CHEMISTRY COMMUNITY

Nina Do 4L

I would remember that zero order would be M/Sec and then divide M each time you go down an order. Here's what I mean: Zero order would be M(molarity)/second so divide by M to get units for first order. Notice how M cancels out. 1st order units: s^-1 Divide M again to get 2nd order units s^-1*M^-1 An...

Nina Do 4L

A zero order reaction does not change if the concentration of the reactants change. It is only based on the k. Therefore, the equation would be rate=k. However, first order reactions do depend on the concentration of reactants and it's equation will be rate =k[A]. When the concentration of reactant ...

Nina Do 4L

Zero order means that the rate constant is independent from the concentration so therefore the concentration is to the zeroth power ie the rate law being rate= k[A]^0 which will just be rate=k.
Same applies for first and second orders.
Hope that helps!

Nina Do 4L

As everyone stated above, it will be M/L per second and as you increase to each order, you're constantly dividing by 1/M so for zero order, the units will be Moles per second. s^-1
1st order will be per second or s^-1M^-1
and so forth

Nina Do 4L

With respect to thermo, it tells us the spontaneity of a reaction and if it will occur. Kinetics allows us to determine the rate at which the reaction occurs, assuming that all reactions happen.

Nina Do 4L

If you are in AAP or are looking to get into AAP, I recommend their PLF sessions and a lot of YouTube videos revolving the topics covered on the syllabus helps a lot. I also would recommend doing the worksheets that flow around Chem Comm. from Karen or other UAs. Best of luck!

Nina Do 4L

Yes, the equations or Gibbs will also be provided to us on the constants and formulas worksheet. The test'll be on Gibbs and Electrochem but not Nernst equation. Here's his official announcement: (2/14/2019) Test 2 covers all topics on Gibbs free energy and electrochemistry up to but not including t...

Nina Do 4L

After you create your two half reactions, you need to balance the amount of electrons in order to combine the reaction into one. In the process of balancing out the amount of electrons, you find your number of moles of electrons. It is the number of e- you used to balance the reaction.

Nina Do 4L

Hi, You would add Pt(s) or C(graphite) if there is no solid acting as an anode or cathode. You still would need to add a solid (either platinum or carbon graphite) in the presence of an aqueous solution because that represents the aqueous solution that the anode/cathodes would be in. So in the absen...

Nina Do 4L

Thank you for this creative mnemonic!

Nina Do 4L

Here is a helpful video for me that I hope will help you: https://www.youtube.com/watch?v=IZ1tKxsqV74

Nina Do 4L

You would use H+ to balance the amount of hydrogens and water to balance the amount of water.

Nina Do 4L

Yes! They are used synonymously.

Nina Do 4L

Reversible reactions always do more work since they are infinitesimal reactions happening so it can do a maximum amount of work.

Nina Do 4L

Right, when pressure increases, the inverse happens to volume. Entropy is the amount of possible positions that thing can have so when there is less space, the options for the different amount of positions it can hold are less so therefore, entropy is decreasing.

Nina Do 4L

A cup of tea can interact with its surroundings, ie release heat from the tea into the universe, making the universe infitesmilly warmer. Closed systems like a hydro flask or thermos cannot do that, it is isolated/insulated.

Nina Do 4L

This Wednesday Professor Lavelle is going to announce various workshops and review sessions located around campus at various times to fit your schedule. I recommend going to those as well as doing some practice problems and finding videos online that relate to the course.

Nina Do 4L

Residual entropy is like a screenshot of the molecules in stationary form but we need to also calculate entropy for its other possible orientations since some molecules can have resonance such as COF2 and BF3. They are both trigonal planar but BF3 is non polar and doesn't have resonance; the three f...

Nina Do 4L

I = initial
C = change in concentration
E = equilibrium, so finding E will give you the concentration of the compound at equilibrium.

Nina Do 4L

I've been told that we can use the quadratic formula when the value is greater than 10^-3. You can also use the 5% rule to double check if the value is eligible to make x equal to zero and to completely ignore it since it's infinitesimal.

Nina Do 4L

pKa is the -log of the Ka value and Ka*kB=1x10^-4 so you can find out the value of Ka/pka if you're given the value of Kb.

pH is the -log of H+ or H3O+.

The 'p' is -log-ing different values, therefore they are different. Hope this helps.

Nina Do 4L

Homework problems from the problem sets from the past two weeks often help a lot and are great practice. Also, if you can go to UAs for some worksheets for more practice, attend office hours and discussions are also a big help. I also do a lot of YouTubing videos on the topics that'll be on quiz 1 (...

