CHEMISTRY COMMUNITY

t_rasul2I

I like handwriting my notes because I feel like there's more flexibility with diagrams and notations and what not. Also, it helps me keep everything together so I can use it later if I take higher level chemistry or need it for future standardized tests.

t_rasul2I

I am taking Chem 14C and Chem 14BL !

t_rasul2I

Based on last quarter, I am thinking next weekend?

t_rasul2I

This was very helpful, thank you!!! Dr. Lavelle also has some notes on his website that go over acid/base redox reactions and that really clarified a lot for me. I'll put the links down below: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Balancing_Redox_Reactions_Acidic_Conditi...

t_rasul2I

I think its just talking about identifying the anode

t_rasul2I

Im getting mixed up with when we use the plan Arrhenius equation separately to solve a problem and when we use the manipulated form. How was it derived? In the solution manual, Dr.Lavelle said to use ln(k/k')=Ea/R( (1/T-1/T') but in some of the other problems, its ln(k/k')=Ea/R( (T-T'/T*T')

t_rasul2I

Dr.Lavelle posted that the -.59 is a typo.

t_rasul2I

a) for how many steps it has, count the number of activation barriers the reaction has which is 3. b) the rate determining step is the slow step, which is the hump with the highest Ea since it has to overcome the most energy. This is the first step. c)the fastest step would be following the opposite...

t_rasul2I

I am really confused by this method as well and my TA said that it doesn't seem correct either.

t_rasul2I

Yes the oxidizing agent is what gets reduced. Essentially, it is helping the other species get oxidized by taking the electrons from them, therefore it is the oxidizing agent/"helper".

t_rasul2I

yes we follow the original equilibrium rules and raise to the power of the coefficients

t_rasul2I

n is the moles being cancelled out as people have mentioned above. It is super simple with practice and is just the number of moles we scaled and cancelled out when we add the half reactions together.

t_rasul2I

We add Pt whenever you can see the metal in question is aqueous in both the reactant and product side. You need to have a metal to have a galvanic cell work when the metals are aqueous so we use platinum.

t_rasul2I

Since this problem was given at 25 degree Celsius, it would be easier to use the other equation Dr.Lavelle gave us. Theres a lot more room for error when you're using the first Nernst equation because there's more numbers to plug in and divide at different points.

t_rasul2I

sapling was being super specific about the states of matter and took a long time to get the formatting right. Luckily the states of matter were already given and if it wasn't, it was either aqueous or liquid in the case of water.

t_rasul2I

I don't think degeneracy has units because the states raised to the number of Molecules is just a integer and ln of that is also an integer. The only unit owe would follow is the kb unit.

t_rasul2I

I think NO2 would have residual entropy and not CO2 is because of the molecular arrangement. I'm not sure fully, but I think the 2 equal double bonds in CO2 makes it harder for molecules to have entropy whereas the hybridized state of the NO2 would have more movement and freedom for the electrons to...

t_rasul2I

residual entropy is the entropy caused by the disorder of a number of molecules. It accounts for how many molecules there are and how many positions they can be arranged in.

t_rasul2I

I've noticed while doing the homework that when in doubt, use the 8.314. However, whenever there's atm in the problem, use the .08206. Worst case scenario is you use the wrong one but you will still have a similar answer with a different power. If it is a multiple choice exam, you can at least narro...

t_rasul2I

m and n are just used to differentiate that there may be different molar coefficient values each of the values. But both represent the molar coefficient and are the same thing!

t_rasul2I

you would simplify out the difference in temperatures multiplied by delta h over R and then take the e of both sides to isolate out the K.

t_rasul2I

Im going to try to enroll in Chem 14BL and Chem 14C next quarter too. I want to keep my memory fresh from Chem 14B going into the lab portion.

t_rasul2I

this was great insight! Especially during online test taking, it is so easy to make silly mistakes and not be in the zone. I think it is important we stay positive!

t_rasul2I

for me I put it straight from the calculator and I got it correct. I think there might be some answers sapling is iffy with.

t_rasul2I

Residual entropy is referring to the entropy caused by the permutations of molecular arrangements. If a molecule is like o-o where there can only be one permutation, there is no residual entropy. Moreover, this is because the number of arrangements to the power of the molecules would equal 1 and ln ...

t_rasul2I

enthalpy can be thought of as an overall energy of the system that is influenced by the molar heat capacity of substances. They are not the same but the molar heat capacity is what influences the overall enthalpy

t_rasul2I

Do ALL the textbook problems. And power through the ones you do not understand because those are the problems that usually have the most similarities with the tests. I recommend starting your studying early. I felt considerably more prepared on the tests that I started studying for earlier than the ...

