CHEMISTRY COMMUNITY

Austin Aldujaili 2D

I agree with what everyone said above, Also would like. to point out that the rate constant k is not like the Equilibrium Constant K. If you were to figure out the rate equation, you only place the [R] in the rate law in comparison to K, where you put [P]/[R].

Austin Aldujaili 2D

Unless it cancels in the problem, It will have to be given. We never learned a way of calculating A for the Arrhenius Equation.

Austin Aldujaili 2D

I believe he only discussed it in class to show us exactly what the A (frequency factor) takes into account and why it is in the Arrhenius Equation.

Austin Aldujaili 2D

Yes, the cathode is conventionally on the right, meaning that the right species are the ones being reduced. Additionally, you have to arrange it with any external metals (ex. Pt) on the outside of the notation diagram and species of the same phases must be next to each other separated by a comma ins...

Austin Aldujaili 2D

Yes, I do have a pdf of the textbook but I don't see a way of sending it on here. If you want to pm me on group me from the 14B Group Chat or drop your email maybe I could send it.

Austin Aldujaili 2D

Kinetics is based on the rate of the rxn vs thermodynamics is based on the stability of products vs. reactants, the bond energies, etc. A rxn that is kinetically driven depends on the things we have been talking about in class (reactant concentration, order, mechanism, etc.)

Austin Aldujaili 2D

The product must be removed any time we are dealing with a period of time that is not instantaneous. For instance, the integrated rate law and an average rate law. This is because the reverse reaction will play a role in the rate of the rxn when we aren't dealing with only an instantaneous time at t...

Austin Aldujaili 2D

Additionally, I believe that it wants you to keep elements/molecules of the same state next to each other in the notation, meaning you would separate them with a comma instead of the line.

Austin Aldujaili 2D

An enzyme is not involved in the chemical reaction and instead just brings down its activation energy. For zero order rxns, all of the enzymes can be saturated (used up) in a reaction that adding more reactant actually doesn't change the rate of the rxn.

Austin Aldujaili 2D

Also, you do not need to convert the H2 from a pressure into a concentration, you can find Q by using the given values for [H+] and PH2

Austin Aldujaili 2D

After balancing, it is considered in basic solution if you have OH- left over or if you have H+ left over for something in acidic solution

Austin Aldujaili 2D

Yes, the anode is on the left (losing electrons), and the cathode is on the right (gaining electrons), meaning electrons always flow from left to right

Austin Aldujaili 2D

This class does cause a lot of stress especially given how hard it is to get an A, but it’s best to take breaks and try and pace yourself. It made a huge difference for me when I stopped cramming for studying the night before and instead did textbook problems throughout the week.

Austin Aldujaili 2D

The way I did it, to include HCl in the half reaction was put Au + HCl —> HAuCl4 and then you can balance the Cl in the reaction as well as the H+ by adding water if needed.

Austin Aldujaili 2D

Whenever it says “agent”, I think of it as the opposite of what it says. So the oxidizing agent would be the reactant being reduced. Another note is that I believe whenever it asks for the agent you want to put in the entire reactant compound as the agent, not just the element/atom that is losing or...

Austin Aldujaili 2D

Yes, I believe that you would still use ∆U = q + w for an open system. Maybe the only other factor you would have to consider that would change the internal energy is an addition of mass/matter since that can be done in an open system; however, I think that may be incorporated into the work (-P∆V or...

Austin Aldujaili 2D

I agree with the comments above, and also work of expansion would be negative because there is an increase in volume do to work being done by the system. In contrast, work of compression is being done on the system since there is a decrease in volume by some external force/pressure.

Austin Aldujaili 2D

I am taking Chem 14C in the spring and taking Chem 14BL over the summer because its. faster and we've already learned the material from that course so I thought that would be best.

Austin Aldujaili 2D

For me, reading the problem and understanding exactly what is happening helps tremendously in determining the sign for heat, work, entropy, etc. Instead of thinking of it as a formulaic type of thing, its best to look at the problem and determine, from the information given, whether a certain value ...

