CHEMISTRY COMMUNITY

SainehaMaddineni_3I

I'm taking 14C in the spring quarter! I wanted to also enroll in 14BL but it was filled up :(

SainehaMaddineni_3I

Your equation for finding Kc is the concentration of the products/concentration of reactants, with each raised to an exponent equal to their corresponding coefficients in the reaction. For example: H2(g) + Br2(g) <--> 2 HBr(g) Kc = [HBr]^2 /[H2][Br2] *Remember that solids and liquids aren't included...

SainehaMaddineni_3I

I plan on dorming in the fall! I'd prefer dorms, but if they don't give an offer due to limited capacity, I'll probably get an apartment. :)

SainehaMaddineni_3I

#Part 5: The generic rate law is: rate=k [A]^a [B]^b [C]^c k is the rate constant. [A], [B], and [C] are the initial concentrations of the reactants. a, b, and c are the orders of the reactants (A, B, and C). You found the order of the reactant A in part 1: 1st order --> a = 1 You found the order o...

SainehaMaddineni_3I

In order for the reaction to proceed, it must pass the transition state. To pass the transition state, the energy needed for the reaction to proceed is known as the activation energy. Activation energy can be thought of as a hindrance to reaction progress. A higher activation energy results in a slo...

SainehaMaddineni_3I

Yes, a catalyst lowers the activation energy in the reverse reaction as well as the forward reaction. A catalyst increases the rate of both the forward and reverse reactions by providing an alternate path of lower activation energy in either direction.

SainehaMaddineni_3I

Here's the solutions: Atkins_7e_SSM_Fundamentals_1-7_9.pdf There are some errors in the solution manual so make sure to refer to the file on the class website that lists the errors --> Here's a link to it: https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14B/Solution_Manual_Errors_7Ed.p...

SainehaMaddineni_3I

If you aren't able to attend some of the sessions, the worksheets from some sessions are usually posted on Chemistry Community, so look out for those too!

SainehaMaddineni_3I

It depends on which way the work is done. If work is done ON the system, then work would be POSITIVE. If work is being done BY the system, then work would be NEGATIVE since the system is exerting energy onto the surroundings by doing the work. Compression of a system usually means work is being done...

SainehaMaddineni_3I

For textbook problems, I'm planning to focus on the more complicated problems from each section. I personally struggled the most on thermodynamics, so I may try to watch more videos from organic chemistry tutor to gain a better understanding and maybe even do some of the basic textbook problems to b...

SainehaMaddineni_3I

You base the rate law on the rate-determining step, which is usually the slowest step. In this case, it is I−(aq) + HClO(aq) → HIO(aq) + Cl−(aq) The rate law for the slowest step would be rate = k [HClO][I−] You also need to note how HClO is an intermediate, so you need to substitute [HClO] with th...

SainehaMaddineni_3I

Yes, brackets are mostly used to indicate concentrations.

SainehaMaddineni_3I

You balanced it right, but I think it's because your (-) for 2OH- and 2ClO2- is after the phase indication. I would put them as 2OH-(aq) and 2ClO2-(aq) instead of 2OH(aq)- and 2ClO2(aq)-

SainehaMaddineni_3I

You balanced it right! Don't put the parenthesis... I did that too and Sapling kept marking me wrong :) It should be: 4Fe+3O2+6H2O⟶2Fe2O3∙3H2O Fe2O3•H2O indicates that it contains water. Any chemical compound that has the dot followed by water indicates that it is a hydrate, which means that it's a ...

SainehaMaddineni_3I

The final is on Sunday, March 14 from 9:30 am – 11:00 am. Unlike the midterms, there aren't separate times for different lectures, but you join your TA proctoring link just like the previous times. I think the only alternative time is for international students from 7:00 am – 8:30 am.

SainehaMaddineni_3I

I would go over the sections you struggled with the most. Since the test questions are similar to the textbook ones, it's good to redo those exercises too. I recommend organic chemistry tutor on Youtube if you want some explanations on topics.

