CHEMISTRY COMMUNITY

SelenaDahabreh1D

I think I actually may have found your mistake, as well as mine. After watching today's lecture, I realized that the denominator for the percentage ionization equation is supposed to be the initial concentration of the acid. Once I acknowledged this, I was able to get the correct answer.

SelenaDahabreh1D

I was wondering the same thing! I am assuming that Dr. Lavelle may include some thermochemistry on midterm 1 because we start that next week, but that is just a guess.

SelenaDahabreh1D

I had trouble with this problem too! I found it helpful to write out the equation for the deprotonation of an acid. HA -> A- + H+. From this step, I used an ICE table to determine the x-value, and thus the concentrations of each element. From then, I plugged those values into the equation for percen...

SelenaDahabreh1D

On that note, does anyone know if we will join a Zoom with our TA during the exam time for the proctoring session?

SelenaDahabreh1D

How do we know whether the change in the ICE tables are +x or -x? I have seen examples and homework problems where it might be +x but I have also seen examples where it might be -x. I am referring to the reactants in the ICE table, by the way.

SelenaDahabreh1D

I kind of like to draw it out and physically visualize a seesaw. When a stronger acid is present, then that means that the acid's conjugate base will be weaker. Vice versa for a stronger base. When a stronger base is present, then that means that the base's conjugate acid will be weaker.

SelenaDahabreh1D

I would re-do your ICE table because when I did it I got x=0.05429. When filling out your ICE table, be sure that for SO3 the change in equilibrium is -2x, and for SO2 it is +2x and for O2 it is +x.

SelenaDahabreh1D

Yes, because equilibrium favors the most stable aspect of the reaction I believe it would favor resonance because resonance also favors the more stable drawing.

SelenaDahabreh1D

The answer to self-test 5G.1A includes water into the equilibrium constant. My TA went over this, but I just wanted to ensure that I fully understand. Water is included into the equilibrium constant because water is in its gaseous phase. I would appreciate it if someone could confirm this.

SelenaDahabreh1D

For anyone trying to get into Michael's UA Workshop, do you know the passcode? When I try to join the Zoom link it asks me to input a Meeting Passcode.

SelenaDahabreh1D

I recommend attending as many as you can! They were super helpful for me last quarter, especially since I was exposed to many different practice problems.

SelenaDahabreh1D

You should be able to choose a free trial on Sapling. I was able to use that option as well.

SelenaDahabreh1D

are we supposed to know how to do problems like J17a (in the textbook) on the final? I am very confused about how this would be solved :/ In each of the following salts, either the cation or the anion is a weak acid or a weak base in water. Write the chemical equation for the proton transfer reacti...

SelenaDahabreh1D

I believe the pi bond composition would just be (C2p, O2p) because p-orbitals are the ones able to take part in pi bonds.

SelenaDahabreh1D

I think if you memorize the trends on the periodic table, then you would be able to decipher which ones are amphoteric compounds.

SelenaDahabreh1D

I think NaCl has a higher boiling point because there is a bigger difference in electronegativity, making the NaCl molecule more ionic than the LiCl molecule. Thus, resulting it NaCl having a higher boiling point.

SelenaDahabreh1D

If you look on Dr. Lavelle's website, there is a file titled "Solution Manual Errors 7th Edition," and there is an error in which one of these chemical reactions is written. So if you were having trouble balancing it, I think this may have contributed to that struggle. Here is the error: i...

SelenaDahabreh1D

This means that strong acids fully deprotonate (meaning they give away a proton) and strong bases protonate (meaning they accept a proton).

SelenaDahabreh1D

I think it is because strong acids and bases fully dissociate in aqueous solutions. So, these strong acids and bases will not be accounted for in equilibrium.

SelenaDahabreh1D

Take a breather. If you are anxious with regards to the final (as I am sure most of us are), I would recommend taking a few moments to write down what you do know and what you don't know. I personally like to take the time to go through each learning objective on all of the outlines and make note of...

SelenaDahabreh1D

I think some questions will come directly from the textbook, but I assume that most of the questions on the final exam will simply be derived from textbook problems.

SelenaDahabreh1D

Darlene Lien 1B wrote:Hi, I think you're supposed to look at Table 6C.2 to get the kB values!

