like with covalent compound...

Anusha 1H

The actual compounds are not covalently bonded to each other.They're ionic.
That's why the lewis structures are drawn so that the electrons are transferred to the anion to complete its octet and the cation is shown with its charge.

Anusha 1H

In some cases, the formal charge will not be equal to zero.
In looking for the most stable lewis structure, though, we should be looking for the structure that has the most individual formal charges close to 0.

Anusha 1H

I think we should be familiar with f orbitals. You should know how to do electron configurations and be aware of the special cases that the TAs have been covering in discussion this week.

Anusha 1H

We didn't receive any handouts.
Gail covered Heisenberg, de Broglie, the photoelectric effect, and electron configurations.
She also went over test questions that were commonly missed like the one about the Alveoli sac where we had to use the uncertainty equation.

Anusha 1H

I don't think black body radiation will be on the midterm because we didn't need it for test 2 and it hasn't been covered in the reviews

Anusha 1H

Both 2d and 4g don't satisfy the requirements for the angular momentum quantum numbers (or l). l can only have the values 0,1,....n-1 In the case of 2d, n=2. This means l can either be 0 or 1. These 2 numbers are l values for s-orbitals and p-orbitals. Because l cannot be equal to 2, it cannot have ...

Anusha 1H

The test covers just quantum material up to Friday's lecture. I think thats about half of section 1D in the 7th edition
This is the outline for what we should know on the website.

https://lavelle.chem.ucla.edu/wp-conten ... _World.pdf

Anusha 1H

My TA said most of the equations we need will be given to us on the front page of the test. For convenience and just to be safe, I would memorize: (frequency)(wavelength) = c (speed of light) Rydberg (or at least remember the Balmer series corresponds to the visible region and Lyman to the UV region...

Anusha 1H

You would use the Heisenberg principle in this problem. 350 pm is your uncertainty in position (delta x) Because the problem is asking for uncertainty in velocity, we have to use (m)(delta v) as opposed to its equivalent delta p The m is the mass of an electron or 9.11 x 10^31 kg. h is always the co...

Anusha 1H

I don't think we're required to know the shapes for this upcoming test specifically, but for the midterm and final understanding the idea behind it and concepts like the probability of finding an electron will probably be important

Anusha 1H

B and C are pretty similar to part A! b) In order to determine the energy emitted, you need to know the number of atoms in the sample. Since we have a 5.00 mg mass we should first convert to grams. From there, use the molar mass to find the number of moles and then multiply that number by Avogadro's...

Anusha 1H

I think we're supposed to use kg. If given g, for the photoelectric effect equation in particular, convert to kg first.
This is because when determining values for things like kinetic energy, the units are kgm2s-2, which is equivalent to Joules.

Anusha 1H

I also think the answer would be yes because otherwise you can't guarantee that the photo electrons maintain their energy.
With the vacuum, you would be able to ensure the frequencies have contact with the sample and aren't influenced by collisions with the air.

Anusha 1H

Yea I thought that was strange too because 1A doesn't address Energy of the photon much apart from Table 1A.1

Anusha 1H

You can use Avogadro's number for atoms, molecules, or formula units.
Formula Units apply to ionic compounds, molecules apply to molecular compounds, and you calculate atoms when in reference to elements.

Anusha 1H

The absorption spectrum basically shows that gases absorb certain wavelengths through the black bars displayed on the spectrum.

Anusha 1H

My TA mentioned the test will be structured a lot like the homework problems, so they are all free response and it's possible to get partial credit.

Anusha 1H

I think we're allowed to submit our questions from either one section or all the sections as long as we have a minimum of 7. The main thing the TAs are grading for is accuracy so any combination of problems should be fine.

Anusha 1H

Do you have an example of a question you were confused on?

Anusha 1H

The modules are basically the material Lavelle has been covering in class. They are pretty helpful if you want to easily follow along with his lecture or if you aren't as familiar with the high school chem material. I also thought they were a good review before attempting the homework.

Anusha 1H

I think the problem is addressing threshold energy. Because the question asks for the energy needed to remove 1 electron, I think you need to convert the kJ mol-1 to J/electron. To do that you can divide the value you calculated (150.6) by Avogadro's number (6.022 x 10^23) because in this case it re...

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