CHEMISTRY COMMUNITY

Andrew Wang 1C

Dear Dr. Lavelle, Thank you so much for all the hard work and resources you put in to help us succeed these past two quarters, especially in this virtual environment! While it certainly was difficult, the opportunities that you, the TA's, and the UA's provided were invaluable and made it so much eas...

Andrew Wang 1C

I don't think there is a direct relation between the forward and reverse rate constant themselves. But they are related in terms of the equilibrium constant, $K=\frac{k_{f}}{k_{r}}$

Andrew Wang 1C

Everything looks good to me!

Andrew Wang 1C

They reversed the order because keeping it as Ag --> Ag+ would have resulted in a negative Ecell, which means the galvanic cell wouldn't be spontaneous.

Andrew Wang 1C

Given that the half-life changes when concentration changes, you can deduce that it isn't a first-order reaction. And I don't think we covered 3rd order reactions in lecture, so that should leave you with 2nd order.

Andrew Wang 1C

Mechanisms aren't numbers, they are the pathway through which a reaction occurs. For this course, my understanding is that we won't necessarily be finding specific mechanisms, just analyzing them.

Andrew Wang 1C

When working with rate laws, we look at the initial rates of the reactants (when there are no products) to keep things simple, so we won't have to worry about products in rate laws. I'm not too sure if this answered your question or not, but I hope it helps!

Andrew Wang 1C

No, the order doesn't matter. Use Pt(s) when you need to add an inert conductor in the electrode (no solid metals present).

Andrew Wang 1C

You'd have to look at the oxidation numbers of the compounds. The easiest one to look at is Fe. It goes from 3+ to 2+, meaning it gained an electron. And since there are 6 Fe's, 6 electrons are moved in the overall reaction, so n=6.

Andrew Wang 1C

For compounds in the same phase, the order doesn't matter. For the overall cell diagram, my understanding was that you would have the inert metals like Pt on the far ends of the cell diagram and fill in the rest like:
anode reactants | anode products || cathode r | cathode p

Andrew Wang 1C

From what I know, you can use the integrated rate law to find the concentration of a substance after some amount of time or use its graph to determine the order of the reaction. There are likely more uses for both the differential and integrated rate laws that others can comment on. But I hope this ...

Andrew Wang 1C

If the standard reduction potential is positive, then it's likely to be reduced. If the standard potential is negative it's likely to be oxidized.

Andrew Wang 1C

Hi everyone, I'm confused about when we should use a comma or | to separate substances when we write cell diagrams. Do we use a comma for compounds in the same phase and | when they aren't? And also when do we need to include Pt (s)?

Thanks!

Andrew Wang 1C

From what I gathered during the lecture, I think you're correct, a concentration cell uses a concentration gradient to produce electric current. It's also helpful to use the concept of a concentration gradient to remember what happens in a concentration cell. Since particles go from high --> low con...

Andrew Wang 1C

The responses above have basically answered your question, but I just wanted to let you know that the agent is the entire compound and not just the element that is reduced or oxidized (that tripped me up when I was doing it lol). For example (this may not be correct in the problem) if Cl was oxidize...

Andrew Wang 1C

I think it may have to do with the fact that the textbook gives you a "skeletal" equation which isn't supposed to be balanced. It just shows you the compounds involved in the reaction, not necessarily the reaction itself. I may be wrong though, since I'm not too confident on skeletal equat...

Andrew Wang 1C

You ignore the products in the first step because the reaction hasn't occurred yet, so the products don't "exist." Same reasoning for the third step; the reaction has occurred and the reactants don't "exist" anymore, so you just calculate q for products and not reactants.

Andrew Wang 1C

The second picture has a negative sign factored out, so they're technically the same.

Andrew Wang 1C

So the three steps are: changing temperature of reactants from 37 C to 200 C, the reaction itself at 200 C, then changing the temperature of the products from 200 C to 37 C. Calculate the \Delta H of each step, then add them all together. This would be a bit easier to explain visually, but I hope th...

Andrew Wang 1C

$\Delta U=q+w$ , so in the first case it's assuming that q=0, so $\Delta U=w$ . The second case assumes w=0 so $\Delta U$ is just equal to q.

Andrew Wang 1C

I think it should start from "Explain the meanings of heat capacity and specific heat capacity."

Andrew Wang 1C

I think one reason is because entropy is a state function, like enthalpy. So we can just take the entropy of the products and subtract the entropy of the reactants to get the change in entropy. \Delta S wouldn't be the same as standard entropy, since standard entropy how much entropy a substance act...