Nina Do 4L

For me, I divided the final volume (.5 L) by the initial (.005 L) and multiplied it by the original concentration of hydroxide (.18) and I got 18 M which is correct according to the solutions manual.

Nina Do 4L

Answer: Refer to the table referenced earlier in the chapter. It tells you that K equals 377 so you would set 377 equal to your K equation (products over reactions, not including solids or liquids), and then determine K from the table again from the other elements associated with the reaction. see t...

Nina Do 4L

No, changing the concentration will not affect K over time. More specifically, at first it will try to balance out the concentrations and eventually and gradually, it will counteract the change and come back to equilibrium.

Nina Do 4L

Pure solids or liquids aren't included in the equilibrium expression because their concentrations stay constant throughout the reaction so we don't include them.

Nina Do 4L

If the reactants are all gaseous, you can use their partial pressures to find Kp. If they are a mixture of gases and aqueous solutions, then your best bet is to use Kc. Very similar equations, just Kp uses the partial pressure of the gases while Kc uses the concentrations. Hope that helps!

Nina Do 4L

I don't believe that you have to since it will be provided for you on the equations worksheet. It is posted on his website and here it is for reference: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/constants_equations.pdf Also, I think you will remember it anyways because of its...

Nina Do 4L

From what was gathered in lecture, the quotient reaction is the values anytime during the reaction, not necessarily at equilibrium. So not only does it determine which way the reaction goes, but if the value of Q is not equal to K then you can tell that it is not at equilibrium. If Q < K, the forwar...

Nina Do 4L

Hi there, I would definitely agree with all of the above suggestions. Personally, I use his website as well as Chemistry Community a lot and the homework questions are the main source of preparation for you so I recommend doing all of them or as many as you can. I also go to the step up sessions and...

Nina Do 4L

Work function, threshold energy, and ionization energy all mean the same thing and can be used interchangeably. The equation you'd use is ephoton - work function (or threshold or ionization energy) = Ekinetic energy, 1/2mv^2. Based on the information of the problem, you can work backwards to find on...

Nina Do 4L

I wrote something along the lines of they are inversely proportional to one another and that it is impossible to know both momentum and position at the same time and got credit for it. Hope that helps!

Nina Do 4L

In the past I have used 3.0 x 10 ^ 8m/s and never saw any problems. However, a coversheet with some equations will be given to us so I'd use that if you're unsure.

Nina Do 4L

On the periodic table, I is in the 7th column meaning that it has 7 valence electrons. We want its formal charge to be zero so we would use the equation FC (Formal charge) = valence - (Lone pair + shared pair/2). If iodine was a double bond, we'd get a formal charge of 1 (7-1) but we want a formal c...

Nina Do 4L

I agree that figure 1 has the largest dipole moment because the most electronegative element is Cl and it is generally going to pull towards the top right direction. The second figure is not as strong of a directional point in electronegativity as the first. And the last figure, they are pulling in ...

Nina Do 4L

Be does not need to fulfill its octet because it is the first four elements on the periodic table and the same exceptions go for Be as it does for H, He, and Li.

Nina Do 4L

I think that it's more reasonable to reason the shape rather than memorizing. From seeing its arrangement like tetrahedral, trigonal pyramidal, or trigonal planar, we can see if the molecule has lone pairs and whittle down from there. We would have to know both its arrangement and shape, like if it'...

Nina Do 4L

Adding on, your hybridization formula should equal the amount of electron density areas. For example, if your hybridization formula is sp3, then you should have four regions of electron density. 1s+ 3 p3 orbitals= 4 hybrid orbitals = 4 regions of electron density.

Nina Do 4L

When writing the VSEPR formula, you begin with A which is the central atom. And then E and the subscript number will depend on the amount of "things" it's attached to, regardless of bonds. Then you'd finish with E and the subscribe number will depend on the amount of LONE PAIRS. We count e...

Nina Do 4L

Professor Lavelle said that we won't need to draw the structures in a 3D form with shading and wedges but we will be asked to name the shape. And regarding resonance, he said that we can use either resonance structure since they do average out to the be the same bond length. Hope that helped! :)

Nina Do 4L

When you determine that the atom is polar or non polar (from the previous comments), the direction of polarity points to the atom that is most electromagnetic! :)

Nina Do 4L

Bond angles are the angles that are created by electrons. Since they are the same negative charges, they repel and that creates a bond. For example, a central atom bonded by two other atoms creates a linear bond. Electrons are so repulsed by one another, they want to remain as far apart from each ot...