Austin Aldujaili 2D

Could someone explain exactly why in an isothermal reversible reaction, the ∆Stot is = 0 and for an isothermal irreversible (free) expansion, ∆Ssurr is = 0.

I'm trying to understand exactly why these are true in those situations

Austin Aldujaili 2D

I always do as many UA worksheets I can about 3 days prior to midterm day. That has always really helped me in getting the score I want (ofc doing that on top of the sapling and bookwork problems).

Austin Aldujaili 2D

To calculate the change in entropy from a temperature change, you don't use that equation. You have to use nC ln(T2/T1). Also, simply multiplying those values together I don't think is an accurate way to show the change in entropy in general because your then taking the natural log of V x K and that...

Austin Aldujaili 2D

(a) Calculate the work that must be done against the atmosphere for the expansion of the gaseous products in the combustion of 1.00 mol C6H6(l) at 25 8C and 1.00 bar. (b) Using data in Appendix 2A, calculate the standard enthalpy of the reaction. (c) Calculate the change in internal energy, DU8, of ...

Austin Aldujaili 2D

I am wondering if we are being expected to know how to calculate Enthalpy at a constant volume using the equipartition theoroem in the textbook. It's something I believe that. Dr. Lavelle did not go over in any of his lectures on molar heat capacity and enthalpy. Also, are we to be expected to memor...

Austin Aldujaili 2D

You. get -3.18 L x atm, not -3.18 J. Using -3.18, multiply that by the conversion ratio, which is 101.325 L x atm to get your answers in J.

Austin Aldujaili 2D

For the heat gained by the ice, there must be a phase transition where energy is put into the system until it becomes liquid water. That is why the problem gives you the standard enthalpy of fusion, allowing you to calculate and accommodate for the energy put into the system that changed the ice int...

Austin Aldujaili 2D

You know that Path B is an irreversible expansion because it states that the external pressure is constant. Also note that the first part of Part B does not effect the energy of the system because there is a constant pressure and no change in volume. If you try plugging that into the work equation w...

Austin Aldujaili 2D

A lot of the equations, once you memorize one variation of them, tie into each other so then it does not become too difficult to memorize a bunch of different equations. For instance, how there are two equations we derived for entropy (S) and they both come from equations we had already previously s...

Austin Aldujaili 2D

Commenting on that last post, entropy is able to decrease (for instance forming a larger molecule from a bunch of smaller one, or forcing gas into a smaller volume as Lavelle showed in lecture); however, for entropy to decrease there needs to be some external force or energy that causes this. In com...

Austin Aldujaili 2D

The internal energy of a system increased by 982 J when it was supplied with 492 J of energy as heat. (a) Was work done by or on the system? (b) How much work was done? I understand that work was done on the system; however, the back of the textbook says that 90 x 10^2 J of w was done and I am not s...

Austin Aldujaili 2D

Changing the pressure of an equilibrium reaction only affects the direction of the reaction because it changes the volume of the reactants/products. By adding an inert gas, you are increasing the pressure however you are not changing the volumes of your molecules and therefore not affecting the dire...

Austin Aldujaili 2D

Q is not neccessarily all of the energy supplied to the system, it is the heat lost or gained by the system.

Austin Aldujaili 2D

A constant pressure calorimeter means that there is probably some work being done on the system by the surroundings, and that is something that needs to be factored into the total internal energy or if you are trying to find the specific heat.

Austin Aldujaili 2D

If there is an aqueous solution in a reaction and the reaction is compressed (pressure increased), then do you not count that when determining which side has more/less moles?

Austin Aldujaili 2D

If the compound in your reactants is able to accept a proton, than you would use Kb. If the compound donates a proton, than you would use Ka. Another way to look at it is that if the reaction produces OH-, then you are dealing with a base and use Kb or vice versa.

