CHEMISTRY COMMUNITY

Chelsea_Guzman_3C

neuroscience undergrad is a good one! don't have to be a neuroscience major, it's really fun and informative overall! Recommend

Chelsea_Guzman_3C

We should have 50 points by Sunday, he might give us more time after the final to post.

Chelsea_Guzman_3C

It's been tough. Life has been hard, but nothing I can't handle. Hopefully.

Chelsea_Guzman_3C

Half life is usually calculated in seconds, but if the question gives you a certain unit, you should probably stick to that.

Chelsea_Guzman_3C

In order to determine which order, you can sum up the exponents in the rate law. However, i find the graphs to be more useful, so I really appreciate whoever posted that!

Chelsea_Guzman_3C

Important to note that NOT all of the elements in the answer bank will be used!

Chelsea_Guzman_3C

The order of reaction refers to the power dependence of the rate on the concentration of each reactant. For a first-order reaction, the rate is dependent on the concentration of a single species. The overall order of the reaction is found by adding up the individual orders.

Chelsea_Guzman_3C

Been listening to a lot of throwback music, especially 2000s R&B. You can never go wrong with that. But when I study, I like to put on classical music to really focus.

Chelsea_Guzman_3C

I usually like to do the problems w/ concepts that I found most challenging in class. Therefore, I could focus on those more and try to understand why I'm having trouble. I like to redo them before exams so they are fresh in my memory, but only on ones that I know will most likely be on the exams. T...

Chelsea_Guzman_3C

I'm part of the Neuroscience Undergrad Club, but I plan on joining more soon.

Chelsea_Guzman_3C

Knives out is a great movie! 11/10 recommend.

Chelsea_Guzman_3C

I'm also having trouble with this problem. I'm not sure what I'm doing wrong.. For the balanced equation, I get ClO- + [Pb(OH)4]2- ---> Cl- + 2OH- + 2PbO2 + H2O

Chelsea_Guzman_3C

Sapling HW is due this upcoming Sunday at midnight, Feb. 28th. Since it covers both week 7 & 8, it was extended a week.

Chelsea_Guzman_3C

The standard reduction potential is the way to go: U should get Au3+, Cu+, Fe2+, Al3+

Chelsea_Guzman_3C

They were released today! They can be accessed in ccle under grades. Lavelle also sent out an email going over specifics.

Chelsea_Guzman_3C

^You are correct, HCl would be included with Au as (HAuCl4)

Chelsea_Guzman_3C

Water does affect the voltage of a cell because it will change the concentration,

Chelsea_Guzman_3C

A handful of people tend to do good in his exams, so curves aren't likely.

Chelsea_Guzman_3C

Just found out Lavelle won't be teaching 14C.. Will be a hard parting, but grateful for the time.

Chelsea_Guzman_3C

The syllabus states that you need at least a 50% to pass the class with a C-

Chelsea_Guzman_3C

Definitely felt a weight get off my shoulder. Week 8, almost done!

Chelsea_Guzman_3C

The values can be found in the equation sheet. What would be most hopefully is to contextually understand how to use/apply these values when needed.

Chelsea_Guzman_3C

K has no units because since it is a ratio, they cancel out in the end.

Chelsea_Guzman_3C

The upcoming midterm should be fairly similar to the first one. Same format and everything. Keep doing textbook problems and getting in the habit of solving similar problems to those. Sapling is helpful as well and if there are any further questions, a lot of other resources are available.

Chelsea_Guzman_3C

To convert Celsius to Kelvin, you just have to add 273. The equation sheet has 273.15 to add more precision.

Chelsea_Guzman_3C

Chem 14C is definitely in my planner!

Chelsea_Guzman_3C

If the total energy required to break bonds in the reactants is more than the total energy released when new bonds are formed in the products, it is an endothermic reaction

Chelsea_Guzman_3C

I believe our TAs count our points, but they should be fixed by the end of this week. Many others have been having issues with this as well, but you should be fine by the end of the quarter!

Chelsea_Guzman_3C

Sapling questions are useful to study, but I prefer the textbook problems because some of them tend to be on the exam, not all the time, but it's still good practice.

Chelsea_Guzman_3C

Keeping an agenda is really helpful to organize and stay on top of things. Whenever you're in lecture, make notes on subjects that may seem difficult or you just need to go over again. It's especially hard to find the motivation to stay on top of things, so make sure you give yourself some time off ...

