CHEMISTRY COMMUNITY

Kristina Krivenko 3I

I'm taking Chem 14C next quarter :)

Kristina Krivenko 3I

The conversion from Torr to atm is very straightforward (1 atm = 760 torr = 760 mm Hg), and it is on our equation sheet. So if units of pressure are expressed in Torr in any of the questions on the exam, you can just use the conversion to convert it to atm :)

Kristina Krivenko 3I

You add Pt(s) when there's no conducting metal present on the cathode/anode part of a cell. For example, when both oxidized and reduced species are in aqueous state or when a solid is a not conducting, such as I2(s).

Kristina Krivenko 3I

Hi! n is the number of moles of electrons transferred in a balanced redox reaction. Hope this helps :)

Kristina Krivenko 3I

We use this equation when the pressure is kept constant, and there is a volume change.

Kristina Krivenko 3I

The slope is equal to negative k for zero and first order reactions, and it is equal to positive k for second order reactions.

Kristina Krivenko 3I

Adding on to what has been said, Dr. Lavelle also has 2 PDF files on his website that explain how to balance redox reactions in acidic and basic conditions. They are listed under "Redox Reactions" closer to the bottom, and I will also link them below. Balancing Redox Reactions: Acidic Cond...

Kristina Krivenko 3I

Temperature and the activation energy can alter a reaction's rate constant. So if you change the temperature or add / remove a catalyst (which would change the activation energy of the reaction), it would also change the rate constant for that reaction.

Kristina Krivenko 3I

I think that a perfectly isolated system is more of a theoretical idea that we can't completely execute in real life. We can get close to it, and a bomb calorimeter would be an example of that, but I believe that Dr. Lavelle mentioned in the lecture that, as time goes on, a bomb calorimeter becomes ...

Kristina Krivenko 3I

Hi, So both graphs would indicate that it is a first order reaction. However, the difference is what you plot on the axes. If you plot concentration of the reactant (y-axis) vs time (x-axis) and get a decreasing exponential graph, it is a first order reaction. If you plot natural log of the concentr...

Kristina Krivenko 3I

Yes, since $Q = \frac{[products]}{[reactants]}= \frac{[Cr^{2+}]}{[Sn^{2+}]}$ (and remember that solids are excluded from the equilibrium constant expression), so Q=0.768/0.0150.

Kristina Krivenko 3I

What ___ agent does:
- an oxidizing agent helps another molecule to be oxidized
- a reducing agent helps another molecule to be reduced

What happens to ___ agent itself:
- an oxidizing agent is reduced
- a reducing agent is oxidized

Kristina Krivenko 3I

If the temperature is not equal to 25 degrees Celsius, you would have to use E = E* - RT/nF(lnQ), because the shortcut form doesn't have a temperature term, which means that it can be only used when temperature is 25 degrees Celsius. However, I believe in cases when the temperature is 25 degrees Cel...

Kristina Krivenko 3I

Yes, you're correct! An oxidizing agent is a molecule that is reduced and that is helping another molecule to be oxidized. Conversely, a reducing agent is a molecule that is oxidized and that is helping another molecule to be reduced.

Kristina Krivenko 3I

I believe that the main difference is that the second form of the Nernst equation can be only used when temperature is 25 degrees Celsius (298.15K), since there is no term for temperature in the equation; in cases when the temperature is 25 degrees Celsius, the equations can be used interchangeably....

Kristina Krivenko 3I

The main difference between delta G and delta G naught is that delta G naught is delta G at standard conditions (1atm and 25 degrees Celsius / 298.15 K).

Kristina Krivenko 3I

To calculate K at a different temperature using Van't Hoff equation, you need to know the value of K at some other temperature as well as the delta H for that reaction.

Kristina Krivenko 3I

We can use Van't Hoff equation to find the value of the equilibrium constant at a different temperature, given that we know the enthalpy of the reaction and the equilibrium constant at some other temperature. We can also use it to find the enthalpy of the reaction, given that we know values of equil...