Austin Aldujaili 2D

I am not 100% sure, but I believe that we have not learned how to calculate K when there is a change in temperature. So for now, all of the chemical equilibrium problems deal with an initial temperature that is the same as the final temperature. If there was a change in temperature, then the equilib...

Austin Aldujaili 2D

When going from liquid to gas, we heated the system and that allows for an increase in enthalpy which ultimately transforms the liquid into its gas phase. Going over the Friday lecture at the end where Dr. Lavelle shows the phase change diagram is a good visual of how the phase changes work.

Austin Aldujaili 2D

Hello,

Does anyone know what book work problems we should be doing to prepare for midterm 1?

Thank you

Austin Aldujaili 2D

Since steam is at a different phase of matter than liquid water, once it touches your skin, it will release all energy it holds until it becomes liquid water (this is an exothermic reaction). From the energy diagram he showed us in lecture, you can see how it takes a lot of energy given off for stea...

Austin Aldujaili 2D

Professor said that if there is ever a situation like that, then x would be small enough to approximate (K < 10^-4). No need to go and practice some crazy trinomial equations and stuff!

Austin Aldujaili 2D

I believe that you have it switched around. When a reaction is endothermic, adding heat will increase product formation given the reaction can only proceed with the addition of heat/energy. If the reaction is exothermic, I just think of it as the reverse reaction then must be endothermic and therefo...

Austin Aldujaili 2D

It is probably best to assume x is negligible when K is < 10^-4 although. sometimes I hear Professor Lavelle say 10^-3.

Austin Aldujaili 2D

The sapling homework problems could have had very had concentrations of a weak acid and therefore that increases the chance of the weak acid ionizing and lowering the pH of the solution. If you see that a weak acid has a low pH, do not immediately assume you did something wrong!

Austin Aldujaili 2D

The solvent is typically denoted in a chemical equation with an (l) beside i, stating that it is a liquid and therefore all other reactants/products simply lie inside that liquid.

Austin Aldujaili 2D

In terms of solving for percent ionization, the terms protonation and ionization would mean the same thing.

Austin Aldujaili 2D

I wouldn't think that there would be any situation where we need to convert to Pascals since Lavelle didn't talk about that unit at all. I think its important to just read the question and figure out what units they want their final answer in.

Austin Aldujaili 2D

I feel as if the textbook goes in more detail on all of the topics Lavelle lectures on; however, the information that is provided in lectures or discussions in my opinion is still enough to succeed on the tests. There's really no info in the book that is important that Lavelle didn't talk about eith...

Austin Aldujaili 2D

Even for scenarios where you start with some amount of product, the products will always need to gain composition/concentration in order to reach equilibrium. Therefore, the products are always +x and the reactants, since they are giving off their concentration to form products, will always be -x.

Austin Aldujaili 2D

I believe later on in the course we will learn how to calculate K when temperature changes. Right now, we just need to know if a change in temp. will favor reactants or products.

Austin Aldujaili 2D

An inert gas is a gas that typically does not react with any other molecules. These are gases in the last column of the periodic table and don't want to give off or gain an electron since their valence shell is already full.

Austin Aldujaili 2D

If you're a big star wars fan, Star Wars the Clone Wars on Netflix is a must watch!!

Austin Aldujaili 2D

I'm not 100% sure, however I do know that the nonhybridized electrons typically are the electrons that are used in the pi bonds and that make molecules more rigid and unable to move around or change their orientation. So maybe there needs to be at least one electron that sits in an atomic orbital ra...

Austin Aldujaili 2D

Not sure, but from the last midterms, a lot of the test questions were exactly the same or at least derived from book problems so to be safe, I would go over as many book problems as possible (that REALLY helped me get through the midterms smoothly as well)

Austin Aldujaili 2D

Personally, using the Bronsted Acid-Base definitions are easier for me to visualize given you can see a proton being accepted or donated by looking at the lewis structures; however, not all acids and bases will be able to be defined by the movement of H+ ions (for example, BF3 can act as an acid in ...

