CHEMISTRY COMMUNITY

IshanModiDis2L

The units go as follows

For Zero-order reaction: M/s
For First-order reaction: 1/s
FOr Second-order reaction: 1/(M*s) or 1*M*s)^-1

IshanModiDis2L

Yes it was best to print out the sheet, as the modified equations did come in handy!

IshanModiDis2L

In lecture, Ea is used to denote activation energy. I find the pun made above very amusing however ^

IshanModiDis2L

E naught is 0 in a concentration cell and at equilibrium

IshanModiDis2L

I think we just had to know, like most things what was covered on the outlines and in lecture.

IshanModiDis2L

There are no units assigned to degeneracy.

IshanModiDis2L

We use this when we have Irreversibile expansion, or when the volume change is occurring under constant pressure.

IshanModiDis2L

If it is zero order, the rate of the reaction is independent of the concentration.

IshanModiDis2L

To restate, the value of Q can be calculated at any point in any given reaction. However, K can only be calculated when the reaction is at equilibrium.

IshanModiDis2L

Basically , a salt bridge helps prevent an excess build up of charge and helps the cell work better.

IshanModiDis2L

A catalyst only lowers the activation energy required. It can speed up the forward reaction by lowering the activation energy, but it would equally speed up the reverse with them ending up cancelling each other out.

IshanModiDis2L

I believe so, as even though an isolated system means can enter or escape , changes can still be made to the system.

IshanModiDis2L

Like mentioned above, the coefficients determine how much of an x value is added or subtracted on each side of the reaction. I would suggest to make sure the reaction is balanced however.

IshanModiDis2L

The irreversible pathway allows for the expansion of gas against a consistent external pressure but a reversible process means that the external must match the internal pressure as mentioned above.

IshanModiDis2L

I feel like I am super late here but I really like the album Malibu by Anderson .Paak. Definitely think its a solid listen all the way through, especially when doing chemistry homework.

IshanModiDis2L

I was confused about this too! Through doing research, I believe you are correct as a bomb calorimeter is an isolated system and thus matter and energy cannot leave the system. I believe this ensures that it is in fact a perfect system.

IshanModiDis2L

A salt bridge basically just allows the two solutions to remain neutral, allowing for the separation and prevention of excessive charge. The process basically ensures the electron transfer from the anode to the cathode to go successful.

IshanModiDis2L

the anode is generally on the left and is losing electrons mwhile the cathode is on the right and gains electrons. This is why they mention that electrons flow from the left hand side to the right hand side.

IshanModiDis2L

I would suggest to just look at the original information given in the problem. Generally, phases are usually the same unless otherwise specified.

IshanModiDis2L

When work is done on the system, the work is a positive value. When the system itself does work, the work is negative.

IshanModiDis2L

I believe that to our knowledge, temperature is the only factor. The way it changes is based on the type of reaction as K increases in an endothermic reaction if the temperature is increased and decreases if its the opposite in exothermic.

IshanModiDis2L

Capital K is the Boltzmann constant and it is roughly 1.38x10^-23 J/K. However, it is on the equation sheet so I believe we are not expected to memorize it.

IshanModiDis2L

Hi! I am currently in Chem 14BL with Casey and currently the workload is not too bad! The workload is manageable as its really only 2 assignments per week but really only one if your group manages to stay on task. I think it definitely could be doable but just remember to really communicate well wi...

IshanModiDis2L

Water molecules have hydrogen bonds between them which provides strong attraction between molecules but when water is boiled, the energy from the heat allows the bonds to break/weaken.

IshanModiDis2L

The state function does not depend on the path taken to the final value or state, only the starting and ending states.

IshanModiDis2L

Residual entropy does not consider movement of molecules due to temperature, more so it is the possible positioning of the atoms like mentioned.

IshanModiDis2L

As mentioned above, I agree that the integral was mostly used to show how the equation was derived conceptually. The equation sheet will be more accessible and useful when it comes to a test.

IshanModiDis2L

Is it updated weekly or is it based off the correct amount at the end of the quarter? I remember last quarter it would be updated towards the end since we would just need to have 50 by the end. Does anyone know if this still applies or should I just communicate with my TA.

IshanModiDis2L

I just was wondering why my chem community points are less than 25 because I have done over the required amount of posts. Is there still a chance TAs are still grading or would it be on an individual basis?

