CHEMISTRY COMMUNITY

charleejohnson1L

the final will be posted on his website i think! we just have to be patient and wait for more information :)

charleejohnson1L

absorption is the process where a fluid thing is dissolved by a liquid or solid thing; adsorption is where the atoms of a substance kinda hang out on top of another thing, creating a surface tension / cover thingy

charleejohnson1L

just wait for the email from your TA! if you're in Lily's discussions, the tests are ready to pick up from her mailbox in young hall. :)

charleejohnson1L

so on test two, there was the question concerning what would happen to the cell potential in a concentration cell if the mass of the cathode electrode is doubled. I figured that since the mass is going up, the molarity will be increasing, the concentration difference will be larger. with a larger co...

charleejohnson1L

the equation sheet is the same and given for every test/exam in lavelle's class!

charleejohnson1L

okay i just understood this like 20 mins ago so here's my rant about this: so the E is the standard reduction potential, so the higher the value is, the more reduction is going to take place. when looking for the strongest reducing agent, since the word "agent" is used, we're really lookin...

charleejohnson1L

in h2o, the oxidation state of o is -2, so that makes the oxidation state of the hydrogen a +2, making it +1 for each hydrogen atom. i hope this helps a little bit :)

charleejohnson1L

if the gibbs free energy was positive, the book might have reversed the equation so that it was a favorable reaction

charleejohnson1L

i'm pretty sure that the standard gibbs free energy is in joules per mole, but i could be wrong. could someone confirm?

charleejohnson1L

that's a good question haha! i've just always remembered that negative delta g means the reverse reaction is spontaneous and the reactants are favored, but i've never really thought about it before. i hope someone can answer! :)

charleejohnson1L

since the standard potential is an intensive property, it won't change depending on how much of a substance is reacting. I hope this helps. :)

charleejohnson1L

if the reaction is spontaneous, chances are that the products will form, so the metal will dissolve into whatever products will be forming. if the reaction isn't spontaneous, then chances are that the products won't form, meaning the metal won't dissolve into any products. I hope this helps! :)

charleejohnson1L

Christineg1G wrote:it might help to see a visual of deriving the Nernst Equation!

this really helped, thank you!!

charleejohnson1L

JesseAuLec1Dis1G wrote:So is test two only on the second page of outline 4 and all of outline 5?

yes! thankfully :/

charleejohnson1L

this has always confused me as well, thank you for asking and thank you to those who answered!
is k the ratio of product concentration to reactant concentration at equilibrium?

charleejohnson1L

to my understanding, Ecell is an intensive property so it doesn't really care how much of the reaction is happening, just that it's happening. so you don't have to multiply it by anything in order for it to correspond to higher reaction coefficients. I hope this helps :)

charleejohnson1L

I'm going to study all of them to make sure I don't get punked, but I'm pretty sure the problems from chapter 5 correspond with the second page of the thermodynamics outline. :)

charleejohnson1L

I think the problems in chapter 5 onward :)

charleejohnson1L

I like to look at what I need my final units to be and choose my R based off of that! so if I'm looking for J/K as my final units, I'd choose to use 8.314 for my R value because when it cancels out with the other units in the equation, J/K will be my final units. I hope this helps. :)

charleejohnson1L

The cathode is the one being reduced and the anode is the one being oxidized. ^^ this! so this is just something else we have to memorize :/ the way that lavelle was explaining it, with the anode being on the left and the cathode being on the right, it's easier to remember because a comes before c ...

charleejohnson1L

MAC 4G wrote:Remember that pressure is inversely proportional to volume (Boyle's Law), so at a constant temperature (isothermal) V2/V1 = P1/P2.

oh! oops thank you for clearing that up for me!

charleejohnson1L

An example to look at would be Mn2+ being reduced to Mn and Al is being oxidized to Al3+. Here, Mn2+ is gaining electrons and Al is donating electrons. So, using Le Chatelier’s principle, an increase in concentration of Mn2+ causes equilibrium to shift the reaction towards the products, which would...

