CHEMISTRY COMMUNITY

Tuong Nguyen 2I

I was gonna tell a chemistry joke, but then I thought "Na... no one would get it."

Tuong Nguyen 2I

For a first order reaction, the half life is the same value regardless of where you are in the reaction. However, in second and zeroth order reactions, the half life value changes depending on the concentration, so it's more difficult to calculate and therefore not as useful.

Tuong Nguyen 2I

A lot of the TA's post a lot of review problems and their weekly worksheets that they hand out in discussion as well!

Tuong Nguyen 2I

Would be acceptable to not round the order and only round the value of the answer, which may be the rate constant, for example? I'd suggest that you round the order rather than the value of the answer, since in the textbook and in lectures, we've only seen reaction orders as whole numbers rather th...

Tuong Nguyen 2I

We round these values because of the possible experimental errors that could occur when we don't round. Plus it just makes calculations easier! Hope this helps!

Tuong Nguyen 2I

No there is no correlation between the order of a reaction and the chemical coefficients.

Tuong Nguyen 2I

It might just be mixed up between the conversion between mmol and mol. But mmol and sec should be in the denominator.

Tuong Nguyen 2I

I just did some Nernst Equation problems and the beginning of the chemical kinetics.

Tuong Nguyen 2I

Reactions rates are always positive and since k is a proportionality constant that relates some given concentration (which are always positive) with the rate (also always positive), it's not possible for k to be negative.

Tuong Nguyen 2I

They will both produce the same value. As long as you have the conversion accounted for in the RT/nF term already.

Tuong Nguyen 2I

In the textbook, it'll be sometimes referred to as just E, so don't get confused with emf and E.

Tuong Nguyen 2I

Yes only free energy can be greatly influenced by a change in temperature.

Tuong Nguyen 2I

As electrons are transferred from the anode to the cathode, there is a salt that builds up which would create a charge gradient. In order to maintain a neutral solution, there needs to be a salt bridge that balances the charges.

Tuong Nguyen 2I

Because you can think of it like the maximum difference in volume between two connected beakers of water. The beaker that has all the water compared to the beaker with no water will show how there is a larger difference. Even at a little flow, the difference in volume will be a large value. This can...

Tuong Nguyen 2I

Going off of what Sophia said, unless there is something in the problem that explicitly states that pressure is constant, you will always use the Cv constant because we can't assume that pressure is constant when nothing within the problem mentions pressure.

Tuong Nguyen 2I

I don't think that there will be curve for the midterm. The only curve that Dr. Lavelle uses is one based on the glass grade by the end of the quarter.

Tuong Nguyen 2I

Well delta U can still have a nonzero value at isothermal, reversible processes depending on what w is. So when we know that the system undergoes an isothermal, reversible process, we know that heat (q) = 0 and therefore the only thing left for delta U to equal is w. So at isothermal, reversible pro...

Tuong Nguyen 2I

I wouldn't be sure about very sure about that. He did state earlier this quarter that making his own problems for midterm reviews is harder for 14B because making the numbers work out is more complicated compared to 14A. So for now, I'd just recommend going over the homework problems and looking at ...

Tuong Nguyen 2I

No it didn't.

Tuong Nguyen 2I

For this problem, my answer for how much work was done was correct according to the solutions manual. However, part c asks for the change in internal energy of the system. My logic was that because there was no heat involved in the scenario, deltaU would just be equal to the work done on the system,...

Tuong Nguyen 2I

I believe the one that is most often used is the 8.314 J.K^-1.mol^-1 one. However, each constant will be on the constants and equations sheet for the midterms/tests, so just pay attention to your units and which ones you need to cancel out, and you'll know which gas constant to use.

Tuong Nguyen 2I

You could also think of solids and liquids as not having much volume change at all, which in result allows us to make the estimation saying that delta V for a solid or liquid is essentially zero.

Tuong Nguyen 2I

I don't think we need to know how to set up a calorimeter experiment specifically, but you should be able to recall that bomb calorimeters/calorimetry in general is an example of an isolation system.

Tuong Nguyen 2I

Yeah I believe that Christina is correct. The overall enthalpy of a chemical reaction should be measured in kJ while other enthalpies for specific chemical species should be measured in kJ/mol unless specified.

Tuong Nguyen 2I

This method is known to be the most inaccurate one of the three however, so I don't think we'll be using this method much, other than in homework problems.

Tuong Nguyen 2I

Those chemical species that are a product of one reaction but also the reactant of another are considered intermediates, so you don't have to worry about them when using Method 1.

Tuong Nguyen 2I

Also remember that the second method is the most inaccurate method of calculating the overall enthalpy of a reaction however. The reason being is that the values that we have for the bond enthalpies are only averages taken over a different number of molecules. The actual bond enthalpy for certain bo...

Tuong Nguyen 2I

Dr. Lavelle said that if K is around 10^-3, then it's a safe range to say that the change x is negligible, but I'd say that 10^-4 is a safer value.