SainehaMaddineni_3I

I just made banana bread today! My most favorite food is Thai food (spicy basil fried rice and pad kee mao) all-day everyday. Favorite dessert would be cinnamon rolls :)

SainehaMaddineni_3I

You need to balance the half-reactions and then add them together to get your overall reaction: First balance the charges in the half‑reactions by adding electrons: In⟶In3+ + 3e− Cd2+ + 2e− ⟶ Cd Then, balance the # of e- transferred in each half-reaction so that it is the same in both half-reactions...

SainehaMaddineni_3I

I'm not really sure, but they may update the professors closer to the summer session. I would advise signing up soon though since most classes seem to be filling up, if they haven't already.

SainehaMaddineni_3I

STP indicates 0 degrees Celcius and 1 atm pressure.

SainehaMaddineni_3I

I binge-read the Lunar Chronicles series!

SainehaMaddineni_3I

I haven't gotten an email about when it would be released, but I would expect it to generally take a week.

SainehaMaddineni_3I

First isolate ln(K) to one side. Then, to get K, you should do e^(whatever is on the other side).

Here's an example of how to do it with some random numbers:
ln(x) = 2.3/4
x = e^(2.3/4)

SainehaMaddineni_3I

I used the same equation: ΔG=ΔG∘+RTln(Q) ΔG∘ = −69.0 kJ/mol and T = 298 K as given in the problem. We first need to calculate Q. Q=(PNO2)^2/(PO2)(PNO)^2=(0.650)^2/(0.250)(0.450)^2= 8.35 Since they want the final answer of ΔG in kJ/mole, make sure all your units are correct . In this case, the R valu...

SainehaMaddineni_3I

I'm currently watching WandaVision. I wasn't into it at first, but it's picking up and dropping a bunch of interesting storylines, so I'm waiting to see where it goes :). I'm also rewatching Cutthroat Kitchen in my downtime. Waiting for Grey's Anatomy to come back in March!

SainehaMaddineni_3I

Yes, you are right! The entropy of a material increases with an increase in the number of microstates. Entropy decreases when going from a freely moving state to one of limited movement where there are fewer possible arrangements of molecules, such as gas to solid. Enthalpy increases as a substance ...

SainehaMaddineni_3I

The two equations seem to be the same. In the equation from the lecture, a (-1) was factored out and put it in front of the rest of the equation. This allowed rearranging of the equation to make 1/T2 positive. In the equation from the textbook, you can see that they didn't factor out a negative, so ...

SainehaMaddineni_3I

I'm planning to take Chem14C, 14BL, LS7B, and Cluster 48CW seminar. I hope I can get my ideal times before they fill up :)

SainehaMaddineni_3I

YES! To calculate the heat of combustion for compound B, qB, multiply the constant C you found earlier by the temperature change: qB=C ×(Tf−Ti) *Since heat is being released, the qB value is supposed to be negative The question asks for the heat of combustion per gram of compound B, so divide qB by ...

SainehaMaddineni_3I

The best thing that helped me was to make a list and schedule out what I would do throughout the day. Breaking up tasks into smaller parts and spreading them throughout the day made it easier for me to accomplish the stuff that I dreaded the most. I start off with a task that I am more motivated to ...

SainehaMaddineni_3I

The heat gained by the cold water is equal to the heat lost by the hot water according to the equation: m (cold water) CΔT (cold water) =−m (hot water) CΔT (hot water) Convert the volume of water to grams using the density conversion of 1.00 g/mL --> The mass of cold ...

SainehaMaddineni_3I

I work part-time at Kumon, but luckily my job is currently over Zoom working with kids. Whenever there is downtime between my sessions, I tend to spend that time getting the simpler class assignments finished, such as article reading. This helps me get ahead in my daily schedule. It's definitely dif...

SainehaMaddineni_3I

You would be using this equation: ΔG = ΔG° + RT*ln(Q) Since you know the partial pressures, you can plug those into the K expression to get Q --> Q = (Pnh3)^2 / ((Pn2)*(Ph2)^3) = (0.600)^2 / (0.500 * (0.200^3))=90 ΔG° is already given: -69.0 kJ/mol R = 8.314 J/mol*K but you want it to be in kJ/mol*K...

