CHEMISTRY COMMUNITY

Ashley Davis 1I

Thank you! Was very confused by this question but this was helpful.

Ashley Davis 1I

Update: I saw there's a chart with values for A depending on the reaction. For this reaction (CO2 + OH- → HCO3-), A would be 1.5 x 10^10 L/mol*s. But even then, ln(1.5 x 10^10) does not equal -0.59. So why is -0.59 included in the solutions manual?

Ashley Davis 1I

Not sure what you mean? You don't need A at all in this problem. Given that ln(k 2 /k 1 )=E a /R * (1/T 1 - 1/T 2 ), you just need to plug in values. Additionally, A is just the pre-exponential constant in the Arrhenius equation, so you could find it. What do you mean we could find it? How? How wou...

Ashley Davis 1I

Oh, okay, I was reading it as "k prime." Thank you.

Ashley Davis 1I

Could someone please clarify why the reaction constant for a reverse reaction is k' (k prime)?

Ashley Davis 1I

My guess is that it's okay to write it in the second step rate law as [HOBr] because that step is not rate-limiting (it's not the slower step). Thus the intermediate [HOBr] is still not in the overall rate law.

Ashley Davis 1I

So because this is a Galvanic cell, we know that we want our Eº(cell) to be positive. We know that we're going to have to reverse one of the reduction half-reactions so that we have one reduction and one oxidation. Since our Eº(cell) needs to be positive, we'll need to make the larger of the two sta...

Ashley Davis 1I

There might be an intermediate step for that last equation, like in the example from lecture on Friday with NO2 + CO → NO + CO2.

Ashley Davis 1I

It's a form of the Arrhenius equation. From the textbook: The Arrhenius equation is used to predict the value of a rate constant at one temperature from its value at another temperature. You would take the standard Arrhenius equation for one temperature and the same for the next temperature, then su...

Ashley Davis 1I

Just to understand what the soln manual is trying to say, how would we determine that increasing the concentration of B by the ratio 3.02/1.25 increases the rate by (3.02/1.25)^2? I'm not sure how we would figure this given how we've gone about solving these kind of problems so far. I personally pre...

Ashley Davis 1I

Do the terms "first-order" and "second-order" only apply to the reaction as a whole and not to the reactants individually? Like if there were two reactants both with coefficients of 1, would we never call the rxn a 2nd-order rxn with two 1st-order reactants?

Ashley Davis 1I

I think Dr. Lavelle will go into this on Friday, or at least in coming lectures, since we haven't worked on any of it yet.

Ashley Davis 1I

The solution manual lists instead of an answer for (c), another answer for (b). And it's solution is +3.3 x 10^-3 instead of -3.3 x 10^-3. Is this another error? Or could someone explain to me why this solution would be positive and not negative, if the formula for unique rate would be -1/a ΔA/Δt.

Ashley Davis 1I

How come the anode rxn's Eº = +1.23 V and not -1.23 V? In the book the reaction is written O2 + 4H+ + 4e- → 2H2O with Eº = +1.23 V. Since they've switched the reaction here, why isn't it -1.23?

Ashley Davis 1I

Still two, it's just that Cl is both the reducing and oxidizing agent.

Ashley Davis 1I

Could someone explain why for 14.13 part c Au^+(aq) \rightarrow Au(s) + Au^{3+}(aq) that one of the half-reactions has Au3+ as a reactant and Au(s) as a product when they are both on the same side in the original equation given? I thought the two half reactions would be Au+(a...

Ashley Davis 1I

Oh, sorry, I actually meant for ΔGº = RTlnK. because then where would the moles cancel out for the R constant?

Ashley Davis 1I

Thank you. Would we have to deduce on our own what kind of work is occurring or would it most likely say in the question what kind of expansion it is?

Ashley Davis 1I

So when using difference of sums to calculate something such as enthalpy, we multiplied enthalpies of formation by however many moles of each molecule were present in the equation. Like if there were 2 moles of CO _2 , we would enter 2 mol(-394 kJ/mol), and the mol would cancel out, leaving us with ...

Ashley Davis 1I

I'm having trouble determining when to use which equation. Could someone summarize when to use which equation to calculate work and internal energy? I know that for work it can be calculated with w = -PΔV or w = -nRT(V_2/V_1) so in what instances would we use these, i.e. how would we know us...

Ashley Davis 1I

It affects the way you balance the equations. For a reaction in an acidic soln, you would balance the O's in the half-reactions using H2O and then balance the H's using H+ ions. For a reaction in a basic soln, you would balance the O's in the half-reactions using H2O and then, to balance the H's, ad...

Ashley Davis 1I

Yeah, basically it's reversed when it comes to the agents: The oxidizing agent is the species that is being reduced. The reducing agent is the species that is being oxidized.

Ashley Davis 1I

For this one, I'm confused how we determine that C was oxidized from -2 to -1? Would we have to draw the Lewis structure of the molecule and calculate formal charge of each atom? I ask because while calculating formal charge of the carbons, I got 0 for both, so how do we know C originally had a char...

Ashley Davis 1I

You don't need the units atm at all because the atm in the ln part of the equation cancel out. For example ln(4 atm/2 atm) is just ln 2, not ln 2 atm. You want your answer to be in J/K. If you were to use 0.0826 L*atm/K*mol, you would have a leftover L and atm in your answer, and no J.

Ashley Davis 1I

Could someone explain why for this problem we assume n = 1.00 mol? Is it because the problem says to assume this is happening under ideal conditions? And as a role of thumb, what do ideal conditions mean for a reaction?

