CHEMISTRY COMMUNITY

Kayla Denton 1A

You're talking about the steady state approach, right? The second video at around 4 mins 30 seconds? For this step, use the forward slow reaction and reverse fast reaction. On the board it says "slow - negligible reverse" which means ignore the reverse of the slow (the reverse is SO slow t...

Kayla Denton 1A

An atom says, "I think I lost an electron." His friend asks, "Are you sure?" and the atom replies, "I'm positive."

Kayla Denton 1A

Oh, sorry! It's there now

Kayla Denton 1A

Hi! I'm a bit confused about how to name this cycloalkane (attached)... why wouldn't it be 1-ethyl-3,3,-dimethyl-2-propylcyclohexane? I thought that the overall sum of the numbers did not matter, and when equivalent numbering was possible we should assign the lowest number to the thing that comes fi...

Kayla Denton 1A

Kayla Denton 1A

Here is an overview of naming organic molecules!

Kayla Denton 1A

Put the equation into y = mx + b form where y is the dependent variable (lnk), x is the independent variable (1/T), b is the y-intercept and m is the slope. You get lnk = (-Ea/R)(1/T) + lnA (I wrote 1/T separately rather than combining it into -Ea/RT to make it easier to visualize). From here you ca...

Kayla Denton 1A

In a highly concentrated solution there is not much water, so as water is used up in the reaction, [H2O] decreases by a measurable amount. Therefore, it is present in the rate law. However, in a dilute solution there is so much water that as water is used up, [H2O] barely changes—so we say that [H2O...

Kayla Denton 1A

By knowing how many moles (per L) of C were consumed, you can figure out how many moles (per L) of A reacted. Once you know the concentration of A that reacted, you subtract this number from your initial concentration of A to get the final concentration of A. Now you have [A] initial, [A] final, and...

Kayla Denton 1A

Catalysts are present as reactants in the very beginning and products at the end. Intermediates, on the other hand, are not present in the initial reaction but are produced within one of the steps and then consumed within another step.

Kayla Denton 1A

Here's a review video Sarah and I made for the upcoming quiz. Feel free to comment below with any questions. Good luck everyone!

(For those of you who can't view the video, here is a screen shot to summarize.)

Kayla Denton 1A

I think that since unique average rate is the rate divided by the coefficients, when the coefficients double the unique average rate will half.

Kayla Denton 1A

Rates are never expressed as negative! Rate of formation of products = delta P / delta t whereas rate of consumption of reactants = -(delta R)/delta t... the negative is there because the change in reactants will be negative as reactants decrease, but rate must be expressed as positive so the - canc...

Kayla Denton 1A

In the question they say that for this certain reaction, Ar has a role (it removes energy as the product forms) and they do include it in the rate expression. Just not the elementary reaction for some reason

Kayla Denton 1A

In question 14.85, why does the textbook put brackets and a symbol around the activated complex? What does this mean & do we have to follow this convention in our answers? Also, for part b, I understand that the reaction is termolecular from the rate law, but why doesn't the solutions manual inc...

Kayla Denton 1A

A quick way to tell the difference between a catalyst and an intermediate is to see whether it is present at the beginning of the reaction. Intermediates are first produced and then consumed in the reaction, so they're not present as reactants in the beginning. Catalysts, on the other hand, are alwa...

Kayla Denton 1A

If the question asks for the "rate" it just wants one number, not the rate law expression. So you'd just do the unique average rate (I guess they just omitted the word average because it was implied).

I don't think in the chapter they ever ask for the "unique rate law."

Kayla Denton 1A

The solution with the lower concentration must be the product in a concentration cell, therefore the Q expression must have the lower concentration (product) on the top. First assign product and reactant, THEN choose anode and cathode. The lower concentration is the product, therefore the anode. The...

Kayla Denton 1A

You need to split the problem into two parts so that you can use two different equations (there's no one equation for both changing volume and changing temperature). 1. heat at constant volume 2. isothermal expansion For part 1, you use delta S = nCvln(T2/T1) and you use 3/2R because volume is const...

Kayla Denton 1A

Thank you so much!!! :) We definitely will!

Kayla Denton 1A

Some last-minute midterm review!

Kayla Denton 1A

Don't omit gases. But F2 isn't present in the OVERALL reaction, that's why it's not in Ka.