Chelsea_Guzman_3C

Going over the outline is very helpful! It leads me to the right direction of which things I should focus on. I then practice the textbook problems so that I know how to tackle those types of problems if they were to appear on an exam. It truly is all about practicing and creating a habit out of it.

Chelsea_Guzman_3C

Changing the temperature towards either a melting or freezing point will change the equilibrium constant K!

Chelsea_Guzman_3C

A phase change is when a substance changes to another state. Ex. liquid to gas. Certain temperatures cause this change, such as melting or boiling point.

Chelsea_Guzman_3C

^To add on, the outlines are truly helpful to review as they show the material that is expected to be on the midterm!

Chelsea_Guzman_3C

Volume expansion work against a constant pressure is
w=-pV
w= -1.013 *10^5 N/m^2 * 6.32*10^-3 m^3 = -640 J

Chelsea_Guzman_3C

The last lecture you should study for is week 3's friday lecture! Week 4 material will not be on this midterm.

Chelsea_Guzman_3C

The temperature remains consistent as the bonds are being broken. Despite heat being added, it is not being conserved.

Chelsea_Guzman_3C

The acid is stronger when Ka is greater.

Chelsea_Guzman_3C

To further add on to this discussion, it makes sense why steam coming from hot water would not cause severe burns because it would have come into contact with cooler air. However, if steam was coming from an enclosed space, such as a pipeline, it would more than likely cause more damage because the ...

Chelsea_Guzman_3C

Tomorrow's lecture will be the cutoff for midterm 1! Therefore, we will have the weekend to look over content that we learned weeks 1-3.

Chelsea_Guzman_3C

For the first midterm in Chem 14A, a lot of the problems were similar to the textbook problems. Hopefully the first midterm for Chem 14B is similar to that one, but all the resources provided should be useful. None of the questions will contain content we haven't already seen.

Chelsea_Guzman_3C

Most acids and bases are weak, meaning that they do not completely dissociate into its ions, when we place them in solution. Only a few acids and bases are strong. Therefore, memorizing key ones could be used to your advantage.

Chelsea_Guzman_3C

Hey! Deprotonation, by definition, is the removal (transfer) of a proton (H+) from a Brønsted–Lowry acid in an acid-base reaction. The complementary process, when a proton is added (transferred) to a Brønsted–Lowry base, is protonation (or hydronation). Percent deprotonation is the percentage of the...

Chelsea_Guzman_3C

Hey! So for the percentage, it says that 60% of the hydrogen has reacted, which means 40% remains. You can find the change in concentration of H2 at equilibrium with (.400)(0.133mol/L-x). If you subtract the original concentration value by 40%, you will end up with the 60%. You want to use the 40%, ...

Chelsea_Guzman_3C

We don't have class tomorrow! Enjoy the day. Get some fresh air.

Chelsea_Guzman_3C

Hey! For the percentage, it says that 60% of the hydrogen has reacted, which means 40% remains. You can find the change in concentration of H2 at equilibrium with (.400)(0.133mol/L-x). If you subtract the original concentration value by 40%, you will end up with the 60%. You want to use the 40%, by ...

Chelsea_Guzman_3C

Kelvins are always used. You can convert Celsius to Kelvin by adding 273.15

Chelsea_Guzman_3C

Read more books and practice more self love.

Chelsea_Guzman_3C

What's the weekly deadline for posting on chemistry community in order to earn your 5 points of the day? Or can we still manage to gain the points during the upcoming week?

Chelsea_Guzman_3C

Everything that goes up, must come down. Therefore, I believe almost every reaction must reach a certain equilibrium in order to be executed properly.

Chelsea_Guzman_3C

Yes, midterms will be administered during lecture time, as opposed to discussion time. Everyone will be taking it at the same time in order to reduce any unfair advantages.

Chelsea_Guzman_3C

The predominant species is neutral because the the pH level is less than the pKa.

Chelsea_Guzman_3C

Therefore, HCIO is considered to be the most acidic, while HIO is the least acidic.

Chelsea_Guzman_3C

Acid rain occurs when compounds like sulfur dioxide and nitrogen oxides are released into the air, from vehicles and factories, and they rise up into the atmosphere to mix with chemicals like water and oxygen to form this acidic pollutant. Typical acid rain has a pH level 4.0.

Chelsea_Guzman_3C

Planning out a schedule of topics to study over helps me feel less overwhelmed. Taking breaks and changing settings also helps reduce my anxiety. Only a few more days left, you got this!

