CHEMISTRY COMMUNITY

Jeannine 1I

I think it's because for each concentration, you need to convert the mmol to mol, because the number gets smaller as you square, and multiply. And it won't get you the same answer if you just convert from mol to mmol at the end, so you have to convert before then.

Jeannine 1I

You have to use the natural log of the Arrhenius equation: ln (K) = lnA - Ea/R (1/T). Then if you subtract, you can get the form: ln(K2/K1) = Ea/R (1/T2 - 1/T1). From this, you can now plug in the given values to get 39 kj per mol Can you explain how to get the second form please? I'm having a diff...

Jeannine 1I

So to clarify from what Dr. Lavelle said today during the review session, can it be said that in cell diagrams, if there are multiple solids on one side, they will always be separated by "|" and never by a comma? Or can this only be applied when one solid is an electrode, and the other is ...

Jeannine 1I

allisoncarr1i wrote:Does that line in the syllabus also mean that anyone who receives over 50% of the points will pass?

Yes, this means that as long as students get 50% (250/500 points), they will pass the class with a C- (:

Jeannine 1I

So the K_{a2} is the second deprotonation of phosphoric acid, which means two H^{+} ions will be gone. The first deprotonation ( K_{a1} ) would be: H_{3}PO_{4} (aq) + H_{2}O (l) \rightleftharpoons H_{2}PO_{4}^{-} (aq) + H_{3}O^{+} (aq) The second deprotonation ( K_{a2...

Jeannine 1I

He mentioned in class that you usually don't write the "1" since it's already implied, just like how in a math equation where x is to the first power, we write it as "x" instead of " $x^{1}$ ".

Jeannine 1I

I believe they will both give you similar answers, the second equation with a number in it is just the simplified version, where the (RT/F) is already calculated, given that the reaction is at the standard temperature 298 K.

Can anyone confirm this?

Jeannine 1I

Yes. When you look at the Arrhenius equation we can see that increasing temperature makes the rate larger, as does lowering the activation energy, which is what a catalyst does, so doing these together should theoretically increase the enhancement of the rate. But isn't the statement saying that th...

Jeannine 1I

I just wanted to ask, when the book or the final asks you to find the reaction rate for a substance, which reaction rate (average, instantaneous, or unique) do we presume to use?

Jeannine 1I

Sometimes there are cases where you simplify the balanced redox reaction, would n be the number of moles before simplification or would it be the number of moles after simplification? In a redox reaction, you would look at the half-reactions you used to cancel out the electrons before you formed th...

Jeannine 1I

I understand that Au (s) is serving as a conducting electrode in this case, but isn't the oxidation reaction that's occurring in this problem Au+(aq) --> Au3+ (aq) + 2e- ? Is it not necessary to include Au+ when writing out the cell diagram for the anode side? You base off your cell diagram from th...

Jeannine 1I

Is it right to say that $H_{2}O$ (l) is never included in cell diagrams, but $H^{+}$ ions and $OH^{-}$ ions are included?

Jeannine 1I

Since K is the concentration of products over the concentration of reactants ( \frac{[P]}{[R]} ) at equilibrium, when K<1, this means there is a higher concentration of reactants compared to products. This means that the *reverse* reaction is favored, which in turn means the forward reaction is NOT ...

Jeannine 1I

I believe the final is cumulative, and I think there might be some basic things we would have to remember from 14A, such as figuring out the limiting reagent given a reaction.

I don't know exactly what else we would be expected to know though:/

Jeannine 1I

And after that, you would add the appropriate number of electrons to cancel out the positive Hydrogen ions?

I think I understand now, thank you(:

Jeannine 1I

How would you figure out what the change in oxidation state is for a reaction where $O_{3} (aq) \rightarrow O_{2} (g)$ ? There's no charge there so I could only assume the oxidation states are both 0, but that would not make sense.

Jeannine 1I

Xingzheng Sun 2K wrote:So which one is the right unit for the equation PV=nRT?

I'd say it depends on which unit the problem uses. Can anyone confirm this?

Jeannine 1I

amogha_koka3I wrote:Do we assume isothermal, reversible, and at equilibrium are all synonymous in this class?

