CHEMISTRY COMMUNITY

Justin Vayakone 1C

Oh nevermind... "The final is 6 questions in 3-4 hrs (still to be decided). Time will not be an issue."

Justin Vayakone 1C

Time limit or due date for the final? I didn't see anything about that in the emails.

Justin Vayakone 1C

If we are just looking at the concentration of [H+], then it would be $(10)^{-64}$
But in the solutions manual, they combined $(10)^{-64}$ with the other concentrations.
This would be $\frac{10^{-2}}{(10^{-4})(10^{-64})} = 10^{66}$

Justin Vayakone 1C

Look up "endgame" to find the worksheets for Lyndon's electrochem/kinetics review.

Justin Vayakone 1C

I know for Lyndon's review session, I can look up endgame to find it, but is there a worksheet for the review session on Thursday, March 12, 6-9pm, Moore 100, Riya Shah, Matthew Tran, Kate Santoso, Chemical Equilibrium, Acids & Bases, Thermochem, Thermodynamics?

Justin Vayakone 1C

I know in class, Lavelle used an example of a reaction that needs an enzyme or catalyst to describe a zero order reaction. Pretty much the catalyst is at max capacity, and the reaction can't go any faster. By adding more reactant, the reaction rate doesn't change, so it is a zero order reaction. I r...

Justin Vayakone 1C

Here's a picture of how to do it: I set up an ICE table. Our initial amounts are 0.245M SO3 and a certain amount of NO that we don't know yet. Since we know that 0.24M SO2 will be formed at equilibrium we can say the change is 0.24. Then by setting up the mass action, we can isolate the initial conc...

Justin Vayakone 1C

Usually for anodes, the solid is the reactant of the half reaction. In this case, the iodine solid is the product. I don't have an explanation why this needs the platinum because I don't know as well, but I think it does have something to do with the solid being a product. Hello, How did you determ...

Justin Vayakone 1C

E⁰ means reduction potential under standard conditions, i.e. concentrations of 1 M in each cell and pressures of 1 atmosphere at 25 C degrees. In a concentration cell, the only thing driving the reaction is the difference in concentration. Under standard conditions, since both half cells have 1M, th...

Justin Vayakone 1C

Usually for anodes, the solid is the reactant of the half reaction. In this case, the iodine solid is the product. I don't have an explanation why this needs the platinum because I don't know as well, but I think it does have something to do with the solid being a product. Hello, How did you determ...

Justin Vayakone 1C

Usually for anodes, the solid is the reactant of the half reaction. In this case, the iodine solid is the product. I don't have an explanation why this needs the platinum because I don't know as well, but I think it does have something to do with the solid being a product.

Justin Vayakone 1C

Write the half-reactions, the balanced equation for the cell reaction, and the cell diagram for each of the following skeletal equations: (d) Au^+ (aq) \rightarrow Au(s) + Au^{ 3+} (aq) Here are the half reactions I used: 2[Au^+(aq)+e^-\rightarrow Au(s)] and ...

Justin Vayakone 1C

Write the half-reactions, the balanced equation for the cell reaction, and the cell diagram for each of the following skeletal equations: (d) Au^+ (aq) \rightarrow Au(s) + Au^{ 3+} (aq) Here are the half reactions I used: 2[Au^+(aq)+e^-\rightarrow Au(s)] and A...

Justin Vayakone 1C

Here is how to balance the oxidation reaction in acidic solution:

Justin Vayakone 1C

Lavelle said that the reaction from diamond to graphite was spontaneous but had a very high activation energy and would occur very slowly. Does this mean that over time, say a few trillions years, the diamond is slowly collecting enough energy to overcome the energy barrier? Or would the reaction ev...

Justin Vayakone 1C

My explanation is sort of modeled off of 4I.9. ΔS of the surr depends on the circumstances. For a reversible process, the system is doing work to push the piston, but the surroundings is giving heat/energy in order to keep internal energy of the system consistent. This means ΔS(surr) = -ΔS(sys) and ...

Justin Vayakone 1C

The units for change in entropy is J/K. That's why we multiply by moles to cancel it. I also think you're confusing state function with intensive property. A state function doesn't necessarily mean it is intensive. Change in entropy definitely depends on how many moles of substance we have.