charleejohnson1L

The question will usually state if the reaction is taking place in a basic or acidic conditions. Usually you will need to use H2O to balance excess oxygen, when you do this you will then add either H+ or OH- to the other side of the reaction to then balance the excess hydrogen from the H2O. If it i...

charleejohnson1L

you can cancel the H+ and OH- ions when you're combining your redox half reactions and making the final full reaction equation. after you're done balancing the two, you flip one of them, most of the time you flip the oxidation half reaction, and then some things will cancel and some won't. you add H...

charleejohnson1L

Transition metals are conducting metals, which is important because it allows for the transfer of electrons in the cell diagram situation. You have the salt bridge to keep the two solutions neutral, and the conducting metals to act as a carrier for the electrons, if that makes sense. I hope this hel...

charleejohnson1L

But don't you need a solid on both sides to avoid Pt? Or does that not apply when it is the same element being oxidized and reduced? Since there's only one element, there will only be one beaker in the cell diagram if you were to draw it, if that makes sense. Everything is happening in one place, a...

charleejohnson1L

I'm pretty sure we're just supposed to balance it with waters and whatever else is in the solution (hydrogen ions for acidic, hydroxide ions for basic). At least, this is what he wants us to prove that we know how to do on exams and stuff, I'm guessing. I hope this helps. :)

charleejohnson1L

Hi I'm reviewing thermodynamics and in the book there's this problem that says "Calculate the change in entropy when the pressure of 1.5 mol Ne(g) is decreased isothermally from 20.00 barr to 5.00 barr. Assume ideal behavior." I got -17.3 J/K, but the book says the answer is supposed to be...

charleejohnson1L

Platinum is introduced into the cell diagram when there are no solid metals as part of the anode or cathode of the cell diagram. It serves as an inert electrode (provides a surface for the transfer of electrons, but does not affect or participate in the redox reaction). ^^ this! also when you write...

charleejohnson1L

cis means same, so for a cis structured molecule, the alike things will be on the same side. I hope this helps :)

charleejohnson1L

it really helped me when my TA like, explicitly stated that you can only add what's already in the solution. so like if it says it's an acidic solution, the only thing in there except for the redox reaction is water and hydrogen ions, and if it says it's a basic solution, the only thing in there exc...

charleejohnson1L

^^ me too oops, thank you in advance! :)

charleejohnson1L

the biggest thing that helped me with this was like, explicitly realizing that you can only add what's in the reaction to your equations. so like in acidic solution, the only things in the solution except for your redox reaction is water and hydrogen ions and in a basic solution, the only things in ...

charleejohnson1L

^^ me too. but I'm pretty sure we can't anymore :(

charleejohnson1L

The salt bridge keeps the two solutions neutral so the systems can exchange electrons. I hope this helps :)

charleejohnson1L

Since you were given a solid thing in the equation, I'm not sure the Pt is needed. You just need to figure out what's being oxidized vs reduced and put them on the correct side of the "salt bridge". I hope this helps :) I'm not entirely sure if I'm correct, but I remember a problem like th...

charleejohnson1L

to easily identify the oxidizing and reducing agents, you just remember that they're backwards! the reducing agent is the thing being oxidized, and the oxidizing agent is the thing being reduced. as long as you can determine the oxidization and reduction half-reactions, determining the oxidizing and...

charleejohnson1L

My TA said to break it into two equations like the one reactant goes to each product in separate equations, then calculate charge and determine oxidizing/reducing agent from there. I hope this helps :)

charleejohnson1L

For those of us who doesn't understand LEO GER, I'm here to offer another acronym that I learned in high school that I find easier to remember: OIL RIG. Oxidation Is Loss, Reduction Is Gain. Happy balancing! :)

charleejohnson1L

I had the same problem :/ but my TA explained that the problem says that the work is done BY the system so that means it will be negative. It's just really important to make sure you pay attention to everything the problem is telling you before you start it, because that's how you forget the importa...