Chelsea_Guzman_3C

In acetic acid, you have CH3 bonded to COOH, whereas in formic acid, only H is bonded to COOH. The CH3 group in acetic acid has electron-donating properties, which means it is less electron withdrawing power than the H attached to COOH in formic acid. Thus, formic acid is slightly stronger than acet...

Chelsea_Guzman_3C

A bronstead acid is a proton donor and a bronstead base is a proton receiver.
a) NH3 has a lone pair of electrons in Nitrogen, which accepts protons.
b) HBr wants to donate its hydrogen
etc....
Hope this helps!

Chelsea_Guzman_3C

^That tip is very helpful, I was able to solve this question in 2 attempts. The subscripts inside the brackets are the only ones to take into consideration.

Chelsea_Guzman_3C

I'm not sure where the .5 came from, but it should definitely be 180.

Chelsea_Guzman_3C

Looking over the textbook problems might be helpful for studying. However, it didn't really help for the last midterm, but hopefully this final goes better!

Chelsea_Guzman_3C

Yes, you would have two take into account the Cl outside of the brackets. In this problem, the oxidative state of [Co (NH3)5 Cl] Cl will be 3+

Chelsea_Guzman_3C

You might have forgotten to add the (-) charge on the three oxygens, not including the one with the double bond.

Chelsea_Guzman_3C

Yes, in the email he sent earlier today, he mentions that chemistry community would still be up and running on Thursday & Friday.

Chelsea_Guzman_3C

I think the shape gives an approximation of the angle, so you are correct. Since it is a trigonal pyramidal, the angles will be slightly less than 109.5

Chelsea_Guzman_3C

The believe the final will have the same structure as the past 2 midterms, just with more questions and more time.

Chelsea_Guzman_3C

I did not do as well as I anticipated, but I guess I have to do my very best on the final! There's still hope, hopefully.

Chelsea_Guzman_3C

The setup will be the same way as midterm 1, with a separate device/camera for zoom and respondus lockdown browser on a computer.

Chelsea_Guzman_3C

In the periodic table, the atomic radius decreases along a row from left to right, but increases in a group from top to bottom. Br is found further down the periodic table than Na, so the atomic radius is larger for Br, despite it being on the further right side of the periodic table.

Chelsea_Guzman_3C

I doubt we will have a video lecture for Friday, but I would just recommend keeping a look out for it. We're almost done!

Chelsea_Guzman_3C

The dipole moment is a measurement of the polarity of a molecule. The larger the difference in electronegativity, the larger the dipole moment. I'm sure you can find more info on youtube videos!

Chelsea_Guzman_3C

I believe you are correct. You can find lone pair # by getting valence electron - 8n / 2.

Chelsea_Guzman_3C

Lone pairs have a stronger repulsion than bonding pairs because the bonding pairs are further away from the central atom, while lone pairs are closer to the nucleus, so it will give off more repulsion.

Chelsea_Guzman_3C

I think the molecule with the highest electronegativity should be in the middle, while the others are around it. This would just make it easier to construct the shape. Or you could just go based how many molecules there are.

Chelsea_Guzman_3C

Tetrachloromethane has a higher boiling point because it has more electrons, so it has a larger molecular surface area which increases the intermolecular interaction strength.

Chelsea_Guzman_3C

To further emphasize what the previous person said, Cl never has a double bond because it is just not stable enough to have that. Oxygen is most likely to carry this, so that it a helpful tip to know! But it would usually go to the element with the least formal charge.

Chelsea_Guzman_3C

Usually before a test I get really anxious because I feel like I won't have enough time, but I take deep breathes and trust in myself. If I come across a problem that I don't understand right away, I move on to the next. Usually when I come back to it in the end, I'm more likely to remember it! Paci...

Chelsea_Guzman_3C

I don't think memorization is required, the periodic table is helpful.
To follow on the previous response, B) indium(III) sulfide, Indium is +3 and sulfide -2, so to balance the charges, you will have In2S3

Chelsea_Guzman_3C

I think being able to recognize or understand what makes an atom an exception is best. But knowing some like Be and B could be just as helpful, but I doubt memorizing all of them is necessary.

Chelsea_Guzman_3C

How to Get Away With Murder! Very tense, but the plot twist are amazing.

Chelsea_Guzman_3C

Drawing them out would be your best bet since you can see the bonds between C and H. This practice will be really helpful for future exams!