SainehaMaddineni_3I

If the conjugate base is strong, it means that pKa would be higher for the weaker acid it comes from (small Ka value for the weak acid).

SainehaMaddineni_3I

Since there are unlimited attempts, I'm assuming you can just re-attempt the questions. I'm not sure you can reset the problems though.

SainehaMaddineni_3I

He hasn't sent out anything yet, but it usually took about a week for the scores to be released. It might be similar this quarter too.

SainehaMaddineni_3I

I plan to take 14C along with 14BL in the spring. This just mostly because it is a long series and I prefer not to take a gap in the series.

SainehaMaddineni_3I

In endothermic reactions , the energy required to break the reactant bonds is more than the energy released during the formation of product bonds. To see which one is endothermic, it should be the reactions that have stronger bonds for the reactants than the product bonds. For example: AB+C⟶AC+B Bas...

SainehaMaddineni_3I

Enthalpy can be thought of as ΔH∘rxn= (bond energy of bonds broken) - (bond energy of bonds formed) For CH4(g)⟶C(g)+4H(g), 4 C−H bonds are broken without the formation of any other bonds. 4-0 = 4. This means that ΔH∘rxn could be equal to 4×bond energy of C−H. C(g)+4H(g)⟶CH4(g) is the reverse reactio...

SainehaMaddineni_3I

I mostly focus on the problems listed on the syllabus, since they seem the most relevant. But, I'm sure the other problems not mentioned can give you more practice if you want to.

SainehaMaddineni_3I

It is based on what's after the decimal when given a pH value. So, in this case, there are 3 sig figs.

SainehaMaddineni_3I

To find the pH, you do: pH = -log[H+]

When you know the pH, you can reverse it to find [H+] by doing 10^(-pH).

Example:
5.6 = -log[H+] ---> [H+] = 10^-5.6 = 2.5 x 10^-6

SainehaMaddineni_3I

Bformal is the formal concentration of B, the amine. You find this by adding the [BH+] and [B].

Bformal = [BH+] + [B]

SainehaMaddineni_3I

Yes, so far that's the correct value for x! You are going to use the Kb expression to evaluate the value of [B] by setting the expression equal to the Kb value and also plugging in your known concentrations (in this case, the products). In the expression, the x value you found earlier is what you pu...

SainehaMaddineni_3I

When it is endothermic, heat is on the reactants side. When it is exothermic, heat is released, so heat is on the products side. So, increasing temperature has an effect depending on if it is an exothermic or endothermic reaction. Increasing the temperature for endothermic reactions shifts the equil...

SainehaMaddineni_3I

The error was in your numbers once you expanded your expression. When you expanded the squared expressions, it should be: 4(0.09 + 0.6x + x^2) = 0.81 - 3.6x + 4x^2 0.36 + 2.4x + 4x^2 = 0.81 - 3.6x + 4x^2 0.36 + 2.4x = 0.81 - 3.6x 6x = 0.45 x = 0.075 When you plug that x value into 0.9 - 2x, you shou...

SainehaMaddineni_3I

I love the Princess Diaries! My go-to movie is Shrek, but I also love to re-watch episodes of Spongebob and The Fairly OddParents for fun.

SainehaMaddineni_3I

So far, you're correct. You have (0.8-2x)^2/(0.1+x)^2 = 25. Take the square root of both sides and solve for x. Once you get the value of x, plug it into 0.8-2x to get the final [NO]

SainehaMaddineni_3I

First, I solved for the Kc value by plugging in the concentrations given in the beginning. (0.4^2)/(0.3)(0.3), which gave the Kc value as 1.778. When you do the ice table, your equilibrium values for the reactants should be (0.3 + x) and for the product is (0.7 - 2x). Plug these into the Kc expressi...

SainehaMaddineni_3I

I don't think you made an error, but it was probably due to sig figs. They did each step with 1 sig fig. So instead of 6.3E-2 for [H3O+] and [Cl2-], their value was 0.06 or 6E-2. This made a difference in the rest of the calculations, which were also within 1 sig fig.

