CHEMISTRY COMMUNITY

Gary Qiao 1D

shouse1f wrote:I know Lavelle mentioned the steady state approach but only went over the pre-equilibrium, so is the pre-equilibrium approach the only one we are expected to know?

Yep, he said that we are only required to know the pre-equilibrium one.

Gary Qiao 1D

The rate constants of the forward and reverse are not always necessarily equal. Go off of the formula K=k/k'. Since K, in that case, is equal to one, we can say 1=k/k', which is the same as k=k'. However, the equilibrium constant K is not always 1.

Gary Qiao 1D

I think adiabatic means something has no heat transfer while isothermic means that there is no change in temperature, but in the overall scheme of things they both refer to q=0.

Gary Qiao 1D

I'm guessing that the question is kind of talking about how relative to the equilibrium concentration, when there is more initial concentration, then the reaction tends to favor the products in order to reach the equilibrium concentrations. So there probably is a case where the product is smaller th...

Gary Qiao 1D

As long as the units are consistent within the problem, any units of atm or bars is alright.

Gary Qiao 1D

Actually, I am not too sure because I put deltaG>0 for photosynthesis also, but it seems like they originally marked me off, but then crossed it out and gave me the points. I would talk with the TA for clarification.

Gary Qiao 1D

Cu was probably the reducing agent, meaning that the reaction occurred in the anode (since the other reaction must've had a larger Ecell value). This would mean that oxidation occurs in the anode, making the reaction the reverse reaction.

Gary Qiao 1D

I usually did balance out the half reactions in order to determine the n. If there are any other ways, I would be glad to know too!

Gary Qiao 1D

The equations should be the same because cathodes are actually always on the right, and the anodes are always on the left.

Gary Qiao 1D

The electrons from the anode are negatively charged, meaning that they want to flow towards the positive charge, which is at the cathode

Gary Qiao 1D

Yes, I do believe so based on Faraday's Laws

Gary Qiao 1D

That is because in the equation ΔG = - nFE, a positive cell potential is associated with a negative Gibbs free energy, which means a reaction is spontaneous. A negative cell potential would make the Gibbs free energy positive, meaning that the reaction is no longer spontaneous.

Gary Qiao 1D

You would use OH- when balancing a basic solution. I would refer to toolbox 6K.1 and example 6K.2 in the 7th edition for a better guideline on the use of OH-.

Gary Qiao 1D

(1)(219.2)(163)+(6)(29.4)(163) only gives you part of the answer as you need to heat it up from 37 to 200, then use the DeltaHrxn then cool it down from 200 to 37. DeltaH1=(1)(219.2)(163)+(6)(29.4)(163) DeltaH2=-2756kJ DeltaH3=(6)(37.1)(-163)+(6)(75.3)(-163) DeltaH1+DeltaH2+DeltaH3 Should give you ...

Gary Qiao 1D

When the gas is expanding in a vacuum, there is nothing to oppose the gas (free expansion), making the work = 0. In a vacuum, there is also no change in temperature or heat change, making delta U = 0, thus q= 0. delta S in free expansion would still be equal to nRln(V2/V1).

Gary Qiao 1D

I'm guessing that because the problem doesn't say that the pressure is independent of the temperature, the change in temperature affects a change in the pressure of the system.

Gary Qiao 1D

I think statistical and residual are the same except that for residual entropy, the system does not account for thermal entropy, meaning T=0. I think the large error in W means that there is a large error in determining the number of states possible? This part I am not too sure.

Gary Qiao 1D

The book is a little inconsistent with their variables, but if you look at the example following the formula, they do use ΔS = nCln(T2/T1). So I'm just guessing they combine the n with C in the book, but both ultimately do use ΔS = nCln(T2/T1).

Gary Qiao 1D

The change in internal energy is ΔU = q + w. The problem gives you q as q = -318 kJ, so we have to find w, or the work done in by the system due to the reaction. w could be found by -PΔV, which means ΔU = q + -PΔV. Because they did not give the volume or pressure, we could convert -PΔV to -ΔnRT beca...

Gary Qiao 1D

Adding on, a reversible and isothermal path should be at equilibrium.

Gary Qiao 1D

I guess that heat capacity will just give you quick information about the energy needed to raise a specific amount of substance, but I do agree that specific heat capacity is more useful in that it could just calculate the heat capacity of that amount of a substance anyway.

Gary Qiao 1D

Extensive means that the properties can change based on the amount of substance (so conditions such as mass and volume), while intensive means that the properties are independent of the amount of substance (so mass and volume would not change an intensive value).

