CHEMISTRY COMMUNITY

Kelly Yun 2I

Anything with the naught sign indicates standard conditions! The good explanations above would also apply to things like E vs E naught :)

Kelly Yun 2I

Kandyce Lance 3E wrote:
Victor Qiu 1C wrote:I think the values are:

1 atm = 101325 Pa = 101.325 kPa = 1.013 bar = 760 torr

What did you use to get those values?

I think those would just be obtained by conversion factors (probably would be given if needed in any future context)!

Kelly Yun 2I

In class I think Dr. Lavelle said that for 14A and B we approximate bar=atm and those two are probably the most common units we will see in the problems.

Kelly Yun 2I

If the equation represents a redox reaction, then I cannot see why there would not be an oxidizing agent so I am not sure why the solution does not include that for reactions that don't favor products...

Kelly Yun 2I

I think that for the final, it would be given in the problem if we need the value for calculations..!

Kelly Yun 2I

A pseudo reaction is when a reaction appears to be/behaves like a first order reaction when it is a second order one. So I think the main difference between the two rate laws would be that they're used for different reactions..?

Kelly Yun 2I

I don't think you should consider the catalyst for the overall reaction because it is neither part of the reactant or product (they're not used up or consumed).

Kelly Yun 2I

Yes, and the reason why it is the peak of the curve is because it has the highest potential energy at that point, where the activated complex is present.

Kelly Yun 2I

I agree that they could be both! I think that the strength would differ depending on the actual element though.

Kelly Yun 2I

Concentration cells have identical cathode and anode compartments, so the concentrations are "equal." This puts the potential equal to 0, since delta G is 0 at equilibrium. Hope this helps :)

Kelly Yun 2I

I agree that it is the general rule--I don't think I've seen a problem with the reverse setting.

Kelly Yun 2I

You should be able to find out the difference in oxidation state for Br by doing BrO- --> BrO3-, and then add the right amount of OH- to balance the electron transfer because we are dealing with a basic solution. Then we would do our normal step of using water to balance the number of oxygen/hydroge...

Kelly Yun 2I

I doubt that we would be given questions like what is the oxidation state of (some random element other than group1/2 or oxygen) without any context. Even then, I think that we'd just be required to know the common ones for the purpose of balancing equations as a part of the problem.

Kelly Yun 2I

To give you a short and succinct explanation, Q is the reaction quotient where it's [products]/[reactants] at any point in the reaction. However, K is the same equation but at equilibrium, so there is just one K value for a given chemical reaction and its conditions.

Kelly Yun 2I

I'm not entirely sure, but I assume that since enthalpy of the dissolution is positive it wouldn't be favored, so then an increase in entropy would drive the dissolution since we were given that it dissolves "readily." Since the dissolved state is less ordered, I think it does make sense f...

Kelly Yun 2I

I do think that just generally knowing how we got the equation as described during lecture helps understand it, but I doubt we would have to actually derive it. I assume we would have to know how to plug numbers into the equation to get a numerical answer for sure, but not the conceptual parts.

Kelly Yun 2I

I'm not exactly sure about like applications in daily life, but I think this would be very important when conducting experiments with chemicals. So maybe like the field of chemistry/biochemistry and related research?

Kelly Yun 2I

I'm assuming you're saying if we can determine q with delta h, since they can't really be "converted," so in that case, for the system with constant pressure like the coffee cup example, there is the equation qp=delta H :)

Kelly Yun 2I

I personally don't think we would have to know quantum content since it's not an objective for 14b but then again, we could get a surprise question so I'm not sure..

Kelly Yun 2I

Yes, you're right because w= p times delta v, and anything times 0 is 0.

Kelly Yun 2I

Adding on, remember to convert J---> kJ and make sure that you multiply the Cv,m value with the number of moles given in the question as well! I didn't do that for my first attempt and had trouble remembering that I had to do that. Hope this helps!

