CHEMISTRY COMMUNITY

Dan M -3E

Dr. Lavelle also posted review questions on his website under "Week 10 Review" that are helpful.

Dan M -3E

Solid.

Dan M -3E

By knowing the change in one of the reactants/products over the reaction, we can use the stoichiometric ratio to calculate the change for the others.

Dan M -3E

I think that way it works out is that each individual test isn't curved, but once the class average is known at the end of the quarter, the grades are curved to reflect that average. But, included in that curve/scaling is that if someone gets at least 250 points out of the 500, they are guaranteed a...

Dan M -3E

Yeah, I've had this problem in the past, too. Unfortunately I can't offer a solution, mine just usually loads when I give it enough time. Any solutions would be helpful.

Dan M -3E

To balance the overall reaction, you have to separate into the redox half reactions, and use H+/OH- and H2O to balance those. After doing that, you just add the two reactions together and get the overall balanced reaction.

Dan M -3E

For a basic reaction, just analyze it and use OH- and H2O on either side and make sure that in the end, it is balanced.

Dan M -3E

Galvanic cells use spontaneous reactions. Electrolytic cells are non-spontaneous, meaning that energy is needed to force the reaction to happen.

Dan M -3E

If the two molecules are in the same phase, you use a comma. But, if the molecules are in different phases, then it's a vertical line, the same as between the different reacting molecules.

Dan M -3E

The biggest thing I try to think that helps to stay calm during tests like the final is just to convince myself right before the test that if I come across a problem that I don't immediately know how to solve, that I am smart enough to solve any problem that could be on the test. We are all smart en...

Dan M -3E

In my opinion, the textbook does a better job explaining what we need to do for the hw problems. I like learning from lecture better, but because the hw problems are in the textbook, they're obviously gonna be based off of the wording and specific equations that are in the textbook readings.

Dan M -3E

You use h2o and it's ions to balance the equations, but in acid you use H+, and in bases you use oh-

Dan M -3E

Yeah I think you're right that we have to take the most stable one / the one that we see most often and use it in this case.

Dan M -3E

They can be on either side cuz you just add them to make sure that everything is balanced

Dan M -3E

Since essentially acid and base reactions are opposite, it would be the other part of the h2o molecule, which is the oh- instead of the h+ in acid solutions.

Dan M -3E

to make sure that the atoms are balanced on each side.

Dan M -3E

This equation takes into account the deltaG(naught) and adds/subtracts what would be different because of the conditions not being standard.

Dan M -3E

I think that the derivations are pretty much all listed on the equation sheet, so we'd only need to be able to know how to use them.

Dan M -3E

If deltaG is equal to 0, than the reaction is at equilibrium, and not proceeding in either direction.

Dan M -3E

Something being a state function means that the pathway by which the value arrived at its final value does not matter in the calculation of the value.

Dan M -3E

A "half reaction" is when you split up the redox reaction into the oxidation reaction and the reduction reaction. So, instead of the electrons being transferred from one atom/molecule to another, we can see just the way the electrons are changed in a single one of the reaction molecules.

Dan M -3E

Gibbs Free Energy is the most accurate way of measuring whether a reaction will be spontaneous. If it's negative, the reaction is spontaneous. If it's positive, the reaction isn't spontaneous.

Dan M -3E

At equilibrium, the energy coming in and being released by work are equal. Basically, the net work being done on and by the system is 0 because the energy of the system is at eq.

Dan M -3E

The temperature is considered "constant" for reversible actions, but not because it doesn't change at all. It's constant because the change is so slow that at an instantaneous moment, it is not really changing. Plus, the system temp and the surrounding temp stay the same as each other.

Dan M -3E

Thanks to everyone that helped out with questions that people were asking here. It was a big help for reviewing for the midterm.

Dan M -3E

Can someone pls explain how to find heat and work in question #6? I get that deltaU and deltaS are 0 because they are state functions, but how/what equations do I use to find work and heat?

Dan M -3E

Energy of a isolated system is always conserved (1st law), so even in an isolated system as big as the entire universe, the overall change in energy is 0

Dan M -3E

Gas always spontaneously expands when put into a larger container, and never contracts into a space smaller than all available.

Dan M -3E

I think that what would decide this is whether separating the HCN molecule into its different elements will occur spontaneously, which is based on whether the reaction is exothermic or endothermic. Because breaking bonds requires energy, I don't think this reaction would occur by itself.

Dan M -3E

If you define a system, and it loses energy to its surroundings, which you don't count as part of the system, then the energy WITHIN the system isn't conserved, but the overall energy of the system+surroundings is still constant.

Dan M -3E

They both mean that the reaction will occur without energy coming from external sources being necessary. I think, as the people said above, there's been a shift toward using the word favorable, because spontaneous comes with the connotation that it will happen quickly, when in reality it just means ...

Dan M -3E

You can also get the amount of work thru the graphs, because the area under the pressure-volume graph is equal to the work. So, when volume is constant, the graph is a vertical line, and there is no area underneath and therefore no work done. When pressure is constant, the area underneath is a squar...

