CHEMISTRY COMMUNITY

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.

IshanModiDis2L

I think rounding to the proper significant figures once you have gotten to the final answer is your best bet.

IshanModiDis2L

It is definitely in your favor to check the coefficients and see that it is balanced. It does not take much time but it is important as the lack of balancing can very much change the outcome of your answer.

IshanModiDis2L

I also believe that your range is pretty on par with what I found as well as it does tend to range from ~400 nm to ~700 nm. For frequencies, I found that violet light has a frequency around 7.5x10^14 Hz and red has a frequency around 4.3x10^14 Hz. I doubt you will have to memorize specific frequenci...

IshanModiDis2L

Like some have mentioned above, I do not believe that we will need to memorize the entire periodic table as that seems a bit excessive. However, it would be in your favor to memorize at least the symbols of common elements but also the weights of some elements such as the elements of life for time p...

IshanModiDis2L

I believe you have the right idea, "in excess" refers that you have more of that element to react in its lowest form. The molecule in excess is one of the reactants and it means the other reactant/molecule involved is the limiting reactant.

IshanModiDis2L

I believe that in most cases, it shouldn't really matter. I think 16.0 would be appropriate but it would also depend on the problem. You should base it off the sig figs available in the problem.

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