CHEMISTRY COMMUNITY

Jasmine Ho 3I

Once you compute the order of each reactant, the rate is equal to k multiplied by the concentration of each reactant to the power of its order. So it would look like rate = k([A]^l)([B]^m)([C]^n) are the l, m, and n the order of the reactants? yes! l is the order for reactant A, m --> B, n --> C

Jasmine Ho 3I

Part 5 isn't asking for a numerical value. You use the order for each reactant (found in parts 1,2,3) to find the rate law for this reaction: rate = k [A]^n [B]^m [C]^l. (A zero-order reactant just becomes a 1 in the rate law.) Using this equation, you can use the rate and concentrations from any o...

Jasmine Ho 3I

Hi! First, I plugged the 4.5 into the 1st order half-life equation to find the value of k. Then you can find the value of [A]naught by dividing mass of phenobarbitol by the mass of the dog. Plug these values into the equation for a 1st order reaction to find ln [A] for time=2hrs. Use stoichiometry t...

Jasmine Ho 3I

Once you compute the order of each reactant, the rate is equal to k multiplied by the concentration of each reactant to the power of its order.
So it would look like rate = k([A]^l)([B]^m)([C]^n)

Jasmine Ho 3I

For this one, you should refer the the appendix for the list of standard reduction potentials.

Jasmine Ho 3I

It helps a lot to draw out the reaction profile! Since deltaH is positive, the reaction is endothermic, and deltaH is the energy difference between the reactants and the products. The activation energy given is for the forward reaction, and you will see that the activation energy for the reverse is ...

Jasmine Ho 3I

There is so much to do this week but being productive is so hard. But I'm keeping myself going by reminding myself that I just have to get through this last week and then I can actually enjoy my free time.

Jasmine Ho 3I

If you press "Hint", Sapling gives a useful version of the Vant Hoff equation. Changing the temperature only affects the k value, so you are looking for a value T2, which makes the ratio of k2/k1 equal to 3.

Jasmine Ho 3I

You can identify the oxidization and reduction half-reactions by comparing the reduction potentials for each. Since the Ti reaction involves 2 ions, you will need to add Pt on that side.

Jasmine Ho 3I

We probably will be asked to figure out half-reactions by ourselves because most of one section of the textbook exercises asks us to write out the two half-reactions in order to find the balanced equation.

Jasmine Ho 3I

I'm gonna sleep a lot and binge some shows I've been wanting to watch since first round of midterms.

Jasmine Ho 3I

Since this is a first order reaction, use the 1st order integrated rate law: ln[A] = -kt + ln[A]naught
Isolate t and plug in all of the given values. This gives you the answer in seconds, so divide by 60.

Jasmine Ho 3I

Part 5 isn't asking for a numerical value. You use the order for each reactant (found in parts 1,2,3) to find the rate law for this reaction: rate = k [A]^n [B]^m [C]^l. (A zero-order reactant just becomes a 1 in the rate law.) Using this equation, you can use the rate and concentrations from any of...

Jasmine Ho 3I

Make sure you are preparing far in advance! The worst thing before an exam is feeling like you haven't studied enough, and that almost always makes me perform worse.

Jasmine Ho 3I

I believe anions flow from the right to the left, and cations have to flow from the left to the right since the oxidized form of the cathode metal carries a positive charge and enters the solution during the reaction.

Jasmine Ho 3I

I believe that Professor Lavelle said that in diagrams the cathode would always be on the right because electrons always flow left to right. If you wanted to make sure, the electrode that is losing material into the solution is the anode, and the electrode gaining material is the cathode.

Jasmine Ho 3I

Go to practice sessions and make sure you're really good with the content we spent a lot of time learning!!

Jasmine Ho 3I

E naught represents the cell potential at standard conditions, while Ecell is the cell potential at non-standard conditions, which someone gave the equation to find above.

Jasmine Ho 3I

They should be; I don't think Professor Lavelle would expect us to memorize any E° values.

Jasmine Ho 3I

Thermodynamics was really tough, but I am glad to have gotten through that midterm and to be able to destress for like a week.

Jasmine Ho 3I

In a setup like that, I believe the anode will always be in the left beaker, and electrons always flow left to right.

Jasmine Ho 3I

Hi! Refer to the table of standard reduction potentials (linked if you click Hint). Professor Lavelle said in class that the half reaction with the lower standard reduction potential will be reversed, and this will end up being the oxidation reaction.

Jasmine Ho 3I

Yes, the cathode has a more positive charge, so electrons will flow to the right, and the reaction on the right will always be a reduction.