IshanModiDis2L

If it is on both sides, we can cancel them out . However, I implore you to make sure that it is the same specific heat constant to ensure your calculations are correct.

IshanModiDis2L

I too did not do well, or rather as well as I would have wanted on MT 1. But here we are a week later, and I can say that my workhabits have slowly changed for the better and I can see positive change. For me , I really feel better when I am ahead so I try to continue to stay that way but UA session...

IshanModiDis2L

I believe we should have 25 posts by the end of tonight. Is it true however, that the points are not in yet or does that vary on the TA. I was curious if we could make up posts like we could last quarter.

IshanModiDis2L

extensive properties are affected/affect the mass whereas intensive properties are independent of the mass and do not change it.

IshanModiDis2L

Like mentioned above bomb calorimeters are really only used to measure the heat of combustion(exothermic) of a certain reaction, in a constrained environment of constant volume. It also measures the heat energy released when food is burned.

IshanModiDis2L

I think next time I am gonna continue what I did well on the preparation, which for me personally was outlining the textbook with the key concepts and sections mentioned in the outline and also continue with textbook problems. Also , UA sessions are amazing for the extra practice and clarification t...

IshanModiDis2L

the entropy is the measure of disorder of a system, and the entropy of the universe is always increasing.

IshanModiDis2L

Yes they will always act this way; exothermic reactions will always be losing heat, meaning that the delta h must be negative and endothermic reactions will always be gaining heat with a positive delta h

IshanModiDis2L

I also think that CO2 should be classified as a linear molecule due to the lack of unpaired electrons.

IshanModiDis2L

Like mentioned above, I agree that Hess's Law problems is pretty much like a puzzle and you just have to try and put the info together. What I like to do first personally is get all the individual components on their correct sides , starting with the ones that only appear once which makes it less co...

IshanModiDis2L

Though I didn't get the best score either, I agree with confidence that the best place to start is the syllabus textbook problems. They are really helpful and similar to the exam questions. What helps me personally is going into the textbook and outlining the sections Lavelle recommends to read ( I ...

IshanModiDis2L

I don't think so but I may bring it up to a couple UAs and respond later with more info

IshanModiDis2L

When water vapor condenses the delta h is negative, so heat is lost. This means it is exothermic as the bonds are formed.

IshanModiDis2L

I think knowing open, closed, and isolated is fundamental for a variety of reason but especially when we are dealing with changes in heat and work. For example, isolated systems do no work done on the system.

IshanModiDis2L

we do not consider solids and liquids when we work with pressure. However, like mentioned previously, there is a net increase in moles of gas, which would do work on its surroundings.

IshanModiDis2L

Steam is also known as water vapor and since condensation forms bonds, energy is released and given off as heat.

IshanModiDis2L

I really feel this one, my confidence was pretty high since the 14A midterms were honestly pretty good but I feel as though this past one was much harder, kind of like how the final was last quarter. I am just hoping the next midterm and the final can help our grade and we'll be good standing in the...

IshanModiDis2L

Hi! I used to also have this problem on exams, especially for this class. It does not help that we're also taking it online which can lead to additional stress. I think that the way I am really able to overcome this is to really go through my initial run of the exam and really try and answer the que...

IshanModiDis2L

I would say you are somewhat right. A vacuum can be an isolated system, but an isolated system does not necessarily have to be in a vacuum for it to be isolated, it just cannot interact with its surroundings.

IshanModiDis2L

Hess's Law is pretty much says that however many state changes in a given reaction may occur, the total enthalpy change is calculated through the summation of all changes , with the proper direction and molar coeffients.

IshanModiDis2L

Like mentioned above, I believe you are correct. Heat is only considered to be a state function if it is under constant pressure which it is here.

IshanModiDis2L

I am also taking it online, and I still think it is possible to get a good grade. The LAs and TAs are very helpful with answering any questions one may have.

IshanModiDis2L

Hi! I am currently in Chem 14BL with Casey and currently the workload is not too bad! The workload is manageable as its really only 2 assignments per week but really only one if your group manages to stay on task. I think it definitely could be doable but just remember to really communicate well wit...

IshanModiDis2L

When work is done by a system, the system is using energy, meaning that energy is being released. This makes work negative. When work is done to a system, the system is gaining energy, so work is positive.

IshanModiDis2L

Yes, it can if a piston compresses or expands the volume. A closed system only means that no matter can exchange with the surroundings.

