CHEMISTRY COMMUNITY

julia_lok_2K

Equations involving Ea are: lnk=-(Ea/RT) + lnA, or k=Ae^(-Ea/RT). You get Ea from the slope of the line (-Ea/R) on the graph plotting 1/T vs lnk.

julia_lok_2K

If standard cell potential is an intensive property, does that mean standard Gibbs free energy is also intensive for a reaction?

julia_lok_2K

It would be bimolecular because two molecules of O3 are involved.

julia_lok_2K

You can find the number of moles of the reactant used up in the reaction from the final concentration of the product by using the stoichiometry coefficients. You then subtract that value from the initial concentration of the reactant to get the final concentration.

Hope this helps!

julia_lok_2K

So it's standard for the units to be in mol/L and not in mmol/L? I thought it didn't matter, as long as you were consistent with your units.

julia_lok_2K

Yes, you would. To make things simple, you can assume that the concentration of SO2Cl2 is 100 M, so the final concentration would be 10 M, thus ln(10). Therefore, with this same principle, if SO2Cl2 concentration were to decrease to 20% its original concentration, you would use ln(20). Generally, I ...

julia_lok_2K

A half life is the time it takes for a substance to fall/decay/be consumed to half its original concentration. Half lives do not depend on the concentration, i.e. the half life of a substance remains the same at any stage in the reaction. We use the t1/2=0.693/k equation to relate half life to the r...

julia_lok_2K

For the rate constant, I keep getting 2.85 instead of 2.85 * 10^14. I don't understand how they got the exponent. Do I have to convert mmol to mol, or do I have to use data from experiment 4, like how the solution manual did it? Am I just making a dumb mistake??

julia_lok_2K

Because K= Ka*Kb, so you need to square K to get only Ka.

julia_lok_2K

Why does the electrode with the higher concentration of the species undergoes reduction and not the electrode with the lower concentration?

julia_lok_2K

The equation from the solutions manual is a simplified version of the original formula, the 0.02568 is what you get when you combine all the constants in the original formula. You can directly plug the numbers into the original formula. I'm a bit confused by exactly what units you're referring to, b...

julia_lok_2K

Will we be given a list of reduction potentials for the test?

julia_lok_2K

Since there is only one reactant in the reaction, would Br2 be involved in both half reactions? Would Br2 also act as both the oxidizing and reducing agent?

julia_lok_2K

I don't think so, since it wasn't covered in lecture. Looking at the material covered in his lectures, it seems like the test covers only up to Section 14.6 (6th edition).

julia_lok_2K

You add the inert conductor when the reactants or products don't involve a conducting solid.

julia_lok_2K

It's because the steam has more heat energy than boiling water, i.e. higher enthalpy.

julia_lok_2K

There wouldn't be any change after time passes because it's isolated--it won't be able to exchange energy or matter with its surroundings.

julia_lok_2K

I don't think you have to. On the constants and equations sheet, they give the conversion between atm and kPa, which is essentially the same conversion between atm L and joules.

julia_lok_2K

It's correct. I guess for this problem you would disregard mass because it's asking for heat capacity and not specific heat capacity.

julia_lok_2K

Why does a change carried out in reversible reactions always does more work than a change carried out in irreversible reactions?

julia_lok_2K

In the solution manual, it says that Ka2 can be ignored. But Ka2 is larger than Ka1/1000, so wouldn't you have to do another ICE table using Ka2?

julia_lok_2K

I don't really understand how the coefficient for compound A is 2. I assumed that since C has twice the pressure of B, the coefficient for C would be 2 and for B, 1. But A isn't twice the pressure of either B or C, so why is the coefficient 2 as well?

julia_lok_2K

In the textbook, it says that if Ka2 is less than Ka1/1000, then you can ignore the second deprotonation.

Hope this helps!

julia_lok_2K

A buffer is a solution that undergoes only slight changes in pH when acids and/or bases are added. I think that you just have to know how to solve problems involving buffers, like the last problem that Professor Lavelle went over during lecture last Friday.

julia_lok_2K

It's because the concentrations of solids and liquids do not change during the reactions, so it is kind of pointless to consider solids and liquids for equilibrium.

julia_lok_2K

Based on the balanced reaction, if the volume decreases and there are more moles of gas on the left, the reaction will shift to the right. If there are more moles of gas on the right, the reaction will then shift to the left. This happens because the side with more moles of gas would have a larger p...

julia_lok_2K

You would put x because you don't know the change in concentration for each molecule. Since the initial concentrations of the products are 0, the system would shift to the right, i.e. make more products, in order to achieve equilibrium, thus you put +x. Conversely, you would know that the concentrat...

julia_lok_2K

The principle quantum number is 5, so the possible l values are 0,1,2,3,4. The total possible ml values for each subshell, therefore, would be 1, 3, 5, 7, and 9, respectively, adding up to a total of 25 orbitals. However, since it involves the g subshell, I don't think you need to know this informat...

julia_lok_2K

I would be safe and just memorize the formulas. You can refer to the "Naming Coordination Compounds" worksheet on the website to help you know which ligands to memorize.

julia_lok_2K

I'm not sure if it will be the same from previous tests since he makes slight changes to the curriculum every quarter, but he probably will use the same one that is posted on the website.

julia_lok_2K

Since we didn't have any homework problems on it but still went over it in lecture, we will probably be tested on at least the concept of second ionization energy.