Andrew Wang 1C

One way to think of a spontaneous reaction is that it does not require an input of energy to proceed, so it can happen "on its own," in a sense. So energy will be released, \Delta G is negative. Since the equation for Gibbs is \Delta G=\Delta H-T\Delta S , this occurs when \Delta S is posi...

Andrew Wang 1C

I think the Sapling homework gives us (3/2)R and (5/2)R as the heat capacity of ideal gases in the problems, so that's probably what we can expect. It wouldn't hurt to memorize them though, they're pretty easy to remember.

Andrew Wang 1C

I just assumed that if they didn't specify another temperature, to use standard conditions (25 C or 298 K)

Andrew Wang 1C

An increase in H will correspond to an increase in S as well (or decrease in H corresponds to decrease in S), since $\Delta S=\Delta H/T$ for phase changes. So $\Delta S$ and $\Delta H$ will have the same sign.

Hope this helps!

Andrew Wang 1C

You'll have to include the enthalpy of fusion in your calculations, to account for the heat absorbed by the ice during its phase change. Then you add that to q=mcΔt for the 0 C water, and set it equal to the negative (because it's losing heat) q=mcΔt of 45 C water.

Hope this helps!

Andrew Wang 1C

So once you solve the first part to get the heat capacity of the calorimeter, you can use that in part 2 to calculate the heat of combustion q for compound B: $q_{b}=c\Delta T$ . Then you'd divide by the mass of compound B to get heat per gram.

Hope this helps!

Andrew Wang 1C

Since we're considering \Delta T in this problem and not the temperature itself, we can ignore the conversion between Celsius and Kelvin because \Delta T of Celsius is equal to \Delta T of Kelvin: K_{f}-K_{i}=(C_{f}+273.15)-(C_{i}+273.15)=C_{f}-C_{i} Note that you can only ignore the...

Andrew Wang 1C

At constant pressure, you'd use the equation \Delta U = q-P_{ex}\Delta V . For constant volume, you'd use the same equation except \Delta V is 0, so it would just be \Delta U = q . I'm not sure about your second question, but I think they'd tell you whether or not the pressure changes. For volume, i...

Andrew Wang 1C

annabelchen2a wrote:Oh, is that because the Ka2 for H2SO4 is rather large/significant (in addition to its already large Ka1 value)?

Yeah, the Ka2 of H2SO4 is large enough to affect the pH so you'd include it in the calculations.

Andrew Wang 1C

Not necessarily. It depends if the reaction is endothermic or exothermic. For exothermic, the reaction shifts to the reactants when increasing temperature, so K decreases. For endothermic, the reaction shifts towards products when increasing temperature, so K increases. The opposite is true for lowe...

Andrew Wang 1C

Since H2SO4 is a polyprotic acid, it'll deprotonate twice. The first time you can directly take the molarity of H+ from the molarity of H2SO4, but the second deprotonation requires the ICE table. Then you add the two values you got together and take the -log of that. Note that H2SO4 is the only poly...

Andrew Wang 1C

Standard reaction enthalpy is the enthalpy of the overall balanced reaction. Standard enthalpy of formation is the enthalpy when one mole of a substance is formed. I'm not too sure about enthalpy of combustion, but I think it's the energy released when one mole of the substance undergoes combustion....

Andrew Wang 1C

Yes, that seems about right! One other connection is that you can use standard enthalpy of formation to calculate the standard reaction enthalpy.

Andrew Wang 1C

When the problem involves a calorimeter the q that you solve for is the q of the surroundings. So to find qsys you'd just take the opposite sign of that.

Hope this helps!

Andrew Wang 1C

When changing the pressure, we only look at the number of moles of gas. In this case there are 6 moles of gas on each side, so there is no change.

Andrew Wang 1C

I assume that's the case, since the sign of delta H is kind of "built in" to the definitions of endothermic and exothermic. Negative delta H means heat is released, so it's exothermic, while positive means heat is absorbed which is endothermic.

Andrew Wang 1C

Yeah they'll most likely be the same as last quarter where we take it in breakout rooms with TA's. And it's fine if your laptop doesn't have a camera, we used our phones for the zoom last quarter too while taking the test on computer.

Andrew Wang 1C

Amphoteric compounds can act as both acids and bases. One example of an amphoteric compound is Al2O3, which can act as a Lewis acid and base despite not having any protons (H+) to donate. Amphiprotic compounds can act as both proton donors and acceptors (you can remember this by the "protic&quo...