Gary Qiao 1D

The new concentrations can be determined by using PV = nRT. If V is halved, so P(V/2) = nRT, making P = (2n/V)RT, the concentrations are doubled, making the new concentrations to be 3.7 M H2, 2.72 M N2, and 5.82 x10^-3 M NH3.

Gary Qiao 1D

It does not really matter since the units will cancel out anyways, and using both bar or atm will get the same Kp value.

Gary Qiao 1D

Protonation is the adding of a proton to an atom. Percent protonated is the percent of the atoms that added a proton to an atom in a reaction since for weaker acids and bases, not all atoms protonate.

Gary Qiao 1D

If you plug in the larger positive x value, you will find that the equilibrium concentrations of the reactants are negative, and that is not possible, meaning that value can be disregarded. Therefore, there would be only one valid x value.

Gary Qiao 1D

if you do not look at the conclusion the solution manual gives, calculating the equilibrium concentration for N2 and O2 would be .114-X, X being equal to 1.8x10^-4. .114-1.8x^10-4 = .1138, which is essentially .114 mol/L.

Gary Qiao 1D

Given a pKa value, you can convert the pKa to Ka using pKa = -logKa. Because Ka = [H+][A-]/[HA], you could find [H+] .
-log[H+] would give the pH.

Gary Qiao 1D

Both essentially mean there are more products than reactants, but I'm assuming "lie to the right" refers to when the reaction has not yet reached equilibrium yet, while "equilibrium shifted to the right" has the equation already at equilibrium. that makes sense! But what does it...

Gary Qiao 1D

Yes, generally when the equilibrium constant deals with an equation with only gases in the formula K= [Producrs]/[Reactants], we use partial pressure as the concentration, labeled as "P". There really isn't a difference in the calculations when writing "P" and without it, but if ...

Gary Qiao 1D

Both essentially mean there are more products than reactants, but I'm assuming "lie to the right" refers to when the reaction has not yet reached equilibrium yet, while "equilibrium shifted to the right" has the equation already at equilibrium.

Gary Qiao 1D

If I interpreted the question correctly, the stoichiometric coefficients don't change in an equation unless if the equation was not balanced in the first place. If "coefficients" refers to the differing concentrations, then yes the ratio of concentrations between products and reactants wou...

Gary Qiao 1D

HClO4 is still neutral because we can look at the formal charges in the Lewis structure to determine the charge of HClO4. The Cl atom would have three double bonds with three oxygen atoms and one single bond with OH, causing the formal charge of Cl to be 0 (7-(0+14/2)). All the oxygens will also hav...

Gary Qiao 1D

I think some strong acids to know are in group 17 ( HCl, HBr, HI) and nitric acid, (HNO3), sulfuric acid (H2SO4), perchloric acid (HCLO4), and chloric acid (HClO3).

Gary Qiao 1D

A neutral ligand would be a molecule that attaches to a central metallic atom, examples being H2O, CO, NH3, etc.

Gary Qiao 1D

The Lewis acid will be the compound that is the electron acceptor while the Lewis base would be the compound that is the electron donor. So in something Like Na+ + Cl- -> NaCl, Na+ will be the Lewis acid and Cl- would be the Lewis base.

Gary Qiao 1D

The bond angles are less than 90 and 120 degrees.

Gary Qiao 1D

Yep, all molecules should have LDF!

Gary Qiao 1D

The bond angles are the following:
See-saw: less than 90 and 120 degrees
T-Shape: less than 90 degrees
Square Pyramidal: less than 90 degrees
Square Planar: 90 degrees

Gary Qiao 1D

I think it is because carbon is on the second row of the periodic table, making it 2.

Gary Qiao 1D

How are delta bonds formed?

Gary Qiao 1D

I am guessing since the lone pairs are not part of a sigma or a pi bond, the 2 is not included in the notation. There might not be a need for the distinguishment for the second row when there are no bonds with lone pairs.

Gary Qiao 1D

I do not think so, pi bonds are formed only after a sigma bond has formed.

Gary Qiao 1D

Hey Nicolette,
I think there actually are 3 lone pairs on the Cl instead of 2, or else the total valence electrons would not match up

Gary Qiao 1D

London forces occur in all molecules (non-polar and polar), while dipole-dipole forces occur only in polar molecules. Dipole-dipole is the result of an attraction of a positive charge on one side of a molecule to the negative charge on another molecule. London forces occur in all molecules because e...

Gary Qiao 1D

Yeah, I think the terms are interchangeable as both point to the shape of the molecule, so both molecular shape and shape do not exclude lone pairs

Gary Qiao 1D

We have kind of gone over the general VSEPR concept, but I do not think we have looked over all the shapes and their bond angles in class yet. For 2E.13, I would just find the number of bonds and lone pairs and then determine the shape and bond angle with that information through figure 2E.1 or any ...