SainehaMaddineni_3I

6CO2(g) + 6H2O(l) → C6H12O6(aq) + 6O2(g) Liquids don't impact equilibrium. Also, pressure relates to gases, so aqueous solutions wouldn't be considered for this scenario. When you focus on the gases, you see that there is an equal amount of moles of gas on both sides. Thus, there would no net impact...

SainehaMaddineni_3I

Pressure and Volume are inversely related. The smaller the volume, the higher the pressure. When the total pressure is increased, the side with less moles is favored. For example, part (a) is 2 moles of gas on the reactants and 3 moles of gas on the products. Since there is a greater amount of moles...

SainehaMaddineni_3I

It comes from the Kc expression. They multiplied the denominator onto the other side. That leaves you with x^2=1.80 (3.00-X). Distribute the 1.80 and you get the expression as x^2= 5.40 - 1.80x. Then, you get all the terms onto one side: x^2 + 1.80x - 5.40. Hope this helps!

SainehaMaddineni_3I

Sophia Kalanski 1A wrote:ok but how do you figure out x from here?

Plug the expressions from the E section of the Ice table into the Kp expression (products/reactants). You know the Kp value, so set that Kp expression equal to that value. Solve for x.

SainehaMaddineni_3I

You need to use an ICE table. You know the Kc value from Table 5G.2. Plug all the expressions from the ICE table into the Kc expression and set it equal to the value from Table 5G.2. Using the quadratic formula, solve for x. Knowing this x value, you will be able to evaluate the concentrations at eq...

SainehaMaddineni_3I

The ratio of initial concentrations refers to products/reactants and their values are what the initial concentrations are when the reaction starts. The ratio of equilibrium concentrations is also products/reactants but the values are measured when the reaction reaches equilibrium. The equilibrium co...

SainehaMaddineni_3I

I'm personally going to take this winter break to relax, especially after a tough quarter. However, I will read over my notes again closer to the beginning of the winter quarter. It might also help to review the textbook problems if you want more practice review.

SainehaMaddineni_3I

A Bronsted acid can donate H+, so there should be a Hydrogen in the formula. A Bronsted base can accept H+, which can be indicated by a negative charge as it allows the substance to accept an H+. If it meets both descriptions, it can be amphoteric.

SainehaMaddineni_3I

Spectator ions don't take part in the chemical reaction to form a precipitate. By looking at the product, you can differ which are the spectator ions. For example, Cu(NO3)2 + NaOH --> Cu(OH)2. In this equation, you can see that Na^+ and NO3^- are not involved in creating the final product of the pre...

SainehaMaddineni_3I

The oxygens have different formal charges, impacting whether they can bind to something else. Two of the oxygen atoms have a formal charge of zero and don't bind to more electrons. The other two oxygens (with the double bonds) have a formal charge of -2. These two have the ability to make bonds. Sin...

SainehaMaddineni_3I

BrO- is the conjugate base of a weak acid (HBrO). The conjugate base of a weak acid is usually relatively strong. Based on the structure, C17H19O3N has a nitrogen with a lone pair of electrons and presents as a weak base. Since BrO- is relatively strong and morphine is weak, BrO- is the stronger base.

SainehaMaddineni_3I

Ok so I understand we should memorize the strong acids/bases but which ones are they... The ones he mentioned in Week 9 Wed. Lecture? I believe these are the strong bases he mentioned (Li2O,NaOH,Na2O,KOH,CaO,Ca(OH2),Mg(OH)2) .. but I cannot find the strong acids. can someone list them? thank you! T...

SainehaMaddineni_3I

A dipole refers to a molecule with opposite charges on both ends. Dipoles tend to be for polar molecules. Induced dipoles are temporary. When an ion or molecule with a dipole comes near a molecule without a true dipole, it will induce the distortion of that molecule's electron cloud, making it have ...

SainehaMaddineni_3I

A Bronsted acid can donate H+, so there should be a Hydrogen in the formula. This leaves H3PO4, HBrO2, H2PO4^- to be possible answers. A Bronsted base can accept H+, which can be indicated by a negative charge as it allows the substance to accept an H+. Out of the 3 possible answers, H2PO4^- has a n...

