CHEMISTRY COMMUNITY

Sapna Ramappa 1J

Ice melting will occur spontaneously without any input of energy, so you can assume the reaction is exergonic/spontaneous, and thus delta G will be negative.

Sapna Ramappa 1J

If you're looking for increasing oxidizing power, you need to see which elements have a higher chance of being reduced. In other words, you'd see which values have the highest standard reduction potential and are the most positive. If you're looking for increasing reducing power, you need to see whi...

Sapna Ramappa 1J

Try to look at the situations given and determine whether or not it'll happen without an input of energy.

Sapna Ramappa 1J

Spontaneity is associated with the delta G values. If you are given delta H and you want to find if something is spontaneous or not, plug in values for T and delta S.

Sapna Ramappa 1J

When work is done on a system, the work has a positive value. When work is done by a system, the work has a negative value.

Sapna Ramappa 1J

It might be helpful to categorize equations based on what circumstances they're used in. Aside from that, it's helpful to see how you can derive useful equations from equations given on the constants/equations sheet.

Sapna Ramappa 1J

I always think about it in regards to endothermic reactions. Increasing temperature will always favor endothermic reaction, and thus decreasing temperature will favor the exothermic reaction. Make sure you pay attention to the delta H, and then you can figure out the effect of a temperature change.

Sapna Ramappa 1J

Introducing noble gases will not change the equilibrium constant.

Sapna Ramappa 1J

The partial pressures of the gases will be in atm or bar, but the Kp will not have units, as it is the equilibrium constant.

Sapna Ramappa 1J

Because the equilibrium constant is large, the products will be more stable than the reactants.

Sapna Ramappa 1J

Yes, Dr. Lavelle told us to not use approximations on the midterm, so you'd just do the ICE tables keeping x in the equations.

Sapna Ramappa 1J

Yes! The equilibrium constant, k, does not have units.

Sapna Ramappa 1J

Don't worry too much about the difference between bar and atm -- the numbers would be relatively similar, so it won't matter too much.

Sapna Ramappa 1J

Just know that strong acids are 100% dissociated, and be comfortable with equilibrium questions regarding weak acids/bases.

Sapna Ramappa 1J

The products will be more stable than the reactants because the equilibrium concentration is large!

Sapna Ramappa 1J

It's all cumulative, and assuming from the 14A final from his class last quarter, there should be a relatively equal distribution of questions. :)

Sapna Ramappa 1J

The Gibbs free energy of the combustion of gas would be negative, as the reaction is exergonic. This is because gas will have more energy than the water and CO2 molecules released after the combustion, and thus the reaction will be spontaneous.

Sapna Ramappa 1J

Use Van't Hoff to find the new kw with the temperature change. Then, take the square root of the new kw to find the concentration of H3O+. Then, you can take -log[H3O+] to find pH, which you can now compare to the pH of 7! Hope this helps!

Sapna Ramappa 1J

In LS7A, we learned that the overall reaction would be exergonic, because it goes from a higher free energy to a lower free energy, thus resulting in a negative delta G. However, it is also true that breaking bonds requires energy.

Sapna Ramappa 1J

Because the inert gases do not really interact with anything else, adding inert gases does not have an astounding effect on the concentrations.

Sapna Ramappa 1J

When you decrease the volume, the reaction will favor the side with fewer moles of gas.

Sapna Ramappa 1J

A change in pressure results in the reaction favoring the side with fewer moles of gas.

Sapna Ramappa 1J

If the K value is less than 10^-3, you can ignore the x in the denominator if it is being added or subtracted to the initial concentration. This is because it'll make such a minute difference in the equilibrium concentration as a whole. Be careful that you don't just substitute all the x values with...

Sapna Ramappa 1J

Homework questions are a great resource for the midterm. Also, I find that the UA sessions throughout the week are really helpful to do some more practice problems.

Sapna Ramappa 1J

Degeneracy is more about the number of different arrangements for the particles, whereas entropy is the chaos which is largely dependent on degeneracy.

Sapna Ramappa 1J

A cup of tea would be an open system because matter and energy can be transferred between the system and the surroundings.

Sapna Ramappa 1J

In a closed system, heat can still be transferred even though the system can't add or remove anything. However, in an isolated system, there is no heat exchange.

Sapna Ramappa 1J

Comparing Q and K allows us to see whether or not the system is at equilibrium. If Q=K, then the system is at equilibrium. Otherwise, we can see if Q>K or if K>Q to determine whether the reaction favors the reactants or the products.

Sapna Ramappa 1J

If both increased by the same amount, K would still remain the same.

Sapna Ramappa 1J

If you're being asked to find pH, make sure to see how many sig figs are in the initial number. The final answer should have the same number of decimal places as the number of sig figs in the initial number.

Sapna Ramappa 1J

K is the generic term for the equilibrium constant, Kc is specifically for concentrations, and Kp is specifically for partial pressures.

Sapna Ramappa 1J

Yes, the reaction will continue for a certain amount of time before it reaches dynamic equilibrium.

Sapna Ramappa 1J

No, they won't be accessible online. They weren't available for 14A last quarter either.

Sapna Ramappa 1J

You can use de Broglie for anything with mass, so that would include electrons. However, you can't use the equation for photons, because photons don't have mass. In this case, you would use an equation with the speed of light.