Kelly Yun 2I

Hi! I think the most significant difference between q and w would be that q is the sum of all heat (enthalpy) trasnsfer into or out of a system whereas w deals with the work done or performed by the system. This isn't like a super long explanation, but I hope this helps!

Kelly Yun 2I

Adding on, I would say that residual entropy is essentially the difference in entropy between a non-equilibrium state and crystal state of a substance independent of temperature and remains present at absolute zero.

Kelly Yun 2I

I'm not sure if this fully answers your question, but basically, gibbs free energy is a measure of how much "potential" a reaction has, so if it equals to 0, that means that there isn't a driving force and there cannot be any more work done.

Kelly Yun 2I

Hi! You have to reverse the second reaction and double it and then add the two reactions to get the enthalpy and entropy values. I think you also have to convert the entropy value from J to kJ and then do ΔG∘=ΔH∘−TΔS to get the answer! Hope this helps :)

Kelly Yun 2I

I think the R value depends on the units and context of the equation, but for this case, I think you would have to use R=8.3145J/(mol·K)

Kelly Yun 2I

going off of this, on exams will we only be given situations where we are only dealing with isothermal reversible expansions so that we can use those equations or is there a chance we'll be exposed to other instances where we'd have to use other ones? if so, what would those equations be? I think a...

Kelly Yun 2I

I assume that it would be given to us, but I think that it would be good to know that the expansion of an ideal gas is irreversible. I doubt we would have to figure this out without information in the question though.

Kelly Yun 2I

I think the first way to approach this would be to use the formula w=-P∆V, and ∆V would be found by finding the volume of the cylinder using the given diameter (cm^3 = L), and since we have the pressure as well, we could plug the values in and get w. Hope this helps..!

Kelly Yun 2I

Higher the entropy--> more disorder, higher the enthalpy--> more heat!

Kelly Yun 2I

Same here--I thought I got rid of this problem in high school but I think doing things online can also cause me to be less focused when doing things and make more mistakes that could easily be avoided if I payed more attention to what I was doing :(

Kelly Yun 2I

The math practice on Sapling doesn't include it so I'm guessing we won't need to know how to solve integrals :)

Kelly Yun 2I

In order for a property to be intensive, it must remain unchanged when the amount is altered, so specific heat is intensive since it accounts for the amount and divides it so that the number represents the amount of heat required to raise the temperature of 1 gram of a substance 1 degree celsius.

Kelly Yun 2I

Ohh I see. Thank you!

Kelly Yun 2I

Ka2 is always less than Ka1, because once it is deprotonated, it will have -1 charge from its original state and will not try to lose more protons.

Kelly Yun 2I

Hi! I feel like the change in pressure should only have an affect on reactions where the moles of gases on either side are different, but would that also apply to reactions where the moles of gases are equal on both sides but one side has more moles of a substance in a different phase? I don't know ...

Kelly Yun 2I

There will be a shift under Le Chatelier principle. In addition, it's important to remember that the K does not change!

Kelly Yun 2I

But I thought that If [H30+]<10^-7 that means that the solution is basic not neutral because the pH would be above 7 and if the pH is above 7 then it is basic? The pH is essentially -log (concentration) and concentrations are usually 10^ (some negative number) , so the smaller the negative exponent...

Kelly Yun 2I

Adding onto the great explanation above, I just wanted to correct the fact that pKa + pKb = pKw = 14 and Ka + Kb = Kw = 10^-14 :)

Kelly Yun 2I

Generally, reactants are to the left of the arrow and products are to the right. When something "shifts" left, that means that the left side is favored (reactants) and vice versa. :)

Kelly Yun 2I

Delta H is essentially change in heat, and endothermic reactions absorb heat (lower H in reactants than products) while exothermic reactions release heat (higher H in reactants than products). When dealing with Q and K, think of heat as part of the reactants/products and apply Le Chatelier's princip...

Kelly Yun 2I

In addition to the good responses above, I think application of quadratic equations could also be used when we deal with pressures and Kp!