Justin Vayakone 1C

My mind was blown when I learned about cellular respiration in LS7A. I remember in high school chem, I merely knew about the overall chemical reaction, glucose and oxygen yielding energy, CO2 and H2O. And now in LS7A, I'm starting to see just how crazy complicated the process really is. The human bo...

Justin Vayakone 1C

Well if it's a neutral solution, that should mean the redox reaction wouldn't be using or creating any H+ or OH-. If we had to balance a redox reaction in neutral solution, we would probably be given something like this where H+ or OH- ions wouldn't be needed: Cu+(aq)+Fe(s)→Fe3+(aq)+Cu(s)

Justin Vayakone 1C

Some people have said that it's a typo, so it should say 2Cl-(aq) as the product rather than Cl2(g)

Justin Vayakone 1C

"You have a system consisting of 0.40 moles of an ideal gas contained in a 100.0L container at 1.0 atm. You just love chemistry to a fault, so you perform a series of steps to the system. First, you perform an isobaric compression of the container to 10.0L. Then, you pressurize the system to 10...

Justin Vayakone 1C

"Initially an ideal gas at 323 K occupies 1.67 L at 4.95 atm. The gas is allowed to expand to 7.33 L by two pathways: (a) isothermal, reversible expansion; (b) isothermal, irreversible free expansion. Calculate ΔStot, ΔS, and ΔSsurr for each pathway." For part b, why is the ΔS of the irrev...

Justin Vayakone 1C

When you use w=P_{ex}\Delta V , external pressure is 0 because free expansion means the system isn't pushing against a pressure. It is freely expanding without any use of energy. This means w=0. For isothermal processes, ∆U = 0 = q + w, so q and w = 0 in this case. I'm not sure why the ∆S is the sam...

Justin Vayakone 1C

"Container B has 1.0 mol of atoms bound together as diatomic molecules that are not vibrationally active. Container C has 1.0 mol of atoms bound together as diatomic molecules that are vibrationally active." I think the wording is just tricky. "1.0 mol of atoms bound together as diato...

Justin Vayakone 1C

A solution of water is neutral, so Ka = Kb. Taking the square root of Kw will give Ka and Kb.

Justin Vayakone 1C

This is how I did the question.

Justin Vayakone 1C

Also getting -138.18, so probably inputting into the calculator wrong.

Justin Vayakone 1C

Yes, endothermic reactions mean heat is a reactant, so in high temperature (lots of heat), the reaction will proceed to the right in order to decrease the temperature/heat.

Justin Vayakone 1C

When work is a positive value, it means the surroundings is doing work on the system while a negative value means the system is doing work on the surroundings. The more negative the value, the more work the system is doing.

Justin Vayakone 1C

For 4E.5) you used the C-C bond enthalpy (-348kJ/mol) when it should be the C-H bond enthalpy which is 518kJ/mol For 4E.7) a) Your work looks right. You should be getting -202 kJ/mol. Maybe you plugged it into your calculator wrong? b) You forgot to include the bond enthalpy of C-C in the products. ...

Justin Vayakone 1C

Heat released = - heat absorbed
Your work here looks right. I'm assuming you used the negative to change (T-100) into (100-T) right?

Justin Vayakone 1C

My answer book shows 28J. Your statement regarding internal energy and work is correct.

Justin Vayakone 1C

I think enthalpy density is a value like regular density or volume, in which the value can't be negative.

Justin Vayakone 1C

Well let's look at the changes for A B and C. A: -10 B: +5 C: +10 The ratio of A/B/C is -10/+5/+10 or simplified -2/1/2. Negative change indicates the reactant while positive indicates products. The ratio above represents the coefficients for the equation. For every 2 A molecules used, 1 B and 2 C m...

Justin Vayakone 1C

Ba(OH)2 is a strong acid, so you can assume all of the reactant turns into products. No need for an ICE table or K value. Just be careful and know that there will be double [OH] compared to [Ba2+].

Justin Vayakone 1C

NH3OH+ is the conjugate acid of hydroxylamine (NH2OH) which is given in Table 6C.2. (CH3)2NH2+ is the conjugate acid of dimethylamine, (CH3)2NH which is also given in Table 6C.2. Just do 14 - pKb to find pkA.