I believe that we can assume (for this class) that isothermal also means reversible. I don't think we can assume it's at equilibrium though.

Jeannine 1I

Does anyone know how you would've figured out the oxidation states of the reactants and products involved?

Jeannine 1I

Aidan Ryan 1B wrote:What are things we have learned that are not state functions?

Basically anything that isn't part of "PD TV HUGS" are not state functions.

Examples of things we've learned now that are NOT state functions would be: work (w), heat (q), and cell potential (E)

Jeannine 1I

Remember that K is a constant, and therefore has no units(:

Jeannine 1I

I made the mistake of using an ICE table when we didn't need to use one. Because of that, I also had to approximate because there ended up being a cubic equation if I didn't. But it was not the right method to answer the question:(

Jeannine 1I

005199302 wrote:for #6, shouldn't the work be -12.4kJ?

w=-PxdV = (-122atm/L)(101.33J/atmL) = -12.4kJ

Were you able to solve 6a? I've tried doing it but keep getting 12.85 kJ, but I don't see where it went wrong.

Jeannine 1I

For Worksheet 1 question 7, are the sig figs correct? Wouldn't it be 3 sig figs, not 4, because the lowest number given is 298 K? Or is it 4 sig figs because we don't use the temperature to solve this problem?

Jeannine 1I

That's a good question! I originally thought it was all in Chapter 8 of the 6th edition, but there are some concepts that I can't find. So, if it isn't in Chapter 8, I guess it would be in Chapter 9.

Jeannine 1I

$\Delta U$ is the overall change in internal energy of a system, and q is heat!

Jeannine 1I

You can imagine that when pressure increases, the gas molecules on both sides of the reaction move faster, and are therefore more likely to bump into each other. The side with more moles of gas will have more molecules that bump into each other, which will drive the reaction to shift towards the sid...

Jeannine 1I

Kenan Kherallah 2C wrote:Whats R?

I believe R refers to the universal gas constant!

Jeannine 1I

Let's say there's a question that asks you if a chemical reaction is at equilibrium, and it gives you a reaction and K-value but doesn't specify if the K is Kc or Kp. Then, it gives you the partial pressure concentrations of the reactants and products. Is it safe to assume then, that the K-value giv...

Jeannine 1I

I am terrible at explaining, but here is the video I watched: https://www.youtube.com/watch?v=54n1XppP-lA and I got confused with the example at 25:46 when the initial concentrations of the products were not 0. The reason why the initial concentrations of the products are not 0, is because they giv...

Jeannine 1I

I understand that when you decrease volume, the side with more moles becomes more crowded and "bumps into each other" more often, which is why the reaction then favors the side with fewer moles. But can someone help me visualize why when you increase volume, the side with more moles is fav...

Jeannine 1I

Since we don't include liquids or solids in K-values, adding water won't change the K-value.The only thing that would affect it is if the concentrations of the reactants or products are increased/decreased, but if the concentrations remain the same, adding water won't affect it.

Jeannine 1I

I'm not sure what that problem is since I have the 6th edition, but I believe it is 10^-14 under lab conditions, which means the temperature is constant. Even then, I'm not sure if it is ALWAYS equal to 10^-14. I'd like to know too, it's a good question!

Jeannine 1I

Remember that when looking at a problem, you use the number with the least amount of sig figs for your answer. So in the problem you're talking about, are you sure there wasn't a number given somewhere that only had 1 significant figure? At least from the problems I've looked at, the solution manual...

Jeannine 1I

I believe that adding a liquid will not affect the K because it will stay the same on both the reactants and products sides.

Jeannine 1I

Yes, you would only include the gases in the $K_{p}$ , and the numerator would be 1 since the equilibrium constant for partial pressure is the concentrations of partial pressure of products over partial pressure of reactants:

$K_{p}=\frac{[P_{products}]}{[P_{reactants}] }$

Jeannine 1I

Thanks that's really helpful. Just wanted to know why would 5/17 be used for the calculation of partial pressure instead of 6/11 from the earlier part B? It's because you're looking for the equilibrium constant K, so you use the flask that's at equilibrium! The flask that's before the one with 5/17...