Kristina Krivenko 3I

No, E (cell potential) is not a state function, which means that we cannot subtract the final state from initial to find the desired value, like we've done for enthalpy, entropy and Gibbs Free Energy.

Kristina Krivenko 3I

Hi, When work is done by the system, work is a negative value. The way you can think of this is that when work is done by the system, it means that the system is dong work and, in doing so, it's losing energy. Therefore, work is negative. When work is done on the system, work is a positive value. Ag...

Kristina Krivenko 3I

In both closed and isolated systems, matter cannot be exchanged with the surroundings. However, in a closed system, energy can be exchanged with the surroundings whereas, in an isolated system, it cannot. An example of a closed system is a cup of coffee with a lid on it or a closed plastic water bot...

Kristina Krivenko 3I

I think units for entropy are usually J/K but you may have to convert it to kJ to perform certain calculations or if the question asks you to give your answer in kJ, so it's best to always look at what the problem gives you and what it is asking for. Hope this helps :)

Kristina Krivenko 3I

Yes, as others have stated, since delta S and delta G are both state properties, just like delta H is, we can apply the same rules, which are: - adding their values when adding the equations - multiplying the value of delta H/S/G by a coefficient if you multiply the equation associated with that val...

Kristina Krivenko 3I

It is uncommon for K value to be equal 1, because it means that neither reactants nor products are favored at equilibrium. Thus, it follows that since K=1 is uncommon, delta G = 0 would be uncommon as well.

Kristina Krivenko 3I

First, you need to find the number of moles, using PV=nRT. Since the pressure is given in kPa, we can use the following value for the gas constant: R = 8.314 $\frac{L*kPa}{K*mol}$ .

Then, you'll use $\Delta S = nC_{v, m}ln(\frac{T_{2}}{T_{1}})$ to find the change in entropy.

Kristina Krivenko 3I

Hi! No, one variable being constant does not mean that others are constant as well. So constant temperature does not imply that pressure is constant, and constant pressure does not imply that pressure is constant. If we look at the ideal gas law, PV=nRT, we can see the direct relationship between pr...

Kristina Krivenko 3I

Since the change in internal energy is the sum of heat and work, whether it is a positive or a negative value depends on the values for heat and work.

Kristina Krivenko 3I

Since there are several different values for the gas constant that differ, depending on the units, look at the units you are dealing with in the problem and choose the appropriate R value (that is, make sure that units match up). The best way to double check whether you chose the correct value for R...

Kristina Krivenko 3I

Hi! Work is denoted by lowercase w :)

Kristina Krivenko 3I

Entropy is a measure of disorder or chaos of a system. In chemistry, it typically refers to molecules, so the more spread out the molecules are, the larger the entropy of a system.

Kristina Krivenko 3I

Hi! Yes, you're correct. An adiabatic process occur without the transfer of heat between the system and its surroundings. This also means that the energy is transferred to the surroundings only as work. Hope this helps :)

Kristina Krivenko 3I

The official definition of phase change is as follows: a phase change is when matter changes to from one state (solid, liquid, gas, plasma) to another. So, essentially, a phase change occurs whenever matter goes from one state to another. There is a diagram ( http://www.chemistry.wustl.edu/~edudev/L...

Kristina Krivenko 3I

Based on 14A last quarter, it took around 1-1.5 weeks for our scores to be released. If there are no mistakes on the test that would need to be fixed to make sure everyone's tests are graded fairly, I'd say we can expect our scores to come out towards the end of Week 5 (so a week after we took the m...

Kristina Krivenko 3I

Adding on to what has been said above, according to the textbook, we can ignore second deprotonation for all polyprotic acids we will encounter in this class, except for H2SO4. While it is still good to compare Ka1 and Ka2 values to double check that we can ignore second deprotonation, I think what ...

Kristina Krivenko 3I

Work is not a state function because it depends on the path taken to get from initial state to the final. More specifically, it is due to the fact that work is proportional to the distance an object is moved.