Kelly Yun 2I

I think it's just a concept that we need to keep in mind and won't be given the value. I do think that remembering that the maximum pH is 14 and p of anything is essentially a -log function would help to remember the value when we need it!

Kelly Yun 2I

Hey, I'm pretty sure that if pH<pKa, that indicates that the acidic form exists in relatively greater amounts than the basic form. Since neutral solutions are preferred, acids would lose protons to become ionized (basic form), and vice versa. Hope this helps!

Kelly Yun 2I

Hi, so I was doing Sapling and I noticed that some questions lead us to an equation for some K value where all terms are squares (e.g. if K=2=[products]^2/[reactants]^2. And when I saw the solution (if I didn't misunderstand) where they said we could take the square root of all the terms and get a s...

Kelly Yun 2I

Sorry--in addition, the fact that there are changes in concentration at equilibrium does not mean that the K value has changed. If we plug the new concentrations into the equilibrium equation, we still get the same K :)

Kelly Yun 2I

For (d), we learned about Le Chatelier's principle where whenever there is a change to the system, the system will want to reverse its effects. So if we have more reactants than we do at equilibrium, the system will use them up to form more products until they reach a new equilibrium. Hope this helps!

Kelly Yun 2I

Yes, I'm pretty sure you can do that since liquids and solids aren't included in Q or K calculations.

Kelly Yun 2I

I was confused on this as well, but I try to remember Q as reaction quotient and K as equillibrium constant and that helps me tell the difference!

Kelly Yun 2I

To be honest, I don't know if there's like a strict guideline when determining how small a value is for it to be negligible, but I have heard of the 5% rule where if x/initial concentration is less than 5% we are able to neglect x. Hope this helps!

Kelly Yun 2I

If products are favored, then the reaction will be driven forward so that more products are formed, and if the reactants are favored, the reaction will be driven backwards (reverse reaction) so that more reactants are present. Hope this helps!

Kelly Yun 2I

I think we could assess pressure change in automobiles mainly by monitoring the amount of air pumped into the tires. As the above reply stated, pressure increases if there are more moles of gas in an enclosed "container".

Kelly Yun 2I

Do we have to know how to find bind order for the final?

Kelly Yun 2I

I think formula units are usually used for solid network compounds (usually ionic) whereas molecular units are for covalent compounds.

Kelly Yun 2I

Hi, since concentration is moles/volume and molar mass is mass/moles, if you multiply the two to volume, moles and volume units cancel out to give us just the mass of the solute in whatever solution we were given. Hope this helps!

Kelly Yun 2I

Oh I see, thank you so much!

Kelly Yun 2I

I was reading through my notes and was confused with the structure of a chelate. Why is it that only polydentates can form chelates and is it just a normal ring shaped structure?

Kelly Yun 2I

The Ka is the dissociation constant that tells you how much of the acid is dissociated in the solution. You can use it to determine the strength of the acid or predict the concentration of the acid or base in the solution.

Kelly Yun 2I

I think you're good to go as long as you can identify which point the stoichiometric/equivalence point lies in (when y axis is pH7). I don't think the rest of the points mean anything, but you should be able to relate those points to acidity of the titration solution by looking at the pH indicated i...

Kelly Yun 2I

pH or pOH is the negative log of the concentration of the molecule, which means that they have an inverse relationship.

Kelly Yun 2I

I don't think that we can label CH3COOH as unstable in its form, especially because it is a weak acid and is usually present in water as whole undissociated molecules (stronger acids are less stable than weaker ones).

Kelly Yun 2I

Even if we place Cl and H so that it looks like the dipole moments may cancel out in our drawing, in a 3-D shape, bond angles are not 90 degrees but are 109.5, so they won't actually cancel out.

Kelly Yun 2I

In lecture, it was mentioned that hemoglobin is made up of four myoglobin-like molecules so we could also infer that myoglobin is at the smaller scale.

