CHEMISTRY COMMUNITY

Eve Gross-Sable 1B

I had the same question, but it just has to do with the concentrations cancelling. They don't end up appearing in the equation visibly, but the concentration is accounted for in k! So you just don't have to factor the initial concentration in again in first order but it is still technically accounte...

Eve Gross-Sable 1B

You can't determine the order just by the chemical equation, so that would be an indicator that you'd need more information!

Eve Gross-Sable 1B

Yes, so the "number" and moles are going to be the same thing, it's the coefficient next to the electron, but it has to be taken from the balanced reaction instead of the half equations.

Eve Gross-Sable 1B

Delta H does not have to do with the activation energy, but rather it is the change in enthalpy from the reactants to products in the most direct path on the graph (excluding the hump). So it's the net, and therefore the activation energy does not affect it. Because the catalyst lowers the activatio...

Eve Gross-Sable 1B

An intermediate is produced and then consumed, while a catalyst is consumed and then produced.

Eve Gross-Sable 1B

The rate of your reaction is going to be dependent on the reactants, so because your equation has coefficients of 1 for NO and Br2, they should both only be raised to the first power in the rate equation.

Eve Gross-Sable 1B

I handwrite my notes for every class because I find that it is much easier to go back and flip through them (specifically for this class) when doing homework, plus it makes it more efficient when you're trying to draw graphs or charts. I try to take notes on the main points of lectures and specific ...

Eve Gross-Sable 1B

Someone also explained it to me that you want to look at k as the factor that will be used to cancel out any of the units that are skewed by the order of the reaction so that the final answer will always be M/s. So whenever you are given one of these, it's helpful to look at what units are present a...

Eve Gross-Sable 1B

^ like the post above mentions, this is applied a lot to enzymes! In the microbio course I'm in right now we're studying enzyme kinetics and zero order reactions occur when the rate is directly proportional to the concentration of the substrate, while the rate therefore has nothing to do with the ac...

Eve Gross-Sable 1B

I'm still very confused on whether you multiply Ered ever when you're calculating the standard cell potential of a reaction. I thought you had to multiply Ered to match the stoichiometric coefficients/whatever corresponds to cancelling out the electrons, but then in a homework problem I attempted li...

Eve Gross-Sable 1B

It looks to me like you may have skipped out on electrons in the sulfur equation or something. I had the same question setup and the answer is 2Mno4- + 3S2- + 4H2O --> 2MnO2 + 8OH- + 3S Remember that the oxidation of S 2- involves the transfer of 2 electrons, while the reduction of MnO4- involves th...

Eve Gross-Sable 1B

I've been wondering this too! Someone said something about it being an property so the coefficients don't matter or something... but I would really appreciate if someone could use an example to show it!

Eve Gross-Sable 1B

I would check to make sure if it was done in a basic or an acidic solution, since this will affect whether you're allowed to have OH- left over.

Eve Gross-Sable 1B

arisawaters2D wrote:Do coefficients matter when calculating Q?

Yes coefficients do matter when calculating Q! They become the exponent that the concentration value is raised to.

Eve Gross-Sable 1B

I didn't know the grades were out, are they on myUCLA? Mine isn't there. Either way, with midterm 2 I was a bit surprised by the amount of conceptual questions and I feel like that means there will probably be more questions on the final that cover midterm 2 calculations.

Eve Gross-Sable 1B

I think what is important is that in a drawn diagram (as opposed to the cell diagram with lines) the anode and cathode may not always be oriented left to right, however in a cell diagram with lines they should ALWAYS be oriented anode on left and cathode on right. So if, for example, you were given ...

Eve Gross-Sable 1B

I'm pretty sure both the cathode and the anode are considered to have an electrode that is initiating the electron transfer, so both sides should acknowledge what is being used to transfer electrons.

Eve Gross-Sable 1B

The appendix from the book is also pretty useful when it comes to deciphering all the rules! https://sites.google.com/site/chempendix/oxidation

Eve Gross-Sable 1B

Yeah, I believe that the standard potentials have to be given to us. We will probably be able to calculate them if he ends up giving us equations that incorporate other functions of reactions, but not at the moment at least.

Eve Gross-Sable 1B

Apparently there is a specific order that the shorthand notation should be written in, in terms of which order the phases are in contact. The Pb/Ag anode and cathode should be on the far left and far right, adjacent to the PbCl 2 and AgCl. The two Cl - s are then placed by the double vertical line,...

Eve Gross-Sable 1B

Hi, I have attempted this question so many times and I can't figure out what I'm doing wrong. I've tried just about everything, at the beginning sapling kept telling me "the cathode should occupy the right‑hand position of the line diagram." and then I moved something and it started saying...