Kayla Denton 1A

No you don't. And yes, omit liquids and solids! Do this for ALL Qs and Ks, not just Ka.

Edit: you want HI in the bottom because it's a reactant!

All you need to remember is that [H+] must be on the top. Then go from there.

Kayla Denton 1A

You want Ka to be EDIT: [H+][F-]/[HI] because the hydrogen concentration should be at the top for the Ka expression. Therefore, the H+ and F- must be the products so you have to write the reaction this way!

Kayla Denton 1A

This video can be summarized in a screenshot, for those of you who can't watch the video but want to have an overview. :)

Kayla Denton 1A

I have a mac computer and both the .mov and the .wmv can be watched without download (but the .mov is a lot better quality). I think most students are mac users so I chose to upload all the KemiStry videos as .mov. I'll upload them all as .wmv later on so they're more accessible to everyone :)

Kayla Denton 1A

K is the equilibrium constant, and whenever we're at equilibrium we have delta G = 0.

So delta G = 0 corresponds to equilibrium concentrations, which can be found by looking at K!

Kayla Denton 1A

W = (# of possible states)^(# of particles). You make the exponent Avogadro's number when there is 1 mole of particles involved. If there are n moles involved, the exponent becomes n x Avogadro's number. In the Boltzmann equation, S = klnW. Since you're taking the ln of W, you have lnW = ln(# of sta...

Kayla Denton 1A

It means that the system can't exchange energy with the surroundings; no heat can be added into the system or taken out of it. In other words, it is an isolated system. In an isolated system, the internal energy remains constant (first law of thermodynamics), and since temperature is a measure of th...

Kayla Denton 1A

In the course reader there's a formula sheet before the practice midterms; I expect it will be the same one on this midterm :)

Kayla Denton 1A

I think it just wants the formula for heat exchange for an isothermal, reversible change in pressure. We know that for such a process delta U = 0 and work = -nRTln(P1/P2). Thus, q = +nRTln(P1/P2).

Kayla Denton 1A

S(gas) >> S(liquid) > S(solid), always. For the same state, to compare entropy, think of the equation S = klnW (where k is Boltzmann's constant). W = (number of states)^(number of particles). With the same amount of moles, monatomic gases have more entropy because there is a greater number of partic...

Kayla Denton 1A

Oh sorry, I meant: reversible, isothermal compression. THEN temperature change at constant volume. You're totally right, I have no idea why I wrote compression at constant volume.

Kayla Denton 1A

You're right about the N2 thing; I think that was a mistake. As for the constant volume, you're separating the problem into two processes since entropy is a state function. 1. reversible, isothermal compression (edit: NOT at constant volume here) 2. temperature change at constant volume. Hope that h...

Kayla Denton 1A

I'm sorry, I didn't see this in time! I will introduce myself soon!

Kayla Denton 1A

Hi! Example 7.6 in the textbook states: Calculate the final temperature and the change in internal energy when 500. J of energy is transferred as heat to 0.900 mol O2(g) at 298 K and 1.00 atm at (a) constant volume; (b) constant pressure. Treat the gas as ideal. My question pertains to calculating t...

Kayla Denton 1A

Here's another format as requested.

Kayla Denton 1A

Of course, here you go!
We've uploaded over 10 clips; do you want me to go through and change them all to WMV?
Anyway, here's this one... enjoy!

Kayla Denton 1A

O2 is found as O=O, not O-O!

Same with other diatomic molecules. Think of the Lewis structures to determine the bonds; they won't all be double. N2 has a triple bond for example.

Kayla Denton 1A

No problem, glad I could help! :)

Kayla Denton 1A

Oh you're right, my bad! Delta U = delta H at constant volume and pressure only when we're working with solids and liquids. For gases, the gas constant changes. Sorry about that :)

Kayla Denton 1A

At constant pressure, delta H = q (heat). You can find q from q = nC(delta T). We know that n = 6.00 moles. And C, since it is a monatomic ideal gas at constant pressure, must be 5/2R, where R is the gas constant. What is the change in temperature? We must find the initial temperature and then do fi...

Kayla Denton 1A

Here is an example in which we use the Nernst equation! Hope you enjoy.

Kayla Denton 1A

1 coulomb = the amount of electrical charge in 6.24 x 10^18 electrons.
Joules = a measure of energy.
Voltage is the amount of energy (J) per unit charge (C).
1 volt is exactly 1 joule of energy done by 1 coulomb of charge (1J/C).