Ahh, okay! Thanks so much!! Do you, by any chance, know if those values will be provided to us on the final exam if we encounter a question similar to this one?

SelenaDahabreh1D

For 6C.17, the question asks us to identify which is the stronger base: BrO- (hypobromite ion) or C17H19O3N (morphine). When I consulted the answer key, it came about answering this question by calculating the pKb values, but the question doesn't offer the Kb values. Are we expected to know the Kb v...

SelenaDahabreh1D

The methyl group (-CH3) in acetic acid inductively donates electron density to the COOH. This makes the O-H bond less likely to break. Formic acid does not have an electron donating group, so its O-H bond is slightly more acidic. So would it be a good rule of thumb that when a methly group (-CH3) i...

SelenaDahabreh1D

Acetic acid is CH3COOH, and formic acid is HCOOH. Comparing these two acids, formic acid would be the stronger acid because it can more easily lose its Hydrogen (H+) to a base. This is because in acetic acid, the CH3 in acetic acid is an electron-donating group, meaning that CH3 inductively donates ...

SelenaDahabreh1D

On almost every UA sheet from Week 9, there was a question about cisplatin. I think Dr. Lavelle spent a significant amount of time explaining cisplatin (and transplatin), so I would expect it to show up on the final in one way or another. On almost every UA sheet from Week 9, there was a question a...

SelenaDahabreh1D

I heard that some of the questions will come from the final, but I believe most (if not, all) will be derived from the textbook problems. I would recommend working through every single textbook problem and ensuring that you understand how to go about answering it, and I think you will be set for the...

SelenaDahabreh1D

I think it would be helpful to know these biological applications:
Cr - assist insulin in control of blood sugar
Fe - hemoglobin, myoglobin, electron transfer in the oxidation of carbohydrates, proteins, and fats (respiratory chain)
Co - vitamin b12
Mn, Ni, Cu, Zn - enzymatic functions

SelenaDahabreh1D

Joshua Eidam 3D wrote:Does anyone know what the format of the test is going to be like as well as how long the test is going to be?

It is going to follow the same format of the previous midterms. It will be 30 multiple choice questions, and it will be 1 hour and 30 minutes (plus 5 extra minutes).

SelenaDahabreh1D

I am having a bit of trouble understanding hybridization orbitals. So, in hybridizations orbitals, are lower level electrons (say in the 2s state, for example) "promoted" to a higher energy level (such as 2p), and is that how they arise from atomic orbitals? Also, how would we go about dra...

SelenaDahabreh1D

Pg. 506 of the textbook goes over this. So, respiratory acidosis is a result when respiration decreases from conditions such as asthma, pneumonia, and emphysema, and even from inhaling smoke. These conditions raise the concentration of CO2 in the blood, and hence the concentration of H2CO3 is raised...

SelenaDahabreh1D

In part (a) you would consider the atomic radius/bond length, since it is a binary acid. However, for the rest of the question, you would consider electronegativity because it is a more complex acid.

SelenaDahabreh1D

I had a similar problem, and another student suggested to me to log into the VPN and it worked perfectly from there.

SelenaDahabreh1D

I think Dr. Lavelle discussed this to highlight the basic concept that the overall Ka value is the product of Ka1 and Ka2. So, I believe that you are going about performing this correctly. But I also think that the biggest takeaway from this was that Ka1>Ka2>Ka3 because it harder to deprotonate a ne...

SelenaDahabreh1D

I believe they just highlight data points, so that when you draw the curve with all of your plotted data you would better be able to identify the stoichiometric point. I think those points you are asking about specifically are just the pH of solution as they correspond to the volume of the solution.

SelenaDahabreh1D

The strength of the acid would decrease as the amount of Oxygen atoms in the acid decreases because there are less Oxygen atoms to stabilize the acid. Thus, the acid would be least comfortable giving up its H+ ion because it has less Oxygen atoms to stabilize the acid after it has given up the H+.

SelenaDahabreh1D

You would typically refer to atomic radius/bond length when looking at binary acids, such as H-F, H-Cl, H-Br, and H-I. However, you would refer to electronegativity for acids that read like HClO4, H2SO4, and HNO3.