charleejohnson1L

The change in the entropy of the surroundings of a system are equal to zero when there is no pressure / the reaction is taking place in a vacuum / when it is a free expansion reaction. I hope this helps! :)

charleejohnson1L

The way he sees it, entropy is representative of the spontaneity of a reaction, like, bottom line. Other people see it as like, the more chaotic it is/ the more disorder there is, the more likely it is that the reaction will occur due to the probability of the atoms hitting at just the right time an...

charleejohnson1L

It has to do with whether you're calculating the entropy of the surroundings or the system and if the reaction is endothermic/exothermic, I'm pretty sure. It'll say in the problem what you're supposed to find. I hope this helps. :)

charleejohnson1L

You use Kp when dealing with partial pressure :) I hope this helps.

charleejohnson1L

If we took that into consideration, we could never use measurements because they'd always be expanding or growing or something like that. Chem's a lot easier if you don't think super deep about it haha. :)

charleejohnson1L

Consider it the inverse of the slope, and it'll be easy to remember. The logic behind it is a little tough for me, too, so I hope someone can answer that! :)

charleejohnson1L

You can only assume that the temperature is 298 K or 25*C if the question says something about ideal situations or standard temperature. I hope this helps! :)

charleejohnson1L

Like said above, Gibbs Free Energy is more stable when negative due to the spontaneity of the reaction. If the reaction isn't favorable/ spontaneous, then forcing it to happen may not result in the most stable products. I hope this helps! :)

charleejohnson1L

a state function is a property of something that depends on the current state, where a state property is a quantity independent of any states. I hope this helps. :)

charleejohnson1L

from what I understand, you can't make up any chemistry community credits. I talked to my TA because I didn't have service the first two weekends of class so I wasn't able to submit any responses and she said there was no way to make up those missed points. she suggested I just do good on future exa...

charleejohnson1L

i think that it's implied that if the system is at equilibrium that the reaction is reversible. i hope this helps :)

charleejohnson1L

I'm also kind of confused on this, but I think that it has something to do with open, closed, and isolated systems? maybe?

charleejohnson1L

the way I think about it is to imagine yourself as the system. if you're doing work, you're doing to be using energy (negative value) and if work is being done on you, that's energy saved (positive value). I hope this helps :)

charleejohnson1L

I think that we'll go over Ka and Kb more in depth in 14B, but that we should know what they are conceptually for the final. So like, we won't be doing calculations right now but we might have to use the conceptual understandings to specify which acids/bases will be strong/weak. I hope this helps. :)

charleejohnson1L

My TA told us that there will be one question per outline, but two for quantum and the other question is free game from the homework. :) I hope this helps.

charleejohnson1L

It's just how the structure is (one O has a double bond, the other is a single bond with an attached hydrogen). I think it might have something to do with its functions, but I could be wrong. I hope this helps a little. :)

charleejohnson1L

When you write the configuration, you write it in the order of lowest to highest energy level. :) I hope this helps, this is what Lavelle said during lecture.

charleejohnson1L

Since there are 5 valence electrons, when you hybridize the orbitals to get sp3, there will be one valence with a lone pair and three orbitals with one electron in it (that's the five valence electrons you're asking about). I hope this answers your question! :)

charleejohnson1L

They normally form acids. :)

charleejohnson1L

Keep in mind that Br is in the group that has 7 valence electrons, so it really only wants one more electron, so if there's more than one atom bonded to it, the octet rule is gonna be a lil wonky. Just always count the areas of electron density and you should be okay. :) I hope this helps.

charleejohnson1L

Lewis acids end up with a new H+ ion, while the Lewis base ends up with a more basic ion. :) Basically acids accept electrons while bases donate them. I hope this helps!

charleejohnson1L

My TA told me that generally if there is at least one atom that is not the same (amongst the outside 4 atoms of a tetrahedral) then the molecule as a whole is polar. This! If the atoms on the tetrahedral aren't all the same, chances are that there will be a dipole moment that doesn't cancel out, ca...