Kayla Denton 1A

Hi! I have a few questions about 13.15 d. The anode reaction is Cd + 2OH- --> Cd(OH)2 + 2e- The cathode reaction is 2Ni(OH)3 + 2e- --> 2Ni(OH)2 + 2OH- In the cell diagram, I understand why they have a Cd(s) electrode, but why do they have an Ni(s) electrode when Ni itself doesn't take part in the re...

Kayla Denton 1A

Hi! I understand that for a cell diagram, the anode is on the left, the cathode is on the right, and the electrodes are at the ends... but within the anode or cathode, does it matter what order you write the species in? e.g. if the anode has 2I-(aq) --> I2(s) + 2e- and the cathode has 2Ce4+ + 2e- --...

Kayla Denton 1A

From the balanced equation, Mn goes from an oxidation state of 0 to an oxidation state of +2 so it loses electrons (oxidation). Therefore, it must be the anode. Ti goes from an ox state of +2 to 0 so it gains electrons (reduction); it must be the cathode. When you're calculating Ecell, you ALWAYS lo...

Kayla Denton 1A

In a concentration cell, for a reaction to be spontaneous, I understand that [products] must be less than [reactants]. I was wondering how this translated into anode and cathode; does the anode always contain the reactants and the cathode contains the products? So should the concentration in the cat...

Kayla Denton 1A

Since lnK = nFEcell/RT, when Ecell = 0, lnK = 0. We know that ln1 = 0, so K must be 1. (Another way to find that is that if lnK = 0, e^0 = K, so 1 = K.) Conceptually: think of a galvanic cell. When Ecell = 0, there is no voltage difference so the two components of the cell have the same concentratio...

Kayla Denton 1A

Yeah this one's weird! So you've set up your table and at equilibrium you get the products CO2(g) and H2(g) as "5.00 + x" and the reactants CO(g) and H2O(g) as "10.00 - x," correct? Well, the equilibrium constant (1.00) equals (5.00 + x)(5.00 + x)/((10.00 - x)(10.00 - x)). Note t...

Kayla Denton 1A

At the beginning of my video here I explain how to assign oxidation numbers! https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=139&t=5147"onclick="window.open(this.href);return false; Also, this is more clear than the textbook—it's from my old IB Chem HL textbook. (Hope this isn't illegal to p...

Kayla Denton 1A

The way I see it is pretty simplistic—the question initially says that the pump "extracts heat from the COLD outdoors and releases it into the WARM interior." It seems to specify an initial temperature difference!

Kayla Denton 1A

In 13.5 a, why is O3 being reduced to O2 when they both have an oxidation state of zero?
I understand why Br- is oxidized to BrO3-, so is it always the case that the other two species not involved in oxidation get reduced?
Thanks!

Kayla Denton 1A

My TA told us in class today that 1 bar is approximately equal to 1 mol/L, so that's probably why the solutions manual is converting partial pressures to bar (by multiplying by 1.01325^2) and then using this value in the Q expression alongside concentration values (as now bar is essentially the same...

Kayla Denton 1A

The key is to not just look at a given reactant or product, but the DIFFERENCE in charge between reactants and products. For iodine, on the reactants side there are 3 moles of I- so a charge of -3. On the products side there is 1 mole of I3- so a charge of -1. The difference in charge is -2, so we n...

Kayla Denton 1A

Given standard conditions, they'll both yield the exact same answer so I think it's up to you which one to use!

Kayla Denton 1A

I'm not sure if this is correct but this is my best guess! So the standard cell potential (Ecell + little circle) is the voltage (potential difference) measured when all of the species are at their standard states (pure solids and liquids, gases at 1 bar, aqueous solutions at 1 mol/L). Ecell is what...

Kayla Denton 1A

In this week's episode we give some rules for assigning oxidation numbers, and do a step-by-step example of how to balance a redox reaction in an acidic solution (part 1) and a basic solution (part 2). We give two 'tricks' that will help you—one lets you check your answer to balanced half-reactions ...

Kayla Denton 1A

Hi! I believe there is an error in the solutions manual for question 10.109 part b. The vapour pressure of bromine is correctly calculated as 0.285 bar, but the textbook converts this to 0.289 atm. However, 1 bar = 0.9869... atm so the correct conversion should give 0.281 atm rather than 0.289. (For...