SainehaMaddineni_3I

Those with higher ph are bases. The strong bases have a higher pH than weaker bases. The weaker bases have higher pH than acids. The strong acids have the lowest pH values. It would be helpful to recognize the strong acids and the strong bases first before you place the weaker acids and bases in ord...

SainehaMaddineni_3I

The website we use is sapling learning. Here is the link: https://www.saplinglearning.com/ibiscms/login/

SainehaMaddineni_3I

One example of an exception I found when doing the week 10 sapling hw was Acetic Acid. It is a weak acid and its formula is CH3COOH.

SainehaMaddineni_3I

So Ka is the acid dissociation constant. That's why the larger it is the more likely it is a strong acid, since strong acids completely dissociate into ions. Ka is essentially the [products]/[reactants]. The [A-] is the concentration of the conjugate base. The [H+] is the concentration of the hydrog...

SainehaMaddineni_3I

I'm not sure if your problem is different but mine was asking to draw AsO4^3-. If that is the same as yours, I put As in the middle and the 4 O's around it. I did one double bond between 1 oxygen and the As. The rest of the bonds between the oxygens and As are single bonds. Then, I just put lone pai...

SainehaMaddineni_3I

Yes! For hybridization, you count the regions of electron density. So, single bonds, double bonds, and triple bonds all count as 1 region of electron density (the type of bond doesn't affect the number of regions of electron density). 5 regions of electron density should be a hybridization of sp3d.

SainehaMaddineni_3I

305311217 wrote:So do we not get credit for attending section?

I don't think so. According to the syllabus, the only things that are worth points are sapling, chemistry community, the 2 midterms, and the final.

SainehaMaddineni_3I

One structure is the linear structure with a triple bond between two of the carbons. Another structure is with double bonds between the carbons. The third structure is a triangular ring of the carbons with 2 carbons double bonded to each other. In this molecule, those two carbons are also bonded to ...

SainehaMaddineni_3I

When determining hybridization, you count the regions of electron density, not the type of bond. Single, double, or triple bonds between atoms count as one region of electron density. That's why there wouldn't be a different hybridization for a molecule with 4 single bonds and a molecule with 3 sing...

SainehaMaddineni_3I

The structure is cyclopropene, with the 3 carbons in a triangular three-membered ring. There's a double bond between two of the carbons. Those same two carbons are also single bonded to 1 hydrogen each. The 3rd carbon atom is bonded to 2 hydrogens. I did my structure like this:

SainehaMaddineni_3I

The total number of electrons in the molecule is 4 + 7(2), which is 18. You would place Sn in the middle and a Cl atom on each side of it. I usually start by placing a single bond between the atoms. Then I fill the octet of the Cl atoms. I then put the remaining electrons on Sn, having 2 electrons a...

SainehaMaddineni_3I

Thank you! Also, would it be okay to take the Chem 14BL Lab section from 11am-1:30pm on Friday even though it goes into Chem 14B's friday lecture at 1pm, since the lecture is recorded? I would avoid overlap with the lecture time because Professor Lavelle said in an email that tests are during lectu...

SainehaMaddineni_3I

I tend to review my notes I wrote again and then watch Organic Chemistry Tutor videos on topics I need more insight on. For the final exam, I'm focusing on understanding the new material first. Then, I'm going through my notes from previous units. After reviewing all my notes, I also redo the Textbo...

SainehaMaddineni_3I

Double bonds consist of one sigma and one pi bond. Single, double, and triple bonds all have one sigma bond and the remaining within that bond are pi bonds. So, for single bonds, it's only one sigma bond. Triple bonds have one sigma and two pi bonds.

SainehaMaddineni_3I

Most of the textbook answers seemed to do this for electron configurations. I think they used the electrons in the 4s orbital to get a half-filled or full d-orbital when possible since those are the most stable.

SainehaMaddineni_3I

You can use formal charge to find the charge of the ion, since if you add all the formal charges, it will give you the molecule's overall charge.

SainehaMaddineni_3I

Shrodinger's equation is essentially a math function where when you square psi you get electron density. Electron density meaning the probability of finding an electron Adding on, the probability of finding an electron varies depending on the orbital, since there may or may not have nodal planes(ar...