Kristina Krivenko 3I

Hi! Based on last quarter, it usually took around 1-1.5 weeks. If there are no mistakes in the test that would require additional time to verify that everyone's scores were graded fairly, I'd say we can expect results to come out in about a week from the day of the exam (so towards the end of Week 5...

Kristina Krivenko 3I

Hi! Splitting 14C and 14D from spring to fall quarters is actually a really good question to pose that I didn't think about before. I feel like it's just a matter of personal preference as well as what classes you're planning to take in the spring / back-up options you have. As of right now (althoug...

Kristina Krivenko 3I

Hi! Your value for Kb is incorrect. Remember that Kw = Ka * Kb, so to find Kb from Ka, you need to do the following: Kb = Kw / Ka = 10^-14 / Ka. In this case, Kb should be equal to 2.5*10^-7. The rest of your set up and solving process seems to be correct, so I think you just need to follow the same...

Kristina Krivenko 3I

You can assume x value is small enough to approximate it when the equilibrium constant (whether it is Ka, Kb, etc) is smaller than or equal to 10^{-4} . You can also double check whether using the approximation in a given case was valid by using the 5% rule. If % ionization or protonation is less th...

Kristina Krivenko 3I

Hi! The general equation for we use for weak acids is: HA_{(aq)} + H_{2}O_{(l)} \rightleftharpoons H_{3}O_{(aq)}^{+} + A_{(aq)}^{-} And the general equation we use for weak bases is: B_{(aq)} + H_{2}O_{(l)} \rightleftharpoons HB_{(aq)}^{+} + OH...

Kristina Krivenko 3I

The midterm will be 50 minutes long, and I believe we will be given 5 additional minutes as a buffer in case there are any problems with the program or submission (at least that's how it was last quarter for 14A). The number of questions as well as the amount of points each question is worth change ...

Kristina Krivenko 3I

Hi! I believe you're correct that we can assume that [H3O+] = [OH-] in that case even if the Kw is not equal to 10^-14 due to a change in temperature. To confirm your assumption, you could set up an ICE table for this reaction, and you will get the same value for [H3O+] and [OH-]. However, I'm not s...

Kristina Krivenko 3I

Hi! Yes, you can use the ICE table when you're given Kp and partial pressures as well as when you're given Kc and molar concentrations. Just make sure that your values are consistent throughout your ICE table (meaning that all of your values given in either partial pressures or molar concentrations,...

Kristina Krivenko 3I

Hi! Yes, you're correct! If we're given Kp and partial pressures, we can use ICE table just like we would if we were given Kc and molar concentrations. And since 1 atm = 1.01325 bar, Dr. Lavelle mentioned in his lecture that, in this course, we can approximate that 1 atm = 1 bar to simplify our appr...

Kristina Krivenko 3I

Hi! The way I think about it is that Kc is a more general one because it can be applied to both aqueous solutions and gases whereas Kp only applies to gases. So you can choose either Kc or Kp for gases, depending whether you're given partial pressures of gases (then you use Kp) or molar concentratio...

Kristina Krivenko 3I

Hi! Midterms will take place during your Friday lecture on weeks 4 and 7.

Here's the link to the Exam Schedule posted on Dr. Lavelle's website that outlines that: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Chem14BTestSchedule.pdf

Kristina Krivenko 3I

I was also wondering what happens if someone originally posts under the wrong topic, and we reply to their post? In that case, would our posts count towards the 5 posts needed for the week or not?

Kristina Krivenko 3I

Hi!

So 1 atm = 1.01325 bar, but since the values are so close, Dr. Lavelle mentioned in his lecture that, in this course, we will just assume that 1 atm = 1 bar to simplify our calculations. Hope this helps :)

Kristina Krivenko 3I

Since midterms will be administered during class time, do you think we will still have a lecture to watch on Fridays of week 4 and week 7?

Kristina Krivenko 3I

I think for this course, we'll mostly use atm as units for pressure, but since 1 bar is approximately equal to 1 atm, we'll use bar as well.