Tuong Nguyen 2I

Yeah I believe that for now, it's safe to assume that the temperature will be 25 degrees C and that pKw is 14.

Tuong Nguyen 2I

Simplifying Cubic Equations is essentially the same process of quadratic equations because once you simplify one of the equilibrium terms, it'll just give you a quadratic equation or simple equation where you solve for x. This will be able to happen if the K constant they give you is small enough to...

Tuong Nguyen 2I

Tony Ong 3K wrote:nope no difference! Hope that helped because I have to sleep now!

Goodnight!

Tuong Nguyen 2I

Like others said, Kp is specifically referring to the equilibrium constant when all relative molecules/atoms are in a gaseous phase, therefore you use partial pressures. Kc refers to the equilibrium constant with regards to the concentrations of reactants and products, so you would use molar concent...

Tuong Nguyen 2I

K is the equilibrium constant that reveals the ratio of products to reactants at the time in which the actual reaction is at equilibrium. Q, however, is the ratio of products to reactants at any other time during the reaction whether or not the reaction has reach equilibrium.

Tuong Nguyen 2I

If you double the moles of a certain molecule/atom in a chemical reaction, you'll have to take that coefficient and use it as the exponent for the concentration of that respective molecule/atom in the equilibrium constant.

Tuong Nguyen 2I

It's helpful to understand the trend of polarizable atoms. The larger the atom is or the more electrons the atom has, the more polarizable it is. This is because there will be more electron clouds that are not as strongly held by its nucleus' protons.

Tuong Nguyen 2I

Chloro and chlorido are interchangeable. I just go with chloro because it's shorter and like Sophia said, Dr. Lavelle used that form as well.

Tuong Nguyen 2I

I'm sure that in the end it doesn't matter what order you write the ligands in, as long as they come after the transition metal within the square brackets. If you want to be safe, I'd suggest just writing the ligands in the order of which you the compound name lists them. Other than that your answer...

Tuong Nguyen 2I

There are also other cases where the ending of certain things change. For example, in a complex the metal should have an "-ate" ending if the overall charge of the complex is negative (ie. cobalt --> colbaltate). Have a good night!

Tuong Nguyen 2I

There can be slight modifications such as one of the atoms making up the tetrahedral that isn't the same as the other atoms being large enough to push the others closer together, which would make some of the bond angles slightly less than 109.5 degrees and the bond angle with the larger atom would b...

Tuong Nguyen 2I

Both are interchangeable, just make sure you know that the lewis definition deals with lone pair donor and acceptor while the bronsted definition views it in terms of proton donor and acceptor.

Tuong Nguyen 2I

The significance of this rule is that it can completely alter the shape of the molecule after drawing the VSEPR model. When we say that the lone pair-lone pair repulsion is the strongest of them all, it means that when we draw a VSEPR model, we try to separate those two as far away as possible, even...

Tuong Nguyen 2I

LeannaPhan14ALec1Dis1F wrote:This applies to the angles?

Yeah this should also apply to the angles as well. Any molecule that is trigonal planar will have 120 degree angles between each pair of bonded atoms.

Tuong Nguyen 2I

It shouldn't matter as long as the overall structure is an equilateral triangle if you were to draw straight lines connecting the atoms surrounding the central atom.

Tuong Nguyen 2I

It's also to keep in mind how lone pair repulsions are the strongest of them all, overpowering a lone pair-bonded pair and bond-bond repulsion. So lone pairs have a significant impact on the 3-D shape of the molecule.

Tuong Nguyen 2I

Professor Lavelle told us that we would be getting the midterm tomorrow during class, and I think the comments will be up to your TA. You could always ask your TA during discussion too.

Tuong Nguyen 2I

I believe that bond angles in VSEPR models are just something that you have to memorize with certain shapes. There's no way to calculate bond angles as of right now.

Tuong Nguyen 2I

You might want to draw out the polarity ends when it comes to drawing multiple molecules that have an effect on each other (ie induced dipole-dipole). It'll just help with understanding what makes a molecule polar and the correlation between electrons and polarity.

Tuong Nguyen 2I

The first four elements (H, He, Li, Be) can have an incomplete octet since their electrons only pertain to the s-orbital which requires two electrons to be stable. Group 13 (B, Al, etc.) can also have an incomplete octet because that is just the exception that we need to know about. Any element in P...

Tuong Nguyen 2I

In some cases, multiple resonance structures can have a formal charge of 0, and if that structure has a central atom, a good way of determining which one is the most stable structure is to find the one where the central atom has a formal charge of 0.

Tuong Nguyen 2I

I agree with Gillian. It's also important to understand that molecules that we consider one or the other will also display characteristics of the other, just to different extents.

Tuong Nguyen 2I

Lyndon Bui (one of our UA's) posted a practice packet that has questions derived from the previous exams as well, so if you just search "Garlic Bread" on the search bar you'll find it. This might give you a good measure on the trickier parts of previous exams. https://lavelle.chem.ucla.edu...

