CHEMISTRY COMMUNITY

Megan Sparrow 1A

Hey there! I'm not sure if this is what you meant, but looking at the photo you provided, the half-reaction (NO3- + 6H+ --> 3NO2 + 3H2O) isn't taking into account the H+ from HNO3 or HCl. Therefore, this question doesn't necessarily ignore the 6H+ ions, they are just not independent ions in an aqueo...

Megan Sparrow 1A

The review sessions tend to be the most useful study tool for me. It forces you to look at the concepts you struggle with. Plus the other students there ask questions that I would not have even thought about asking prior. Last quarter these were great resources!

Megan Sparrow 1A

Homogeneous catalysts are those that occupy the same phase as the reaction mixture (typically liquid or gas), while heterogeneous catalysts occupy a different phase. In general, heterogeneous catalysts are solids that are added into gas or liquid reaction mixtures. In heterogeneous catalysis, the re...

Megan Sparrow 1A

The sum of each elementary step in a reaction mechanism must yield the overall reaction equation. The rate law of the rate-determining step must agree with the experimentally determined rate law. This refers to the slow step representing the rate law. If this is different than experimental data, you...

Megan Sparrow 1A

The elementary step that has a rate law that matches the overall reaction's rate law is the slow rate-determining step. It is also characterized by its high activation energy. The higher the activation energy in an energy diagram, the longer it takes to reach that threshold value.

Megan Sparrow 1A

Thermodynamics is not about things moving and changing but instead about how stable they are in one state versus another, while the kinetics is about how quickly or slowly species react. To clarify, thermodynamics describes the overall properties, behavior, and equilibrium composition of a system. K...

Megan Sparrow 1A

I was also wondering this! I'm assuming the E* values will be provided for us in this scenario unless the questions give us the tools to calculate it.

Megan Sparrow 1A

What Sophia said above is true. Conversely, it is important to know basic solutions have a different approach. For basic solutions, you would add water to the other side to balance the oxygen molecules. From there, you have extra H+ molecules, so you add OH- to counteract these molecules due to the ...

Megan Sparrow 1A

I was worried about the same thing. I usually see how Sapling sets up these problems as well because recently they have been similar to the test. They have been giving use the values for E or at least a table to look it up.

Megan Sparrow 1A

I had the same question. This is very helpful, thank you! I was still confused in terms of cell diagrams which are the anode and which is the cathode because in some problems I've seen the cathode on the left side of the cell diagram. Was this just an exception, or would we need the E* values to ens...

Megan Sparrow 1A

Is there any way oxygen can have a different oxidation number of -2 in a compound? I think in H2O2, oxygen could have an oxidation number of -1 instead, but I don't know if this is the only exception.

Megan Sparrow 1A

In order to do this problem, you break up the reaction into two separate half reactions: Ag+(aq) + e- --> Ag(s) E*(cathode)= +0.80V AgBr(s) + e- --> Ag(s) + Br-(aq) E*(anode)= +0.07 The cathode shows the reaction where reduction occurs, and the anode shows where oxidation occurs. You should then rev...

Megan Sparrow 1A

This such a hard thing to balance, but I find that forming a study group really helps. Not only can they help study, but they remind you that you need to take a break when you need it. Also when you are studying with them, it doesn't feel like a hassle because you can actively take breaks that you p...

Megan Sparrow 1A

The struggle is real with these next results. Just remember that you did all you could to prep and know that there's nothing you can do to change what happens next. I usually focus on doing other homework (I had a lot backed up in order to study for this test), which distracts you for a few days unt...

Megan Sparrow 1A

I completely agree with the post above. Remember that we are all here in this class for a reason, and with hard work, we can pull through! Good luck with the next exam.

Megan Sparrow 1A

The textbook questions are VERY similar, if not the same, to a handful of the questions on the last final. I felt as if there was more time to do the problems on the final compared to the midterms, but I generally don't consider time a major issue when taking these tests. For these tests, you either...