IshanModiDis2L

I assume we do not have to memorize molar entropies because like as mentioned above, they should be given to us.

IshanModiDis2L

HF is not a strong acid because the bond between H and F is too strong. Essentially the bond length between H and F is relatively short, making it harder for the bond to break when placed into a solution. This makes the dissociation harder and overall lowers the Ka.

IshanModiDis2L

I would agree with the others, monoprotic just means the acid has one donatable proton to give off whereas polyprotic acids can give off multiple.

IshanModiDis2L

I believe that the most stable structure will have the lowest formal charge, as it tends to be most stable. If it does have a charge, it tends to be that the most electronegative atom holds the most negative charge.

IshanModiDis2L

I believe charge takes precedence and then size but it may be situational as well.

IshanModiDis2L

I don't think there is something like a curve in this class but we never know due to it being a multiple choice final and what I can assume is Dr. Lavelle's first final of this sort.

IshanModiDis2L

I believe the charge usually will come after the ion but I do not think it rly does matter too much. Only time I would really consider it is when drawing a structure.

IshanModiDis2L

I think practice calms my nerves down but breaks definitely are essential too.

IshanModiDis2L

Strong acids are characterized by their ability to dissociate completely into H+ ions. F is so electronegative with a short bond length to the Hydrogen so it will not want to fully dissociate that easily.

IshanModiDis2L

Like mentioned above, The -dentate state represents how many lone pair of electrons it can give or how many bonds can be formed with a central atom. For instance, polydentates are ligands that can form more than one bond with the same central atom, whether it be two bonds (bidentate), three bonds (t...

IshanModiDis2L

Yes, like the posts above mentioned, the dipoles wont ever cancel out so there will always be an unequal sharing of electrons as the lone pairs cause an asymmetrical shape where the dipoles still exist.

IshanModiDis2L

I think it just refers to the fact that they are on the same plane/surface and an even number of carbon atoms leads to coplanar hydrogen atoms as the C-C bonds are parallel to each other

IshanModiDis2L

I believe you are correct. A bonding region and a region with a lone pair are considered regions of electron density. Having more than one bond does not change the amount of electron density.

IshanModiDis2L

Like mentioned before, I expect that we should be able to take ions from their verbally expressed name and transform that to a chemical formula, with balancing and such considered as well. I do not think it should be a major topic of focus since it is not on the list of topics on the outline but the...

IshanModiDis2L

Hydrogen bonding is a consequence of a hydrogen atom bonding to a very electronegative atom specifically F, N, O. Hydrogen is able to form a hydrogen bond with these atoms because the difference in electronegativity is large enough unlike it is with carbon. The small difference between the hydrogen ...

IshanModiDis2L

I would say that when considering if a bond is polar or non-polar, I would compare the electronegativity of the two atoms as the more electronegative atom is more likely to pull the shared pair. Id also consider the shape of the molecule because polar bonds arising from differing electronegativities...

IshanModiDis2L

I don't think that we are expected to know or calculate the exact angles unless its a generally specific bond angle. I asked my TA and he agreed that looking at a VESPR chart would be helpful to determine the approximate size of the angle and that we may not to know some of them but I checked with h...

IshanModiDis2L

If you do not see anything stating otherwise, it is safe to assume the system is at equilibrium.

IshanModiDis2L

If the formal charge is not zero, it is in your favor to include the charge regardless if it is positive or negative.

IshanModiDis2L

Bent and angular are the same thing and produce the same shape so I think we will not have to worry too much about it for the midterm.

IshanModiDis2L

It has 3 sig figs since the decimal is after the 0 in 350 and if there were not a decimal then 350 would only have two sig figs.

IshanModiDis2L

When determining the most stable structure, we must consider the individual charges of each element and choose the one with the least formal charges with the most electronegative element having the most negative formal charge and the less electronegative atoms to have positive formal charges.

IshanModiDis2L

Hydrogen Bonding only works for certain elements (Nitrogen, Oxygen, Fluorine) because those elements have a strong/noticeable amount of electronegativity difference to develop attraction and thus form hydrogen bonds. Hydrocarbons would only happen in certain instances as described above.

IshanModiDis2L

Polar molecules have unequal charge distribution due to an unequal sharing of electrons in a bond that creates a net dipole. Nonpolar molecules have no net dipole/unequal sharing of electrons due to an equal sharing of electrons and thus creates symmetry.