Nina Do 4L

At 0, 0=s At 1, 1=p At 2, 2=d At 3, 3=f S has to have one orbital and one orbital can contain 2 electrons. P has three orbitals (the orbitals increase in increments of 2), containing 6 electrons. D has five orbitals, containing 10 electrons. F has seven orbitals which can hold 14 electrons. The s or...

Nina Do 4L

I agree with what Tony said! A combustion reaction will always be (some compound) + O2 --> CO2 + H2O And according to that compound, that will help you balance the equation. Sometimes, there will be more things produced but the question will have to tell you the extra molecules that were produced.

Nina Do 4L

Start off with what you have and convert it to grams of NO using the chemical equation. 21.1 g NH3x(1 mol NH3/ 17.031 g NH3)x(4 mol NO/4 mol NH3) [because we are trying to get grams of NO, use the chemical equation ratio] x(30.006 g NO/1mol NO) That will give you 37.2 grams of NO produced using 21.1...

Nina Do 4L

From my understanding, we will use the smallest amount of significant figures that is used in the problem for example, if the question gives us a velocity of 3.00x10^3 and energy of 1.2x10^-10, the smallest amount of sig figs would be 2 because 1.2 has two significant figures. Therefore, our answer ...

Nina Do 4L

First, convert all that was given in the equation to moles. .561 g CO2*(1mol Co2/44.01 g CO2) Then, use the mole ratio of how many carbons there are in CO2. So your entire table should look something like: .561 g CO2*(1mol Co2/44.01 g CO2)*(1 mol C/1 mol CO2) CO2 is on the Botton to cancel out, leav...

Nina Do 4L

Covalent bonds tend to share electrons to complete the octet rule which makes them anions.

Nina Do 4L

The first 4 elements are the ones that are an exception (along with others). The elements H, He, Li, and Be are exceptions to the octet rule because they are in the 2s orbital, meaning they can only have most 2 electrons. Here's more info on them in case you are confused: https://lavelle.chem.ucla.e...

Nina Do 4L

Since in D orbitals l=2, that means that l can be 1 and 2 so it breaks the octet rule

Nina Do 4L

This is what I needed to read! Thanks so much for sharing. During the gap year, aside from studying for the MCAT what do you recommend we do in that free time?

Nina Do 4L

Hi, First I converted the given keV to eV. I did that by multiplying 140.511 by 1000 and got 140511 eV. Then I used that and multiplied it by 1.602*10^-19 J because that is how many joules per eV. That gave me 2.251x10^-14 J for my energy. Then I set that equal to the (E=hc/lambda) equation to find ...

Nina Do 4L

First, I converted my given wavelength of 420 nm to m (420 x 10^-9) and multiplied the given 32 W to joules (32 J/s). Since the problem gave us a time, I multiplied the time with the Joules/sec to give me 64 total Joules. That is what I would call the TOTAL Energy, not the energy for one photon. So ...

Nina Do 4L

In peer learning sessions, we use 6.626*10^-34.

Nina Do 4L

An electron will only be launched/ejected if the threshold energy is greater than or equal to the amount of work needed to do so.

Nina Do 4L

The photoelectric effect should be the answer because during the experiment, light (which is also electromagnetic radiation) was shined on a metal in attempt to launch an electron. In trying to do so, they tried amping up the intensity but that still did not eject the electron. Upon more attempts, t...

Nina Do 4L

A watt is also J/seconds so you can multiply 32 by 2 to get the total energy, 64 J. Then, use E=(hc)/lambda which will give you the energy of only 1 photon. I believe you then multiply the energy of 1 photon. You can then proceed to use 6.022*10^23 (Avogadro's) to find the number of moles.

Nina Do 4L

I began the question by turning the given 4.14 g P into moles. (4.14 g P)/(30.97g P) = .13367 moles P. (I got 30.97 g from its molar mass on the Periodic Table). Then I went back to the question and subtracted the total mass of what the equation produced, 27.8 g, and subtracted it with the original ...

Nina Do 4L

You would look at the element first and then see its atomic mass on the periodic table. Ex: O=~16g/mol. And since you're asking about O2, we'd multiply that by 2 because we have two of them so it would be 32g/mol.

Nina Do 4L

Hi! Even though they aren't required and it is no way identifiable, I found them extremely useful and great as a tool to refresh what happened in the lecture. There are also a lot of good example problems that might help for future quizzes and a great way to review for the final. I plan on doing the...

Nina Do 4L

When you are adding or dividing or multiplying or subtracting with significant figures, your answer will have the least amount of sig figs because your answer can not be more precise than the least precise measurement. In this case, yes, I also agree that the answer would be 4.56 because 3 is less t...

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