Andrew Wang 1C

The previous answers all explained it very well! I just wanted to add that you can ignore Ka2 for most acids except H2SO4 (this shows up in a textbook problem)

Andrew Wang 1C

Using the pH, you can find the current (diluted) concentration of OH-. Then you can use M1V1=M2V2 to find the concentration of the original solution. Using the concentration you found, you can figure out the moles of NaOH, then use molar ratios and stoichiometry to find the mass of Na2O added. Hope ...

Andrew Wang 1C

How I did it was by starting with an ICE table and filling in values for what we know, and using variables for what we don't know. Although it's a bit different for this problem since the variable isn't in the "C" row. This should leave you with one unknown variable, which you can solve fo...

Andrew Wang 1C

I'm pretty sure Kc will always be small on these questions since Dr. Lavelle said we wouldn't be able to algebraically solve cubic equations, so we have to just approximate.

Andrew Wang 1C

It depends on the value of Q relative to K. If Q is greater than K, then the reaction proceeds towards the reactants, so you would put "-x" on the products side and "+x" on the reactants side. If Q is less than K, then the reaction proceeds towards products and you would put &quo...

Andrew Wang 1C

Use table 5G.2 to find two or more reactions that add together to result in the desired reaction 2BrCl(g+H2(g)⇌Br2(g)+2HCl(g). Then use the equilibrium constants for the reactions that you found in the table to calculate K for the overall reaction by multiplying them together.

Hope this helps!

Andrew Wang 1C

For (c), the equilibrium constant will not change due to an increase in pressure (or concentration) on one side of the reaction. Equilibrium constants for reactions only change if the temperature is changed. For (d), according to Le Chatlier's principle, if you start with a higher concentration on o...

Andrew Wang 1C

Use table 5G.2 to find two reactions that, when added together, result in the desired reaction 2BrCl(g+H2(g)⇌Br2(g)+2HCl(g). Then you'd use the equilibrium constants for the two reactions that you found in the table to calculate K for the overall reaction by multiplying them together. Hope this helps!

Andrew Wang 1C

You would use PV=nRT when you are converting from partial pressure to concentration, or vice versa. So if the problem asks for Kc, but gives you a measurement in atm, you would use PV=nRT to convert that value to the concentration and then plug it into Kc=[P]/[R]. For problems that ask for Kp, you'd...

Andrew Wang 1C

While Kc and Kp are calculated the same way, I don't think they are equal in value. Kp can be used if the reaction involves only gases, while Kc can be used for gases and aqueous solutions. The problem will usually tell you which one to solve for by looking at the units (moles for Kc, atm or bar for...

Andrew Wang 1C

It would depend if we are given the values of reactants/products in concentration or bars. I remember some problems in the Audio-Visual modules had us calculate Kc when there were only gases in the reaction.

Andrew Wang 1C

Yes, you can get the Rydberg equation by setting E=hv equal to the difference between energy levels -(hR)/n2^2 - (-(hR)/n1^2). then you cancel h from both sides to get the equation in the equation sheet. If they were asking you to calculate the energy of a specific level you'd probably use -(hR)/n^2...

Andrew Wang 1C

I was having this problem on Monday with lecture 28, and just kept refreshing the page until it played again. It shouldn't take too long for it to work again, though.

Andrew Wang 1C

Yes, when you combine (add) the two reactions together you would multiply their individual K's together to get the K of the overall reaction. I think the equilibrium constants will be covered more in-depth in 14B, and for this class just know K A decreases as you remove more protons, and that a high...

Andrew Wang 1C

I tend to first look at the bond that is being broken when an acid loses a H+, and comparing them to each other. The longer that bond is, the stronger the acid. If the bond being broken is the same across the compounds being compared, I would look at which compound has the more stable anion. There a...

Andrew Wang 1C

If the reaction involves a strong acid or base, the concentration of H3O+ or OH- will be equal to the concentration of the acid or base, respectively. If it's a weak acid or base then we need to use equilibrium which won't be tested in this course.

Andrew Wang 1C

The oxygen atom is common across all three compounds, so you wouldn't use it to compare them to each other. Instead, we use the halogens because they are different in each compound and therefore comparable.

Andrew Wang 1C

My TA gave us some examples of when to use the Latin name for transition metals: Fe (ferrate), Cu (cuprate), Au (aurate). Looks like there's maybe a pattern here to use the Latin name when the atomic symbol isn't an exact abbreviation of its name?

Andrew Wang 1C

So would it be true to say that the weaker the bond, the stronger the acid since it dissociates more easily?