Justin Vayakone 1C

That equation given by your TA comes from this equation (PV=CRT) given by Lavelle. Here is an example I made illustrating where the difference in moles comes from.
Edit: accidentally put exponent inside B brackets twice

Justin Vayakone 1C

NaCl doesn't affect the pH of the solution because the Na+ and Cl- ions don't react with anything. This means we completely ignore NaCl. We are given 0.20 M of NH2NH2, which is a weak base. Create an ICE table and look up the Kb for NH2NH2 in one of the tables in the textbook (6C.1 , 6C.2 and 6E.1)....

Justin Vayakone 1C

When doing the approximation, you have to double check the ionization percentage. Lavelle told us in lecture today that if this ionization percentage is above 5%, then the approximation isn't accurate enough, and we should use other methods like quadratic formula. To find this percentage, do the amo...

Justin Vayakone 1C

Yeah it looks like the answer key is wrong. The pH should be -0.176. With this pH, the pOH would now be 14.176 and $[OH^-]$ = $6.67 \cdot 10^{-15}$

Justin Vayakone 1C

For each mole of Ba(OH)2 that dissociates, 2 moles of OH are created, so 0.0092 M Ba(OH)2 turns into 0.0184 M OH.

Justin Vayakone 1C

My solutions manual only has odds and not even numbered problems. Could you post a picture of it so I could see?

Justin Vayakone 1C

Being thermodynamically stable means being in a low energy state. The less energy or enthalpy a molecule has, the more stable it is. Now to the problem, we are asked which reactant (Cl2 or F2) is more stable. The more stable a reactant is, the less it wants to dissociate into product. Comparing 1.1x...

Justin Vayakone 1C

The photoelectric experiment (photons shined on a metal surface which ejected electrons) showcases the particle property of electrons. For the wave property of electrons, I don't know the name, but I think the experiment involved particles passing through a hole in a wall, then a detector at the end...

Justin Vayakone 1C

This is the answer to part b: (in attachment). I understand the arrangement of the equilibrium constant and where the values (5, 10, and 18) come from, but I don't understand why these values are over 100. Is it because of units? The graph uses P/pKa, but what does that mean? Pressure per kiloPascal?

Justin Vayakone 1C

5G.4 is all about Gibbs free energy which doesn't have anything to do with question 5G.11. Read through 5G.2 and you should be able to solve this problem. It's all partial pressures and mentions "Pure liquids and solids do not appear in K" (5G.2)

Justin Vayakone 1C

Exothermic means energy is released while endothermic means energy is taken in, so exothermic:energy is a product, and endothermic: energy is a reactant. What I use to help me determine which is which is to think that endo sounds like "in", so it takes in energy. In exothermic reactions, e...

Justin Vayakone 1C

If you can solve the problem, without the x^2 approximation, then do it that way. In this case, making x^2 equal to 0 is too much of an approximation. This is how I solved for x: (in file attachment)
Edit: Click on the image to see it right side up

Justin Vayakone 1C

A very large K value means the reaction at equilibrium will heavily lean towards the products. The equilibrium constant is basically concentration of products over concentration of reactants, so for K to be a large value, the numerator (product) should be big while denominator (reactant) should be s...

Justin Vayakone 1C

My solutions manual uses partial pressures. Is your's seventh edition?

Justin Vayakone 1C

Use this equation: $\Delta G^o = -RTlnK$

Justin Vayakone 1C

Doing log[.88] means the concentration was reduced by 12%, but the problem is asking for when it decreases to 12%. The equation for the initial pH is: pH = -log[HCl]_i Final pH: pH = -log(0.12[HCl]_i) To find the change, just do final pH - initial pH. Through some algebra, you should get -lo...

Justin Vayakone 1C

Wait there was a worksheet he emailed us? When did he email it? I don't see it.

Justin Vayakone 1C

When figuring out whether a ligand is monodenate or polydentate, we have to look at the lone pairs. Both of the lone pairs of water are on the oxygen right next to each other. When lone pairs are on the same atom, only one can bond because the other lone pair will point away from the bonding metal a...

Justin Vayakone 1C

When it comes to boiling point, we have to look at the intermolecular forces (e.g. dipole-dipole and london dispersion forces) not the intramolecular strength (like bond strength within a molecule). Since both CH4 and CCl4 are nonpolar, they both only have london dispersion forces for intermolecular...

Justin Vayakone 1C

In this lecture, what information do we have to know since Lavelle said some of the information was 14B material? Like for example, would we have to know stoichiometric point and the polyprotic acids and bases for the final?