t_rasul2I

So this problem can be done intuitively. We multiply the masses with their respective temperatures and add them together. Then we divide by the sum of the two masses. We are essentially getting the average temperature so we would want to find the total heat and see, on average, how much of that heat...

t_rasul2I

You might need to multiply the -pressure times volume by 101 J to convert to joules.

t_rasul2I

We know that the work value is negative because the work is done by the system, meaning the system is what puts out the energy.

t_rasul2I

the constant-volume molar specific heat is Cp minus R. Im not sure if this is the case 100 percent of the time but that is usually the process.

t_rasul2I

Chem is that one class where I don't realize the amount of time that has passed by (watching lectures, going to UA sessions, Sapling, Syllabus problems, etc). I'd say I sped 5-20 hours a week depending on how long the concept that week is.

t_rasul2I

I am planning on taking 14BL next quarter because I heard it is easier when you already have the knowledge of 14B.

t_rasul2I

If the bonds start out strong and go to weaker bonds, you know you need an input of energy to break the bonds (endothermic). If the bonds are weak to strong, you know energy is released with the formation of the stronger bond and is therefore exothermic.

t_rasul2I

X2 would require energy to break the bond since bonds need energy to be broken, where as 2 X molecules are already broken. This means that it is an endothermic reaction. And you know if it is endothermic and temperature is increased, it favors the products.

t_rasul2I

Hess's Law draws on the idea that enthalpies are a state function. Therefore it does not matter how the final enthalpy was obtained. With that being said, this means that the enthalpies are additive and can be manipulated based on molar amounts etc.

t_rasul2I

update: my TA confirmed that the 5 percent rule also works with partial pressure!

t_rasul2I

I'm not sure if this rule would apply to partial pressure but I would think if the partial pressure was significantly small, it wouldn't affect the equation when solving it out anyways. I will double check with my TA though.

t_rasul2I

you would leave out solids and liquids in their pure form because their concentration is not over a volume. Therefore, their concentration is exponentially larger than other species in the equation and that will throw off your answer.

t_rasul2I

you seem to be doing it correctly and set it up right. I would try holding onto as many significant figures as possible when completing the problems because sapling can be a little picky

t_rasul2I

like people have answered, you take the -log of the H+. However, to make it easier and know when to use pOH and pH, convert everything to the ka, pka, or pH and convert the answer after you have solved for the problem (assuming the problem doesn't have any intermediate steps requiring pOH)

t_rasul2I

stronger acids have a higher acidity dissociation constant (ka) which is because the ratio of products over ratio are higher when the acid is able to disociate more. This is why strong acids don't have a ka. They have an exponentially larger number of products to reactants and so it looks like the [...

t_rasul2I

you always compare the pka to its environment. if the environment has a lower ph than pka, then the environment is already as protanated as it is and so the pka species will not want to give off more protons (therefore it is going to stay neutral. If the pH is higher, then it is not protanated enoug...

t_rasul2I

If Q is less than K the reaction is favored to the products and to the right. There are more reactants in this scenario. If Q is greater than K, the reaction is favored to the left and reactant side. There are more products in this scenario.

t_rasul2I

You would solve for kc first and then equal that to the ice box equation containing that is set up using the second NO concentration. Then you find x and you multiply it by 2 because there are 2 NOs and subtract it from that second NO concentration.

t_rasul2I

I think you will typically see mols over liter since it is standard. However, I could also expect some other concentration metrics and I believe that would be fine as long as you stay consistent and the question of the problem asks for it in that metric

t_rasul2I

we inverse kc when we are flipping the reaction or in other words using the reverse reaction. A + B -> C w a k of 2 would have a k of 1/2 if the reaction was flipped and C -> A + B

t_rasul2I

like others have said in the post, the temperature definitely impacts which side is favored and this depends on if the reaction is endothermic or exothermic. Basically you have to see which side is something (in this case temperature) added and that will push reaction to the other side.

t_rasul2I

If you refer back to 14A, you will remember we consider molarity or concentration with mols divided by volume. n/V is the algebraic representation of mols divided by volume.

t_rasul2I

Like the comments said before, you will most likely use the standard R constant 8.206 x 10^-2. However if you have the other variables to solve for R or have different units, you would reevaluate how you find R. I don't think we have to worry too much about R in general.