Andrew Wang 1C

I'm not too sure of this myself, but I think one way to do it is to see how many atoms in the ligand have lone pairs, and use that number to determine if it is bidentate, monodentate, etc. Also consider if the ligand can rotate or not (pi bonds), since double bonds may affect how many sites a metal ...

Andrew Wang 1C

Hi! In today's lecture, Lavelle told us it was fine to use either, but he used -o in his lectures. -ido is a new naming convention that not many people know about yet.

Andrew Wang 1C

Transition metals often don't follow the octet rule since the octet rule only considers the s and p orbitals. Transition metals are located in the d block so they often use their d-orbitals when bonding.

Andrew Wang 1C

Finding the hybridization of the oxygen is similar to finding the hybridization of the carbon. You'd look at how many regions of electron density the oxygen atom has (# of bonds + lone pairs), and use that to find the hybridization.

Hope this helps!

Andrew Wang 1C

I'm not sure if this is what you mean by "general" or not, but it could be that cisplatin isn't specific to just cancer cells since it can also affect healthy cells. What you said about it treating many types of cancer also makes sense though!

Andrew Wang 1C

I think it will be, but Dr. Lavelle doesn't recommend overlapping 14B with another class since the exams might take place during the lecture time.

Andrew Wang 1C

Do all double bonds involve unhybridized p orbitals? What other examples of the unhybridized p orbital should we need to know?

Andrew Wang 1C

A triple bond between the N and O is the only way for both atoms to have a full octet.

Andrew Wang 1C

I'm not sure about the specific details of how they are determined, but I remember reading somewhere that scientists can use x-ray or electron diffraction to measure bond angles in different substances.

Andrew Wang 1C

If an orbital only has one electron and is in the ground state, I'm pretty sure the spin should be +½. It wouldn't be in the ground state if it had spin -½.

Andrew Wang 1C

Boiling and melting points can be related in the sense that both are higher when the substance has stronger intermolecular forces. So yes, a higher melting (and boiling) point means stronger intermolecular forces, since more energy (heat) will be required to overcome the interactions to change phase.

Andrew Wang 1C

Yes, I think the other Group 16 elements are also exceptions to the trend and have lower ionization energies, since they all have 4 electrons in their outermost p orbitals. This makes them have more repulsive interactions, so it takes less energy to remove an electron.

Andrew Wang 1C

Lone pairs are only attracted by one nucleus, compared to a bonding pair which is attracted to two. This makes it so that lone pairs have a greater negative charge and thus more repulsion, which accounts for the extra "volume."

Andrew Wang 1C

From Dr. Lavelle's most recent lecture, we can determine how viscous a substance is compared to another by analyzing their IMFs. The stronger IMF's something has, the more viscous it will be.

Andrew Wang 1C

How I did it was I drew the Lewis structure for CH3CHO and used that to determine which interactions were present. I'm not sure if there's a method to determine by just using the formula, but drawing the Lewis structure should be helpful.

Andrew Wang 1C

Yes, I believe that is the case. Since diethyl ether is nonpolar, it only has induced dipole-induced dipole interactions (LDF) between its molecules. Butanol also has LDF, but will also have dipole-dipole interactions because it is polar. So because of this extra interaction, it requires more energy...

Andrew Wang 1C

The 4s orbital is filled first before the 3d due to its lower energy. Although this isn't shown in our textbook's electron configurations, there is another way to order them called the "filling order" that shows this more clearly.

Andrew Wang 1C

Oh I completely missed that the second spin in 2p was pointed down. Thank you!

Andrew Wang 1C

In your example, the electron in the 3d orbital can act as a valence electron and determine how the element reacts, which is why it is important to include it in the noble gas configuration.

Andrew Wang 1C

I'm confused about how 7b (Nitrogen) is in an excited state. I said that it was in the ground state since it has the electrons in the correct orbitals, but the answer key says otherwise.

Andrew Wang 1C

Each of the single-bonded oxygens has 3 lone pairs, which combined with the single bond gives them a full 8 electron octet. I'm not sure about the "-" in NO3-, but I hope this helps!

Andrew Wang 1C

When drawing Lewis structures, how do we know when to use the formal charge or when to use the octet rule for each element?

Andrew Wang 1C

Generally when shortening electron configuration, we would substitute the preceding noble gas, instead of the immediately preceding element. This allows us to see the electrons in the highest energy level (valence electrons).