IshanModiDis2L

1/2 and -1/2 denotes whether or not the electron spins clockwise or counterclockwise or rather the orientation of the spin momentum of the electron in an orbital. When electrons are paired up in the subshell orbitals, they have opposing spin directions, which is also important in terms of bonding be...

IshanModiDis2L

The formal charges with the lowest energy/energy that is closest to zero would be more stable. Also, the most electronegative element will be more stable with a negative charge.

IshanModiDis2L

I would agree with what was said above, the way to tell that they are exceptions is that if there is no way for the middle atom to achieve a full octet. You do not need to memorize all of them, I would say its more favorable if you recognize them instead.

IshanModiDis2L

When you are configuring an electron configuration, you would write it in the order of lowest to highest energy level/shell number.

IshanModiDis2L

Yes you definitely do because you need to use both the number of lone pair electrons and the number of bonded electrons when considering the formal charge on the element.

IshanModiDis2L

Hi, as mentioned above the spin state is the fourth quantum number and it refers to the orientation/specific spin an electron in an orbital may have. If an electron spin state is positive 1/2, it means the electron is spinning facing up or clockwise on its axis whereas if the spin state is -1/2, the...

IshanModiDis2L

I believe it depends on what the question is asking but usually if it does not ask specifically in reference to the nearest noble gas, I would just write it all the way out with the entire configuration provided by you.

IshanModiDis2L

You would first start off by using the equation c= wavelength*frequency. you convert 2.4 μm to meters using unit analysis into meters and then solve for frequency. Once calculated, you would then plug your frequency value into E=HV, where h is represented by Plank's constant and V is the frequency a...

IshanModiDis2L

Empirical formulas do need to be whole numbers as you cannot have a fraction of an atom. That is why we multiply to get a whole number ratio (ex. 1.5x2=3) but it is key to make sure that it is in its lowest whole number ratio, otherwise it will be considered a molecular formula.

IshanModiDis2L

For periodic trends, I find it easy and useful to remember a few key things. As you go across a period horizontally, your ionization energy and electron affinity increase but your atomic radius decreases. As you go down a group, the atomic radius increases but electron affinity and ionization energy...

IshanModiDis2L

I agree with the person who just commented above me, I believe in a previous lecture Dr. Lavelle mentioned that it is best to not really give much attention to sig figs until after you are done with your final calculation. However, you should just round your numbers to a proper decimal to ensure per...

IshanModiDis2L

C is a constant that will not change and it is 3.00x^8

IshanModiDis2L

It should be okay, just make it clear what you are using and display it well on your webcam.

IshanModiDis2L

That should be correct but I would also suggest using the units to check your work when doing it.

IshanModiDis2L

I would prefer you think about it as initial energy level and final energy level rather than just mathematically as light operates at discrete energy levels. It is basically n(final)- n(initial) and n2 should be the initial shell where it begins and n1 where it ends up.

IshanModiDis2L

To think of its simply, one mole of a something is equal to 6.02x10^23 units of that specific thing. It could be cars, bananas, footballs, etc, It really does not matter. A mole of something is basically the avagadro's number of that thing and that's why it can be universally used for any atom or su...

IshanModiDis2L

I believe there are many ways to do this but the best way to do this is to do the least common element first because it will set you up the best when you do have to do the tricky balancing at the end usually with Hydrogen or Oxygen atoms. There is however one real answer and you wanna make sure to a...

IshanModiDis2L

Hello! Ok so I talked to my TA about this and he said that we do not necessarily have to memorize them as Dr. Lavelle will provide them to us but it is in our favor to be comfortable with the names in case of time constraints for example.

IshanModiDis2L

With your calculations, energy is measured in Joules (J). However, it can also be written as kg*(m^2/s^2) so it helps to be aware of that as well.

IshanModiDis2L

From what I can understand, it depends on the question and if it does not specify then it comes to your judgement and knowledge of chemistry on what would look more appropriate.

IshanModiDis2L

I do the pre and post assessments which help me quite a bit, and really just try to get ahead of the topics of what Dr. Lavelle will be teaching us in the next week. I find great use in the textbook problems and all the additional office hours resources as well so I would definitely recommend those ...

IshanModiDis2L

I do not believe so, the homework got pushed back and all but it may be helpful to do review and the modules as well.

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