Dan M -3E

Basically what they said above, the volume of a closed system can change when its container expands or contracts, but the amount of mass will not change.

Dan M -3E

An open system can gain/lose energy from the environment. A closed system will not lose energy between the internal parts of the system, but its surroundings can add/take energy from the entire system. An isolated system does not gain or lose energy from the surroundings.

Dan M -3E

On a closed system, you can add energy by adding it to the entire system, which is different from an open system where you can pick where to add work/energy, or an isolated system where you can't add energy, I think.

Dan M -3E

I'd probably say the drop-in sections are better if you have specific questions about the material, but if it's more than you want a review of an entire topic, the 2-hour workshops have been good in my opinion.

Dan M -3E

Use ICE when you are given K and need to find the equilibrium concentrations of the molecules.

Dan M -3E

It's the non-h30/oh molecule of the product divided by the non-water molecule of the reactants.

Dan M -3E

Yeah, it being less than 5% just means u can neglect it the x that is subtracted when calculating the Kc using x

Dan M -3E

Would we use this if given a volume of gas and not the number of moles?

Dan M -3E

You can only use pvnrt for gases, so in the calculations, use the moles of gas.

Dan M -3E

Inert gases, usually just the noble gases including helium, don't react with other substances. Because of that, we can add it to a solution and the volume will increase without altering the reaction.

Dan M -3E

R= 8.314 J/mol*K if the units we're using is atms, which is usually is.

Dan M -3E

I think the most important way we can use this formula in this class is to convert between partial pressure and concentration when we need to. Other than that it is used more for physics than chemistry.

Dan M -3E

P is partial pressure, n/V is concentration because it's the number of moles divided by the volume of the solution, so using PV=nRT, you can convert between the two.

Dan M -3E

There's a longer explanation that I'm sure is in the textbook, but like was said above, when pressure increases, the reaction will shift to the side with less moles.

Dan M -3E

You can use PV=nRT to convert between partial pressure and concentration for a gas so that you make sure all the units are the same.

Dan M -3E

K is the equilibrium constant. Q is the same calculation as K, but when the reaction isn't at equilibrium. From the value of Q, you can tell if the reaction is making more products or reactants.

Dan M -3E

Also make sure you look at the syllabus and which chapters of the textbook are in each topic that we've learned about so you can make sure you're studying the right things.

Dan M -3E

The main inert gases are the ones in the far-right column of the periodic table because they have full octets. (He too because it only takes 2 electrons and has both)

Dan M -3E

I think you should prob be find doing some of them as equilibrium but should do some as the next topic too. Last quarter in 14a my TA gave credit as long as the 5 problems were about pretty recent topics, and being that it'll be week 2, I feel like it would be ok.

Dan M -3E

I doubt Lavelle will give us a problem where the pH is negative, but if that's what the calculation comes out to, than yeah it means that the molarity of hydrogen ions >1

Dan M -3E

Calculate in the Ka and Kb aren't hard though if he gives us the concentrations of the reactants and products. Other than that, I dont think we'll have to calculate it because its in 14b

Dan M -3E

For strong acids, you just calculate pH, not Ka or Kb.

Ka is equal to the concentration of the products divided by the concentration of the reactants, so unless you are given those, you won't have to calculate it. Then, finding pKa is the negative log of it. pKa + pKb = 14

Dan M -3E

An acid accepts an electron pair, a base donates an electron pair.

Dan M -3E

There are only a few amphoteric compounds, so it probably wouldn't be hard to memorize them just in case, or at least be familiar with them

Dan M -3E

Pretty much, the more ionic character a molecule has (EN difference, bond length), the stronger acid it is

Dan M -3E

The only way to calculate the Ka or Kb would be if you are given the concentrations of both ions and the molecule in the water. The equation is that Ka=[products]/[reactants], or the multiplied concentrations of the ions divided by the concentration of the full molecule. That gives you the amount th...

Dan M -3E

Also, if they give you the Ka, then you know it's a weak acid/base because strong acids/bases ionize completely

Dan M -3E

It don't think they would keep going after the formal charge reaches 0 because then the molecule is becoming less stable than it was when the FC was 0, and molecules try to stay as stable as possible.

Dan M -3E

I don't think it matters cuz its clear what you mean by both.

Dan M -3E

There is also a bigger electronegativity difference in the HO bond, which means it is has more ionic character and is therefore stronger.

Dan M -3E

I think that we might have to test the formal charges of the differently drawn lewis structures to find our wether its a chain or a ring, but hopefully we won't be given one of those, or there will be parts in parenthesis so that we can tell.

Dan M -3E

It wasn't on that test but I'd guess we might have to know it for the final.

Dan M -3E

We don't determine bond lengths ourselves, we are given the lengths because they have been experimentally found from repeated experiments of different molecules involving the same atom.

Dan M -3E

Rod shaped molecules interact more easily because of the increased surface area, and also because when they line up next to each other, their nucleus's and electrons can get closer together because in the short side of the rod, the outside of the molecule is closer to the nucleus's.

Dan M -3E

The bonds in ions are always ionically stronger than the induced dipole bond, because it all depends on the electronegativity difference in between the atoms. When they reach a certain point, it's considered an induced dipole, and at an increased point in EN difference, it is called an ion.