Kelly Yun 2I

VSEPR helps determine the molecular shape. It is based on the fact that molecules tend to achieve a geometry that minimizes the repulsion of valence electrons. On the other hand, hybridization refers to atomic orbitals mixing to reach the lowest overall energy states. It explains why atomic orbital...

Kelly Yun 2I

Hi, so I was rewatching lecture from last week and was wondering if there are cases where hybridization and VSEPR suggest different shapes for a molecule because Professor Lavelle mentioned that we always use VSEPR over hybridization when determining shape.

Kelly Yun 2I

Kelly Ha 2L wrote:
Kelly Yun 3K wrote:Oh, I see so size of cations and polarizing power and size of anions and polarizability are the opposite right? Thanks!

Yes, larger anions have greater polarizability and smaller cations have greater polarizing power.

Okay, thank you so much :) Happy Thanksgiving!

Kelly Yun 2I

I'm not sure if we need to know exceptions, but I think we will be notified which exceptions to know once we go over it in lecture :)

Kelly Yun 2I

Hi! I'm not entirely sure, but I think the lone pairs from nitrogen are reactive and will be donated and used with the two guanines to form coordinate bonds. Happy Thanksgiving :)

Kelly Yun 2I

Hi, so is it safe to assume that the number of "bonding areas" like a lone pair or bonding pair can determine hybridization for all molecules? Or are there any exceptions?

Kelly Yun 2I

It simply means that there are no bonds produced in the molecule because there are an equal number of electrons in bonding and antibonding orbitals.

Kelly Yun 2I

Hi, in a more simple sense, you would first need to see where the polar bonds are located and see if they are symmetrical and therefore can be canceled out. If they cancel out, you have a nonpolar molecule despite the polar bonds, but if they do not cancel out, the molecule would be polar.

Kelly Yun 2I

Oh, I see so size of cations and polarizing power and size of anions and polarizability are the opposite right? Thanks!

Kelly Yun 2I

When looking at polarity of the molecule as whole, it is important to look at the geometry of the molecule. Any case where polarity cancels out (trigonal planar would also cancel out the polar bonds) and polar bonds are symmetric would lead to a nonpolar molecule, no matter how polar the bonds are.

Kelly Yun 2I

I like to think of hybridization as the number of groups so like one lone pair would be a group and one bonding pair would be another group. In this case, there are four groups so the answer would be sp3.

Kelly Yun 2I

I thought that larger anion size meant higher polarizability and more covalent character, so wouldn't NaBr have more ionic character? Please correct me if I'm wrong.

Kelly Yun 2I

Hi, can someone explain the relationship between cation size and covalent character (solubility)? I only remember taking notes on the changes of halide anions affecting solubility.

Kelly Yun 2I

I was also confused about this because I searched up online to find that some say that the smaller the ion the stronger the interaction...Which was not what we learned in class. Would it be safe to generally follow the rule that smaller anions are more soluble and that higher covalent character mean...

Kelly Yun 2I

Polarizability can be seen as a measure of the molecule's ability to acquire a dipole moment when exposed to an electric field, so I would say the consequence of polarizability would essentially be stronger IMFs.

Kelly Yun 2I

I think we were just looking at the rule where formal charge of each molecule is closest to 0 so that we have a lot of atoms with a formal charge of 0. I was confused as well, but I think that since the question was looking at resonance in different steps using different aspects, we just had to cons...

Kelly Yun 2I

I thought that polarizability was more related to intermolecular forces like dipole-dipole moments than intramolecular forces (bonds)...Can anyone correct me if I'm wrong? Thanks

Kelly Yun 2I

Yes, for any "one mole of interaction" I believe it would be one mole of each, since one of each make up one interaction.

Kelly Yun 2I

I think you were right in looking at how close the molecules were to each other, but because the negative and positive charges align in only II, it would be II. I think that if we were given three structures all looking like II but being different lengths apart, we would choose the closest ones.

Kelly Yun 2I

Oxidation numbers basically tell us how many electrons an atom would gain or lose when forming a bond with ionic character. We can see the overall charge of the molecule and use our basic knowledge (which I was told to memorize in HS, but I'm sure there would be a way to understand it otherwise) in ...