t_rasul2I

anything y2k like Mean Girls, Clueless, or Legally Blonde or Disney teen movies are so comforting and remind you of the good ol' days.

t_rasul2I

last quarter I was behind a couple of chem community points and I received the full 50 in the end by making them up throughout the quarter. I wouldn't recommend waiting on chem community points, however, to make sure you are updated on what's happening in the class.

t_rasul2I

Partial pressure can be broken down into its words: partial because each gas exerts a fraction of the total pressure and pressure to represent pressure

t_rasul2I

I think watching the lectures and doing the textbook problems ultra thoroughly is the best way to go. The material itself is not much but its the application that can make it confusing and so practicing the application is very important

t_rasul2I

I felt rushed as well, especially because transferring the question from the screen to my paper was time consuming. But I am grateful Dr.Lavelle is giving points based on that.

t_rasul2I

Dr.Lavelle said grades should be in by Saturday!

t_rasul2I

They don't really affect pH because the interaction they have with a water molecule is not strong enough to change the water molecule and generate H+

t_rasul2I

I was wondering if hydrogen bonding impacts the acidic or basic nature of a compound. Like if HF vs. HCl is dissolved in water, would the hydrogen bonding in HF impact the Ph? I know the bond lengths of the two would make HCl more likely to donate a proton. But wouldn't the hydrogen bonding in HF ma...

t_rasul2I

Is it just me or are the recorded lectures glitching again?

t_rasul2I

I am blessed to have a designated desk in my room so I lock myself in my room, hide my phone in a drawer, and grind. Sitting at a designated study space makes me feel like I'm productive even though I am not. Moreover having that space is nice for me because I often have my mom basically screaming o...

t_rasul2I

I once emailed Dr.Lavelle about Chem Community and he verified that if you are missing a couple of points here or there from CC, you can still make up for those by the end of the quarter.

t_rasul2I

I was so lazy in week 8 because of thanksgiving and all. But I did get some nice retail therapy in to treat myself for doing the bare minimum. *Sigh* I'm already exhausted from finals week and I haven't even done anything but think about it.

t_rasul2I

negative poles are toward the most electronegative regions. it's kind of similar to how negative formal charges like to stay with the most electronegative atoms

t_rasul2I

the type of bonds do not affect the shape of these models, only the strength of the bond. What does affect shape are the lone pairs and their repulsions.

t_rasul2I

Since its a resonance structure, all the bond lengths and charges will distribute in real life and so you will be left with even distributions of bonds and charges to form a tetrahedron shape. Moreover, since the elements bonding to the central atom are the same, you know it has dipole moments that ...

t_rasul2I

I think the one you might be missing is using a ring of a triangle to connect the carbons together. This should help you get that last resonance structure.

t_rasul2I

Hybridization, as mentioned above, is the mixing of orbitals. To find the hybridization symbol (which is essentially what the homework asks for most of the time), count how many areas of electron density there is and divide it into orbitals. 1 area of electron density is s and 4 areas of electron de...

t_rasul2I

Hey David, Yea it's kind of weird because for the textbook problems, it seemed like formal charge wasn't what determined which atom has the unpaired electron. However, from what I've seen it seems like the most electronegative atom will want the full octet, so the less electronegative species will ...

t_rasul2I

the carbon atom has the negative formal charge so I think that's why iron binds to it. The same principle applies with CN or cyanide which is why both of these compounds are deadly.

t_rasul2I

im not going to lie, but I forgot everything about debroglies wavelength or e=hv or stuff like that.

t_rasul2I

London dispersion forces are always present and in every molecule including non-polar forces. There are more dispersion forces in bigger molecules because there is more surface area essentially. Therefore, bigger molecules have higher boiling points.

t_rasul2I

You would have to consider the points in the class. 100 points are from weekly homework stuff, 220 are from midterms and 180 is from the final. You can miss 50 points to get an A- (3.7) in the class and 35 points to get a A (4.0) in the class. Calculate all the points you have right now out of avail...

t_rasul2I

I run about 5 miles a day. Ok so originally I started working out and running this January for the sole reason of getting ready for college and looking sNaTcHeD and literally no intentions of being healthy. It was literally for all the wrong reasons. But then I liked it eventually because exercising...

t_rasul2I

I totally feel you. Smh I feel like I mastered most of the concepts but then when I take the tests, I make such silly errors and feel kind of bummed out since each point has so much worth in our grade. But we will get through this!!!!!

t_rasul2I

Dipole forces are intermolecular forces (between different atoms) and ionic forces are intramolecular forces (between on atom).