SelenaDahabreh1D

Heather Szeszulski 1I wrote:what is pOH? i am a little confused

pOH is a measure of the concentration of the hydroxide ion [OH-].
Just know that [OH-}*[H+]=1.00*10^-14, and pOH+pH=14.00
Here is a site that explains it in more detail: https://courses.lumenlearning.com/chemi ... h-concept/

SelenaDahabreh1D

When trying to watch today's lecture, I was prompted with Error Code: 232001. Has this happened to anyone else? If so, is there another way we can access today's lecture?

SelenaDahabreh1D

In Friday's lecture, Dr. Lavelle noted that the alkali metals and alkaline earth metals (excluding Beryllium) typically give rise to bases (or basic oxides), whereas elements on the right hand side of the periodic table typically give rise to acids (or acidic oxides). However, amphoteric elements a...

SelenaDahabreh1D

The prefixes do not apply to elements outside of the coordination sphere (or outside of the brackets). I think the only way to indicate that it is dichloride is to look at the charge of the coordination compound.

SelenaDahabreh1D

To be considered a strong acid, the acid must completely deprotonate. Thus, CH3COOH is considered to be a weak acid because it doesn't completely give up its protons, as it only gives up one H+ ion. However, HCl is considered to be a strong acid because it completely deprotonates in an aqueous solut...

SelenaDahabreh1D

In Friday's lecture, Dr. Lavelle noted that the alkali metals and alkaline earth metals (excluding Beryllium) typically give rise to bases (or basic oxides), whereas elements on the right hand side of the periodic table typically give rise to acids (or acidic oxides). However, amphoteric elements ar...

SelenaDahabreh1D

I believe you are right with respect that Lewis acids accept electrons. With regards to the proton, I believe that acids produce hydrogen ions (which would be the protons) in solution because it reacts with the water molecules by giving a proton to them. This site speaks more on this: https://www.ch...

SelenaDahabreh1D

Sigma bonds are single bonds and pi bonds are the additional bonds that build on top of sigma bonds. So, for example, a double bound would have 1 sigma bond and 1 pi bond. Also, a triple bond would have 1 sigma bond and 3 pi bonds. It helps to think of it as the sigma bond being the initial first bo...

SelenaDahabreh1D

If you give up on a question, you will lose the points available for that question. I'd recommend just attempting the question as many times as you need. Not only will you keep those easy points, but you'd also exercise your thinking which would eventually be beneficial when studying for the final e...

SelenaDahabreh1D

I find that this one organizes it well: https://mmstcchemistry.weebly.com/uploa ... ometry.pdf

SelenaDahabreh1D

I would expect all outlines to be covered on the final. Best of luck studying! :)

SelenaDahabreh1D

I believe that since there are lone pairs present on Br the dipole moments don't entirely cancel out, so it would be a polar molecule.

SelenaDahabreh1D

I think it would help to restart your computer. Or maybe you could try using a different browser. I found that using Chrome is better than using Safari, if you have a Macbook.

SelenaDahabreh1D

Electron density refers to the probability that an atom would be found in any given region of an orbital. And to answer your second question, I believe that any area where an electron could be found would be considered a region of electron density.

SelenaDahabreh1D

Since F is the largest element, it would have a greater atomic radius, and thus greater bond length, so the bond would be weaker.

SelenaDahabreh1D

Also, it is helpful to look at lone pairs. I remember there was a problem in the textbook in which we were asked to draw the figure of AsF3 and AsF5. And since the dipole moments cancelled out in AsF3, it was considered to have a lower boiling point because it was no longer polar. On the other hand,...

SelenaDahabreh1D

Sigma bonds show up only in s- orbitals and pi bonds show up in p- and d- orbitals, and so on. I'm not entirely sure if this is correct, but I distinctly remember reading this somewhere.

SelenaDahabreh1D

Lewis acids accept electrons, whereas Lewis bases donate electrons. It helps to draw out the Lewis structures to determine whether a molecule is a Lewis acid or base.

SelenaDahabreh1D

I believe the 4s subshell would be occupied first because it has a lower energy. But when it is written, you would write 3d before 4s. This link provides a nice visual for understanding which order subshells are filled: https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Map%3A_Organic_Chemist...