Megan Sparrow 1A

Although he will ask conceptual questions, it doesn't help to memorize them! I found this online, which helps me. 1. You can’t win. / Energy can neither be created nor destroyed. It can only change forms. 2. You can’t break even. / The entropy of an isolated system not in equilibrium will tend to in...

Megan Sparrow 1A

In the free expansion, Q = 0, so the entropy of the surroundings does not change and the change in entropy of the universe is equal to ΔS for the system.

Megan Sparrow 1A

In order to have a reversible process, net heat transfer occurs between the system and its surroundings that are arbitrarily close to thermal equilibrium. A reversible process does not have to be a constant-temperature process. There should be no trace on the universe that the process ever occurred....

Megan Sparrow 1A

Hey there! From my understanding, positional disorder refers to the change in entropy from the rearranging of molecules. Thermal disorder refers to the thermal motion of molecules, which increases with temperature. In this case, the positional disorder is dominant because KNO3, when dissolved in wat...

Megan Sparrow 1A

I feel the exact same way! Textbook problems are always the go-to for me, I find they are very representative of the tests and midterms. Wishing luck to you all :)

Megan Sparrow 1A

A reversible process is one in which both the system and its environment can return to exactly the states they were in by following the reverse path. An irreversible process is one in which the system and its environment cannot return together to exactly the states that they were in. Most people con...

Megan Sparrow 1A

Maybe the molar values were already accounted for in the specific problems you are referencing. On Dr. Lavelle's equation sheet, moles should be included in the equation you mentioned. Do you mind sharing which questions specifically ask for this?

Megan Sparrow 1A

These all seem correct to me. If you are more of a visual learner, these diagrams may help!

Megan Sparrow 1A

Entropy is the amount of disorder in a system. Negative entropy means that something is becoming less disordered. In order for something to become less disordered, energy must be used. So when something is in a state of negative entropy, something else must be in a state of positive entropy to balan...

Megan Sparrow 1A

When a phase change occurs - that is, a substance moves from one region of its phase diagram to another - the entropy of that substance changes even as its temperature stays the same. A substance in the solid-phase has low entropy; in the liquid phase, it has medium entropy; in the gas phase, it has...

Megan Sparrow 1A

Summer enrollments opened, so I decided to do BL during the summer in case the spring sessions fill up quickly.

Megan Sparrow 1A

Also, a comparison between the two values can be shown by cp = cv + R. These values indicate the quantity of heat that can increase the temperature of unit mass by 1°C. Hence, the specific heat at constant pressure is more than the specific heat at a constant volume, Cp > Cv.

Megan Sparrow 1A

Thank you for the response, I had the same question. Further, when finding the energy required for phase change, did you leave it as joules or kJ. This was in relation to adding this product to the final equation you wrote.

Megan Sparrow 1A

I'm not sure if we use work as always negative for this course, but if the sign of the work is positive, then energy enters the system. An equivalent statement is that a positive value of w means that work is done on the system by the surroundings. When w is negative, we say that the system does wor...

Megan Sparrow 1A

Although this isn't referenced in the outlines, Dr. Lavelle mentioned the difference between an open system, a closed system, and an isolated system. The open system can be represented by a beaker where matter and energy can exchange with the surroundings. A closed system would be represented by a s...

Megan Sparrow 1A

I had a similar question. Could someone show the general set up for this problem. For some reason, my q is just not adding up properly.

Megan Sparrow 1A

Thank you for the help because I had this same question. On another note, if the question said that the volume was held constant, would deltaU=q ?

Megan Sparrow 1A

Hi! I'm not sure if this has already been asked, but does anyone know which bullet points from Outline 3 (Thermochemistry and The First Law of Thermodynamics) will be covered on midterm 1? I know Dr. Lavelle sent an email reviewing textbook questions and general concepts from the Thermo chem unit th...