Kristina Krivenko 3I

Yes, all of the three phrases express the same idea, which is that more products will be formed. And, as Dr. Lavelle mentioned in the lecture, there may be some confusion with the term "shift," because "shift" typically means "change," but K value is not changing. For t...

Kristina Krivenko 3I

Hi! I also found UA sessions to be extremely helpful last quarter. I'd suggest going to many different UAs in the first couple weeks to see whose style / worksheets you like and then stick going to those UAs for the rest of the quarter. In addition, I try to read the textbook and take notes on it (w...

Kristina Krivenko 3I

You can generalize the "Quick Way" shown in the lecture to any reaction as the following: - when volume is increased, the side with more moles of gas will be favored - when volume is decreased, the side with less moles of gas will be favored This way, reactions adjust to minimize the effec...

Kristina Krivenko 3I

The way I memorized it was:
- Endothermic: as T increases, K increases (up up)
- Exothermic: as T increases, K decreases (up down)

Kristina Krivenko 3I

In both molecules, the hydrogen atom is bonded to oxygen, so O-H bond is the same length. In that case, we don't look at size of the atom and rather figure out the anion of which acid is more stable since the more stable the anion, the stronger the acid. Since both molecules have equal number of oxy...

Kristina Krivenko 3I

Yes, they are used interchangeably. It is because $H^{_{+}}$ has no electrons and only 1 proton.

Kristina Krivenko 3I

In one of the UA sessions, it was explained that, at first, 4s has lower energy, which is why it gets filled up first. However, once electrons occupy the 3d orbital, the energy "settles down," and 3d now has lower energy that 4s, so we write 3d and then 4s.

Kristina Krivenko 3I

Since Dr. Lavelle finished covering the material today (Wednesday), the review on Friday will be during our "lecture" time, so it will be recorded and posted on CCLE.

Kristina Krivenko 3I

Ni is in period 4 and is a transition metal, so it has 3d orbitals that it can use to have expanded octet.

Also, Ni does not have f-orbitals. Remember that f-orbitals lag by 2, so 4f orbitals do not appear until period 6.

Hope this helps :)

Kristina Krivenko 3I

Both are correct, but I'd stick to the 1.0 x10^6 one as opposed to the shorthand one because 1.0 x10^6 is more commonly used, and 1.0 E6 is more associated with calculations on a calculator and such. However, I don't think it will matter for this class because our final, just like our midterms, is m...

Kristina Krivenko 3I

Our final is on December 13th.

You can find this information on his website, it's under "Test and Exam Schedule," and here's the link: https://lavelle.chem.ucla.edu/wp-conten ... hedule.pdf

Kristina Krivenko 3I

I think memorizing where metalloids are located is the most important because if you know where metalloids are, you also know that metals are to the left of it and nonmetals are to the right. The way I memorized the location of metalloids on the periodic table is that I imagine a mental staircase th...

Kristina Krivenko 3I

A little throwback to when classes were in-person and we did labs :"

Kristina Krivenko 3I

No, we don't write the equilibrium sign when writing reactions for the dissociation of a strong acid/strong base because the assume that they dissociate completely.

Kristina Krivenko 3I

Hi! No, the order doesn't matter. And the questions usually simply ask us to list how many sigma and pi bonds are there :)

Kristina Krivenko 3I

You are correct that you only consider the bonds when determining the molecular geometry. And the shape of the molecule is just another name for molecular geometry. As for electron geometry, Dr. Lavelle often referred to it as "the arrangement of electron density," but they both mean the s...

Kristina Krivenko 3I

Hi! It's important to differentiate between the arrangement of electron densities and the shape of a molecule.

H2O has 4 regions of electron densities. Therefore, it has tetrahedral arrangement of electron densities.

However, because 2 of them are lone pairs, the shape of the molecule is bent.

Kristina Krivenko 3I

How I remembered it is you just count the "number of things" attached to the central atom, where single/double/triple bonds & lone pairs count as a single region on electron density (so "1 thing"). P has 3 single bonds & 1 lone pair, so it has 4 things attached. Therefore...

