CHEMISTRY COMMUNITY

Amy_Bugwadia_3I

You put a catalyst in the overall rate law only if the catalyst is part of the reactants side of the rate-limiting step.

Amy_Bugwadia_3I

As stated above, there is no lowest sum rule for IUPAC naming. Thus, we must try to number the carbons in order to give us the lowest numbers possible. In that respect, we simply look at the numbers individually. 1-ETHYL-3-Methyl-4-PROPYLCYCLOHEXANE and 4-ETHYL-2-METHYL-1-PROPYLCYCLOHEXANE both have...

Amy_Bugwadia_3I

Generally, Professor Lavelle's quizzes cover everything from where we left off last page (so, this quiz's "fair game material" starts on page 74) and ends on the Friday the week before the quiz, aka tomorrow's lecture.

Amy_Bugwadia_3I

I said that the rate = k * [A] * [B]^2 (didn't include [C] in the rate equation), but I did explicitly state that the reaction is zero order with respect to C. I think if you put [C]^0 into the rate equation it wouldn't make much of a difference as long as you stated that it's zero order with respec...

Amy_Bugwadia_3I

The order of B according to the solutions at the back of the book is 2 (the reaction is second order with respect to B). The way I solved for B is (rate of expmt. 3)/(rate of expmt. 2) = 16/4.0 = (200/100)^b. Solving for b, I got b = 2. Thus, in the rate equation, the concentration of B is raised to...

Amy_Bugwadia_3I

The units for second order reactions is M^(-1)* s^(-1), or L/(mol*s), so you are correct!

Amy_Bugwadia_3I

Yes, for this question I simply ignored the temperature, because I think the only reason it was given is just to give context as to why it's a decomposition reaction. Since it's a first order reaction, the rate will be k * [A]. We are given k, but have to solve for concentration using the given gram...

Amy_Bugwadia_3I

Yes, the quiz will only cover up to page 73 (so does not include info about temperature taught at the very end of Friday's lecture). The quiz will not be cumulative, however, so it only covers Kinetics. It'll be 6 questions long according to Dr. Lavelle.

Amy_Bugwadia_3I

If you're given the piece of information that the temperature is 0 K, then you are looking for the residual entropy. Residual entropy has to do with degeneracy, W, which essentially means the number of microstates/atomic positions that a compound can exist in. Because the residual entropy is given b...

Amy_Bugwadia_3I

Yes, that is correct! For second order reactions, the units are L^2/(mol^2 * seconds).

Amy_Bugwadia_3I

Christopher is right about the residual entropy vs. the standard molar entropy. Since the question mentioned 0 K, we know that we have to use the S = Kb * lnW equation because at this temperature, we are looking at the residual entropy, which has to do with degeneracy (W). Degeneracy is essentially ...

Amy_Bugwadia_3I

Exactly what Christina said! Essentially, the salt bridge is to keep both solutions neutral (allows for the transfer of ions between solutions). A porous disk can also function the exact same way as a salt bridge. However, the difference is that with a porous disk, it's essentially a membrane that s...

Amy_Bugwadia_3I

Whenever a problem gives you something about 0K, it means that they're asking about residual entropy instead of standard molar entropy. Residual entropy has to do with degeneracy (W), specifically the following equation: S = Kb * ln(W). This equation shows that S is directly proportional to W (that ...

Amy_Bugwadia_3I

As Maggie stated above, a positive cell potential is not always required (it's only for when you want a spontaneous reaction). The reason why we know we want HF on the reactant side and the H+ and F- ions on the product side is because we are asked for the Ka value, which implies that the acid (in t...

Amy_Bugwadia_3I

Generally speaking, we need to add Pt(s) (in order to conduct electricity) for reactions with only aqueous or gases on one side of the equation (you would add Pt to that side). However, there are two exceptions to this rule: if you have solid Iodine, you still need to add in Pt (because Iodine is no...

Amy_Bugwadia_3I

Because the question is asking for the value of Ka, we know that the question implies that the acid (in this case, HF) is dissociating into ions. Thus, the reaction equation will be HF --> H+ (aq) + F- (aq). Because the equilibrium constant K is always defined as the concentration of the reactants (...

Amy_Bugwadia_3I

We are given the equilibrium constant K for both the temperatures. Since it's 1.7 x 10^(-3) at 600K but 7.8 x 10^(-5) at 700K, we know that K is smaller at the higher temperature. When we increase the temperature of ammonia (NH3, on the product side), the reaction will form more reactants due to Le ...

Amy_Bugwadia_3I

I read the textbook to prepare for Quiz 1, and what I found to be very helpful was to make my own formula sheet separated by subject (based on each section of the book). For example, I had one section devoted to enthalpy, one to entropy, one to Gibbs, etc. I also find highlighting very useful, but I...