Dan M -3E

When we talk about the intermolecular forces, it's just between the different molecules, in between atoms of the same molecule is different and has different affects on the structure of molecules and substances.

Dan M -3E

Yeah, when two atoms that have an electronegativity difference, a dipole moment occurs between those two atoms.

Dan M -3E

I think that, in general, covalent bonds are considered stronger than ionic bonds.

Dan M -3E

Yes, they are essentially opposites, so they increase in opposite directions.

Dan M -3E

Yeah this was a tricky question on the midterm but because of the extra (4th) electron in the 2p subshell, oxygen has a slightly lower IE than nitrogen.

Dan M -3E

I think for a single atom that has 5 electrons, it would still be considered nonpolar because the electrons move around the nucleus and end up being equally distributed, on average, around the because they move so fast around it.

Dan M -3E

In case you can't read the above responses, Lavelle said we'll get the midterms back on wednesday in lecture.

Dan M -3E

oxygen has the higher electronegativity.

Dan M -3E

And ml can be any number between -l and l, so it's range would depend on the value of l, but if l=11, than ml can be anything btwn -11 and 11.

Dan M -3E

It's based off of which subshell the electron is in. For s, there is only one orbital. In p, there are three degenerate orbitals (px,py,pz). In d there are 5 and in f there are 7.

Dan M -3E

Shells are whether its 1/2/3/etc, the subshells is the s/p/d/f, and the orbitals are more specific for the subshells because all except for s have multiple trajectories that the electrons travel through. So for example, you could classify an electron as being in 1px, which covers all three levels- 1...

Dan M -3E

As long as the units are the same on each side, it doesn't really matter because you can always convert btwn the different ^10 levels. The only trouble is for units like Joules where its specifically kg, and as long as you know that (or the entirety of the joules unit breakdown), you're fine.

Dan M -3E

You just have to memorize the periodic trends, with IE increasing from left to right and from top to bottom. So, Helium will have the highest ionization energy, meaning it is the least willing to lend an electron, while the bottom left atom will have the lowest, meaning it is the most willing to giv...

Dan M -3E

They are essentially opposites, being that electronegativity pulls on other atom's electrons vs. effective nuclear charge being the nucleus pulling in that atoms own electrons.

Dan M -3E

They are essentially opposites, being that electronegativity pulls on other atom's electrons vs. effective nuclear charge being the nucleus pulling in that atoms own electrons.

Dan M -3E

Yeah, the size of the atoms is different because of the periodic trends so the orbital location would be different even tho the shape is the same because its in the 1s orbital.

Dan M -3E

Yes, they will have the same wavelength because all the variables in the equation are the same between protons and nuetrons if they are moving at the same velocity.

Dan M -3E

Yup did the same thing and got the same answer as you guys.

Dan M -3E

You add an electron with each atom going left to right, and if you memorize how many electrons go in each orbital, its easy to tell when you add a new level of orbital. Going from top to bottom, you just add a full set of orbitals

Dan M -3E

Yeah its something that we just memorize because were in the basic level of chemistry, but I think that its based off the number of workable answers to the psi function for each orbital.

Dan M -3E

For all but two of the atoms in that area and row of the periodic table (silver and one other in the first row), 4s just happens to be the lower energy orbital so its filled first. When both are full 3d has lower energy, but with empty orbitals, 4s is a lower energy state than 3d.

Dan M -3E

Lavelle said in class that it is arbitrary which axis is which, because like you said it depends on the point of view. We put it in the x first out of convenience and its standard, but no there isn't a defined x, y, and z axis.

Dan M -3E

Not totally sure, but I think that the probability is uniform because of the energy of the electrons going around the orbital, and at that point the pull from the nucleus doesn't make a difference in the location of the electron.

Dan M -3E

The first one (a) is asking about the wavelength of the electron, and I think that (c) is asking about the wavelength of the photon that caused the electron to be ejected

Dan M -3E

What they said, no each photon interacts with just one electron with the excess energy being released, not applied to other electrons

Dan M -3E

If you go to your own profile it shows you how many times you've posted on here

Dan M -3E

It's always the same.

Dan M -3E

By that it means the proportion of the proton's wavelength to the neutron's. As in divide the proton's by the neutrons

Dan M -3E

Does anyone know if we have to know the english names for chemical formulas on the test? (ex ethene) .

Dan M -3E

It's probably not essential, but it can't hurt to have them on there in case. Its a little bit of an inconvenience but it doesn't make the problem any harder

Dan M -3E

Yes, I think you guys are correct that we will get the important formulas given to us on the test so we don't have to memorize all of them, but it prob helps to know at least the most important ones just in case.

Dan M -3E

Do we have to work with sig figs on the test? Because in discussion we just skipped over them because they're kinda unnecessary, but we learned them so and theyre in the textbook

Dan M -3E

There are technically prefixes for every different level, but I think we usually work in every three because it's a lot easier to remember. Having a number be to three places or two decimals isnt the end of the world, so I think that is easier than memorizing every single place.

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