Eve Gross-Sable 1B

Hi! Here is a super useful chart for this concept. Screen Shot 2021-02-10 at 2.24.49 PM.png The bottom two rows explore the temperature-dependent scenarios for Gibbs free energy. In these scenarios, a temperature can be above or below (or at, if in equilibrium and delta G is 0) a certain point of mu...

Eve Gross-Sable 1B

Hi! I'm pretty sure thermal equilibrium just refers to the state at which all components of the system possess the same amount of thermal energy, so there is no net exchange of thermal energy between them. They could still exchange energy, but not to facilitate a net gain or loss.

Eve Gross-Sable 1B

Hi! The "spontaneous" labels go with the arrows on top, while the "Q<K" and "Q>K" labels go on the bottom spaces. You have it all correct. Q<K is spontaneous forward and Q>K is spontaneous reversely, while Q=K is when the reaction is at equilibrium.

Eve Gross-Sable 1B

I find this concept kind of confusing too. Based on my understanding, I think we are looking at it theoretically within a given state of matter. I think the number of possible microstates (degeneracy) has to do with the number/identities of different atoms (since this determines the number of possib...

Eve Gross-Sable 1B

Hi! The answer is that you actually are using deltaS = deltaH/(T). Using deltaS = deltaH/(T) will give you the same answer because at equilibrium, deltaG is 0. So I think this question wanted us to make the connection between the two equations by using the Gibbs free energy equation. deltaG = deltaH...

Eve Gross-Sable 1B

Hi! After calculating q I wrote the equation deltaU = q +w, and that w = -PextdeltaV Knowing I would need the final and initial volumes, I used the ideal gas equation two separate times to generate the final and initial volume values based on the values provided in the question. Then, I used those t...

Eve Gross-Sable 1B

Hi! I first converted the mass of NaN3 to moles because that is the only thing that the question provides a mass for. Then, I used stoichiometric coefficients to get to what I was looking for: the moles of N2. I did this by multiplying the moles of NaN3 by 3/2, because that is the ratio of N2 to NaN...

Eve Gross-Sable 1B

Hi! You can find q by using the equation q = nCsubS*deltaT, but because what we are looking for is the final temperature, we manipulate the equation to show final temperature as a variable. The problem involves changes of state, so you want your equation to show that heat to melt ice + heat to bring...

Eve Gross-Sable 1B

Yeah, the question says the calorimeter was calibrated by carrying out a reaction known to release heat and then gives you the negative q of that reaction. When you calculate the heat capacity of the calorimeter, you're using that q without the negative sign because heat capacity is always positive,...

Eve Gross-Sable 1B

I like the explanation above! I also find that it helps to work backwards in terms of making it easier to understand, at least when you first start doing these problems. So you know deltaU is 7545 J, which means that 7545J = -346J + q It looks like you tried to use 16000c as a sub for q to get C fro...

Eve Gross-Sable 1B

I think the explanation above is really good! Sometimes with these problems it can be hard to figure out what you're looking for, but I just try to focus on the variable that it's asking for and the variables that I already have. For this question they only give you grams of NaN3, but you're looking...

Eve Gross-Sable 1B

I started by converting the mass of NaN3 to moles. The reason you want to do this is because NaN3 is the only compound for which you're given a mass -- it's important to mention that what you're ultimately looking for is the quantity of N2, so if the problem gave you the mass of N2, you would just g...

Eve Gross-Sable 1B

what's the difference between standard enthalpy of formation and standard enthalpy of a reaction? Standard enthalpy of formation is the kj/mol quantity for the formation of an element/compound, but the standard enthalpy of a reaction is the kj or J value of the entire reaction, which includes react...

Eve Gross-Sable 1B

You can rearrange the equation as long as you account for sign changes. Typically with Hess's law problems they give you a few different reactions and the enthalpies of those reactions, and you might also be given the standard enthalpy of formation of one/some of the molecules. So if you're rearrang...

Eve Gross-Sable 1B

Hi, in the book example 4A.2, it says that "Provided the initial and final states are the same, a change carried out reversibly always does more work than a change carried out irreversibly, so you should expect the second path to produce more work and therefore correspond to a more negative val...

Eve Gross-Sable 1B

This was really hard for me too, but I find it helps to visualize what is actually happening when you put it in water. If you use the charts in the book that someone mentioned above, you'll be able to see the character of the cations and anions when they're in water. If the anion is from a weak acid...