Austin Aldujaili 2D

They are the same thing, the only difference is that, experimentally, H30+ is technically more accurate given that is what is happening at a molecular level (the water molecule gains a proton, lowering the pH of the solution rather than just a H+ ion floating around in solution).

Austin Aldujaili 2D

Atoms in the same row increase in electronegativity towards the right because more protons = stronger pull on the e- = higher affinity/ability to attract e-

Austin Aldujaili 2D

I believe because CO3(2-) has resonance and that can change where the two - formal charge oxygens are. If the single bonded oxygens with a - formal charge are near one another, it can form a bidentate and if they are far apart in the trigonal planar shape, then only one O can bind with a TM and that...

Austin Aldujaili 2D

Instead of Iron they are using the latin name based off of Ferrum (which is why it is called Fe). I'm not sure if there are other exceptions like that one but it's probably not something you have to worry too much about.

Austin Aldujaili 2D

I'm a little confused on drawing the aufbau diagrams for hybridized orbitals because I am not sure if we always move one of the electrons to the higher atomic orbital (ex. e- in 2sp3 moves to 2p) to prevent electron repulsion or if we sometimes do this and it depends on circumstances... If someone c...

Austin Aldujaili 2D

From what Lavelle said in the lecture about how ligands can only form a coordinate bond with one of its lone pairs to a TM, does this mean that ligands cannot form double bonds?

Austin Aldujaili 2D

When figuring out hybridization, I try to look specifically to how many atoms or lone pairs are around the central atom rather than looking at the type of bonds for instance. This is because all the bond types (single, double, triple) are treated equally in hybridization.

Austin Aldujaili 2D

Hybridization treats single, double, triple bonds, and lone pairs all as one region of electron density. Instead of looking for bonds, look for how many atoms and lone pairs are around/bonded to the central atom to figure out its hybridization.

Austin Aldujaili 2D

Due to differences in electronegativity between N and O, the bonds in NO2 are polar. However, if NO2 did not have a lone pair, then these polar bonds would cancel each other out in a linear structure. With the lone pair, the bonds are not directly opposite of one another, meaning the dipole moments ...

Austin Aldujaili 2D

When figuring out hybridization, you only need to count the number of electron densities around the central atom (including lone pairs). there is only one possible s orbital, then 3 possible p orbitals, 5 possible d orbitals, etc.

Austin Aldujaili 2D

Remember when trying to figure out molecular shape, you only want to consider the amount of atoms present, not the amount of e- densities. So H20 has only two atoms branching from the central atom, making it bent with its lone pairs.

Austin Aldujaili 2D

The smaller bond will typically be the strongest so in this case the CF4 bond is strongest. I think it is talking about the covalent vs ionic character of the bond where out of all of those options, CF4 has the least covalent character. Their wording was a little confusing.

Austin Aldujaili 2D

When looking at molecules such as the tetrahedral shape or the trigonal bipyramidal, the equatorial atoms are the ones in the same plane versus the axial atoms are the ones that lie above or below that plane. Although it seems that if you rotate the molecule the perspective changes, I think that it ...

Austin Aldujaili 2D

When determining polarity, it is important to look at not only the difference in electronegativity but also the shape of the molecule. When there are two or more polar bonds opposite of one another, I have always thought of it as like two people playing tug of war with electrons that have the same a...

Austin Aldujaili 2D

I believe the question is trying to get you to look at the molecular shapes of each molecule where in answer b, COFH has higher polarity (meaning higher dipole moment) because there is no F on the other side of the molecule to cancel out the electronegativity of the other F. In the other two molecul...

Austin Aldujaili 2D

SO2 can have an expanded octate since some of its valence electrons are in the 3d orbital. That means it will have a lone pair to satisfy all the electrons in the molecule. Lone pair makes it polar versus CO2 has only 4 bonds with the two 0xygen molecules, no lone pairs.

