CHEMISTRY COMMUNITY

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.

SainehaMaddineni_3I

As per Lavelle's email, it should be until 1D.2 in the textbook, but the problems in section 1D don't seem to apply to what we expect to be on the midterm.

SainehaMaddineni_3I

In Atomic spectroscopy, if it says a "blue light is emitted" or any other color, it would indicate it deals with the Balmer series, which is the visible light region. Through this, you would be able to know n1=2 for Rhydberg's equation when calculating the wavelength of a spectral line.

SainehaMaddineni_3I

The size of an ion depends on how far the outermost electrons are from the nucleus. As effective nuclear charge increases, the electrons are pulled closer to the nucleus and the radius decreases. Since they have the same number of electrons, the ions are considered isoelectronic. Therefore, the effe...

SainehaMaddineni_3I

Perfectly full shells are the most stable and half-filled shells are mostly stable. Since d^9 elements are neither half-filled subshell nor perfectly full subshell, they are considered unstable. To fix the instability of the d^9 and to complete the d subshell, an electron is borrowed from the 5s orb...

SainehaMaddineni_3I

how would we go about calculating the wavelength range? The range of wavelengths would of the highest energy transition (n=7 to n=1) to lowest energy transition (n=7 to n=6). Use the Rydberg's equation to solve for the wavelengths. The n1 would be the final low energy level. The n2 should be the in...

SainehaMaddineni_3I

The amount of spectral lines can be found depending on the amount of energy level transitions. In your example, the hydrogen atom is excited to the n=7 shell, so it can transition to 6,5,4,3,2, and 1 energy levels. There is a possibility of 6 energy level transitions, so there should be 6 spectral l...

SainehaMaddineni_3I

It is true that each element unique amount of energy and colors that are emitted. The amount of spectral lines/emission lines depends on the amount of electrons. The more electrons an element has, the more emission lines there will be as there is more possibility of energy transitions.

SainehaMaddineni_3I

Each transition of energy level is one spectral line. So based on that, you can calculate the total number of spectral lines by subtract the two given energy levels.

SainehaMaddineni_3I

To find wavelength, you can use the c = λv equation and solve for λ (wavelength). Plug in the frequency you found as v and use 3.00*10^8 m*s^-1 as your c (speed of light).

SainehaMaddineni_3I

Hi! Energy's unit is Joules.Your answer right now is in J*mol^-1. You need to convert further by dividing by Avogadro's number (6.022*10^23 mol^-1) in order to get to Joules. I hope that helps!

SainehaMaddineni_3I

Hi! Electron configurations are generally written in increasing order of energy level. Since 3 is less than 4, it is placed before 4s even though 4s orbital would be filled up first. By placing 4s after 3d, it is more easier to picture what order electrons are lost. Electrons lost first come from th...

SainehaMaddineni_3I

BoparaiAdesh2I wrote:What do you mean as a straight scale?

A straight scale means that your grade is dependent on what you score on the assignments and exams (the points you can get is provided on the syllabus). The class isn't curved, so your grade isn't dependent on the class average or how others do in the class.

SainehaMaddineni_3I

So based on the responses, I gather the understanding that in order to find the amount of excess, we need to subtract the mols of product formed by the limiting reagent from the mols of product formed by the excess? Then we convert that number into grams of the excess reactant? Can someone clarify ...

SainehaMaddineni_3I

In general, to find the limiting reactant, you compare the amount of product produced by each reactant using the stoichemtric coefficients. First, you convert the reactants' masses to moles. After, based on the balanced equation, you use the mole ratio of that reactant's stoichiometric coefficient t...

SainehaMaddineni_3I

Hi! Stoichiometric coefficients are preferred to be an integer as the smallest possible whole number. Balanced equations would show the ratio of each product. It is much easier to utilize these ratios in calculation if the equation is balanced with whole numbers as the coefficients.

SainehaMaddineni_3I

Your final answers should be in the units they ask for. However, if they don't state specifically, you could keep it in the units they gave or convert to SI units. SI units are generally preferred. It is ok to use common units, such as mL, when working through problems as long as it is appropriate a...

SainehaMaddineni_3I

Hi! It usually depends on the question and whether they tell specifically what units should be used. If they ask to keep it in mL, then keep the answer in mL. Liters are the usually most preferred unit for measurement. However, unless they state otherwise in the problem, it shouldn't matter whether ...

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