CHEMISTRY COMMUNITY

David Liu 1E

yes, the anode is always oxidizing and the cathode is always reduced

David Liu 1E

I agree that we need metals with a lower reduction potential as they would work better

David Liu 1E

it means that we need no reactant, and we would have a 0 poewr exponent because the ratio of the two (if calculating this way) would be equal to one

David Liu 1E

we use the equation specifically for half life for zero order reactions! which is the equation everyone else has stated

David Liu 1E

like everyone else said, you look at the sum of the rate law exponents!

David Liu 1E

you can figure it out by looking at how the half life changes related to molarity, and then look at the half life equations and find which one fits the proportions of molarity related to the time of the half life

David Liu 1E

The slow step should have the highest activation energy, and you would also find the products of the slow step in the final reaction!

David Liu 1E

Dear Dr. Lavelle, Thank you so much for teaching us for 2 quarters (14a and b) especially during quarantine! Thank you for putting in so much effort recording lecture videos and making them really enjoyable and happy during this time. The transition to online learning has been hard, but you've made ...

David Liu 1E

i agree with above statements! It states in the question that it's a first order reaction, so we take the equation for a half life first order reaction.

David Liu 1E

because we're looking at the slow reaction, we use that one and then replace hclo with the equation from the first as it's an intermediate product!

David Liu 1E

potassium just helps with the concentration and pH of the solution to help with ion movement i believe

David Liu 1E

i agree with the person above! you half reaction sshould be Zn2+(aq) + 2e– → Zn(s) and Au3+(aq) + 3e– → Au(s), which you can then use to calculate the E value

David Liu 1E

anodes are for oxidized substances while cathodes are generally for reduced substances; as the person above said~

David Liu 1E

the anode is always on the left and the cathode is always on the right!

David Liu 1E

these two would have the same values if the number of moles were the same!

David Liu 1E

leo is just for substances that are oxidized, it doesn't mean anything beyond that!

David Liu 1E

it would be 0 alone, but -2 in another compound!

David Liu 1E

I use oil rig! (oxidation is lost and reduction is gained) and like the comments above lavelle also uses leo the lion goes grr (which I don't use as much)

David Liu 1E

the coefficient doesn't matter because it doesn't affect the oxidation numbers for the elements

David Liu 1E

you should first look at the oxidation numbers of all the parts!

David Liu 1E

You have to find the change in entropy of both the volume and pressure, and then add them together, (making sure to keep the signs in check!)

David Liu 1E

you generally solve it through the equation, but it's good to check first based on concept whether or not it changes due to an increase or decrease in entropy

David Liu 1E

d is for derivatives and delta is always for change!

David Liu 1E

^^sorry to add on you don't use the neutralization reaction at all, so it's not something to worry about!

David Liu 1E

all you do is calculate the specific heat of the calorimeter and then plug that in to find the q of the second

David Liu 1E

you first find q with nCdeltaT and then you use the equation delta E/U = w+ q

David Liu 1E

also the reverse is true, which is Cv=Cp-R which is what you use in the sapling question!

David Liu 1E

The equation for delta u is the same except you use cv instead of cp

David Liu 1E

Yea it just looks at VESPR structure and drawing out the molecule would help determine if it's linear or not

David Liu 1E

The setup looks good, it's probably just the temperature difference mentioned above^

David Liu 1E

I don't think it was mentioned in lecture either (at least not the ones associated with this homework), but Lavelle did mention the difference between Cp and Cv when he first explained them in Lecture 9, and I just assume that means we should probably know this formula. It's also the one used in the...

David Liu 1E

Because it stated that it was an ideal gas, we first find the number of moles of the gas with the ideal gas law!

David Liu 1E

Also, check to make sure yours is CO2, mine was NO2 so I used the nonlinear ratio!

David Liu 1E

Delta U can also be calculated with the equation: ΔU=nCVΔT

Essentially you solve it the same way, but you would change Cp to Cv (constant volume molar specific heat). I don't know if it was mentioned in lecture, but I looked it up for this question and Cp and Cv are related with the equation Cv=Cp-R

David Liu 1E

I agree that I solved it using the equation people talked about above, and my answer to both parts were the same because delta E = w + q, but in this case we don't have work because the volume stays constant so you just solve for q

David Liu 1E

you have to add 0.000018 and 0.32 i'm pretty sure because both of those numbers would be part of the initial molarity. So it would be 0.000018/0.000018 + 0.32

David Liu 1E

pKa is the negative log of Ka!

David Liu 1E

he said less than 10^-4 during lecture you can ignore the bottom x, and like the person said above, once you check the ionization/deprotonated species % that also has to be less than 5% once you ignore 10^-4

David Liu 1E

it's based off the coefficients!