Kristina Krivenko 3I

Yes, you're correct. Because when there are only single bonds (and, therefore, only sigma bonds), there can't be resonance since all bonds are the same. Thus, resonance can only exist when a molecule has at least one double bond (for example, CO_{3}^{2-} ), and, as we learned, a double bond consists...

Kristina Krivenko 3I

Hi! Single/double/triple bonds & lone pairs all count as a single region on electron density.

Thus, a molecule with 2 single bonds and a double bond would have sp^2 hybridization because the "number of things" attached to the central atom is 3.

Kristina Krivenko 3I

Hi! An easy way to remember this is: when determining hybridization, count everything that's attached to the atom.

Thus, this includes single/double/triple bonds & lone pairs, and they all count as a single region of electron density.

Kristina Krivenko 3I

Yes, Dr. Lavelle mentioned in one of the lectures that single electrons are also counted as a single region of electron density, along with lone pairs of electrons & bonding pairs.

Kristina Krivenko 3I

Hi! Yes, I believe the final is cumulative :)

Kristina Krivenko 3I

Here's the list of intermolecular forces in the order of increasing strength (from weakest to strongest):
- LDF (induced dipole-induced dipole)
- dipole-induced dipole
- dipole-dipole
- ion-dipole
- hydrogen bonding
- ion-ion

Kristina Krivenko 3I

I think it should be released some time next week, like it was with MT1, or possibly sooner since there is no partial credit this time.

Kristina Krivenko 3I

I'm not sure by how much but G-C pair is definitely stronger since, as you mentioned, G-C pair can form 3 hydrogen bonds whereas A-T pair can form only 2. In his lectures, Dr. Lavelle mentioned that the effect of IMFs adds up, and, thus, G-C pairs are more stable.

Kristina Krivenko 3I

The boiling point is the temperature at which a liquid evaporates and turns into gas whereas the melting point is the temperature at which a solid melts to become a liquid. For example, at the melting point of water, ice turns into liquid water; and at its boiling point, water evaporates and turns i...

Kristina Krivenko 3I

Similar to boron, aluminum typically forms 3 bonds. For example, AlCl₃.

Kristina Krivenko 3I

It doesn't matter which name you use since all of them (induced dipole-induced dipole, LDF, Van der Waals) are used interchangeably. Hope this helps :)

Kristina Krivenko 3I

My question didn't have CH4 as one of the choices, but CH4 only exhibits LDF because it is nonpolar due to its molecular shape. When you draw the Lewis structure of CH4, you can see that all C-H bonds will cancel each other out and, thus, the molecule as a whole will be nonpolar.

Kristina Krivenko 3I

The strength of dipole-dipole forces depends on the molecule's polarity. Thus, as the more polar the molecule is, the stronger the dipole-dipole interactions it forms.

Kristina Krivenko 3I

Hi! Both of your Lewis structures are correct, except you incorrectly identified formal charges on the second one (where there is a double bond between C and N). N does have a positive formal charge. However, the formal charge on C is 0, and the formal change on both O is -1.

Kristina Krivenko 3I

For this question, you need to identify the number of sites where there could be a potential hydrogen bond. Thus, you need to count N/O/F with a lone pair (note that if an atom has 2 lone pairs, it counts as 2 hydrogen bond sites) and number of H atoms bonded to N/O/F. For example, on the left side,...

Kristina Krivenko 3I

If the molecule you need to draw a Lewis structure for is a hydrocarbon/similar organic compound, write down a chain of C atoms first. From there, you know that each carbon can form up to 4 bonds and that C-H bonds are singular bonds and C-O bonds are double bonds. Based on this information, put dow...

Kristina Krivenko 3I

We don't have to draw double / triple bonds shorter than single bonds when drawing Lewis structures. Lewis structures are just a way to represent what molecule structure looks like, so they don't have to drawn to scale. Just know that the bond length decreases as the number of bonds increases :)

Kristina Krivenko 3I

While noble gases are very stable due to them having a complete octet and don't form compounds under ordinary conditions, some heavier noble gases will form compounds with highly reactive ions (ex. halogens) when ionized and under pressure. It is due to the fact that heavier noble gases have more el...