Jasmine Ho 3I

I've been eating ramen and watching youtube and my weekend has been very nice so far<3

Jasmine Ho 3I

Yes, in all of the equations Professor Lavelle has given us with a capital K, the K is the equilibrium constant.

Jasmine Ho 3I

If a system is acting against a constant external pressure, the reaction is irreversible. I think the problem would most likely tell you if you had to find work for an irreversible reaction since they do not occur in the real world. The equation for work for an irreversible reaction is w = -PΔV. The...

Jasmine Ho 3I

I would say it's not necessary to know how to derive the equation since exams are multiple choice.

Jasmine Ho 3I

If you have q(system) = q(surroundings), I believe it depends on if the reaction is exothermic or endothermic. If it is exothermic, q(sys) is negative because the system releases heat into the surroundings.

Jasmine Ho 3I

I'm taking it next quarter hopefully along with LS7B and 14C.

Jasmine Ho 3I

I know this feeling so well, and lately I've been trying to set timers for myself to make sure I am productive for some period of time, and giving myself breaks frequently. It worked pretty well when I was preparing for midterms!!

Jasmine Ho 3I

Enthalpy is the total energy in a system, and entropy is a measure of disorder. One thing that helped me understand entropy is the more ways things can be arranged, the higher the entropy. For example, 1 glucose molecule has lower entropy than 6 CO2 and 6 H2O.

Jasmine Ho 3I

It is because water forms so many hydrogen bonds, which are a relatively strong intermolecular force (compared to something like dispersion forces). When water reaches its boiling point and is being converted to a gas, the energy is being used to break the hydrogen bonds in the water so that the wat...

Jasmine Ho 3I

For this all I can really say is writing out your work clearly and including units. Also double checking often.

Jasmine Ho 3I

All three examples that we went over in class allow you to find the change in enthalpy for a given reaction. However (I'm not sure if you had this problem on the MT), you could also find a certain bond enthalpy using the 2nd method if you are given the right information.

Jasmine Ho 3I

I think I prepared as much as I could have for the exam but my nerves really got to me. Overall it was mostly what I expected though.

Jasmine Ho 3I

I believe heat capacity is referring to the heat required to raise the temperature of an object of any mass by 1 degree C, and has the units kJ.K^-1. To get specific heat capacity from heat capacity, you would divide the heat capacity by the # grams of substance present. This gives you the units kJ....

Jasmine Ho 3I

I'm hoping to take it next quarter with 14C but I'm a little worried because I heard enrollment is pretty competitive.

Jasmine Ho 3I

treating myself by binging netflix for the entire weekend woooo

Jasmine Ho 3I

I believe they're converting kPa to atm because that is what the formula for K requires. 1 atm is 101.325 kPa.

Jasmine Ho 3I

I think that they gave us the concentration of NH4HS so we could recognize the fact that it has no effect on the equilibrium constant. And I believe we would still use +x for all of the products because it is always true that if x of one product is formed, then x of the other product is formed. The ...

Jasmine Ho 3I

Actually I believe the answer key did use the value for 1000 K. Make sure you're looking at the equilibrium constant for gases!

Jasmine Ho 3I

I think that if you've done a lot of practice with the calculations then you should be fine with time.

Jasmine Ho 3I

It's 10^(-pH).

Jasmine Ho 3I

For 14A we only needed 50 by the end of the quarter, so I think if you have 100 by the end of the quarter it should be fine. I don't think the TA's keep track of how many posts we started with.

Jasmine Ho 3I

I set study goals for myself every day in order to avoid having to cram everything before exams.

Jasmine Ho 3I

From the information given, you can compute the equilibrium concentration of O2. Using stoichiometry, you can also get the equilibrium concentration of SO2 (2 moles of SO2 produced per 1 mole of O2) and the equilibrium concentration of SO3 (2 moles of SO3 consumed for every mole of O2 produced, and ...

Jasmine Ho 3I

I just watched Mulan last night for probably like the 50th time!!

Jasmine Ho 3I

I want a regular sleep schedule and to move around more.

Jasmine Ho 3I

I remembering seeing fractions as coefficients in my high school chemistry class so I think it's still correct. I assume you would plug everything into the equation the same way, or you could make all of the coefficients whole numbers.

Jasmine Ho 3I

Warm drinks and netflix are my favorite!!

Jasmine Ho 3I

increase/decrease of pressure does not necessarily mean the volume changes. Changing the volume of a (gaseous) solution is just one way to change its pressure.

Jasmine Ho 3I

When the pressure of a gas changes (because its volume changes), the K value stays the same for that reaction. However, the concentration of the solution changes because c = n/v. When you put the new concentration values into the equation Q = [P]/[R], you see that the reaction quotient Q is unequal ...