Chelsea_Guzman_3C

Hey! I believe high polarizability refers to a cations ability to distort an anion while an anion is highly polarizable to become polarized by the cation... If that makes sense. And I believe your other statement is correct, since they have a lower polarizability, that means their electrons are all ...

Chelsea_Guzman_3C

A higher frequency light would emit electrons with higher kinetic energy if and only if the frequency is already high enough for an electron to be ejected. If not, you can't make that assumption, therefore it would be false.

Chelsea_Guzman_3C

4s contains several peaks within the space of the 3d orbital. As a result, it experiences penetration of the shielding the 3d would otherwise provide, and is pulled closer and thus its energy level is lower.

Chelsea_Guzman_3C

An element enters their excited state when an electron temporarily occupies an energy level greater than its ground state. This occurs when the outer electrons observe energy and jump to a higher orbital. Eventually it can release this energy and fall back to a lower state. During this process, the ...

Chelsea_Guzman_3C

Yes, this trend does apply to elements in different rows. As you move down the rows, the first ionization energy decreases. When you move across the periodic table, the first ionization energy increases.

Chelsea_Guzman_3C

I believe it is important to know that the s-block elements are made out of metal because that is one of the properties that they all share. Collectively, they always lose their valance electron that's in the s-orbital. Because of this, they are considered to be the most reactive elements in the per...

Chelsea_Guzman_3C

I believe they should be converted to nanometers, but you could look at the answer choices to see which units you need to convert the answer to!

Chelsea_Guzman_3C

I refers to the number of orbitals in a subshells. I = 1 corresponds to the p-orbital, I=2 corresponds to the d orbital and I=3 to the f-orbital.

Chelsea_Guzman_3C

The momentum of an object is = to the mass of an object x the velocity. Momentum is a vector, which has the same direction as the velocity.

Chelsea_Guzman_3C

The problems on Sapling give me a basic foundation of what to expect on the midterm, so I make sure I'm familiar with how to solve these kind of problems. I also look through my notes and review terms that I highlighted because those were emphasized during Dr. Lavelle's lectures. Doing practice prob...

Chelsea_Guzman_3C

If i'm not mistaken , the vacuum permittivity is the force between two electrons. The value is 8.854 x 10^-12

Chelsea_Guzman_3C

I believe it is mandated by Dr. Lavelle in order to reduce any unfair advantages during the exams!

Chelsea_Guzman_3C

Hey!
The possible number of spectral lines can be found (n2 - n1)(n2-n1 +1) / 2
so: (5-1)(5-1+1)/ 2 = 10

And I'm actually not quite sure how to find the wavelengths, so I would also appreciate the help!

Chelsea_Guzman_3C

Hey!
E(n) represents the energy of n energy level.. I believe n could stand for either initial and final, depending on the value (such as n=3 went to n =2).. Someone correct me if I'm wrong :)

Chelsea_Guzman_3C

Hey!
It's important to remember that energy is proportional to the number of photons, therefore, the wavelength would be λ, with a frequency of v and Energy of 100.
Someone might want to check my answer or further explain it!

Chelsea_Guzman_3C

I believe we will have access to this equation sheet during our midterm in week 4.. I may be wrong tho, might have to ask my TA during my discussion tomorrow

Chelsea_Guzman_3C

Hey!
I believe there is an error in the problem because when I worked it out, I got ∆v = 10.0 m/s instead of 5.0 as the minimum uncertainty of the position is the spread of the values.. I may be off somewhere, someone might have to check my answer!

Chelsea_Guzman_3C

First start off by figuring out the formula mass of MgSO4 * 7H2O, which is 246.48 g/mol Then, divide the amount of grams of the sample by the molar mass After finding out the amount of moles of the sample, you can multiply the mole ratio to find the number of moles of oxygen Lastly, multiply by Avog...

Chelsea_Guzman_3C

Your answer is correct Jenny!
The balanced equation I got : 2 C6H6 + 15 O2 --> 12 CO2 + 6 H2O
shows that 15 moles of O2 are required for 2 moles of C6H6 to combust into CO2 and H20

- If it were just 1 mole of C6H6, you would simply need 7.5 moles (15/2) of O2!

Chelsea_Guzman_3C

I had the same issue! Could it just be for more precision? Maybe, but my answer for d) was

O = 4(16.00 g / mol) / (206.53 g/mol) = 0.3099
*Total mass of sample due to oxygen*

Search found 102 matches