SelenaDahabreh1D

My TA said that we won't need to memorize the values. Rather, he suggested using and applying the trends in the periodic table to determine the difference in electronegativity.

SelenaDahabreh1D

Polarizability refers to a measure of how easily an electron cloud is distorted by an electric field. And I believe polarizability and electronegativity are inversely related.

SelenaDahabreh1D

For elements in the transition metals group, the elements occupy the d-block, which is why 3d^10 came before 4s^2. For example, for elements in period 4 would typically have this electron configuration: [Ar] 3d^10 4s^2. It helps to visualize it on a periodic table. Here's a link to check out: https:...

SelenaDahabreh1D

In simple terms, the octet rules refers to the tendency of atoms to prefer to have a full valence shell (8 electrons). Thus, atoms tend to form compounds in a way that allows them to have a complete valence shell. This website explains it very well: https://courses.lumenlearning.com/cheminter/chapte...

SelenaDahabreh1D

It helps to change locations while studying at home! Sometimes I'd find myself unable to focus in my bedroom, but I've found that it helps to go to a different area in my house to study. Also, I recently read an article that said studying in various locations helps you retain the information you're ...

SelenaDahabreh1D

I think Nitrogen, in this case, would be an exception because while drawing Lewis structures, symmetry also plays a role in determining the shape.

SelenaDahabreh1D

Formal charge would typically be used for determining the correct structure for the molecule. Thus, I believe you would most likely calculate formal charge when drawing structures that have resonance.

SelenaDahabreh1D

In simple terms, the inert-pair effect is refers to the tendency of valence electrons to remain unshared in the compounds of post-transition metals. Here is a website that explains it well: https://courses.lumenlearning.com/chemistryformajors/chapter/periodicity-2/#:~:text=This%20phenomenon%2C%20the...

SelenaDahabreh1D

Simply put, electron affinity increases as you move up a group and across a period. So, for example, Oxygen would have a higher electron affinity than Nitrogen and Sulfur.

SelenaDahabreh1D

So firstly, what I would do is I would convert the masses of carbon dioxide, water, and nitrogen to the moles of Carbon in carbon dioxide, Hydrogen in water, and simply Nitrogen. To do this for carbon dioxide, you would essentially perform this calculation: 0.682gCO2 x (1molCO2/44.009gCO2) x (1molC/...

SelenaDahabreh1D

There isn't really a set-in-stone cut off value for rounding moles to a whole number. But, in some of the UA sessions I've attended, they said to just use your logic in rounding and typically round down or up is the value is less than 0.1 or greater than 0.9. Hope this helps!

SelenaDahabreh1D

The UAs don't typically send out an answer key, in hopes of encouraging students to attend the Workshop/Step-Up/Drop-In sessions. I am sure you could post any lingering questions you may have about certain problems to Chemistry Community, and the students (including myself) would be more than happy ...

SelenaDahabreh1D

The photoelectric effect needs to be done under a vacuum so that the experimental conditions are as controlled as possible. If the photoelectric effect weren't done under a vacuum, then the electrons would interact with the atoms in the atmosphere, which would have an influence on the experimental r...

SelenaDahabreh1D

Yes, typically when writing the molar mass, it is always assumed that its units are g/mol.

SelenaDahabreh1D

I've been going to a few each week, and I find them super helpful! The UA workshops and step-up sessions are where UAs would give you a worksheet to work on, and it really great extra practice. I haven't gone to any drop-in sessions, but I hear that those are most like office hours, and you can go t...

SelenaDahabreh1D

I think we will only have to be able to understand the relationship between Schrodinger's equation (H ψ = E ψ), wave functions, and orbitals. But not necessarily know how to apply them to particular questions. I think he just wants us to understand Schrodinger's equation from a fundamental and conce...

SelenaDahabreh1D

We will not be tested on wave functions and atomic orbitals. In Dr. Lavelle's email, he stated that we wouldn't need to be responsible for the last 8 topics on the Outline 2: The Quantum World. Hope this clears up any confusion!

SelenaDahabreh1D

Hi! Here is a link to the webcam I purchased: https://www.amazon.com/gp/product/B088F ... UTF8&psc=1. I also have a MacBook Pro, and this camera works excellently.