julia_lok_2K

Cl and CN would be first by alphabetical order.

julia_lok_2K

Would carbonato also be considered a polydentate then?

julia_lok_2K

If you look at Ka = ([A-][H+])/[AH], a higher Ka value denotes higher concentrations of [A-] and [H+], which means the acid has become more ionized/dissociated. Therefore, a higher Ka value makes a stronger acid.

julia_lok_2K

Why is water a monodentate, even though it has 2 lone pairs of electrons? Wouldn't water also possibly be a bidentate?

julia_lok_2K

Ligands are atoms, ions, or molecules that are attached to a central metal atom in a coordinate compound. They typically have at least one lone pair of electrons to form a coordinate covalent bond with the central atom; thus the ligand acts as a Lewis base while the central atom acts as a Lewis acid...

julia_lok_2K

Also, atoms with dipole moments typically result in a polar covalent bond, unless the atoms in a molecule are arranged in a way where the dipole moments cancel out, such as CCl4, in which case the molecule becomes nonpolar.

julia_lok_2K

You would first find the element with that ground state configuration, which would be As. Then, since you are looking for the ion of +1 charge, that means the ion lost an electron, so you would look at the element that is to the right of As, which is Se. Therefore, the answer is Se+.

Hope this helps!

julia_lok_2K

I think it's because instead of four bonding pairs in a tetrahedral, you have three bonding pairs and a lone pair of electrons in a trigonal pyramidal molecule. That lone pair of electrons wants to take up more space, thereby pushing down on the molecule. This causes the bond angle to be less than t...

julia_lok_2K

VSEPR model explains how to draw the 3D shape of molecules. It stands for Valence Shell Electron Pairs Repel, which summarizes how to draw them--you want to draw the models in a way that will minimize electron repulsion. From the VSEPR model, you can determine the molecular shape and its name. I hop...

julia_lok_2K

Sorry for the typos, but I meant to say Valence Shell Electron Pairs Repel.

julia_lok_2K

It stands for Valence Electron Shell Pairs Reel--this name conceptualizes how to draw the 3D model of molecules. It denotes that you want to draw the model in a way that will minimize electron repulsion.

Hope this helps!

julia_lok_2K

I think it complies with the octet rule, as in taking away electrons or adding more electrons until you get the previous or next noble gas configuration. So to find the minimum oxidation number, you take away electrons from the atom's outer shell until you reach the preceding noble gas. In this case...

julia_lok_2K

O is double bonded with Xe because that way, the formal charges for both will be zero, making the structure more stable. If O and Xe had a single bond, then O would have a formal charge of -1, and Xe would have a charge of +1, which is less favorable.

Hope this helps!

julia_lok_2K

I don't think the textbook explains this exception, and I tried looking it up on other websites as well but they didn't really explain it either, and Lavelle didn't give a reason too. So my guess is that you only have to memorize this exception and won't have to explain this trend on the exam. Sorry...

julia_lok_2K

Honestly I never remember which one to use, but how I think of it is whichever order will get you a positive value. And if it's asking for energy of an electron going from a higher level to a lower level, I just get the positive value first then stick the negative sign in front of it, or say "t...

julia_lok_2K

I know there's the exception for electron configurations where if the d orbital is almost half full or almost full, one of the electrons from the outermost s orbital will go to that d orbital to make it half full or full. I know that Cr, Cu, and Ag are exceptions, so wouldn't tungsten also be an exc...

julia_lok_2K

One of the UAs said that all you have to know is that the solutions you get from Schrodinger's equation are the quantum numbers. And an orbital is a wave function that relates those quantum numbers, specifically n, l, and ml, i.e. denotes the locations of electrons.

Hopefully this helps!

julia_lok_2K

So far, we have learned about the trends for atomic size, ionization energy, and electron affinity. 1. Atomic size: atomic radii increase as you go down a group because additional shells are farther from the nucleus atomic radii decrease as you go across a period because while more electrons are add...

julia_lok_2K

I think a nodule plane is the area where the electron probability density is zero.

I don't know if I answered your question well, but I hope this helps!

julia_lok_2K

I think this is because a smaller amount of energy is emitted when an electron goes from n=4 to n=2 compared to going from n=5 to n=1 since the energy difference is smaller. This energy can be represented by E=hv, from which the equation E=(ch)/wavelength, or wavelength=(ch)/E, can be derived. From ...

julia_lok_2K

I would definitely show work on the test. In addition to getting partial points, it also decreases your chances of making minor mistakes instead of doing the work in your head.

julia_lok_2K

I agree, homework problems definitely help with solidifying your understanding of the concepts that will appear on the test. In fact, I think that the quiz was easier than the homework, so if you did all the homework then you should be fine. And it's definitely possible to finish earlier--some peopl...

julia_lok_2K

I suppose so. Can you give a problem as an example?

julia_lok_2K

I didn't see this in the list of homework problems? Anyway, how I did it was that since they give you the mass percent of M, you can figure out the mass percent of O by subtracting 88.8 from 100 to get 11.2. You can then set this number out of 100g to get 11.2 g, then find the number of moles of O, ...

julia_lok_2K

So you know how many moles of copper(II) chloride tetrahydrate you have. You multiply that number by the molar ratio of water and the molecule to get the number of moles of water. In this case, since there are 4 molecules of water in in one molecule of copper(II) chloride tetrahydrate, you would mul...

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