t_rasul2I

No there are definitely other Lewis acid/bases reactors like people have mentioned above but group 13 elements are kind of like a guarantee that they are involved in this acid base reactions.

t_rasul2I

coordinate covalent bonds are when a lone element like F comes and donates both of their electrons to create covalent bond. Basically, the electrons are not shared and one of the compounds donates their two electrons to create a bond. This happens where there are partial positive and negative charges.

t_rasul2I

Hi Kaitlyn. You can identify radicals by normally drawing out a Lewis structure and assigning formal charges to find the lowest formal charge. An example is laughing gas or nitric oxide (NO) and seeing that the lowest formal charge is when N has a radical.

t_rasul2I

Like the other posts have mentioned above, dipole dipole forces happen between two polar molecules because their charges are not distributed evenly and so one positive end of a molecule will bond to one negative molecule of another molecule. A dipole induced force means that one strong nearby force ...

t_rasul2I

rod structures have more surface area (because they are longer and stretched out) and so they will have more areas to bond whereas a spherical structure does not have as many areas to bond. As a result, 2 rod structures are more strongly held together than two spherical structures.

t_rasul2I

Based on the lecture on Wednesday, I think the main point of talking about radical was to show that they are compounds with unpaired electrons in an unstable, reactive state and that there are compounds in real life (like Vitamin C) that minimize radicals.

t_rasul2I

Electronegativity does not directly affect atom size but there is correlation. Typically cations are smaller and they have less electronegativity and anions are larger and they have less electronegativity. Anions are smaller because they have more e- (more electronegativity) and so they have a large...

t_rasul2I

atomic radii increase down a group and decrease across a period. cations are typically smaller and anions are larger. Ionization energy decreases down a group and increases accross a period. Electron affinity increases across a period.

t_rasul2I

Another simplified way to calculate formal charge is to subtract the number of "things" (symbols) around the element of interest from the number of valence electrons. Its not a precise way and I recommend using Dr.Lavelle's method to ensure you know the reasoning behind the formal charge, ...

t_rasul2I

The L stands for the number of lone electrons, or dots on the Lewis Structure.

t_rasul2I

There is not really a way to determine if an electron has a plus or minus 1/2 spin. Its just a numerical representation of the Pauli Exclusion principle where two orbitals in the same orbital will have opposite spins.

t_rasul2I

I believe it is applied to all the other subsequent shells!

t_rasul2I

I just remember two things: the farther right it is on the periodic table, the smaller it is. And, the more protons there is to the electrons, the tighter and smaller the atom is.

t_rasul2I

The ionic radius is the distance from the nucleus to the outermost electron. Because electrons orbit super fast, there is a little cloud around the nucleus. The farther away the electrons are from the nucleus, the bigger their "cloud".

t_rasul2I

Ground state refers to having it at a state that is expected of the electrons, meaning they are not exited or anything. Moreover, just count the electrons given despite the configuration to determine the specific electron.

t_rasul2I

I understand that h bar is the same set up as h/4pi. But why do we have a complicated/different set up of the equation. Is there something it was derived from?

t_rasul2I

Hi! I think you might not have converted your units and/or may have plugged your answer into the calculator incorrectly like the previous comments said. You set up the initial problem variables correctly though.

t_rasul2I

How does the particle in the box concept relate to what was on that lecture? I remember it was mentioned after the electron constructive and destructive phases.

t_rasul2I

I think for the midterm, since it is multiple choice, we can use either value to calculate because we should still get a very close answer to the correct answer.

t_rasul2I

Yes we will be allowed to have scratch paper lol. Because I do not know how we would be able to do work without it.

t_rasul2I

"The transition from n = 4 to n = 2 emits radiation of longer wavelength than the transition from n = 5 to n = 1." is the correct answer because n = 4 to n = 2 emits less energy and less energy means longer wavelengths

t_rasul2I

Black body radiation is essentially a "theoretical or model body which absorbs all radiation falling on it, reflecting or transmitting none back. it has something to do with temperature. it won't be covered on the midterm though I believe

t_rasul2I

I think we kind of end up memorizing the equation sheet as we do the HW because there's only a handful of equations and constants.

t_rasul2I

In Dr.Lavelle's lecture, he mentioned some electrons only jump down to a few levels. This was seen in the hydrogen spectrum example where some electrons jumped all the way down to n=1 and some were at n=3

t_rasul2I

intensity is the amplitude of the wave. And even if you read the threshold with the proper intensity, one photon can only interact with one electron.

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