CHEMISTRY COMMUNITY

Salman Azfar 1K

Probably the simplest way to explain this is that the first step is so fast that you end up with overflow. Imagine I'm baking a huge number of cakes for a bunch of birthday parties. It takes me 5 minutes to prepare batter but an hour to make the cake. I'm going to end up with a ton of batter (which ...

Salman Azfar 1K

The key is that when you put them into the formula (integrated second order rate law) I believe you end up using reciprocals due to the way the law is formatted. Just remember that half life problems for second order problems usually require use of the rate law; they aren't usually as simple as addi...

Salman Azfar 1K

The organic chem we covered (I think) was more to just give us a bit of a preview of organic chemistry. That being said it's only like one or two small things to remember so it probably wouldn't hurt to glance at it.

Salman Azfar 1K

I doubt it matters. As long as the answer makes sense it would probably be accepted as an answer (though moles is used a lot more). Rate units will usually stay the same aside from small changes like converting between stuff like seconds to minutes.

Salman Azfar 1K

As the post above me pointed out, it isn't an equation. It is the derivative for concentration of A, which as you said is essentially change in concentration; however it is not over dt, so it is not with respect to time. The difference... well, that probably has to do more with derivations and integ...

Salman Azfar 1K

I don't recall ever going over that equation in class and Dr. Lavelle usually emphasizes that we understand concepts rather than memorize stuff, so I highly doubt it.

Salman Azfar 1K

I think for a lot of these thinking about products involves getting into the reverse reaction which can get a little messy so we focus on reactants.

Salman Azfar 1K

Usually a half life is a time value for how long it takes half of a substance to be converted or something along those lines. Because of that it is pretty much always notated T1/2 but often you need to use multiple half lives to get a smaller amount. Like it I wanted to know the amount of time for s...

Salman Azfar 1K

I don't have my book on me right this minute but I would just check the section headings for 15.1-6 and then look for similar headings in the homework problems. Plus look for problems similar to the example problems.

Salman Azfar 1K

Not sure about that term explicitly but I haven't heard that specific word come up to much and people have started taking the test already, so probably not a huge amount. Now if you are looking at a cell diagram where the reaction is electrolysis, you need to know it (but then you need to know every...

Salman Azfar 1K

Usually writing a cell diagram is something like this (item)|(item)||(item)|(item)
In each slot you fill in the elements of the cell setup. One side is the cathode and one is the anode; the book's examples will tell you exactly what goes where though.

Salman Azfar 1K

I would assume you just use the smallest number of given sig figs from the problem. Still, if you see one consistently, that is probably fine; either way I don't think Professor Lavelle is super strict about that as long as you understand the topic.

Salman Azfar 1K

Definitely. It was only excluded from the first test since we hadn't done much practice, but we have definitely covered it enough now and most TA and LA workshops have had problems involving those topics on practice worksheets. It is safe to assume that we need to know them.

Salman Azfar 1K

I don't think we need to understand each in extreme depth but knowing each one is probably helpful for the conceptual parts of the exam. I'd just check the book to see what types of molecules exhibit which types of energy.

Salman Azfar 1K

An isolated system means no energy or matter can flow in or out, so often that will be the case. So yea, in isolated systems w and q will both be zero and delta U is zero.

Salman Azfar 1K

The reason is that the equation we usually use is a version derived from what the Vant Hoff equation originally was; the actual scientist discovered an equation that was expanded on a bit further for most formula sheets.

Salman Azfar 1K

Gibbs free energy is determined in part by delta H. Delta H is determined by the amount of energy invested in bonds, and that energy is somewhat related to the forces between elements and molecules.

Salman Azfar 1K

I'd assume that there are more complex systems where things like temperature can affect stuff; however, with the level that we are working at I would pretty much assume an increase in moles of gas is an entropy increase.

Salman Azfar 1K

I'm pretty sure the point was to show gas expanding against a piston with a vaccum being on the other end of the piston. It was just an alternative way to show gas expansion.

Salman Azfar 1K

You may want to confirm with Dr. Lavelle but I think that the ratio depends on a number of things. For one, different ratios are used for different types of molecules (like molecules that are linear vs non-linear, etc.). There are other factors that come into play too, but that's sort of an example.

Salman Azfar 1K

With regards to what has been discussed in lecture recently, irreversible pathways are usually what is done in biological systems where the internal and external pressures differ a lot. In a reversible path the pressures are about equal and change very slowly. That's the key way that they work diffe...

Salman Azfar 1K

Just remember that forming bonds is an exothermic process (it releases energy) so delta H for bonds formed is negative. That means that breaking them is a positive delta H, so subtract formed from broken.

Salman Azfar 1K

We may need to know how to solve for them in some cases but lots of questions about phase changes usually give them to us to calculate the overall heat change. Check out the textbook problems, they should show whether or not you need them.

Salman Azfar 1K

It may depend though. If they are the only numbers we are given, we may need to default to using them for sig figs. Plus some emthalpies (I think bond enthalpies in particular) aren't always exact.

Salman Azfar 1K

Pretty much what the above post said. A large number of calorimeter reactions do involve heat transfer and will use that equation, but do keep an eye out for any extra factors (like maybe it leads into a question about enthalpy).

Salman Azfar 1K

Not sure about your calculations, but make sure to look at it logically. You are forming bonds, so you release energy. Energy is lost, so your answer will be negative.