Jeannine 1I

I think using either will be fine! I have heard before that Lavelle will be using the old convention. Can someone confirm this please?

Jeannine 1I

Yes, I think that's a good way to think about it! The H+ counts as a proton.

Jeannine 1I

It helps me to know what the trend of sizes are for cations. They generally decrease across a period, and increase down a group (due to the addition of another shell). From there, you can tell that the smaller the cation, the greater its polarizing power. Hope that helps(:

Jeannine 1I

what were the other names we had to remember besides bis when the other prefixes are already used in the name? We need to remember to use: bis- instead of di- tris- instead of tri- tetrakis- instead of tetra- pentakis- instead of penta- We use those names when the ligand already has di-, tri-, or t...

Jeannine 1I

My TA told us that usually a central atom will at most form 6 bonds, with a total of 12 valence electrons in its outermost shell, so since pentagonal bipyramidal has 7 bonds to its central atom, I think it is a rarity.

Jeannine 1I

Rebeca_Ruiz_3D wrote:So just to be clear the test won’t cover anything from Monday’s lecture? Like coordination compounds?

Yes, the test will not cover anything from Monday's lecture!

Jeannine 1I

sp and sp2 and so on are the general hybridization of atoms. When it says 2sp2 that refers to the second period orbitals. For example, if we are asked to label the bond between C and H in CH4, we would label it as \sigma (C2sp^{3}, H1s) . This is because Carbon has valence electrons in perio...

Jeannine 1I

You do have to be careful when looking at dipole moments because some nonpolar molecules can have dipole moments, they just cancel out against other dipole moments. For a molecule to be polar it must have polar bonds with dipoles that don't cancel. Professor Lavelle also said it would be helpful to...

Jeannine 1I

They could also ask us to label all existing bonds in a molecule. For example, in $CH_{4}$ we would label the bond $\sigma$ (C2 $sp^{3}$ , H1s).

Jeannine 1I

According to electronegativity trends, Cs has a lower electronegativity value than K, so the difference between Cs and Cl is greater than the EN difference between K and Cl. In this case though, the difference is so great that both KCl and CsCl will be ionic, with CsCl having greater ionic characte...

Jeannine 1I

How does electronegativity relate to greater covalent character when comparing 2 molecules?
For example, between KCl and CsCl, which one has the greater covalent character? And why? Does electronegativity explain it?

Jeannine 1I

The VSEPR formula refers to AX(sub n)E(sub m), where A is the central atom, X is the atoms attached to the central atom (so n would be the number of atoms attached), and E refers to electron lone pairs (so m would be the number of lone pairs attached to the central atom). For example, in H20, the VS...

Jeannine 1I

Do the bond angles of the octahedral shape (90 degrees) ever change (because of things like electron pairs)? I believe if there is only one electron pair (AX5E), it will slightly push the other 4 bonded atoms further away from it, so that the angle between the central atom, the opposite atom from t...

Jeannine 1I

Yes, if you were to draw the VSEPR model of NH3, given that it has a lone pair, you would draw N as the central atom with a lone electron pair attached, one H with a regular line, one H with the stripes, and one H with the triangular wedge! The striped lines means the H atom is seen in the back, whi...

Jeannine 1I

I think knowing their 3-Dimensional geometry definitely helps determine their bond angles! I would generally say though, that the most common angles are 90, 109.5,120, and 180. It's usually either those angles, or slightly less than those angles (when lone pairs are involved). Thinking about what th...

Jeannine 1I

How would you determine these trends for diagonal (non-adjacent) elements? For example, between Carbon and Sulfur. I think between those two, carbon would have a higher ionization energy because sulfur is in period 3, which means it has another shell around the nucleus. The extra shell shields vale...

Jeannine 1I

I know the general rule for categorizing ionic vs covalent bonds, is knowing the difference in electronegativities of the atoms involved, and if it's less than 1.5 it's covalent, and if it's greater than 2 it's ionic. What exactly is the bonding type, or what happens between the atoms, when the elec...

Jeannine 1I

Yes! Because electrons in the real world are 3-dimensional, the electrons maximize their repulsions to each other, and do their best to have as much space between each other as possible. This results in the tetrahedral shape Dr. Lavelle mentioned in class. As for drawing it out, I'm not too sure, bu...