SelenaDahabreh1D

Hi! So I just saw Dr. Lavelle's email regarding the Equipment Request form, and I noticed that a webcam was listed. Does this mean we need to purchase an additional webcam, even if our laptop already has a built-in one? Thanks!

SelenaDahabreh1D

The order would be: Gamma Rays > X-rays > Ultraviolet Rays. On the electromagnetic spectrum, energy increases as frequency increases and wavelength decreases. Thus, the rays with the higher frequency and smaller wavelengths would generally be the strongest.

SelenaDahabreh1D

If I remember correctly, I believe he said that on exams he would provide the name of the compounds with the molecular formula in parenthesis. I assume it would look similar to this: "carbon monoxide (CO)."

SelenaDahabreh1D

The quizzes are just to test the lockdown browser for when we use Respondus for our midterms and final. On a separate note, does anyone know how we would be able to remain on Zoom during the duration of our exam if Respondus closes the Zoom application once we login?

SelenaDahabreh1D

keV stands for 'kiloelectric volts.' This unit could be converted to Joules using this conversion factor: 1 eV = 1.602 × 10^-19 J.

SelenaDahabreh1D

Hi! If there is not enough energy to emit an electron, then that energy will simply be absorbed by the metal.

SelenaDahabreh1D

Hi! I agree with the post above, Khan Academy is a life saver! Also, if you look up "Chemistry Lumen Learning" on Google, it should take you to this website: https://courses.lumenlearning.com/introchem/. This is an excellent resource that explains topics in a a very understandable manner! ...

SelenaDahabreh1D

When expressing a number in scientific notation, the number being expressed should be between 1 and 10. For example, when expressing 10000.00 in scientific notation, I would express it as 1.00 * 10^4 rather than 100.00*10^2 because in the first one 1.00 is between 1 and 10. At least this is how I le...

SelenaDahabreh1D

Here's how I like to think of it (and this is a great example provided by Dr. Lavelle in today's lecture): Imagine you have a vitamin C (C6H8O6) tablet. When you add water to that tablet and that tablet dissolves, have the moles in that tablet changed? The answer to that would be no. Thus, in these ...

SelenaDahabreh1D

When balancing a chemical equation, you would typically want it in its simplest form for nicer numbers when doing calculations related to that chemical equation. I think there are definitely instances in which you balance a chemical equation and realize that it is not in its simplest form. In that c...

SelenaDahabreh1D

I'm not entirely sure if each UA session goes over the same exact material, but I'd assume that they review the material that we are learning in lectures for that week. You don't need to attend every single UA session. I would recommend going to the sessions, as necessary. So if you feel that you ne...

SelenaDahabreh1D

Hi! An email was sent out to a student in this course who had the same issue. Turns out, if you bought your code from the UCLA store, there's a chance you may not get it for a few more weeks. The reason being that the UCLA store has run out of Access Codes and they are still waiting for more from th...

SelenaDahabreh1D

So for this one, you are going to determine what percentage of C (Carbon) is present in each of these compounds. So, for example, in part (a) you will be determining what percentage of C (Carbon) is in ethene, whose molecular formula is C2H4. To do this, you will first find the molar mass of ethene....

SelenaDahabreh1D

I had a similar issue, and I was able to resolve it by accessing Sapling Learning through Dr. Lavelle's page on CCLE. Just click on Dr. Lavelle's site and at the very bottom of the 'Site Info' tab there is a hyperlink that says "Chemistry Principles/Sapling Learning." Just click on that an...

SelenaDahabreh1D

The mass percent is essential for finding empirical and molecular formulas because it tells you the amount of each element that is present in any given compound. Then by treating each mass percent as a portion of the compound with a sample of 100g, you can more easily find the amount of moles in eac...

SelenaDahabreh1D

This is how I went about this question: Skeletal Second Stage Chemical Equation - Fe3O4 + CO --> Fe + CO2 So for balancing the chemical equation for the second stage, I'd start by first balancing Fe. Thus, I'm left with this chemical equation: Fe3O4 + CO--> 3Fe + CO2. Next, I'd experiment to see how...

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