David Liu 1E

Percent ionization is the calculated x value (A-) divided by the initial molarity (AH) x 100%

David Liu 1E

we mainly need to know atm and torr!

David Liu 1E

you multiply them to add them together!

David Liu 1E

sapling gives everyone different values, but we're graded on completion so you should always try your answer if you think you solved it!

David Liu 1E

even though both are positive, only one should make sense given the conditions of the question

David Liu 1E

it would shift right because according to the equilibrium there would now be too much reactant and it shifts to the products in order to achieve equilibrium again

David Liu 1E

k is only affected by temperature!

David Liu 1E

make sure to factor/remember to use inverse reactions for these types of questions! It was something I forgot about when I first did it too

David Liu 1E

A large k value is anything over 10^3 and a small value would be anything lower than 10^-3!

David Liu 1E

should be bars and atm for units that we've seen so far!

David Liu 1E

for the questions that we've discussed, I think that they mean the same thing for calculations (though it'll get harder and we'd need to know the difference later)

David Liu 1E

you should look at what accepts/loses an electron/proton and that will tell you which ones act as an acid or base! You could also memorize certain patterns like how anything that starts with h is generally an acid and anything ending with oh is usually a base (and the group 1 and group 2 patterns)

David Liu 1E

a ligand is just a lewis base (donates electrons) to a metal usually to form a coordinate compound

David Liu 1E

lewis acids/bases focus on electrons, while bronsted focuses on the transfer of h+ ions

David Liu 1E

I think it's normally j/photon, but you would check with the question to make sure because sometimes it can change to j/atom

David Liu 1E

paired electrons and parallel spin are different! paired electrons are the 2 that would be in the ml orbital with opposite spins (-1/2 and 1/2) and parallel electrons are ones that have the same spin meaning they're definitely not paired

David Liu 1E

the more electronegative an atom is the stronger acid it'll be generally

David Liu 1E

I agree with a lot of the responses mentioning that you most likely have to look at the definitions of what you want but a rule of thumb would be to look at the H and OH groups at the ends of the compound

David Liu 1E

it can act as either depending on the question and the other compounds

David Liu 1E

the carboxyl groups shows that it's an acid!

David Liu 1E

do you guys know if we'll need to know this by memory for this class/series? wondering if it's something I should look at before the exam in case it shows up

David Liu 1E

I would definitely first look at the bond angles, because not every molecule with lone pairs would be polar as they could be evenly spaced from each other

David Liu 1E

for this class c is always the constant value for speed of light!

David Liu 1E

My ta mentioned that you would get the best answer by being as exact during calculations and rounding at the end only for the best/safest results

David Liu 1E

definitely should be 180! I checked my textbook and it says 180 if that helps

David Liu 1E

they have different amounts of electron dense areas - trigonal planar has 3 electron dense (paired) areas while t-shaped has 5 electron dense areas while 2 are lone pairs (aka there should be a different amount of electrons on the molecule that will help you tell)

David Liu 1E

You would first count the valence electrons and then create the dot structure around it based on bonds that you need to create and lone pairs

David Liu 1E

I would just focus on each one individually, as the structures of each tend to be self-explanatory. If the overall structure affects it, I would focus on that after looking at each individual carbon

David Liu 1E

lewis acids are compounds/elements that accept electrons, and bases will donate electrons. acids and bases donate/accept in lone pairs, so it's a matter of seeing which ones can donate a lone pair or give one up.

David Liu 1E

like lavelle said in lecture, I feel like it's important to note that the "normal" elements we use generally follow the trend, but any element that has access to the d orbital past period 3 can generally overcome the octet rule

David Liu 1E

it should be false because intensity is based on how many photons there are, while energy is focused on the individual photon!

David Liu 1E

the homework is due tonight! it's always the same time as chem community too~ :)

David Liu 1E

with ice it forms a lattice structure that actually makes the structure have less density, because water is more dense in liquid form as it flows freely and forms/reforms bonds while in ice all the molecules are structured

David Liu 1E

you locate the amount of N,O,and F atoms that a hydrogen bond can form, and then you check how many lone pairs each one of these have for a hydrogen bond to form with!