Kayla Denton 1A

Hi! The way I approached question 10.23 in the textbook (written below) is different than the approach in the solutions manual. I was wondering if my way was valid as well?
Thanks!

Kayla Denton 1A

You're right, the solutions manual is wrong! There are other mistakes as well, like the units for specific heat capacity of air. It's posted on Lavelle's website here: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Solution_Manual_Errors_5Ed.pdf"onclick="window.open(this.href);ret...

Kayla Denton 1A

The key is that in S = q/T, the temperature is constant. Your previous conceptual argument applies to a CHANGE in temperature. To explain S = q/T conceptually: If the (constant) temperature of the system is low, then even a tiny amount of heat added will greatly increase the entropy (due to signific...

Kayla Denton 1A

Yes! The water and metal will reach thermal equilibrium and have the same final temperature.

Kayla Denton 1A

Hi guys! Here is this week's episode of KemiStry with Kayla and Sarah. We explain how to calculate the change in entropy when both volume and temperature change.

By: Kayla Denton and Sarah Brown

Kayla Denton 1A

It's easy to visualize on the graph! If you plot the two pathways with pressure on the y axis and volume on the x axis, the work is simply the area under the graph. If you're having trouble with that I'll try to explain it in words too. So w = -P x delta V. So you want to find a volume CHANGE and th...

Kayla Denton 1A

Thank you so much!! :)

Kayla Denton 1A

Here is part two! As noted in the previous thread: At the end of the video I found the heat transferred to air (q), but there’s one more step: finding the mass of octane that needs to be burned. The “q” of the combustion of octane is –q of air, so -884 kJ. We know that delta H for the combustion of ...

Kayla Denton 1A

I'm pretty sure this file will work! It will appear small so you'll have to click the rightmost part and play it in a different screen. Here are the instructions for how to run it on google chrome. http://support.ultranet.co.nz/entries/266304-FAQ-How-do-you-play-WMV-files-in-Google-Chrome-"onclick="...

Kayla Denton 1A

Hey there! That's my video, I'll upload it as a different format and tell me if it works for you! :)

(I'm so glad someone's actually watching it, yay!)

Kayla Denton 1A

The solutions manual is not using the equation "bonds breaking - bonds forming", it's using the equation "delta H reaction = delta H formation products - delta H formation reactants"! Edit: I think the reason they don't use the equation "bonds breaking - bonds forming" ...

Kayla Denton 1A

It's the delta H of combustion divided by the number of moles of CO2 gas produced for each substance (which can be found by the balanced equation of combustion for each substance—methane, ethanol, and octane).

Kayla Denton 1A

Here is part two—entropy.

Note: we simplified the question on the board; it should really read 100. g, 1.00 atm, -10.0 degrees C and 150.0 degrees C.

Kayla Denton 1A

Yes! Bond forming is an exothermic process because it is the reverse of bond breaking, which is endothermic (as heat is required to break bonds).

Kayla Denton 1A

You don't need it! :) If you use the equation w = -nRTln(Vf/Vi), all you need to know is that the final volume is half of the original volume, so the ratio Vf/Vi is 1/2. You have n, R, and T, so you can calculate w. Hope this answers your question! The rest of the question is below just in case. w =...

Kayla Denton 1A

Yes, you're right. The number is the same due to sig fig rules: when you add or subtract two numbers, you round your answer to the least number of decimal places present in the numbers you are adding or subtracting. Since 5500 has 0 decimal places, you round your answer to the nearest whole number.

Kayla Denton 1A

In a combustion reaction, X + O2(g) --> CO2(g) + H2O(?). I was taught that H2O was gaseous (water vapour). However, in the solutions manual, H2O is written as H2O(l). This affects the answer to question 7.105, in which the change in moles of gas must be calculated for the reaction C6H6(l) + 15/2 O2(...

Kayla Denton 1A

Hi! I noticed that for question 7.99, the solutions manual says that the molar mass of CO2 is 28.01 g/mol when it is really 44.01 g/mol. They seem to have forgotten an oxygen. :)

Kayla Denton 1A

I think we just have to assume (unless otherwise stated) that a reaction occurs at room temperature, 25 degrees C or 298K.