Eve Gross-Sable 1B

The numbers next to Xs are just the coefficients and do not have to do with the molarity. Molarity is the concentration, which you end up plugging in later.

Eve Gross-Sable 1B

The other day I was in Justin's step-up session and he asked a question about equilibrium that didn't mention delta H at all, but then the entire mode of solving it was around delta H? Can someone explain the relationship between enthalpy and equilibrium and what the implications could be if there i...

Eve Gross-Sable 1B

Hi! It looks like my problem has different values for the masses which is strange but otherwise it's the same question. I used the word equation from the hint and then plugged in the formulas corresponding to what it was describing. So for "heat to melt ice" I used deltaHf which is 333.6J/...

Eve Gross-Sable 1B

Hi! I was wondering if someone could help me understand how to do this problem, I've been able to do all the problems fine until #9 and I have no idea where to even begin with it, I looked in the book and it didn't help me. "If you combine 300.0 mL of water at 25.00 ∘C and 130.0 mL of water at ...

Eve Gross-Sable 1B

Hi! I know everyone has their own way of doing things, but I personally will go over the outline at the end of the section and look at the textbook problems at the end of the week. If there are things we didn't cover in lecture/homework, people will usually ask about them on chemistry community or i...

Eve Gross-Sable 1B

You can subtract the pH from 14 to get the pOH. Then you can do 10^(-pOH) to get [OH-]

Eve Gross-Sable 1B

Hi! To do this, you have to look at how the pH compares to the pKa. For acids, if the pH is higher (more basic) than the pKa, then the charged, deprotonated species of A− will become more predominant in solution. If the pH is lower (more acidic) than the pKa, the neutral, protonated species of HA wi...

Eve Gross-Sable 1B

Hi! I am not sure what you mean by writing a new equation or making a second ICE table, typically there should just be one for these problems. But aside from that, when dissociation occurs, it means that the strong acid or base completely separates into its ions when it is in water. Therefore, H+ (f...

Eve Gross-Sable 1B

My ICE table was initial: N2 [.3] O2 [.3] 2NO [.8] change: N2 [+x] O2 [+x] 2NO [-2x] equilibrium: N2 [.3+x] O2 [.3+x] 2NO [.8-2x] The wording of these questions can be tricky and confusing, but I find it helps to focus on figuring out what the question is asking, which in this case is, what is the c...

Eve Gross-Sable 1B

If you were able to use your x value to calculate the equilibrium concentration of NO2 you're probably on the right track. Because NO2 is 2.87-2x, but N2O4 is 3.11+x, did you make sure to add x (as opposed to subtract x and/or 2x)? Hope this helps or you're able to find the issue!

Eve Gross-Sable 1B

Hi! I first calculated Q by plugging in the concentrations to [NH3]^2/[N2][H2]^3. Then I compared Q to K to answer the second part.

Eve Gross-Sable 1B

Hi! Sapling problems are considered separate from "textbook problems," the textbook problems are the book problems recommended by Lavelle in the outlines, doing them won't get you points but he includes textbook problems on exams and they're good practice. Sapling is required (if you want ...

Eve Gross-Sable 1B

Yeah Q would be affected, I think to know if it would increase or decrease you have to know which side has more moles but I may be wrong.

Eve Gross-Sable 1B

Hi! My understanding is that when you have an endothermic reaction, composition shifts towards the products resulting in an increase in heat absorption that helps offset the increase in temperature. K increases in endothermic reactions, but the addition of heat is not going to make Q decrease becaus...

Eve Gross-Sable 1B

An amphoteric substance must have - hydrogen present (so it can act as a Bronsted acid to donate H+) - a feature that allows it to accept an H+ (typically this feature is a negative charge) so it can act as a Bronsted base So by looking at the formulas and seeing if they fit those bullet points, you...

Eve Gross-Sable 1B

From what I know, since C17H19O3N has a N with a lone pair of electrons, it would be an ammine. Ammines are generally considered weak bases. BrO- is the conjugate base of HBrO, a weak acid. Therefore, it would most likely be considered a stronger base than morphine (the weaker the acid, the stronge...

Eve Gross-Sable 1B

I agree, I think just make sure you are familiar with the sheet from the website and know what the prefixes mean. I assume that the examples of terms you used in your question will be defined on the final if they are included since that is not emphasized on the outline/in lectures.

Eve Gross-Sable 1B

Although XeF2 has 3 electron pairs because it is arranged symmetrically around Xe, so the bond angles will still be 180 degrees, making the molecule linear. Do the lone pairs not count in the bond angle? And I am assuming the 180 is between the bond of Xe and F, what happens to the lone pairs and w...