Andrew Wang 1C

I saw other explanations saying that extent of change refers to amplitude. Can this also be true? I don't think so, since the amplitude corresponds to the electric field. The extent of change refers to the slope from a trough to the next crest on the wave, or vice versa. So while changing the ampli...

Andrew Wang 1C

I think the answer key had the energy of the incident light equal to the work function because the kinetic energy of the ejected electron is so small that it's basically negligible, so adding it to the work function wouldn't change anything. I'm pretty sure either way is fine and gives the same answ...

Andrew Wang 1C

I think the amplitude of the wave is 0 at the ends of the box because that's where the box's borders are, so there is no chance that the particle will actually be at that location (since it's already taken by the edges of the box). The wavefunction represents the probability that the particle will b...

Andrew Wang 1C

I'm not 100% sure, but I think there are a couple concepts in that learning outcome: one is to understand that you can analyze the spectroscopy of both atoms and molecules, and the second is to understand the differences between Lyman and Balmer series, like how they each correspond to one region an...

Andrew Wang 1C

I think you would use the diatomic form in calculations when the diatomic elements are alone (not bonded to anything else). Otherwise, you should use what is given in the chemical equation.

Andrew Wang 1C

The Heisenberg indeterminacy equation applies to all objects, but is only noticeable and mostly affects objects on an atomic scale.

Andrew Wang 1C

Spectral lines represent the wavelength of the energy absorbed or emitted by an electron when it transitions from one energy level to another. The energy absorbed or emitted is equal to the difference in energy between the two levels that the electron transitions between, and you can use E=\frac{hc}...

Andrew Wang 1C

The uncertainty of position or momentum is like standard deviation of either the position or momentum (velocity) of an object. For example, if an object's position was 3\pm2 m, then \Delta x would be equal to 4, as the object could be anywhere from 1 m to 5 m, and 5-1=4. Once you know the uncertaint...

Andrew Wang 1C

The answer should be D. The photoelectric effect is one of the experiments that shows the particle properties of a photon. If photons acted only like waves, then increasing the intensity (amplitude) should result in ejected electrons (ike a wave at the beach, the higher it is, the stronger it is). H...

Andrew Wang 1C

In addition to what Sid posted, the spectral lines in the UV region are part of the Lyman series, which has n₁=1 for its lower energy level and n₂=2 as its higher energy level, which is why they have higher energy and a shorter wavelength.

Andrew Wang 1C

A helpful way to identify when to use each series is to remember that the Balmer series occurs within the visible light range on the EM spectrum, so it'll have a longer wavelength and less energy, which makes sense since the energy difference is smaller between the higher levels and n=2 for the Balm...

Andrew Wang 1C

I was going over the solution to #15 in Focus 1A of the textbook, which said that each spectral series has its own lower energy level (Lyman is n=1, Balmer n=2, etc.). So this got me curious if each of the spectral series also corresponds with one region of the electromagnetic spectrum?

Andrew Wang 1C

While it is true that c=λv describes the relationship betwen the wavelength and frequency of light, I don't think this equation allows you to figure out the exact graph. You would be able to conclude that the shape would be oscillatory however, since wavelength and frequency are both qualities of a ...

Andrew Wang 1C

Φ is the amount of energy (unique to each element) that a photon needs to have for an electron to be ejected from a metal when it is hit with the proton. E = hv is the energy of a photon when it hits the metal (hv = Φ + E k ). So Φ would be equal to hv only if the electron is ejected with no excess ...

Andrew Wang 1C

Hi there! E = pc is derived from E = mc 2 . Since c is the speed of light, we can think of E = mc 2 as E = mv 2 , or E = mvv. And since momentum p = mv, we can plug in momentum to get E = pv. After substituting c back in, we end up with E = pc. You can connect this with the equation from photoelectr...

Andrew Wang 1C

I also thought it was E at first, but Arielle is correct that the answer is D. My reasoning was that although λv = c is used in photoelectric effect problems, it isn't specific to the photoelectric effect itself, while the other three equations are.

Hope this helps!

Andrew Wang 1C

Hi!
Since oxygen is one of the diatomic elements (along with Br, I, N, Cl, H, and F), it exists as O2. So when the question is talking about oxygen gas (O2), we should use 32.00 g/mol as a conversion factor.

Hope this helps!

Andrew Wang 1C

Hi! The moles of each respective molecule is equal to the stoichiometric coefficient. The moles of each individual element in a molecule is equal to the stoichiometric coefficient times the subscript. So for your question, there would be 4 moles of hydrogen atoms, but 2 moles of water molecule. Hope...

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