Megan Sparrow 1A

The molar heat capacity of a chemical substance is the amount of energy that must be added, in the form of heat, to one mole of the substance in order to cause an increase of one unit in its temperature. In order to find this, you should use Cv or Cp. Use Cv when there's constant volume. This is the...

Megan Sparrow 1A

Reactions that have a negative ∆G release free energy and are called exergonic reactions. A negative ∆G means that the reactants, or initial state, have more free energy than the products, or final state. This in turn relates to K being larger, which favors product formation.

Megan Sparrow 1A

Usually, the water in which the solids have been dissolved in the surroundings, while the dissolved substances are the system. The temperature change that is measured is the temperature change that is occurring in the surroundings. For example, if the temperature of the water increases as the reacti...

Megan Sparrow 1A

K in general is an expression to describe the composition of the reaction at equilibrium. Specifically, Kc is the equilibrium constant when it is found through the use of concentrations, while Kp is the equilibrium constant when it is found through the use of partial pressures.

Megan Sparrow 1A

The partial pressure is a portion of the total pressure exerted by the presence of a single gas molecule. The sum of the partial pressures of all the gases in a mixture equals the total pressure.

Megan Sparrow 1A

Thank you for reviewing this, I had the same question too! To avoid mistakes with calculations like this, I found it useful to put your scientific calculator in "sci" mode instead of "deg". This shows the number in scientific notation.

Megan Sparrow 1A

Atm and torr are the most important to know, but I have seen threads where people say we may need to know pascals and Barr. Honestly, it is best to just follow the course outlines because Dr. Lavelle includes units we should know. If not there, check the new equations sheet because maybe there are c...

Megan Sparrow 1A

Okay, thank you for letting us know! Does anyone know if for the test if we are expected to know how to convert from Barr to atm or atm to pascals? I'm not sure if it was mentioned in the section outline or if this was just for practice on the homework and such.

Megan Sparrow 1A

Usually, the problem will either tell you which way the reaction is proceeding (or which way is more favorable). If K > Q, a reaction will proceed forward, converting reactants into products. If K < Q, the reaction will proceed in the reverse direction, converting products into reactants. If Q = K t...

Megan Sparrow 1A

To start, whenever you see a "p" in front of a value, like pH, pKa, and pKb, it means you're dealing with a -log of the value "p". For example, pKa is the -log of Ka. pH is the -log of the concentration of H3O+ ions in solution (M), and the pOH is the -log of OH- ions in solution...

Megan Sparrow 1A

Yes, thank you everyone for clarifying the rest of the dates!

Megan Sparrow 1A

Thank you all so much! I see where I got a bit mixed up when using the ICE table.

Megan Sparrow 1A

Based on research I found online, a superbase is an extremely basic compound or substance that has a high affinity for protons. The hydroxide ion is the strongest base possible in aqueous solutions, but bases exist with much greater strengths than can exist in water. In contrast, superacids have the...

Megan Sparrow 1A

To clarify, these midterms are held in class this quarter. Last quarter they were held in discussion section times. They are still within the same time frame of 50 minutes for each midterm. You can also look at alternate times if you are an international student.

Megan Sparrow 1A

In addition to this, even if it seems that some of the questions are harder, I found it valuable to ask for sapling homework help here. Sometimes I try to do them as warm-ups before learning the information in a lecture to remind myself of the application. Plus, there is no penalty for getting the q...

Megan Sparrow 1A

I haven't seen any posts for this discussion section, but if you check the big Group Me chat for 14B, someone may have posted a link there.

Megan Sparrow 1A

In this case, the equilibrium constant would be less than 1. A small equilibrium constant (when Keq is less than one) means that the chemical reaction will favor the reactants and the reaction will proceed in the opposite direction.

Megan Sparrow 1A

In addition to this, at a constant temperature, changing the equilibrium concentration does not affect Kc because the rate constants are not affected by the concentration changes. When the concentration of one of the participants is changed, the concentration of the others varies in such a way as to...