Justin Vayakone 1C

The answer is electron-electron repulsion. There is more repulsion when two electrons share an orbital making it less favorable and unstable. Therefore, to avoid this, one of the electrons in the s-orbital jumps to fill an empty p-orbital. Hope this helps! I'm still confused. In the first photo, th...

Justin Vayakone 1C

Yeah generally, molecules with OH will indicate a bronsted base and H for bronsted acids. For part a though, NH3 (ammonia) is a base. It will gain a proton to become NH4 (ammonium). Memorizing some of the common acids and bases will help in figuring out these molecules. B and D are the acids, while ...

Justin Vayakone 1C

Why is it that when we use sp^2 orbitals for Carbon, it involves three 2sp^2 and one 2p^1 orbital? Why not three 2sp^2 orbitals and one 2s^1 orbital?

Justin Vayakone 1C

It depends on what we are given. If a weak acid is involved, then we need to use the equilibrium constant (Ka) along with the given concentrations to make an ice table and calculate [H+] or [OH-], but I don't think we'll be facing these types of problems in this course. I haven't seen the practice p...

Justin Vayakone 1C

The problem is asking for equations involving a cation that is a weak acid reacting with water or an anion that is a weak base reacting with water, so they will look something like this: \textup{A}^++\textup{H}_2\textup{O}\rightleftharpoons \textup{A}+\textup{H}_3\textup{O}^+ or \textup{B}^-+\textup...

Justin Vayakone 1C

Yes each hybridized orbital represents bonds in the molecule. Let's take methane (CH4) for example. Because there are four regions of electron density around the Carbon atom, the s and p orbitals combine to create sp^3 orbitals: https://ibb.co/DGWMx8v Each of these sp^3 orbitals has an unpaired elec...

Justin Vayakone 1C

Do all molecules use hybridized orbitals? No. Hybridized orbitals are only considered when necessary. For example, H2 molecules use the 1s orbital and don't need hybridized orbitals. You will have to look at the electron configuration of the atom to figure out if there are enough unpaired electrons...

Justin Vayakone 1C

Image of Lewis Structure: https://ibb.co/FbB3TwM There are 2 electron densities around the right carbon, so sp hybridized orbitals are used for the triple bond. Here is the electron configuration shown by my epic MS paint skills: https://ibb.co/kQTh67k For the left carbon atom, sp^3 orbitals are use...

Justin Vayakone 1C

Hybridization is just a model. It's used whenever necessary. If the current orbitals of a molecule don't have enough unpaired electrons to make all of the necessary bonds, then we use hybridized orbitals to explain for the lack of unpaired electrons. For example, a molecule like H2 won't have hybrid...

Justin Vayakone 1C

Hybridization is just another addition to the bonding model to fix the issue that there weren't enough unpaired electrons to make all of the bonds, so I don't think energy of the molecule changes. But when it comes to the hybridized orbitals, I thought Lavelle said the energy of these orbitals were ...

Justin Vayakone 1C

https://imgur.com/a/AUAeA2Z Out of these three structures, the top right one is the best structure. Negative formal charge should be placed on the most electronegative atom if possible. In this case, O is more electronegative than N, so O should get the negative formal charge over N. Edit: Is the i...

Justin Vayakone 1C

Nonpolar molecules can't have dipole-dipole interactions. Within each molecule, all of the dipoles cancel out, so there should be no dipole interaction between molecules. The only intermolecular force of nonpolar molecules is london dispersion forces.

Justin Vayakone 1C

I think double/triple bonds can distort the molecule geometry slightly, but the VSEPR model ignores that for simplicity I guess. It's just a model, so there are probably limitations and exceptions.

Justin Vayakone 1C

Imagine a closed container half filled with liquid. The temperature is then increased to the point where some of the liquid evaporated into gas. The gas particles colliding with the liquid creates what's called vapor pressure, pressure exerted by gas against it's solid or liquid phase. Vapor pressur...

Justin Vayakone 1C

Lavelle hasn't taught us hybridized orbitals yet, but it's pretty much combining orbitals into one type. For example, the two 2s electrons and two 2p electrons could combine to create 2sp^3 orbitals. Here is a table I made back from AP chem to help determine what hybridized orbitals are used. I woul...

Justin Vayakone 1C

Alright, I'll pretty much paraphrase the paragraph above the figure. In the gas phase, molecules are rotating rapidly. An assumption you can make is that there is no net interaction between the molecules because the attraction between opposite charges are canceled by the repulsion of like charges. T...