Eve Gross-Sable 1B

Hi! So SO2 has 3 possible structures, one has 2 double bonds (one between Sulfur and each oxygen) and the other two have a double bond between sulfur and one oxygen and a single bond between sulfur and the other oxygen. Since the second two structures are essentially mirrors of each other they have ...

Eve Gross-Sable 1B

I think coordination compound is an umbrella term for a central atom surrounded by ligands, if it does not have a net charge then it is neutral and if it has a net charge then it is a complex ion.

Eve Gross-Sable 1B

^ and to elaborate, bidentate means that the ligand has two atoms that coordinate to the central atom, as opposed to one, which would be monodentate.

Eve Gross-Sable 1B

This link has a good diagram for electronegativity and acidity. Acid strength increases with increasing electronegativity of the central atom because the OH bond is more polarized. https://groups.chem.ubc.ca/courseware/pH/section10/index.html#:~:text=Acid%20strength%20increases%20with%20%3A,for%20th...

Eve Gross-Sable 1B

The coordination number is equal to how many atoms the central atom is bonded to. You use it to determine the name of the coordination complex

Eve Gross-Sable 1B

Just in the home stretch and excited for this quarter to be over! Can't wait to have some time to relax during break.

Eve Gross-Sable 1B

Shapes like that can be really intimidating, but just remember to look at the atom the question is referring to and focus on that. I think those shapes are meant to distract, but it really is all the same rules for hybridization despite the confusing shape.

Eve Gross-Sable 1B

The explanation above is really great! I would also just say that overall, orbitals hybridize when chemical bonds are forming so that electrons from the atoms can be paired. I usually imagine it like a venn diagram in terms of shape overlap

Eve Gross-Sable 1B

How do you know when orbitals overlap at the ends vs at the sides? They look the same in the diagram

Eve Gross-Sable 1B

Hi! I think the primary tool we are supposed to use for determining bond angles are the shapes from the formulas that Lavelle has been talking about. Based on those formulas, there is a pretty definitive set for bond angles, which are still approximations but it's my those angular approximations can...

Eve Gross-Sable 1B

How do you determine if the overlap is end to end vs side to side? They look basically the same in the diagrams. Thanks!

Eve Gross-Sable 1B

A high viscosity substance has high resistance to deformation due to strong intermolecular forces. This means it resists flow, so water is low viscosity.

Eve Gross-Sable 1B

For me, it helps to look at the outlines and check off what I know and what I need to work on. Once I know that I develop the knowledge by doing the practice problems/watching videos if I must. It also helps that you can always post questions on here!

Eve Gross-Sable 1B

You know that it is an electron acceptor if it needs to take on an electron to fill the shell. So this would be figured out by counting the valence electrons and then seeing what shell of the orbital that is and how many electrons it would have if it were full

Eve Gross-Sable 1B

It is for sure hard to study at home and I totally get you. Personally, I have struggled a lot to discipline myself and actually sit down and do stuff, especially when there isn't always a quiet place with enough room. I've found it very useful to set a timer when I'm studying or doing work and to s...

Eve Gross-Sable 1B

I go on bike rides every day which as been such a good source of exercise and entertainment for me!

Eve Gross-Sable 1B

can someone revisit the octet rule conceptually again? The octet rule says that atoms prefer to have eight electrons in their valence shell, and if they don't have eight, they will react with others to form stable compounds. I found this link gives a pretty good explanation https://chem.libretexts....

Eve Gross-Sable 1B

I agree with lots of these responses, think of it like if there is more pull = more power = closer together.

Eve Gross-Sable 1B

Yeah, lewis basis typically have lone pairs on the central atom so that they can donate to compounds that will accept electrons.

Eve Gross-Sable 1B

Also on that note, since it seems like we are all quite confused about oxidation numbers, can anyone explain what the concept is? I also noticed on the sapling problems that the explanations were not very clear and it seems like a relevant concept.

Eve Gross-Sable 1B

Their electronegativities are too similar. Because hydrogen bonds are formed from electron via a difference in electronegativity, it won't happen when a hydrogen and a carbon are near each other that level of pulling does not occur.

Eve Gross-Sable 1B

My TA told us that if we want to go over the problems we can do it in office hours, they just can't post the test material. So if you want to know why you got stuff wrong before the midterm (even though it is not cumulative) I would go to office hours!

Eve Gross-Sable 1B

I believe most replies above have the definition of London dispersion forces. Here is the order of bond strengths from strongest to weakest 1. Ionic bonds 2. Covalent bonds 3. Dipole Dipole bonds 4. Van der Waals (London dispersion) Are London Disperson/Van der Waals considered a form of bond? Or a...