Jasmine Ho 3I

This exam definitely did feel more rushed than the midterms :// For both midterms I had time to go over all of my answers with time left and I ended up submitting this final in the last 10 seconds.

Jasmine Ho 3I

Not sure about the grade, but Professor Lavelle sent an email out today saying we have until Tuesday 9am to get our 50 posts. Also, Chem community will be closed to posts from 11:30am to 10:30pm today during finals.

Jasmine Ho 3I

I believe ferrate is only used when the coordination compound has an overall negative charge; otherwise you can just use iron. I also didn't know about this until I did the textbook problems so I'm not sure if this will show up on the exam. And from what I know, there are a few other similar excepti...

Jasmine Ho 3I

You can look at this entire compound as an ionic compound, where K+ is the cation and the coordination compound is the anion. Since the whole compound is neutral, the coordination compound must have a charge of -1. So, since each CN has a charge of -1, Cu has a charge of +1.

Jasmine Ho 3I

Hi Dr. Lavelle!!
I genuinely enjoyed your lectures so much and wish so badly I could have taken this class in person. I am so grateful for all the resources you set up for us, as well as the TAs and UAs who worked so hard, and I am looking forward to 14B next quarter!!

Jasmine Ho 3I

You can get the coordination number by looking at how many atoms/compunds are inside the brackets (aside from the transition metal), because those are directly bonded to the transition metal.

Jasmine Ho 3I

a bit stressed, very tired of being in my house but overall okay thank you for asking!!

Jasmine Ho 3I

Don't think Dr. Lavelle ever mentioned anything like that, so probably not, but it would be very nice.

Jasmine Ho 3I

Yes, I think proctoring is the same, but the exam is 30 questions, 6 pts each.

Jasmine Ho 3I

With 6 bonds, a planar structure would never be ideal because the regions of electron density want to be as far from each other as possible.

Jasmine Ho 3I

Try to visualize an H2O molecule in 3D. The regions of e- density want to be as far from each other as possible, and have a square planar configuration with only 4 regions of e- density leaves 2 sides of the central atom empty. Professor Lavelle showed us a representation of what an atom with a tetr...

Jasmine Ho 3I

If you look at the two models, you will see that the first one has 6 regions of electron density (with 2 lone pairs) while the second one only has 5 regions of electron density (with 1 lone pair). :)

Jasmine Ho 3I

When looking for a cyclic structure, I would just start by forming a simple ring (in this case, a triangle) with the carbons and adding all of the hydrogens after. Then, add bonds wherever necessary so that each carbon has 4 bonds.

Jasmine Ho 3I

Yes, because the tetrahedral shape makes it impossible for all of the dipoles to cancel out unless all 4 are the same magnitude.

Jasmine Ho 3I

I believe you can only tell if something can be an acid or base by looking at the Lewis structure. A base has an extra lone pair and an acid can accept a lone pair because it is electron deficient (ex. H+ ion or BF3). As for covalent vs. ionic, you would have to have an idea of their relative electr...

Jasmine Ho 3I

Yes, you will lose points.

Jasmine Ho 3I

Yes, all you really have to look at is the electronegativity difference between the two bonded atoms. The smaller the difference, the more covalent character.

Jasmine Ho 3I

If all of the dipole moments cancel out (by being of the same magnitude and in exact opposite directions), then the molecule will be nonpolar.

Jasmine Ho 3I

If you look at the Lewis structure for this molecule, you can see that there are 5 regions of e- density, and 3 of them are occupied by atoms. The regions of e- density are arranged in a trigonal bipyramidal shape, and the two lone pairs occupy 2 of regions in the equatorial plane. So, if you were t...

Jasmine Ho 3I

It is nonpolar because the two dipoles are going in exact opposite directions. When you add two vectors of equal magnitude and that are opposite direction, they cancel each other out. In the case of cis-dichloroethene, you can see that both vectors are facing right (one to the top-right and one to t...

Jasmine Ho 3I

I believe there's a table in the book or you can look up a VSEPR chart online and you should find everything you need!

Jasmine Ho 3I

Since the observed bond length is very close to the expected length of a Cl=O bond, you know that the bond has greater Cl=O bond character than Cl--O bond character. So I just chose the models that had more Cl=O bonds than Cl--O bonds.

Jasmine Ho 3I

The ones that contribute the most are the ones that have formal charges closer to zero for all of the atoms in the molecule because neutral atoms require the least energy and therefore are more stable.

Jasmine Ho 3I

Essentially that question was just asking how close the observed bond lengths matched up with the expected lengths for a single or double bond. You could say it has ample bond character for all of the expected bonds because the observed lengths were between both and not very noticeably closer to one...