Tuong Nguyen 2I

You can check this by analyzing the ionization energy of each of them. Each element wants to become stable by either losing or gaining electrons so that their outermost shell follows the octet rule. For example, the element Sulfur has relatively high ionization energy and has 6 valence electrons. Th...

Tuong Nguyen 2I

Yes elements within the 3rd period/row can have more electrons than the octet rule allows. Also, the first four elements (Hydrogen, Helium, Lithium and Beryllium) and elements in column 14 can have incomplete octets (less that 8 electrons).

Tuong Nguyen 2I

Yeah this would be another example of that exception in electron configurations. The transition metals from Copper and and Chromium take one electron from the 4s orbital to either have a half filled 3d orbital (in the case of Chromium) or a fully filled 3d orbital (in the case of silver).

Tuong Nguyen 2I

We aren't sure based on what we learned in lecture today. All Dr. Lavelle said was that we wouldn't be able to tell just by looking at the periodic table, rather we know that 3d has a lower energy than 4s because it was measured and given to us as a scientific fact.

Tuong Nguyen 2I

Yeah that looks about right, since 2.45 and forward is all the stuff we learned in lecture today and will learn in the upcoming lectures.

Tuong Nguyen 2I

We don't need to know electron configurations for this upcoming test. Dr. Lavelle said Test 2 consists of all the first half of quantum up to the quantum number material we learned about on last Friday's lecture.

Tuong Nguyen 2I

My TA personally requested that we did the first test in pen but I guess it's up to the TA. However, on the syllabus, Dr. Lavelle states that "All tests and exams are closed book and must be written in pen." So get used to it by using pen on homework if you feel like it.

Tuong Nguyen 2I

I'm not sure if a question like this has been asked before but do posts on Sunday count as posts for Week 3 or Week 2? I responded to a couple posts on Sunday night (October 14) and was wondering what academic week it would apply for, since at the time the CCLE front page still said Week 2 at the ti...

Tuong Nguyen 2I

As the temperature of the black body increases, the frequency of the light that causes that temperature increases as well. In other words, frequency is proportional to energy of a certain radiation. However, wavelength and frequency have an inverse proportional relationship, so as the frequency of t...

Tuong Nguyen 2I

Based off of the way that our first quiz went, I don't think you need to know too much about the history of those experiments. Many of the problems assigned for homework are more mathematical and application-based so just study the formulas and steps needed for certain types of problems and you'll b...

Tuong Nguyen 2I

Besides doing all the homework problems, you've probably also heard about going to your TA hours or Dr. Lavelle's office hours for help. I would recommend going to the UA Peer Learning Sessions that have the different levels like step by step reviewing/extra practice or the more fast paced sessions ...

Tuong Nguyen 2I

This might sound a little unnecessary but when it comes to the detailed notes and jotting down key concepts, you could type them so that when you go back to study them, you'll have clean, neat words that fully explain the concept without having to use so many abbreviations and shortened versions of ...

Tuong Nguyen 2I

For the most part, you should get partial credit because the TA grading your test should be able to see that you were on the right track and interpreted the question correctly. If your main problem is being off by a hundredth from the correct answer, it could be your use of sig figs and rounding too...

Tuong Nguyen 2I

I made sure to ask Dr. Lavelle this same question! For this quiz, we don't have to know the formulas of anions or cations like the ones you exemplified. However, expect to have to know some in the future for later quizzes!

Tuong Nguyen 2I

Can there ever be an unidentified solid/gas and you would do the same thing Or willl we only ever come across unidentified metals? Yeah there are possibilities that this same type of question could appear but with gases instead of metals. It's just that in terms of the periodic table, there are man...

Tuong Nguyen 2I

You could always leave the element that is by itself for last. In this case, the Fe is pure iron so just focus on balancing the carbon and oxygen first, then look at the Fe which is easy since there is only one mole of it.

Tuong Nguyen 2I

Does anyone know when our first homework assignment is due? Is it due on the Friday of week 1 during lecture or during our discussion in week 1? Also, I remember Professor Lavelle saying that we can "choose" our homework. Does that mean we can go through our textbook and choose the questi...

Tuong Nguyen 2I

Since you know the moles of each reactant in the equation, along with the molar ratio between the two reactants (being 4 moles of NH3 for every 5 moles of O2) you would just have to multiply it by 4/5. If you want to know how many moles of NH3 you need for 0.266 moles of O2, you would multiply the 0...

Tuong Nguyen 2I

I'm pretty sure you don't have to write out the question. Just make sure to write the chapter, question number, and the edition of the textbook that that certain question is from, since Dr. Lavelle allows us to use the 6th and 7th edition for homework!

Tuong Nguyen 2I

Yes the charge of Ag would be +1 in this case. Transition metals are stated to have an oxidation state of 0 (having no charge) so when combined with another element like Cl which usually has a -1 charge in chemistry problems, you assume that the transition metal takes on a charge that would stabiliz...

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