Eve Gross-Sable 1B

305614361 wrote:When are we taking the second midterm?

The second midterm is during week 7 discussion

Eve Gross-Sable 1B

This is such a nice thread. I didn't realize how hard it would be not to be able to collaborate in person for classes like this. It really affects my entire understanding of the material and I think I took that for granted. But I really appreciate all the effort that has been made by the instructors...

Eve Gross-Sable 1B

Can someone elaborate on the step-up sessions, I'm kind of confused about what happens in them. Are they worthwhile? Thanks

Eve Gross-Sable 1B

Is there a definite answer for when to use de Broglie and when not to? I thought that it was straightforward but then it seemed like it was being applied to whatever they wanted it to be applied to in the sapling problems. Like you are able to calculate using de broglie for electrons but then also n...

Eve Gross-Sable 1B

Hi! Based on my understanding, the "2p electrons in fluorine are experiencing a higher effective nuclear charge than those in oxygen" is the opposite of the answer about shielding. So if the electrons in fluorine were more highly shielded, that would mean that they would experience a lower...

Eve Gross-Sable 1B

There will be multiple l values for an n value unless the letter of the orbital is specified, unless it's n=1 because there is only one subshell. FOr example if all you're given is that n=2, l could be 0 or 1 depending on the subshell letter. Hope that helps!

Eve Gross-Sable 1B

Hi! I remember having a good understanding of orbital shells and quantum numbers in high school but I cannot seem to get the same grip on it that I used to have. Does anyone have any tips or good videos for understanding it and being able to apply it to problems? Thanks

Eve Gross-Sable 1B

Hey Monica! So for that one you'd have to use DeBroglie's wavelength. (Wavelength=planck's constant/mass x velocity) You're already given the velocity so now you'd just find the mass of a Chlorine atom. In this case, since they said it's a diatomic molecule so you'd multiple 2(the molar mass of chl...

Eve Gross-Sable 1B

He said in the lecture that we do not need to know about black body radiation for this class, but if it's just a curiosity thing I found this link pretty explanatory
http://galileo.phys.virginia.edu/classe ... ation.html

Eve Gross-Sable 1B

Definitely just an application thing, but i don't think there would be a question where you'd have to differentiate between definitions or anything like that.

Eve Gross-Sable 1B

Definitely just an application thing, but i don't think there would be a question where you'd have to differentiate between definitions or anything like that.

Eve Gross-Sable 1B

I do not think there will be very many since it's only a 50 minute period, but on a related note, does anyone know the format?

Eve Gross-Sable 1B

I love riding my bike around Westwood and other parts of LA! it's so nice to get fresh air and fun to do with friends.

Eve Gross-Sable 1B

Avogadro's number is just the number of "something" (units) in a mole. A "dozen" is 12 of "something" -- that could be 12 pounds, 12 kilograms, 12 grams etc. Think of the mole in the same way regarding avogadro's number. A mole is [avogadro's number] of an object.

Eve Gross-Sable 1B

Relating to this, I am struggling to understand conceptually how this all relates to the bigger picture. The last time I took chemistry was tenth grade and while we did go in-depth, it included no discussion of photons. So in terms of quantum, are we just using photons as our prime example regarding...

Eve Gross-Sable 1B

Based on what I know, there is a correlation between photon energy level and color of light emitted. The energy level of the photon absorbed by the element can excite the atom's electron and make it jump and then fall back down. So based on my understanding the number of energy levels it falls will ...

Eve Gross-Sable 1B

I get what you are asking and I am also confused by it. I found what seems to be some pretty in-depth explanations at this link https://physics.stackexchange.com/questions/110844/why-electrons-have-less-energy-than-photons-with-the-same-wavelength But I am having a hard time understanding it, based ...

Eve Gross-Sable 1B

Atomic radius increases as group level increases and atomic radius decreases as you move across a period

Eve Gross-Sable 1B

Are the practice problems that people are using in the book or from elsewhere? Typically I do the discussion section problems and the sapling hw but I'm not sure where else I should go for more problems because I feel like I am not that good at the application.

Eve Gross-Sable 1B

Moles will stay the same as long as you do not add more solute. However, say you're adding more water, which will increase the volume, that would decrease the molarity because adding more solvent decreases the concentration. So despite there being the same number of moles, the molarity is less.

Eve Gross-Sable 1B

^ Yes try capitals, that's what worked for me

Eve Gross-Sable 1B

I was not told about any quizzes or respondus but if it's written in print anywhere I'd love to see!

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