Sapna Ramappa 1J

If the electron is ejected, then increasing the intensity of the light will increase the number of ejected photons, but this does not have an effect on the speed of the ejected electron.

Sapna Ramappa 1J

Because you would only be able to find the exact bond angle through an experiment, it's totally fine to put <120 degrees for the test!

Sapna Ramappa 1J

The end-to-end overlap of the sigma bond is far more significant than the side-to-side overlap of the pi bond. As a result, the sigma bond is stronger.

Sapna Ramappa 1J

In a triple bond, there are two pi bonds!

Sapna Ramappa 1J

It typically goes on the element like Nitrogen, in which the placement of the radical ensures stability of the atom. See whichever placement gives you a formal charge of 0 for each atom.

Sapna Ramappa 1J

Double bonds would be shorter (and thus stronger) because there are more electrons that are bonding, and as a result, the atoms are pulled closer together.

Sapna Ramappa 1J

Yes, the pi bond would break first, because it has less strength (the electron density is more spread out within pi bonds).

Sapna Ramappa 1J

Because there is resonance here, the bond length is the average of the bond lengths of the two structures.

Sapna Ramappa 1J

Unfortunately, I think we just have to purely memorize the different shapes and the names of them. However, if you're a visual learner, there are a bunch of helpful charts online that helped me! Here's one: https://chemstuff.co.uk/academic-work/a ... molecules/

Sapna Ramappa 1J

Because lone pairs have a high electron density, they take up more space. As a result, lone pairs are able to push the angles of the other bonds away from the lone pairs, thus changing the shape.

Sapna Ramappa 1J

Personally, I think that visualizing the shapes really allowed me to memorize the names of them. If you're a visual learner, there are many charts online that might help elucidate the connection between number of bonds, number of lone pairs, shapes, and the names of the shapes. One example is this c...

Sapna Ramappa 1J

Due to resonance blending, the bond length will be 130 pm (or really any value in between 120pm and 140pm)!

Sapna Ramappa 1J

So far, we've primarily just looked at resonance structures in the context of delocalized bonding pairs :)

Sapna Ramappa 1J

Yes, the molecules can still have resonance as long as the number of electrons still works in the Lewis structure! The lowest formal charge just indicates the stability of the structure.

Sapna Ramappa 1J

I believe we just need to put the most stable resonance structure. To find this, use the formal charge equation to see which structure has the formal charge closest to zero.

Sapna Ramappa 1J

You can look at whatever has the lowest ionization energy, the lowest electronegativity, and whichever shows up the least in the formula.

Sapna Ramappa 1J

In order to calculate formal charge, use this equation:
FC= V - (L+ (S/2)), where V is the number of valence electrons, L is the number of electrons in lone pairs, and S is the number of shared electrons.

Hope this helps! :)

Sapna Ramappa 1J

Question: If the uncertainty in the momentum increases for a given particle what does this mean for the uncertainty in its wavelength? I was wondering what the relationship between these two uncertainties are and how one can potentially affect the other? Because there's an inverse relationship betw...

Sapna Ramappa 1J

Z is the atomic number for an element, and thus, it is also the number of protons!

Sapna Ramappa 1J

Is there any type of indicator that you may have not converted right or messed up somewhere in the equation when calculating velocity? For example should you ever get like 3.4 x 10 ^ -12 m/s as your velocity? Is it possible to be that small? Yes, it is possible for the number to be that small, beca...

Sapna Ramappa 1J

There are other levels that exist, but we'll just need to focus on s, p, d, and f, because those are the four main orbital types.

Sapna Ramappa 1J

Nope! It would probably be a good idea to know the general order of the lights' wavelengths, but you won't need to know the specific ranges for anything other than visible light! :)

Sapna Ramappa 1J

Either option would work for tests and/or homework, but 2.998 * 10^8 is more accurate! :)

Sapna Ramappa 1J

The de Broglie equation is used to calculate the wavelength of any moving particle. In most of the worked examples completed during lecture, some questions have asked specifically for "the de Broglie" wavelength, whereas other questions might just mention components of momentum and/or wave...

Sapna Ramappa 1J

The intensity of a light source can be determined by looking at the amplitude of the wave, whereas the energy of the light wave can be determined using the equation E = h*v (E is energy in Joules, h is Planck's constant, and v is velocity of the wave).

Sapna Ramappa 1J

The De Broglie equation enables us to see the wave-like properties of moving objects. We can use the equation ( λ=h/(m*v)) to calculate the wavelength of moving objects/particles (λ is wavelength, h is Planck's constant, m is the mass of a particle, moving at a velocity v).

Sapna Ramappa 1J

How do we know when to write solutions using scientific notation? I usually do so as a default but when checking the solutions manual, I noticed that sometimes answers like (for example) .0623 is left just like that and not written as 6.23 * 10^2. Is there a reason for that? Are there specific rule...

Sapna Ramappa 1J

I know that in a chemical equation the delta symbol above the arrow means that heat is added to the equation. However, I cannot remember whether the heat changes any of the compounds in the reaction or not. Do you add any products or reactants to the equation if you add heat? The delta sign indicat...

Sapna Ramappa 1J

Hello! I had a question regarding atomic masses. Do we round the atomic masses on the periodic table or leave them as is? We can calculate everything with the atomic masses given on the periodic table, and then we can round after we consider significant figures at the end of the problem! This will ...

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