Gary Qiao 1D

Generally, anything below 10^-18 m in wavelength is undetectable making that the shortest.

Gary Qiao 1D

The nitrogen cannot have an expanded octet, thus it cannot form five bonds. There are no available d-orbitals for nitrogen to have that expanded octet for two double bonds with the oxygen and one single bond.

Gary Qiao 1D

I think Dr. Lavelle meant for the central atom of sulfur to have a formal charge of 0, but the overall formal charge to stay at 2-. For sulfur here, with 6 valence electrons, in the formula FC = V - (l + s/2), FC = 6-(0 +12/2), FC=0.

Gary Qiao 1D

You only count the s and p block electrons. You do not count the d block electrons because the d block would not be part of the outermost shell. For instance, with gallium, [3d104s24p1], the s and p are in the same larger shell than d (4 compared to 3). For transition metals, groups 3-11, they are a...

Gary Qiao 1D

Valence electrons refer to the outermost electrons in the outermost shells, so usually when an atom has a d orbital, you would just count the electrons in the s and p shells. If you are referring to how many electrons can be in each orbital of the d shell, then yes there can be one or two electrons ...

Gary Qiao 1D

The h constant would be given on the test

Gary Qiao 1D

The h in the heisenberg equation actually has a slash through the h, which symbolizes h/2π. This means that the "h" variable, the planck constant, 6.63 x 10^-34 Js is the same in both the wavelength = hp and heisenberg equation.

Gary Qiao 1D

I'm sure that the value 1.22 x10^-7 is correct, but you have to also be aware that the negative that you got from the calculations earlier comes from the hydrogen atom emitting energy. This means that when the problem asks for the wavelength, the wavelength must have a positive value because a negat...

Gary Qiao 1D

I think it depends on the problem at hand, most likely the problem would state if you need to use the weight of a neutron or proton or a car. However, the weight of a neutron and proton are very similar and almost essentially the same at 1.67 x 10^-27 kg.

Gary Qiao 1D

HI, I don't think so, the professor said that we do not have to go over it.

Gary Qiao 1D

I think anything generally shorter than 10^-18 m for wavelength

Gary Qiao 1D

Hi, the Rydberg Equation is just a derivation of ΔEn = Efinal - Einitial, with En = -hR/n^2. In other words, it is just the simplified form to find the change in energy from one energy level to another for the frequency or 1/λ. The professor wanted to show how this formula was derived, but I think i...

Gary Qiao 1D

I'm not a hundred percent sure, but I think it would be best to look over the names of the polyatomic ions again and refresh your memory since on the tests there might be word problems where only the chemical name is given and you would have to derive the chemical formula yourself. It doesn't hurt t...

Gary Qiao 1D

So first, you have the balanced equation: Cu(NO3)2 + 2NaOH --> Cu(OH)2 + 2NaNO3 The next step after this is to find the moles of the reactants in order to find the limiting reactant. So for Cu(NO3)2: n/L =M n/.08L = .500M n = .04 moles Cu(NO3)2 And for NaOH: molar mass: 22.99g/mol Na + 16 g/mol O + ...

Gary Qiao 1D

What I usually do is first make sure that the chemical equation is balanced, and then I try to convert the grams of the reactants to moles (unless the problem already gives the number of moles of each reactant). Then, I would take these mole values and divide them by the stoichiometric coefficients ...

Gary Qiao 1D

BrF_3 is the excess because although there are 12 moles of ClO_2 present to react with only 5 moles of BrF_3, the chemical equation (6 ClO_2(g) + 2 BrF_3(l) --> 6 ClO_2F(s) + Br_2(l)) shows that 6 moles of ClO_2 would react with 2 of the moles of BrF_3 for one reaction. After two reactions, 12 moles...

Gary Qiao 1D

I think the solutions manual skips finding the mass percentage of each element in this problem because the mass percentage, in this case, would only be needed to find the mass of each element and the problem already gives the mass of phosphorus (4.14g) and chlorine (27.8-4.14g). With this, you can l...

Gary Qiao 1D

Yea usually I would just look it up since we have not really gone over determining the states of the atom in class. The states of molecules will generally be gas (g), liquid (l), solid (s), or aqueous (aq) (when the molecule is dissolved in water). Often times the word problem may give hints to what...

Gary Qiao 1D

The coefficients in a chemical equation are pretty much the stoichiometric coefficients, which means that the coefficients really represent the ratio between the substances in a reaction, so it is not necessarily always the exact molar amount. Expanding to your hydrogen gas example, for the equation...

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