Kayla Denton 1A

Hi! Those numbers are the enthalpy of formation, you were right! Make sure you look up Zn2+(aq), not Zn(s) (in appendix 2A of the textbook). As for HCl (aq), you don't need to find the number; just know that the enthalpies of formation cancel out. And you're right about the second part too; 800 g is...

Kayla Denton 1A

My video doesn't explain part b so I'll explain it here. You need to look up the conversion between gal to mL. 1.0 gal = 3.785 x 10^3 mL. Also, note that the molar mass of octane is 114.232 g/mol. Next: 1.0 gal x (3.785 x 10^3 mL/gal) x (0.70 g octane / mL) x (mol octane/114.232 g) x (-5471 kJ/mol [...

Kayla Denton 1A

Hi! I posted a video explaining this question here: https://lavelle.chem.ucla.edu/forum/viewtopic.php?f=76&t=4968"onclick="window.open(this.href);return false; You can just scroll through to find the answers to your questions! To quickly answer your first two questions: 277.6 K is the initial te...

Kayla Denton 1A

Here is part two to the video above!

Kayla Denton 1A

I think so, because it comes up in the homework problems (which are an indication of what's going to be on the tests).

Kayla Denton 1A

"Heat" refers to a transfer of energy due to a temperature difference. Heat flows from hot objects to cold objects. "Thermal energy" is a property of a system; it is the sum of the potential and kinetic energies of the atoms/molecules/ions in the system.

Kayla Denton 1A

Here is part 2! At the end of the video I found the heat transferred to air (q), but there’s one more step: finding the mass of octane that needs to be burned. The “q” of the combustion of octane is –q of air, so -884 kJ. We know that delta H for the combustion of octane is -5471 kJ (given in the or...

Kayla Denton 1A

Here is part 1 out of 2!

Thanks to Jasmin for filming!

Kayla Denton 1A

Hi! I am going to upload a video explaining this question.

Kayla Denton 1A

To expand on the previous answer: The "m" in Cv,m stands for "molar" because Cv,m is the heat capacity per mole (in JK^-1mol^-1) (at constant volume). Similarly, the m is Cv,m is the molar heat capacity at constant pressure. Section 7.10 in the textbook explains how to predict th...

Kayla Denton 1A

Here's a video I made explaining question 7.20 in the textbook! "A piece of metal of mass 20.0 g at 100.0°C is placed in a calorimeter containing 50.7 g of water at 22.0°C. The final temperature of the mixture is 25.7°C. What is the specific heat capacity of the metal? Assume that there is no e...

Kayla Denton 1A

I got a different answer than the textbook for self test 7.1 A. The question was, "Water expands when it freezes. How much work does 100. g of water do when it freezes at 0°C and pushes back the metal wall of a pipe that exerts an opposing pressure of 1070 atm? The densities of water and ice at...

Kayla Denton 1A

In question 11.61, the concentration of codeine's conjugate acid = concentration of OH- = 1.1 x 10^-4 mol/L. The initial concentration of codeine is 0.0073 mol/L. The % protonation is therefore 1.1 x 10^-4 / 0.0073 x 100 = 1.5%. The textbook does this calculation correctly but writes the answer as 2...

Kayla Denton 1A

When this question says "0.250 N2" does it mean to say "0.250 M N2"?
Thanks!

Kayla Denton 1A

No problem! :)

Kayla Denton 1A

You are given K! K = 4.0 :)

Kayla Denton 1A

My TA said that either one is fine, as long as you stick with one or the other!

Kayla Denton 1A

Hope this helps :)

Kayla Denton 1A

It has to do with spectator ions; I will upload a drawing in a minute explaining it!

Kayla Denton 1A

Hi! I think question 2.61 d) "Write the Lewis structure for TeCl4" has an error in the solutions manual. Te should have 10 electrons around it but it's only shown with 8. Shouldn't there be a lone pair as well?

Thanks!
-Kayla

Kayla Denton 1A

I thought that at first too! But in the question it says it is a "hydrocarbon," which means that it only consists of hydrogen and carbon.

Kayla Denton 1A

To the first part of your question, yes, L = 0 corresponds to s, L = 1 corresponds to p etc, and different 'mL's correspond to different orientations of the subshells (like px, py, pz). (I'm using capital Ls for clarity.) When it says mL = L, L-1... -L, this is kind of confusing, but the important p...

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