Megan Sparrow 1A

Hello everyone! I was having an issue calculating the sapling problem below. I understand that you have to use the molar concentration of the products over the reactants (with their appropriate powers) in order to find the Kc value. Can someone walk me through this or show me how they found the mola...

Megan Sparrow 1A

This happened last quarter as well, and my roommate had this same issue. I use Chrome to access it and then you have to turn on the pop-up blockers in settings. This fixed her issue. Hope this helps!

Megan Sparrow 1A

There can be many ways to approach this, but here are a few. When the forward reaction is favored, the concentrations of products increase, while the concentrations of reactants decrease. When the reverse reaction is favored, the concentrations of the products decrease, while the concentrations of r...

Megan Sparrow 1A

In addition to everything above, there was another post on a study tips section that had a google doc with recorded UA meetings and worksheets. I also looked at Khan Academy for practice tests on concepts I was struggling with. Overall, textbook problems are life savers though!

Megan Sparrow 1A

Hey guys! I was hoping someone could help me work through this problem. I understand the concepts but keep getting the value incorrect. Specifically, help with finding the pOH would be great. Any insight is appreciated!

Megan Sparrow 1A

Thank you for posting these! Did you find certain recordings or UA sessions to be most valuable to you?

Megan Sparrow 1A

Hey! I was hoping someone could help guide me through this problem. For some reason, I keep having trouble classifying this. Any insight would be great!

Megan Sparrow 1A

To add on to this, resonance may delocalize the electron pair that the base might use to form the new bond with the proton. This delocalization increases the stability of the base. Greater stability results in lower reactivity. A base that has resonance delocalization of the electron pair that is sh...

Megan Sparrow 1A

I don't believe we will need to know very specific aspects of the concepts you listed. Those seem to be highlighted more in Chemistry 14b. Plus in the lecture, Dr. Lavelle mentioned that we wouldn't be going in-depth into equilibrium concepts. Below, I attached the outline for the Acids&Bases un...

Megan Sparrow 1A

Make sure to write down or check when your finals are on MyUcla. They won't be during normal discussion times or lectures, so it is important to check your email for more information given the whole only learning shift. I attached when ours is, and it should be longer than the midterm times.

Megan Sparrow 1A

Perfect, that makes more sense. Thank you all for the help!

Megan Sparrow 1A

Based on what I have been doing and hearing, the textbook problems and workshops are highly effective. Sapling homework is very useful in terms of learning the basics, but I've noticed that the difficultly of the midterms matches that of the textbook work. Khan Academy videos on YouTube are also liv...

Megan Sparrow 1A

Hello! I was unsure of how to approach the first problem. If you have any insight, that would be greatly appreciated.

Megan Sparrow 1A

Hi, I was having trouble with Sapling problem #6 on week 9 Homework. If anyone could give insight, that would be great. Thank you in advance :)

Megan Sparrow 1A

In terms of ligands, NH3 and H2O are both neutral. In this example, there are 2 chlorines to take into account, each with a -1 charge (as with the rest of the halogens). Since this entire compound is neutral, the oxidation number for Co must be +2, hence the Cobalt(II).

Megan Sparrow 1A

You are correct, light does not have mass. It does, however, have momentum. Since photons (particles of light) have no mass, they must obey E = pc and therefore get all of their energy from their momentum. This is why the DeBroglie equation works in this instance. Hope this helps!

Megan Sparrow 1A

I haven't heard about Chem BL in terms of the waitlist, but since it is considered a lab, I think it would be more difficult to get in. If you don't get in this quarter, I know a lot of people are trying to do 14BL with Chem 14C for the spring quarter, since you already learned the topics from the p...

Megan Sparrow 1A

In addition to what Katarina said, when a double bond breaks, pi bonds will always break first. Pi bonds are weaker than sigma bonds. This is because sigma bonds have head-on overlapping while pi bonds only have lateral overlapping. On the other hand, the electron density is more spread out in pi bo...