Ashley Davis 1I

Could someone explain to me why to find change in entropy we would calculate negative q/T as opposed to positive q/T for this problem? I'm also confused because even though the calculation is –(100. J.s^-1)/(293 K), the solutions manual still has a positive number as the solution. Why would it be po...

Ashley Davis 1I

yes, in this case ΔS would be inversely proportional to T.

Ashley Davis 1I

for the second part of your question, i believe w = -PΔV when work is done BY the system and w = +PΔV when work is done ON the system.

Ashley Davis 1I

Do all phase changes require energy? Or just the "backward" ones like condensation, solidification, etc?

Ashley Davis 1I

For this one, I thought we only had to determine how to heat up the water, but it looks like we need to determine the same for the copper kettle itself. Will we always need to do this whenever it is a question concerning heating up something in a container? For example, would we need to do the same ...

Ashley Davis 1I

I'm confused about work in general for a couple reasons. Why is it that when a system does work, its work value is negative? Also during condensation, how is it that work is done ON the system rather than by it?

Ashley Davis 1I

I think it's just the standard temperature because that's the temperature of most rooms. Like it's the typical room temperature.

Ashley Davis 1I

The answer is positive because you can't have negative values for density. Like an object like TNT can't be -1 g/mL, just as you can't have negative grams of something.

Ashley Davis 1I

I think it depends on your TA! I know tests have to be written in pen.

Ashley Davis 1I

Could someone explain the difference between intensive and extensive values/properties? We went over this briefly in my discussion but I was a little lost in how it pertained to our course...

Ashley Davis 1I

Could someone explain why we know for sure that the concentration of OH- will be twice the amount of Ba(OH)2? Like I get that it will be twice an amount because of the coefficient of 2, but why do we assume it's twice the concentration of Ba(OH)2?

Ashley Davis 1I

i think right now we're allowed to do both, but if it's strong, i would just put the forward one, and if it's weak, do the double arrows.

Ashley Davis 1I

As far as I know, the orbitals being hybridized has no effect on bond angles. For example a tetrahedral with sp3 hybridization still has 109.5 bond angles. I wouldn't think of hybridization as a separate entity from molecular shape. The shape is the same. Hybridization just goes into WHY it is the s...

Ashley Davis 1I

That would be my guess as to why it's like that, since it has more than one bonding site.

Ashley Davis 1I

1) Can we use something like cyano- and cyanido- interchangeably? I've seen both used in the solutions manual.

2) For the test, are we expected to know the charge of various polyatomic ions like sulfate, oxalate, etc.?

Ashley Davis 1I

If the coordination compound has an overall charge like -1 or +1, it would then be an ion. For example, hexacyanoferrate(II) ion is an ionic compound because it has a charge of -4. But a compound like potassium hexacyanidochromate(III) is a neutral compound: three potassiums with a charge of -1 boun...

Ashley Davis 1I

I think for the ligands we're meant to know what charge they have? I'm not sure if we need to memorize polyatomic ions... And as far as I know I don't know an easier way to compute it :/

Ashley Davis 1I

There's a difference between molecular geometry and electron geometry. Lone pairs are included in electron geometry but not in molecular geometry. For example for something like NH3, the electron geometry is tetrahedral (including the lone pair) and the molecular geometry (how the molecule itself wi...

Ashley Davis 1I

Could someone explain to me why 1) you wouldn't arrange CH2Cl2 with H's and Cl's opposite to each other and 2) it's a polar molecule? I thought it might be nonpolar if you arrange them symmetrically...

Ashley Davis 1I

No it's not trigonal planar cuz there's only 2 things bound to the central atom, rather than three. It would be linear.

Ashley Davis 1I

I think it's just drawn this way because that's what it says white phosphorous is, four phosphorouses each connected to three others. I wouldn't worry too much about it.

Ashley Davis 1I

Could someone explain how to do this problem. Like how would we know which Period 3 element it would be...

Ashley Davis 1I

Could someone please explain how to do this problem? I'm confused on how to determine the compound formula when it comes to ions with 2+ and 3- charge, and vice versa.

Ashley Davis 1I

If this is true then why, for elements like tungsten, would it be 4f14 5d4 6s2 instead of 4f14 5s2 5p6 5d4 6s2?

Ashley Davis 1I

For anyone who has done question #2.43 on the homework: Why would tungsten's configuration be [Xe]4f^14 5d^4 6s^2 instead of [Xe]4f^14 5s^2 5p^6 5p^5 5d^4 6s^2?

Ashley Davis 1I

Could someone help me with 1.39? My units don't cancel out properly. I thought J = kg * m^2 * / s^2? But in my equation I have one kg on top, two meters, and only one second. This is what I entered:

wavelength = (6.626 x 10^-34 kg • m^2 • s^-2) / (0.15 kg)(41 m)

Ashley Davis 1I

I get that the photoelectric effect proves light is a particle, but how exactly does it prove that? Like how does it prove that it is a particle?

Ashley Davis 1I

will we be given the mass of an e- and planck's constant in a problem or will we have to know them by heart?

Ashley Davis 1I

Could someone explain to me how units cancel out in equations like E = hv and E = 1/2mv^2? Do I even need to worry about it?

Ashley Davis 1I

We've basically been calculating theoretical yields rather than actual yields this whole time, so we would only know the actual yield if it were given to us in the problem. And then you would take the actual yield that is given and divide that over the theoretical yield calculated times 100 to find...

Ashley Davis 1I

We've basically been calculating theoretical yields rather than actual yields this whole time, so we would only know the actual yield if it were given to us in the problem.

Ashley Davis 1I

Add the coefficients of the reactants together and then subtract the sum of the coefficients of the products from that. For example, if the reaction were 2K + 10W --> 5L + 4R, then the net moles produced would be 12 - 9 = 3 net moles.

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