Austin Aldujaili 2D

A dipole moment in a molecule occurs when electrons are unequally shared between atoms. It can technically occur between all atoms in a molecule regardless of polarity. So no it does not determine if a molecule is polar or nonpolar.

Austin Aldujaili 2D

A dipole-dipole bond and dipole-induced have relatively the same energy in them

Austin Aldujaili 2D

Going back to the last unit/outline, microwaves have very long wavelengths and low frequency, meaning they do not have high enough energy to excite an electron. The movement of electrons in water molecules however will increase the temperature and produce heat.

Austin Aldujaili 2D

A logical way to view if an atom is more likely to become an anion or cation is to look at electron affinity trends in the periodic table and ionization energy trends. So yes, metals will mainly be cations and nonmetals anions

Austin Aldujaili 2D

I am a little confused on what the rules are for electron configuration for some of the transition metals when to break those rules for example in Cu. I know that one of the electrons in the Cu electron configuration is transferred from the 4s orbital to the 3d orbital to make 3d104s1 which is more ...

Austin Aldujaili 2D

Over time, developing test taking strategies as well as an understanding of what a professor wants during an exam or what he/she focuses on is very helpful. I also believe in the idea of studying very lightly before the final (given you have a solid understanding of everything before).

Austin Aldujaili 2D

Remember there is a difference between an ion, an atom with a charge, and an ionic bond, the complete transfer of electrons between two atoms. Because it states there are polyatomic ions does not mean there are ionic bonds. Lewis Structures will be used to show covalent bonds.

Austin Aldujaili 2D

For electron spin, you do not need to worry about figuring it out if one is upward or downward spin (+1/2 or -1/2). The idea of electron spin as the 4th quantum number shows that there will never be electrons with the same set of quantum numbers, they will always have individual electron configurati...

Austin Aldujaili 2D

The numerator is the number of shared e- in the hydrogen atom. Remember that one bond contains 2 shared e-, so the one covalent bond between the N. and the H gives you the 2 in the numerator.
Hope that makes sense.

Austin Aldujaili 2D

I am wondering if, when drawing the valence electrons of an atom, if we follow the same principle that we used when writing out electron configuration: Hund's Rule states that due to e- repulsion, e-in the same subshell occupy different orbitals. As in, we would rather have. lone electrons before we...

Austin Aldujaili 2D

In the diffraction experiment, the wave-like properties are shown in the constructive and destructive interference of the radiation when they go through the slits in the wall. If light solely acted as a particle, then the light would have only showed up right behind the openings in the wall, yet it ...

Austin Aldujaili 2D

Remember if you ever get stuck on problems that ask for an answer in specific units to go through it step by step through dimensional analysis. There you can see how the J will cancel out and leave you with photons/sec and you can convert photons to moles using Avagrado's number.

Austin Aldujaili 2D

Atomic Spectroscopy would support the particle model of light because atoms are only able to absorb a specific set of wavelengths or frequencies given it must match the change in energy between shells like n=1 and n=2. By showing that light is absorbed in only a set wavelength or frequency, it prove...

Austin Aldujaili 2D

Just as they described above, you want to solve for the number of photons that the metal absorbed because, given the photoelectric effect and the particle model of light, one photon will only be able to eject one electron. Therefore, solving for the number of photons will result in the same value as...

Austin Aldujaili 2D

We only know that the molecular formula of sodium carbonate is Na2CO3 by understanding charges in the molecules. CO3 has a -2 charge when by itself and Na has a +1 charge when isolated. Therefore, the formula would be Na2CO3 in order to result in a neutral charge.

Austin Aldujaili 2D

Since electrons have a mass, using E=hv to calculate the energy of an electron would be incorrect. As everyone else has said, use DeBroglie's Equation to calculate speed, then KE of an electron.