Kristina Krivenko 3I

I think the best you can do when it comes to determining a bond length in a resonance structure is to approximate that it would be the average of all bonds. For example, if there are 2 single and 1 double bonds in a resonance structure, the length of bonds would be the sum of the three bonds divided...

Kristina Krivenko 3I

Writing things down helps me to remember the information better, so I try to take thorough notes on the text from the textbook (and I also took notes on the videos from the modules). I highlighted and looked over these notes a couple days and then a day before the midterm. To be honest, taking notes...

Kristina Krivenko 3I

Since covalent compounds share electrons, we draw a line between the 2 atoms indicating that. However, in an ionic compound, that doesn't happen, and one atom donates its electron to another, more electronegative atom. As a result, we don't draw a line between them but rather use brackets ([]) to si...

Kristina Krivenko 3I

I don't think we need to memorize them, so I believe if we would need them to solve a problem, they'd be given in the question. However, it's useful to know the general correlation that as the number of bonds increases, the length of the bonds decreases. For example, a double bond is shorter than a ...

Kristina Krivenko 3I

Hi! It is a typo in the answer key. So you're correct, $\Delta v = 10$ m/s. You can find the correct solution in Solution Manual Errors (https://lavelle.chem.ucla.edu/wp-conten ... rs_7Ed.pdf), which is also posted on Dr. Lavelle's website. Hope this helps :)

Kristina Krivenko 3I

Yes, you're correct. So a wave function ( \psi _{n, l, m_{l}} ) that is a solution with quantum numbers n, l, m_{l} is an orbital, and wave function squared is different from that; \psi ^{2} is probability density, which means that the probability of finding particle in a region is proportional to t...

Kristina Krivenko 3I

Hi! There are 2 things you need to look at to determine whether a wave is compatible or not. First, if you look at the given options, you'll notice that some of the waves are irregular, meaning that they have different frequencies and amplitudes within one wave. Irregular waves are not compatible. S...

Kristina Krivenko 3I

I found them very helpful as well! However, I believe Dr. Lavelle mentioned that they take a long time to make, so while he might create more in the coming years, it's highly unlikely that more modules would be added this quarter.

Kristina Krivenko 3I

Rydberg's equation is a modified version of $E_{n}=-\frac{hR}{n^{2}}$ , which comes empirical observations rather that being derived, so it only works for H-atoms. Therefore, Rydberg's equation also works only for H-atoms.

Kristina Krivenko 3I

I think it's best to convert it to meters, so it is consistent with units of velocity (m/s) and momentum (kg*m/s) and you don't make mistakes when calculating the final answer. Then, if the problem asks you to give your final answer in nm, just convert it. Hope it helps :)

Kristina Krivenko 3I

Choice A isn't the right answer because the speed of radiation is always constant and equal to the speed of light ( $2.99*10^{8}$ m/s). Therefore, as frequency decreases, the wavelength increases, so that their product would always be equal to the speed of light. Hope this helps :)

Kristina Krivenko 3I

Hi! Lyman and Balmer series are the only ones that have been emphasized in this course so far, and I haven't seen any problems that would require you to know the other series. So I believe knowing these two should be enough. However, if you want to know the third series as well, and it's called Pasc...

Kristina Krivenko 3I

You are correct. A nodal plane is a region where there is zero probability of finding an electron in that area. Thus, the fact that p-, d- and f-orbitals have nodal planes means that there are areas in those orbitals (the area is the nodal plane) where the likelihood of finding electrons is zero.

Kristina Krivenko 3I

Hi! Since the midterm is all multiple choice, I don't think we'll have to necessarily show our work / how we derived equations. However, since Dr. Lavelle went over the derivation of De Broglie equation, and it is also one of the bullet points on the "Quantum World" outline, I believe we d...

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