Megan Sparrow 1A

Based on Dr. Lavelle's Outline 4 (Molecular Shape and Structure), the only part that specifically references bond angles is below: - Explain why lone pairs are more likely to found in certain locations around a central atom and how and why they affect the bond angles in a molecule, cation, or anion....

Megan Sparrow 1A

Sigma and pi bonds are chemical covalent bonds that are formed by the overlap of atomic orbitals. Sigma bonds are formed by end-to-end overlapping and Pi bonds are when the lobe of one atomic orbital overlaps another. Double and triple bonds are comprised of sigma bonds between hybridized orbitals a...

Megan Sparrow 1A

During hybridization, the C-C sigma bond is formed when one sp orbital overlaps from each of the carbons, and two C-H bonds are created when the second sp orbital on each carbon overlaps with 1s orbital of hydrogen. In this, the carbon atom will have two half-filled 2p orbitals. These two pairs of p...

Megan Sparrow 1A

To answer the first part of your questions, in inorganic chemistry, π bonds can be made from both p- and/or d-orbitals. As long as there is a general overlapping of the electron-dense regions, a pi bond can be formed. The diagram below shows different examples.

Megan Sparrow 1A

After hybridization, all four hybrid orbitals have the same energy, lower than p orbitals, but higher than s orbitals. In terms of d orbitals, this website might help explain a bit more. http://butane.chem.uiuc.edu/cyerkes/Chem102AEFa07/Lecture_Notes_102/Lecture%2015'-102.htm#:~:text=Hybridization%2...

Megan Sparrow 1A

All molecules are polarizable to some extent. Polarizability affects dispersion forces in the following ways: As polarizability increases, the dispersion forces also become stronger. The relationship between polarizability and the factors of electron density, atomic radii, and molecular orientation....

Megan Sparrow 1A

The general order you listed is correct. I believe for this course and the tests, this is what we need to know. It is based on general trends we can find using a periodic table and given information.

Megan Sparrow 1A

The other group 16 elements that follow this trend include O, S, and Se. The other group 16 elements get too large for the ionization energy to reflect this change going down. This photo shows the energies of the first ionization energy.

Megan Sparrow 1A

Hello! If you are comparing the bond strength of diatomic halogens, we can compare F2 to Cl2. The bond dissociation energy of F2 is less than Cl2 because fluorine has a high electronegativity, resulting in F−F repelling each other and having lesser bond dissociation enthalpy than Cl2 since Cl is les...

Megan Sparrow 1A

The most plausible resonance structure is the one that is most stable. For example, the greater the number of covalent bonds, the greater the stability since more atoms will have complete octets. Also, the structure with the least number of formal charges and the least separation of formal charge is...

Megan Sparrow 1A

In this case, the reason that SO3^-2 is a trigonal pyramidal shape is due to only 3 of the 4 bonding positions are being occupied by atoms, when using 3 single bonds and a lone pair. The main difference is the bond angles. The reason it couldn't be trigonal planar is that a molecule with that format...

Megan Sparrow 1A

I'm not sure if this is what you meant, but the energy released is found experimentally. It will always be given on Dr.Lavelle's test. In today's lecture (11/09), he noted that he found these values online and didn't personally calculate it. You could experimentally find the energy via calorimetry r...

Megan Sparrow 1A

For anions, the larger the ionic radius of an anion, the greater its likelihood of being polarised is (the greater its polarisability). This is because the outer electrons are more loosely held by the nucleus (weaker electrostatic force of attraction due to greater distance from the nucleus). In ter...

Megan Sparrow 1A

I believe you are on the right path in terms of 4s orbital filling first since it has lower energy compared to 3d orbitals. I'm not too sure about the covalent bonding aspect, but this problem and explanation I found online might help explain a bit.

Megan Sparrow 1A

Hey! To answer your question, I believe that all resonance structures technically contribute to the overall structure, but the major resonance contributor has the lowest energy. When the formal charge is 0 or close to 0, then the structure is most stable. In general, the most stable structures (lowe...