Salman Azfar 1K

One method is to say, evaporate water in an open beaker. Basically some of the substance leaves the system.

Salman Azfar 1K

A quick and easy way to put it is that the temperature is technically changing but tech is being used to keep it constant anyway.

Salman Azfar 1K

If I had to wager a guess, interactions between molecules are determined by which molecules are present, and nothing changes there. However, their arrangement and ability to move freely does change.

Salman Azfar 1K

I don't really think that there is much of a connection. I'm guessing you're thinking that PV=nRT so if everything is constant increasing temperature could increase volume, etc. However, most of what we work with (if I had to guess) will either be in a rigid container or we simply won't care much ab...

Salman Azfar 1K

Correct me if I'm wrong but is it not just a way of saying acid-base reaction at equilibrium? Since the reaction is basically just proton transfer.

Salman Azfar 1K

I don't really recall Dr. Lavelle ever saying that we needed to memorize biological examples (I assume you mean coordination compounds). However, it would be nice if he could confirm this in class; I would ask him just to be sure. The one thing that will definitely be good to know though is which li...

Salman Azfar 1K

I don't know about the individual smaller gas laws but at the very least, it may not hurt to be familiar with the ideal gas law since it encompasses the smaller ones. Plus, it can be used to switch between Kc and Kp; not sure if that may be on the test, but it could be useful.

Salman Azfar 1K

I am not sure as to whether or not we will need to know how to do this, but basically you need to use the ideal gas law. If given pressure to get Kc, you need to put pressure into the ideal gas law and calculate the unknown values (most likely moles). Then use those moles to get concentrations and c...

Salman Azfar 1K

Alternatively, if the entire equation has only gases, you can just leave them all as pressures and get Kp instead.

Salman Azfar 1K

With regards to what you asked about the equation changing, it literally will not change at all. You pretty much just switch the sides the reactants and products are on and that's it, and as mentioned above, the new equilibrium constant is just the inverse of the old one.

Salman Azfar 1K

It's also fairly important to consider that this isn't necessarily a trend. It's not like elements grow less reactive moving to the right towards the p-block or anything; in fact, the halogens are often very reactive as they very badly want to gain that one electron to fill an octet.

Salman Azfar 1K

It implies what side of the double bond something is located on. Two things are on the same side for cis and opposite for trans. It doesn't directly define where the dipoles are but the location of the atoms around the double bond affects the dipole directions so it is quite important.

Salman Azfar 1K

The key is the lone pair. Because one of the electrons is a lone pair, the shape will be a seesaw regardless of what VSEPR predicts for electron geometry.

Salman Azfar 1K

A good thing to remember is that F has the highest electronegativity. That means the stuff past it on the right (Noble gases) has lower values, and Noble gases have their octets and actually really don't want electrons.

Salman Azfar 1K

I think with regards to where the electrons are added, usually only one or two of the atoms in the structure (if I had to guess) will actually be able to handle an expanded octet. We have not really covered any extremely complex structures, though if you are personally interested in the topic Dr. La...

Salman Azfar 1K

So while there are ways to measure this sort of thing, we have not really covered those in class. Generally, the most important information will be the trends across the periodic table. Other than that, I personally find that it helps to know that Fluorine with its value of 4 is the most electronega...

Salman Azfar 1K

Like the answer above said, I'm pretty sure changing them up would be okay. At the same time however, I think we are trying to imply viewing things in three dimensions in which case the most logical variables to use are x, y, and z as you would for a 3 dimensional set of coordinates.

Salman Azfar 1K

So the shell is the quantum number n, the general energy level. The subshell consists of the levels within that quantum number, so within n=2 you have the s orbitals and p orbitals corresponding to l=0 and l=1. The ones I'm not entirely sure about are the orbitals, but here is my best answer: orbita...

Salman Azfar 1K

With regards to delta X and delta P, my TA gave us a little info. Basically, the more resources you put into knowing an electron's position, the less you know about its velocity and vice versa. So you can get pretty accurate results on one but adversely affect what you know about the other. That's w...

Salman Azfar 1K

For the purposes of this topic, it generally means that the energy levels are going to be whole numbers n=1, n=2 etc. You cannot send in a photon with enough energy to move an electron to say, n=2.5 or else the photon will pass through rather than being absorbed by a sample.

Salman Azfar 1K

It's essentially what the reply before me said. Amplitude has its purposes but I've looked through the entire reading and I haven't seen any mention of it in equations unless I missed something. At the very least, it doesn't impact our current calculations with wavelength and frequency with respect ...

Salman Azfar 1K

I haven't seen any specific guidelines but more significant figures will usually lead to the more accurate answer, so I wouldn't really recommend using 3.0 unless you're in a rush or something.

Salman Azfar 1K

Okay, here's a general list of steps to follow. Convert the given values from grams of the various compounds to grams of H, C, and N. Use those gram values to subtract for the mass of O. Once you have that, take all the gram values from above and convert to moles. Divide by the smallest one to get t...

Salman Azfar 1K

This one can be tricky. You need to use the initial molarity and volume values they give you to calculate the number of moles, and from that, use Avogadro's number to get the number of atoms/molecules. Once you have that, critical thinking comes in. Keep in mind that when you double the volume, you ...

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