Jeannine 1I

The ground state electron configuration for Cu would be [Ar]4s^1 3d^10. You would think it would be [Ar]4s^2 3d^9, but a full d-orbital is more favorable than a full s-orbital. This means one electron in the s-orbital will jump to the d-orbital in order for that orbital to be full. So in Cu+, the el...

Jeannine 1I

When you remove an electron from an atom that has a p-orbital (and no d-orbital), you remove it first from p and then from s. However, if you remove an electron from an atom that has a d-orbital, you remove it first from the s-orbital.

Jeannine 1I

I was also very confused by the book definition, but my TA explained electron affinity by saying it's basically how badly an atom wants an electron. The atoms that need only a few more electrons to have a full shell, want electrons much more than those that need many more electrons to fill in their ...

Jeannine 1I

I think it's true that technically all matter has wavelike properties, but I believe in one of his lectures, Mr. Lavelle mentioned that anything smaller than 10^-18 would be so small that it most likely would not display wavelike properties.

Jeannine 1I

Yes, I would think so! Since the equation involves a greater than or equal to sign, it would make sense that they ask for minimum of something, and not maximum.

Jeannine 1I

If you answer all the questions on the post-modules, the whole thing is scored, and although it doesn't tell you what the correct answers are (for the ones you got wrong), it's still useful to know! And there's always the option of retaking it and choosing different answers until you get 100% of the...

Jeannine 1I

Question 1.3 asks, "What happens when frequency of electromagnetic radiation decreases?" and the answer is "the extent of the change in the electrical field at a given point decreases." Would someone be able to explain to me what the answer to this means, or how this relationship...

Jeannine 1I

Yes, for every test we will be given a sheet of equations and the periodic table!

Jeannine 1I

I believe he mentioned in class that we don't need to know about black bodies, so I wouldn't worry too much about studying them. It's also not on the course outline!

Jeannine 1I

why is the energy level of 102.6 nm 1? My reasoning for why n1=1, is that the 102.6 nm is a part of the Lyman series, and a general rule in the Lyman series is the first energy level (n1) is always equal to 1, so n1=1. I'm not sure if that line where 102.6 nm is is n1, but that was just my reasonin...

Jeannine 1I

A "Joule" is a unit of energy, and is named after the man who discovered it. I think we just use the term "joule" as a shorter way to say 1 kg*m^2/s^2, and it lets people know automatically that we're referring to energy. I hope that helps at least a bit haha(: !

Jeannine 1I

What I end up doing to show my work for balancing equations, is count the number of moles on each side and write them out. For example, if there are 12 total carbon atoms on both sides, I'd write C: 12 on each side! I hoped it would show them that I checked my work.

Jeannine 1I

I got the same answer of 4.23 x 10^-7, and since nanometers is in 10^-9, I just moved over the decimal so that it would be 10^-9. The answer then becomes 423 x 10^-9, which is the same thing as 423 nanometers. I hope this helps!(:

Jeannine 1I

You're right, it should be 4 L of water as the final volume! Did you check to see if the final answer is correct? Maybe it was a typo, or they really did solve it incorrectly. The answer should be 1.0 x 10^-2 mol of N!

Jeannine 1I

I attended a chemistry study session where I was told there would only be about 3-4 questions, but most likely with multiple parts. I was also told that really studying and understanding what is listed in Outline 1 would help us on the test! I don't know how strict they'll be on grading though:/

Jeannine 1I

In general, you use the given information to figure out how many sig figs you need in your answer. So yes, in this case you base it off the 25.2 kg in the question, which means your answer will also include 3 sig figs!

Jeannine 1I

My chem TA told us to include our Bruin ID number and our discussion session! I'm not sure if that applies to all classes though.

Jeannine 1I

In this equation, we need to balance the equation C10H15N + O2 -----> CO2 + H2O + CH4N2O, and the answer to that is 2 C10H15N + 26 O2 -----> 19 CO2 + 13 H2O + CH4N2O. I ended up with the correct answer, but it took me a while to figure it out. Does anyone know a faster way to balance out equations, ...

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