Megan Sparrow 1A

Generally, polarizability increases as the volume occupied by electrons increases. On rows of the periodic table, polarizability decreases from left to right (larger molecules are more polarizable than smaller ones). Polarizability increases down on columns of the periodic table. I hope this helps!

Megan Sparrow 1A

An excited state means that, typically, the valence electron has moved from its ground state orbital ( lowest available energy) to some other higher energy orbital. So any electron configuration in which the last electron (the valence electron) is in a higher energy orbital, this element is said to ...

Megan Sparrow 1A

Silver would have 47 electrons as long as it is in a neutral state. It is easy to assume that the configuration is [Kr]5s^2 2d^9, however you have to be careful with silver because it is a transition metal. This implies that the occupied d-orbitals are actually lower in energy than the s-orbitals th...

Megan Sparrow 1A

In terms of bond length and the stability of a molecule, the measures are directly proportional. When the bond order is higher, the bond length is shorter. A shorter bond length means that the bond is held tighter to the other atoms in a molecule, and it would require more energy to remove them comp...

Megan Sparrow 1A

For the Heisenberg equation, we haven't used psi I believe. Psi is used in Schrodinger's equation and represents the wave function. Psi squared represents the probability density of an electron. That tells us where the electron is most likely to be found in the space around the nucleus. If you want ...

Megan Sparrow 1A

I believe you can position the resonance structure in both ways. You can have the oxygen atoms on opposite sides, or as it was shown in the textbook. When I was looking online, I found this photo below. Either way is a correct representation of the resonance structure. In addition, during Dr. Lavell...

Megan Sparrow 1A

Another good tip for scientific calculators is to set it SCI mode. Usually, the default is on FLO, but SCI shows results in scientific notation which saves time from homework and test questions. I believe our week 0 sapling optional homework noted this.

Megan Sparrow 1A

Nodes in general are locations where electrons are not present. Therefore, angular nodes are either x, y, and z planes where electrons aren't present while radial nodes are sections of these axes that are closed off to electrons. This website helped explain more of this concept to me. https://chemis...

Megan Sparrow 1A

Since we haven't reviewed the specifics in a class of the information you noted, I believe he will give us the information. I would double-check the learning outcome sheet (https://lavelle.chem.ucla.edu/wp-content/supporting-files/Chem14A/Chem14A_Outline2_The_Quantum_World.pdf) and the equations/con...

Megan Sparrow 1A

I believe they are very similar in terms of each element and each molecule has its own individual "fingerprint". However, there may be differences in terms of energy from the light being used to promote an electron from a bonding or non-bonding orbital into one of the empty anti-bonding or...

Megan Sparrow 1A

The main difference between shell subshell and orbital is that shells are made up of electrons that share the same principal quantum number. Subshells are composed of electrons that share the same angular momentum quantum number (l), whereas orbitals are made up of electrons that are in the same ene...

Megan Sparrow 1A

To help clarify the second part, the theoretical yield is the maximum amount of product calculated that could be produced if the reaction is precise given the limiting reagent. The reason you couldn't always add the product grams from both reactants is that you could be factoring in an excess reacta...

Megan Sparrow 1A

Start by multiplying the grams of ice by the enthalpy of fusion in order to get the total energy required (in J) to melt the ice. Then substitute the given wavelength into the equation "E=(hc)/lambda" in order to find the energy per photon. Remember to convert nanometers to meters when usi...

Megan Sparrow 1A

I'm assuming you are referring to the information that Dr. Lavelle mentioned in his last slide when going over the controversy of the equation. De Broglie hypothesized that 1) this equation works for ANY particle with momentum (which includes photons that have momentum) and 2) that it exhibits wave-...

Megan Sparrow 1A

I was having an issue solving the parts b and c of the textbook problem B.15. The original question is: 1B.15 The velocity of an electron that is emitted from a metallic surface by a photon is 3.6×10^3 km*s^−1 (a) What is the wavelength of the ejected electron? (b) No electrons are emitted from the ...

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