David Liu 1E

it doesn't need to be attached to a hydrogen, but you need ot make sure there's a partial charge on the molecule that it's attached to, which is why in the sapling homework it generally can't attach to a carbon that has 4 bonds (because it becomes nonpolar)

David Liu 1E

I think it's really helpful to draw out the structure because LD forces can only be used with nonpolar molecules, so drawing out the structures generally gives you an idea of how they should work, besides a few exceptions

David Liu 1E

thank you for organizing this! hope we all do well :')

David Liu 1E

double bonds hold electrons together more closely, which is shown as it takes more to break a double bond over a single bond. Because they're more tightly bound, they tend to be smaller

David Liu 1E

flourine only needs one electron, which means that it has a high electronegativity as it doesn't need a lot of electrons to fill it's outer shell

David Liu 1E

sulfate/sulfur can have expanded d- orbitals; I'm not entirely sure why it happens, but because it has something to do with the d-orbitals elements in the second period can't have expanded octets, which is why sulfur is an example

David Liu 1E

covalent bonds generally share electrons in a pair, while ionic bonds generally result in ionization where one atom donates electrons to another

David Liu 1E

the max l value is always one less than the n value, and you can tell what it's in by the orbital that's stated. s=0, p=1, d=2, f=3 and so on

David Liu 1E

Do we need to know how to identify intramolecular forces? I'm pretty sure there are only 3 intramolecular forces (ionic, covalent, and metallic bonding) and it's not too hard to remember so I would just make sure to know the difference between the two. Intramolecular forces are bonds because it all...

David Liu 1E

electronegativity is how much an electron is drawn to an atom, and we can see the trends on a periodic table (as we probably won't have to know specific electronegativity numbers for this class)

David Liu 1E

N, O, and F have the strongest attraction to a hydrogen because they are all really close to their octet, and it was mentioned that in most of the time we would see hydrogen primarily bind to these elements

David Liu 1E

I'm pretty sure ionic bonds are stronger for this class, but as comments above have said, it seems that covalent bonds are stronger for chemistry

David Liu 1E

does anyone know if we just use the equation sheet from his website, or will he email another sheet out? I remember someone telling me that their TA said Lavelle would email one for us to print for this test specifically

David Liu 1E

Avogadro's number is just a set number of anything, and we then assign to units to it based on what we're looking at. Essentially it's like the number 10 or 5, and we assign units to this number to represent what it is (i.e. molecules of water, eggs, test questions, etc.)

David Liu 1E

Hi this is probably another dumb question but why does electron affinity increase as we go up and to the right? I understand how electronegativity goes up but not affinity

David Liu 1E

it's always n-1, and I'm sure where in lecture he stated that, but it should only be for the 0 and 1 orbitals, and not the second!

David Liu 1E

everything seems right, so you should probably check the parenthesis of your calculator like the last person said! I've found that adding extra parenthesis is always better (better safe than sorry) and just as a trick you only really need the first parenthesis and you don't need to work on closing i...

David Liu 1E

I think that the levels are more of a concept to help us understand electrons, and as previous posts above have stated with the uncertainty principle, electrons probably travel without stopping to levels as they "jump", as an electron is only in one space for a split second. There's probab...

David Liu 1E

electrons have negligible mass and are mostly there for the charge/energy, which is when we use them for calculations! Protons and neutrons are around the same mass and are larger than electrons

David Liu 1E

from people i've talked to the getting a 50% is a c- until like 73% and then it's a normal grading scale

David Liu 1E

I'm taking notes/outlining the lectures and sample problems bc they're really useful for understanding the concepts first! I've been doing sapling problems for hw after and then reviewing a few textbook questions on subjects I'm struggling in! I think that starting from the lectures and working your...

David Liu 1E

There isn't one way to do it, but I definitely find it easier to first try to balance out a molecule that has only one product with it on the right side because by attempting to balance it, a lot of the times the rest of the numbers will balance each other out through trial and error!

David Liu 1E

Actual yield should be able to be calculated given certain information from the problem, though the questions that we've done so far mostly have them given. I think that in the future we might get questions that ask us to calculate both given idealized information but I think all the questions this ...

David Liu 1E

There would be 3 sig figs on this because it specified 7.00 and all trailing zeros behind a decimal are significant. If the question had written 700 it would probably be one though, so I'm fairly certain it's based on how the question is worded.

David Liu 1E

If you're using molecular mass, you should get the empirical formula already in it's simplest state without decimals after calculating. You'll use whole numbers for the molecular formula too, so I think that even if you were to somehow get decimals it would be easier to work with whole numbers for t...

David Liu 1E

what I did on the homework problem was round the really close ones like oxygen (15.999) and selenium (78.96), but like ones that don't round as well I keep to one decimal point. I've gotten all the right answers with sig figs according to the textbook, so it's what I've been doing since it works.

David Liu 1E

The person who posted above is right! And I'm also kind of rusty on sig figs but I'm pretty sure 0.340 has 3 sig figs, because the sig fig rule is that trailing zeroes to the right of the decimal are significant, so it should count?

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