Austin Aldujaili 2D

An easy way to remember the varying wavelengths on the EM spectrum is that infrared has longer wavelengths and is close to red in the visible light region. As the wavelengths get shorter, you get to violet in the visible light then you move into ultraviolet, and so on into x rays and gamma rays.

Austin Aldujaili 2D

There is no specification but I recommend using as many as possible ESPECIALLY when dealing with very small numbers like moles for instance. I have gone through some molecular formula problems and got varying answers solely because I did not use exact amounts for the molar masses.

Austin Aldujaili 2D

Personally, I feel that for chemistry doing practice problems is a great way to ensure that you are on top of the information. Sometimes, I would make a short outline or "cheat sheet" of the information I know as a refresher and to make sure I understand all of the concepts.

Austin Aldujaili 2D

I just watched the lecture on De Broglie's Equation and am curious as to whether there is a set value on when an object has a detectable wavelength (not insignificant due to too large of a mass) or if this is something that scientists deem as subjective. Is everything going to be seen as not having ...

Austin Aldujaili 2D

Lavelle uses the idea of the free electron, an electron not bound to the nucleus, to explain that why the Emperical Equation is negative. Using a free electron as a reference point, once the electron moves into lower and lower energy levels approaching an atom's nucleus, it is going to be going into...

Austin Aldujaili 2D

An important thing to note here is that from n=1 to n=2 has the highest energy difference and is therefore the hardest to overcome. This is why UV radiation (Lyman Series) always drops down to n=1 from some higher energy level when emitting electromagnetic radiation given UV light has more energy pe...

Austin Aldujaili 2D

Q: In the ultraviolet spectrum of atomic hydrogen, a line is observed at 102.6 nm. Determine the values of n for the initial and final energy levels of the electron during the emission of energy that leads to this spectral line. I am really confused on how to use the different series (Balmer, Lyman,...

Austin Aldujaili 2D

Ever since I was a child, I have been told to always stand away from the microwave when it is heating food because it is "dangerous" or bad for me. However, according to the electromagnetic spectrum, microwaves have a lower frequency (and therefore less energy) than visible light. Does thi...

Austin Aldujaili 2D

Another way to think about it is to utilize both equations (E=hv and c= v) and combine them into one larger equation that only has the values E, c, h, and \lambda . Therefore you would get E = ch/ \lambda Knowing your constant values for c and h as well as the Energy of a photon (which is equal to t...

Austin Aldujaili 2D

Since you do not know the mystery metal "M" in the metal hydroxide, that is what we need to figure out before we can substitute the hydroxide with sulfide. To do so, subtract the molar mass of the hydroxide (OH)2 by the entire molar mass of the compound: 74.10 g/mol - 33.996 g/mol = 40.104...

Austin Aldujaili 2D

Another way to think about it is to utilize both equations (E=hv and c= \lambda v) and combine them into one larger equation that only has the values E, c, h, and \lambda . Therefore you would get E = ch/ \lambda Using 1.97 x 10^-7 m and your constants for c and h, you can find the energy of the pho...

Austin Aldujaili 2D

First, convert the 5.00 g of glucose and the 5.00 g of Oxygen to moles using their respective molar masses. Glucose: 5.00 g x (1 mol/180.156 g) = 0.028 mol Oxygen: 5.00 g x (1 mol/31.998 g) = 0.156 mol Then, we use the molar ratios to determine how many moles of oxygen we would need to complement th...

Austin Aldujaili 2D

Here's the question for reference: Which sample in each of the following pairs contains the greater number of moles of atoms? (a) 75 g of indium or 80 g of tellurium; (b) 15.0 g of P or 15.0 g of S; (c) 7.36 3 1027 atoms of Ru or 7.36 3 1027 atoms of Fe. Correct, it doesn't matter what element you h...

Austin Aldujaili 2D

It's also important to always make a quick check as to whether the chemical equation is balanced or not. In this case it already was; however, this should be one of the first things you do or else your molar ratios you use during calculations may end up being wrong and throwing your values off. I kn...

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