SainehaMaddineni_3I

Under my posts, they're just grouped by forum, so you won't see each post you made on the same forum. However, when you're in a forum, under your name it shows the number of posts you've made. If you click on that number, it shows all of the comments, even the ones on the same forum. So, I think you...

SainehaMaddineni_3I

Ionization energy increases across a period because there is a higher effective nuclear charge, so the electrons are held more tightly, and thus requires more energy to remove an electron. Ionization energy also decreases down a group, since the outer electrons are farther from the nucleus and are h...

SainehaMaddineni_3I

How do you find the differences in the electronegativities? Is it based on where they are on the periodic table? Since we probably won't get the values of electronegativities of the elements for the midterm, you can use the periodic table as a measure. Electronegativity increases up and to the righ...

SainehaMaddineni_3I

Thorium is another exception. In this case, the energy level 6d is lower than 5f, so it will be filled first. You probably don't need to focus much on this element since it isn't one of the exceptions we should know for this class.

SainehaMaddineni_3I

It's mostly up to you. Some decide to take it sequentially so that they will have learned the material and have a basis before taking 14BL. If you think you want to take them together, you can!

SainehaMaddineni_3I

For formal charge, electrons are assigned assuming that they are equally shared without concern for which atom is most electronegative. On the other hand, oxidation numbers assign electrons to the most electronegative atom in the bond.

SainehaMaddineni_3I

At least for this Midterm 2, the questions are worth fewer points each unlike Midterm 1. We just need to try our best by preparing and understanding the material as well as we can!

SainehaMaddineni_3I

No, it should be up until Wednesday's lecture.

SainehaMaddineni_3I

Yes, the midterm covers material up to Wednesday's lecture. It covers the last 8 points of the Quantum World (Outline 2) and Outline 3.

SainehaMaddineni_3I

Polyatomic ions are just ions composed of multiple atoms. The atoms in polyatomic ions are held together by covalent bonds. It is only ionic when a polyatomic ion bonds with a metal. The elements in NH4 are nonmetals because they possess the properties of nonmetals. Also, they are under the nonmetal...

SainehaMaddineni_3I

The grade breakdown isn't available on his syllabus. But, if it is similar to other classes, I assume 90% is an A- (450/500).

SainehaMaddineni_3I

Yes! You are correct in your calculations. Anions gain electrons and become negatively charged since they have more electrons than protons. So, you do add an electron (or more depending on the charge of the anion). Cations lose electrons as they become more positively charged. So for protons, you su...

SainehaMaddineni_3I

David He wrote:When will the score released, and where can we see them?

Dr. Lavelle sent an email earlier today that said Midterm scores should be updated by tomorrow(Thursday) midday. You can check on CCLE once they are available.

SainehaMaddineni_3I

I'm not sure if this question got answered earlier in the thread, but do posts have to be strictly chem related (like asking or answering a lecture question) to earn points, or would a post like this one count towards the weekly total? Since it's participation points, I don't think it has to be str...

SainehaMaddineni_3I

I think it's good to know the general range of wavelengths for visible light and UV since it's in our notes. Knowing the general trend of increasing wavelength sizes might help for the rest of the light spectrum (which are shorter and which are longer).

Here's an EM Spectrum that might help!

SainehaMaddineni_3I

In regards to molecular spectroscopy, it is more complex because it is composed of multiple atoms and will have a unique spectra (unlike in atomic spectroscopy which deals with one atom). When electrons transition in a molecule, there will be more spectral lines because each individual atom in the m...

SainehaMaddineni_3I

Yes, Cl2 is the correct way if it says chlorine gas. Oxygen would also be O2. This is because these elements exist as diatomic molecules.

SainehaMaddineni_3I

I agree that they helped a lot to better understand topics with more practice examples especially when studying! Unfortunately, I think the ones posted are it. There doesn't seem to be any mention of more audio-visual focus topics in other outlines.

SainehaMaddineni_3I

As a trend, yes. Group 15 elements generally have lower ionization energy than group 16 elements.

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