charleejohnson1L

I'm not sure, but I think we do need to know that the pH is the negative log base 10 of the concentration of H+ ions and the same for pOH and OH-. I hope this helps! :)

charleejohnson1L

So if there's two di-methyl ethane or something like that and they're bonding together, then you'd say bismethyl ethane or whatever molecule it is. I hope this helps! :)

charleejohnson1L

After drawing the lewis structure, the answers are drawn! sigma bonds are single bonds, I remember it with the alliteration "sigma single". pi bonds are double bonds, and I remember that one with the two lines down when you draw a pi symbol. :) I hope this helps.

charleejohnson1L

After drawing the lewis structure, the answers are drawn! sigma bonds are single bonds, I remember it with the alliteration "sigma single". pi bonds are double bonds, and I remember that one with the two lines down when you draw a pi symbol. :) I hope this helps.

charleejohnson1L

From what I remember from high school biomedical classes, sickle cell anemia is caused by a point mutation in DNA, that when translated into RNA and then coded for proteins, accidentally makes a different protein that causes the structure to change and make the weird banana shape. I may be wrong, bu...

charleejohnson1L

Google says that a ligand is an ion or molecule that binds to a central metal atom to form a coordination complex, but in actual english, a ligand is just like, a smaller atom that binds to a larger atom or molecule.

charleejohnson1L

Something starts boiling when the external pressure is the same as the internal pressure, but something melts when the energy is being absorbed by the solid from to become the melted, liquid form of the substance (the temperature which solid's vapor pressure is equal to the liquid form's vapor press...

charleejohnson1L

A linear molecule can have no lone pairs or it can have lone pairs. Since the angle remains at 180 degrees, these lone pairs must be oriented in such a way that they cancel out, permitting the bond angle to remain at 180. This often found on central atoms possessing three or four lone pairs of elec...

charleejohnson1L

Thank you for this! I've been struggling to recite my flashcards, but this helped a lot haha. :)

charleejohnson1L

To be considered T-shaped, the VSEPR label thing (whatever it's called) is AX3E2. It has to have a lone pair on both the top and the bottom of the "T-shape" in order to take that form. With just the one lone pair, the bonds will be repelling from the lone pair, rather than getting caught i...

charleejohnson1L

The geometry is based purely on electron dense areas, but the VSEPR model only uses the bonded parts, as the repulsion will cause all electron-dense areas to move as far away from each other. :) I hope this helps.

charleejohnson1L

Test 2 is going to cover everything that we started learning after the midterm, so like. Outline 3 and 4 on the website. Basically just chemical bonds and VSEPR. :) Hope this helps.

charleejohnson1L

It's during week 8 and it'll only cover the material after the midterm to week 8! :) I hope this helps.

charleejohnson1L

There's a hydrogen between an oxygen on thymine and a hydrogen on adenine, and a hydrogen on thymine and a nitrogen on adenine. it just has to do with the lone pairs on the oxygen and nitrogen. :) I hope this helps.

charleejohnson1L

Basically, in really broad terms, a dipole moment is found between two atoms who are sharing an electron. Due to one's higher electron affinity, it might hold the electron closer to it than its partner. The increase in negativity as the electron is closer to an atom is what the dipole moment is, if ...

charleejohnson1L

I'm not sure what you mean, but the H in H2O has hydrogen boding with the oxygen atoms within the molecule, if that helps answer at all. :/ For there to be hydrogen bonding, there needs to be either a nitrogen, oxygen, or fluorine atom with at least one lone pair bonding to a hydrogen. I hope that h...

charleejohnson1L

I had this question, too! Thank you everyone for posting the explanations :))

charleejohnson1L

No :) The formal charge on each element helps determine which lewis structure is "more accurate", as when the formal charge on each element is zero, the structure is more stable. So the formal charges will always be different depending on how many lone pairs, bonding electrons, and valence...