Jasmine Ho 3I

Dipole-dipole interactions occur between two polar molecules (which have permanent dipoles). Dipole-induced dipole interactions occur between a polar molecule and a nonpolar molecule, where, say, if the positively charged end of the polar molecule comes close to one end of the nonpolar molecule, tha...

Jasmine Ho 3I

I think that since same charge particles repel each other, it would be lower energy if the charges were separated but I'm not completely sure.

Jasmine Ho 3I

They are the three orbitals in the p (n=1) subshell.

Jasmine Ho 3I

The structure with the lowest energy would have formal charges closer to zero. However, I think it is also important to consider what charges are next to each other because two negatively charged atoms repel each other = more energy.

Jasmine Ho 3I

The number of valence electrons corresponds to that element's column in the periodic table, ignoring the entire d-block. So, any element in the 1st column has 1 valence electron; any element in the "5th" column (the column with N at the top) has 5 valence electrons, etc.

Jasmine Ho 3I

For H and He, it's because those only have electrons in the 1s orbital (H has 1, He has 2), so they cannot have a full octet. Not really sure about the other 2 to be honest, but thanks for reminding me to look into that!

Jasmine Ho 3I

I believe neighboring atoms just means 2 of the same atoms that are directly next to each other with no space in between.

Jasmine Ho 3I

Essentially the physical pull on the electron by the positively charged nucleus is weaker on electrons in the outer layers. This means that the outermost electrons are more easily removed from an atom. For example, Ca has 2 electrons in the n=2 subshell, so those are shielded by the inner electrons ...

Jasmine Ho 3I

You can use the de Broglie equation to get the velocity of an electron in the beam and then use the formula KE = (1/2)mv^2 and convert to eV.

Jasmine Ho 3I

Ionization energy is just the amount of energy to remove an electron from an atom in the gas phase. First ionization energy refers specifically to the amount of energy to remove the first electron to create a cation with a +1 charge, and second ionization energy is the amount needed to remove a seco...

Jasmine Ho 3I

I like to watch Netflix and listen to playlists and sometimes just stare at my ceiling for a little bit when I've been thinking for too long.

Jasmine Ho 3I

Just wondering, but how come the uncertainty value is just 2x4 from 14.3 +/-4? If I were to write it out the long way, would it be since that the range is between 14.3-4 = 10 and 14.3 +4 = 18.3, and that the difference between the two is 8? I just need some clarification! I’m not sure where you got...

Jasmine Ho 3I

For this problem, just know that the electron can move from n=6 to any lower energy level, meaning n=5, 4,3,2,1, which then gives 5 spectral lines. Then you find the energy of the n=6 to n=5 transition and get the wavelength for that photon, and do the same for the n=6 to n=1transition. Subtract the...

Jasmine Ho 3I

Uncertainty is mainly about the range of values that the real value could be in. The actual value of the velocity doesn't really matter. For the question you specified, the uncertainty in velocity would be 4 m/s. Thank you for the clarification! I have a better understanding of it now. Do you know ...

Jasmine Ho 3I

The steps you listed sound correct to me. Did you convert the wavelength from m to nm? That's all I can think of unless you maybe mixed up some variables.

Jasmine Ho 3I

I believe you made an error when you converted g of H2O to g of H because there are two H atoms in and H2O molecule. The mass of H should be double.

Jasmine Ho 3I

Yes it is since the mass of the empirical formula adds up to 135.21.

Jasmine Ho 3I

yes you did :)

Jasmine Ho 3I

Yes, the intensity of light is related to the number of photons! More photons = brighter light

Jasmine Ho 3I

Jasmine Ho 3I

You were correct in finding the energy of the photon, but I think you made an error when subtracting E-K. Since it is approx. (8*10^-19 - 3*10^-19) you should get an answer closer to 5*10^-19 as your work function. Maybe you plugged in the wrong value for excess energy?

Jasmine Ho 3I

You are correct that moles of solute don't change when diluting a solution. The original 10 mL of solution contains .0010 mol of solute, and that number of moles stays the same throughout all 90 doublings of the volume. To find the final volume, you calculate 10*(2^90) , since one doubling is 10*2, ...

Jasmine Ho 3I

I make a chart of every element for both reactants and products to keep track of how many of each element I have. I don't think I necessarily follow an order; I just go with what seems the most obvious at first, I think.

Jasmine Ho 3I

You can multiply the volume of solution used in the reaction by the concentration of the solution (given in g/mL) to find out how many grams of 2-butanone was reacted, and then convert that mass to moles using the molar mass of 2-butanone. I assumed the equation was balanced, but you could check to ...

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