Kelly Yun 2I

Sometimes there can be multiple structures that are favored so we come up with resonance structures. In these structures, electrons are delocalized and occupy a larger volume, which lowers the energy of a molecule and stabilizes it.

Kelly Yun 2I

I'm not sure if I understood your question, but I'm assuming you're asking how to determine the number of electrons given the ml value. If ml is given, we already know the sublevel, which contains two electrons with a spin up and spin down. If we were only given the n value like n=2, we wouldn't rea...

Kelly Yun 2I

I believe s,p,d,f is all we need to know for this course, even though there may be angular momentum #s that have an absolute value greater than 3.

Kelly Yun 2I

The d orbital can hold up to 10 electrons, but the nth s orbital may hold 1 or 2 electrons, so I think it's easiest to think of the number of valence electrons in the d block as equal to the group number.

Kelly Yun 2I

When we use a wave function (math) to determine the probability that we will find an electron at some place, we are representing electron orbitals in terms of mathematical concepts. The three quantum numbers tell us which energy level and orbital (s,p,d,f) to further help us conceptualize electron o...

Kelly Yun 2I

I believe the groups have more in common than the periods. Each vertical column has its own name and the atoms in the column act in a somewhat similar way (i.e. noble gases).

Kelly Yun 2I

Effective nuclear charge is basically the atomic number - # of shielding electrons. So I think we can generally assume that the more inner electrons an atom has compared to the number of protons, the lower the effective nuclear charge.

Kelly Yun 2I

Just to add on, I think there was a post-module question that addressed this on the website where the answer was that atomic spectroscopy helps us determine characteristics of a certain element while molecular spectroscopy does the same from a molecule.

Kelly Yun 2I

Hi! We will not be tested on the particle in a box on the midterm. Dr Lavelle said in his lecture today that our midterm will only go up to the last slide he did which introduced the different atomic orbitals and did not discuss a particle in a box. In general, our midterm will only contain all the...

Kelly Yun 2I

I think some things to remember would be that the wavefunction gives us the energy of the electron, Ψ represents the height of a wave at a position, and (Ψ)^2 represents the probability of finding an electron. Also, the H (hamiltonian) gives us a double derivative of the sin/cos wave function.

Kelly Yun 2I

I'm not entirely sure but I remember someone saying in the Groupme that they asked their TA and she said the midterm questions for this chapter should only include material from 1A and 1B but to read 1C. I have no idea why the email said we needed to know up to 1D2, but I hope this can clarify thing...

Kelly Yun 2I

According to the question, a bowling ball of mass 8.00 kg is rolled down a bowling alley lane at 5.00+-5.0m/s, then since the position is + or - 5 from 5, isn't the spread/delta x 10.0? The solution manual used 5.0 so I was confused because I thought that delta x was the spread of uncertainty (2x th...

Kelly Yun 2I

Yes I believe that quantum numbers or s,p,d orbitals won't be on MT1 according to the email, so I don't think orbitals will take up much of the test? I thought he stopped towards the end of lecture #9 to say that what he has covered so far will be it for MT1 and then went on to discuss orbitals in m...

Kelly Yun 2I

Hi, because Sapling is still randomizing questions for me I just skipped the ones that weren't covered up to Wednesday week 3's lecture but I ended up solving 18/30 instead of the 19 questions that Professor Lavelle told us. I skipped all the electron configuration, periodic table trends, and orbita...

Kelly Yun 2I

I don't know about other people's saplings but mine is very randomized :( I have a Ephoton and Eelectron question for #25 and an orbital question for #11.

Kelly Yun 2I

Adding on to Sydney's explanation, I think delta v had to do more with the fact that the electron could change paths due to its entanglement with the photon which leads to varying distances of the path that the electron takes.

Kelly Yun 2I

Sorry, just to clarify, do free electrons have more energy than bound electrons?

Search found 117 matches