charleejohnson1L

To answer 2A13, you only need to name the orbital that the electron would be removed from, which you got correct. :) However, on your other question, there is only one electron removed to form an ion. It is rare for Cl to lose electrons in general, as it normally gains one to form Cl-. I hope this h...

charleejohnson1L

Okay, so when the 3d shell starts to gain electrons, its energy level drops and it ends up having less energy than the 4s energy level. Since the electrons are removed from the orbitals that are farthest away from the nucleus (and that encounter the most shielding), the electron would be removed fro...

charleejohnson1L

From what I understand, you always count the subshells when finding ground-state configurations. Even when finding an ionic configuration, you still count the subshells, just some of the electrons have moved to other energy levels so it may seem as if you "aren't counting" them. For Mn in ...

charleejohnson1L

In the ground state configuration, the number of protons matches the number of electrons. Helium has two protons, so it will have two electrons in its ground state! :) I hope this helps.

charleejohnson1L

Elements like neon have a full valence shell and meet the octet rule! They don't really have a need for more electrons, so they're less likely to form bonds. :) I hope this helps!

charleejohnson1L

This would be an important topic to know for the midterm! So when you look at the periodic table, electrons fill orbitals in that order. So since the elements that have 4s orbitals are in front of those that have valence electrons in the 3d state, they fill the 4s first. :) Over all, it's important ...

charleejohnson1L

I would memorize which series starts at n=0 vs n=1 and like, the scale of wavelengths (?, I'm talking about like, gamma rays having higher wavelengths than ultraviolet rays), but other than that, I don't think there's much of a use to memorize all these things that you most likely won't use. :) Hope...

charleejohnson1L

The negative sign is always present in the equation! :) Sometimes people don't use it when calculating because it makes sense that the final answer would end up being negative in any case. Hope this helps!

charleejohnson1L

The way I learned it (and I know this isn't very representative of how it actually works, but it's easy to understand like this), some electrons spin clockwise, and others spin counter-clockwise. In order to "have a full set," or "see all angles of the clock," you would need one ...

charleejohnson1L

My TA did this problem in discussion. He used v = R ( (1/n1^2) - (1/n2^2) to get n2. To get n1, just determine whether to use the Lyman series (for ultraviolet, starts n1 = 1) or the Balmer series (for visible, starts n1 = 2), as each starts on a different energy level. R is Ryderberg's constant. I ...

charleejohnson1L

Lavelle said that we'd get all equations and constants on every test, I'm pretty sure. :) He most likely won't provide all of the derivations, but those should be easy to get from the original.

charleejohnson1L

E= h (c(lambda^-1) and E = (hc) / (lamda) are the same equation. :) A negative exponent, in the simplest terms, just means that it's supposed to be in a denominator. I hope this helps.

charleejohnson1L

In a vacuum, there is less of a possibility of interference, so the data from the experiment is more accurate. :)

charleejohnson1L

When a photon is emitted, light is produced. When a photon is absorbed, the electron gets excited and moves up an energy level. I hope this helps!

charleejohnson1L

I'm pretty positive it's because they're unstable in singularity and/or they naturally occur in nature as diatomic. :)

charleejohnson1L

My AP Chemistry teacher told me to always do chem work in pen, so it's kind of a habit for me. I'm sure it'll be accepted either way, since it's the first assignment due and all. I wouldn't worry too much. :)

charleejohnson1L

Avogadro's number represents the amount of things inside one mole! Fun fact: it's dimensionless, as it represents different things. I say things because Avogadro's number can represent the amount of particles, formula units, atoms, etc. in one mole. :)

charleejohnson1L

When calculating molarity, you should convert it into the units given in the standard equation (moles and liters). From experience though, if you have a method to keeping track of your units, just go for it. Either way, you're still calculating the concentration of a solute in a solution. :)

charleejohnson1L

I've been basing my significant figures off of the molar masses indicated on the periodic table. If there is a given mass in the problem, match your percentages to that amount of sig figs. :)

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