Amy_Bugwadia_3I

The way I prepped for Dr. Lavelle's quizzes last quarter was to first go through the course reader and notes, writing down all important equations/formulas/definitions, just as the previous poster did. I would then do homework problems, noting down which topics were of difficulty for me. I'd next lo...

Amy_Bugwadia_3I

Like Ara mentioned, Dr. Lavelle will let us know if and when we need to start bringing it to class. I don't think he'd expect us to study it on our own time yet, because currently we've been working on Thermodynamics and Thermochemistry, both of which are Ch. 8 topics. The recommended readings and h...

Amy_Bugwadia_3I

Because we do not have a workbook this quarter, I don't think there will be any replacement quiz (also because nothing is mentioned in the syllabus about it). However, there are a few practice quizzes to help us prepare in the back of the course reader.

Amy_Bugwadia_3I

Attending the UA office hours was definitely beneficial to me last quarter, especially before the quizzes. My TA's office hours were also very useful, because he was able to provide me with often one-on-one help in solving certain homework problems step-by-step or anything else related to the course.

Amy_Bugwadia_3I

Today in class (the 3 pm lecture), Dr. Lavelle mentioned that even though there is no such thing as a perfectly isolated system (other than the universe), bomb calorimeters can be considered to be isolated systems.

Amy_Bugwadia_3I

Generally, just look at the K value and as the above students mentioned, if i'ts less than 10^-4, you can assume that x is negligible. However, if the question asks you to test that assumption, after you solve for x, make sure to calculate the percentage deprotonation or protonation by dividing x by...

Amy_Bugwadia_3I

A "small" Ka value is when Ka = 10^-5 or smaller. However, as good practice, it's advised to calculate the percent error, and if it's less than 5% then using the approximation is okay. The approximation is essentially when we disregard the " - x" in the denominator, so we are lef...

Amy_Bugwadia_3I

H3O+ and OH- are in equilibrium with water. 2H20<-->H3O+ + OH-. Therefore their Kc(written as Kw)=[H3O+][OH-]=1x10^-14. Therefore, in any aqueous system, their concentrations will equal 1x10^-14. Therefore, obtaining the concentration of OH- in the equations above will allow you to obtain the concen...

Amy_Bugwadia_3I

Adding onto what the previous students said, I like to remember the order (HOMO to LUMO instead of the other way around) by the fact that H comes before L in the alphabet and therefore comes first here too. Essentially, HOMO stands for Highest Occupied Molecular Orbital, and LUMO stands for Lowest U...

Amy_Bugwadia_3I

Not sure about the use of hyphens specifically, but the basic protocol for naming is as follows: 1) identify the ligand and numerical prefix 2) add an "o" to the end of the anion name 3) put the cation/Lewis acid name after the anion. Include the oxidation state in parentheses and Roman nu...

Amy_Bugwadia_3I

That's helpful, thank you! Good luck on the quiz :)

Amy_Bugwadia_3I

Are we allowed to use fractions when balancing chemical equations as long as we put the coefficient at the front of the equation? I think that if you do use fractions as coefficients in balancing equations, that should only be an intermediate step. Your final answer should always be whole number co...

Amy_Bugwadia_3I

I've been making flashcards to memorize these. As the previous poster mentioned, it really helps to visualize the shape that corresponds to each term, because otherwise it's really easy to get the terms confused since they're all pretty similar. It also really helps to go through each term and try t...

Amy_Bugwadia_3I

You are correct! Maybe you were looking at the wrong problem's answer? At the back of my book, it says the following for the answer for 2.93: "A Cl; B Na; C Cl- ; D Na+", which is the answer that fits your initial explanation. You're right about the fact that Cl has a smaller ato...

Amy_Bugwadia_3I

Like the other posters mentioned, in order to find the number of molecules in a compound, you would need to use Avogadro's number. If you're given the amount of the compound in grams and you want to find how many molecules of that compound there are, convert grams to moles, and then go from moles to...

Amy_Bugwadia_3I

I was wondering the same thing! However, when I was working in my workbook on the Quiz 1 Prep Fall 2014 quiz, problem number 1 says "How many grams of ammonium chloride..." without giving us the chemical formula for the compound. However, no names of polyatomic or other ions are provided o...

Amy_Bugwadia_3I

I am not positive about this, but I think that if you aren't given the mass of the second reactant, you should assume that the reactant with the given mass (in this case, PCl3) is your limiting reactant. Then, solve it as you would any other limiting reactant problem by first always remembering to b...

Amy_Bugwadia_3I

I would tell y'all another chemistry joke, but all the good ones argon.

Amy_Bugwadia_3I

The reason why we have to balance equations is due to the law of conservation of mass, which states that the mass of the reactants has to equal the mass of the products. From my understanding, a hydrogen atom, for example, has the same mass regardless of state of matter (which is why we only have on...

CHEMISTRY COMMUNITY

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Re: writing reaction rate laws [ENDORSED]

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