Justin Vayakone 1C

Vapor pressure is relative to boiling point. If the boiling point of one molecule is lower than a different molecule, then the first molecule will have more gas molecules, leading to more vapor pressure. To tie all of this in terms of IMFs, you can say a weaker IMF makes the molecules easier to brea...

Justin Vayakone 1C

Yes, Mo and Ag want to have the half filled d-orbital over the full s-orbital, just like Cr and Cu above them.

Justin Vayakone 1C

Do we use the De Broglie equation for the midterm review question 13a? Yes we do. We know this because we are using the wavelength of a potassium ion, which has mass. The speed of light equation (\textup{c}=\lambda \upsilon ) on the other hand would be used for situations involving electrom...

Justin Vayakone 1C

Well for visible light, I use (ROY G BIV) to remember colors from weakest to strongest energy. Then to the left of visible light is infraRED light and to the right of visible light is ultraVIOLET light. Before infrared light and the start of the EM spectrum is radio waves followed by microwaves. I r...

Justin Vayakone 1C

The electron configuration of Ag is: [\textup{Kr}] 4\textup{d}^{10} 4\textup{s}^1 . Silver is one of the exceptions of the Aufbau principle because it is more stable to have a full d-subshell and one s-electron rather than a full s-subshell and 9 d-electrons. Since the question is asking for \textup...

Justin Vayakone 1C

Do we use the De Broglie equation for the midterm review question 13a? Yes we do. We know this because we are using the wavelength of a potassium ion, which has mass. The speed of light equation (\textup{c}=\lambda \upsilon ) on the other hand would be used for situations involving electrom...

Justin Vayakone 1C

I think if the cation's radius is small, the electrons of the anion would be closer to the cation nucleus, creating a stronger pull and polarizing the electrons more easily.

Justin Vayakone 1C

Actually, when we add electrons, the attractive pull between the nucleus and electrons are weaker. With more electrons to pull in, the force is weaker. There is also added repulsion between the electrons that cause them to be further apart, creating a larger radius.

Justin Vayakone 1C

So for number 9 in the Dino Nuggets Midterm Review, the UA's explained that oxygen actually has less ionization energy than nitrogen despite being on the right of nitrogen. I know it's because of half-filled orbitals being pretty stable, but does this concept apply to the periods below nitrogen and ...

Justin Vayakone 1C

Well d-subshells can hold 10 electrons, so I guess 18 could be the max? I don't think we need to worry about a max because we won't ever get to that point.

Justin Vayakone 1C

If we were looking at \textup{M}^{2+} , then your answer of Iron and Manganese would be correct, but we are looking at metal cations with a 3+ charge or in other words elements that lose 3 electrons. Let's take a look at Iron. The electron configuration of Iron is [\textup{Ar}]3\textup{d}^64\textup{...

Justin Vayakone 1C

The size of the atoms largely contribute to the melting point of molecules. Fluorine may have stronger dipole-dipole attraction than Iodine, but that difference in electronegativity is trumped by the difference in size. Iodine is bigger than Fluorine, meaning more electrons and stronger London Dispe...

Justin Vayakone 1C

In my solution manual, the answer to part c is ns^2 (n-1)d^3. That's the answer you said in your post right? ns^2 is the 2 valence electrons, and the (n-1)d^3 is the outermost d-orbital electrons.

Justin Vayakone 1C

For part a and b, keep in mind that the 4s orbital is higher in energy than the 3d orbital. The question is asking for metal cations with a 2+ charge, meaning the metal element will lose 2 electrons. Part a and b involve electron configurations with electrons in the 3d and 4s orbitals. Since 4s is h...

Justin Vayakone 1C

To solve this problem, we can use the \textup{M}_1{}\textup{V}_1 = \textup{M}_2{}\textup{V}_2 equation. To find the original molarity, we have to first convert the 5.00 g of KMnO4 into moles using KMnO4 molar mass ( 5.00\textup{g} \textup{KMnO}_4\div158.04\textup{g/mol}= 0.0316 \textup{mol KMnO}_4 )...

Justin Vayakone 1C

BrF3 has 28 electrons. This molecule is one of the exceptions to the octet rule. Br can have an expanded octet. Br will be the central atom with single bonds to each F and 2 lone pairs to itself.

Justin Vayakone 1C

I guess the increase in nuclear charge trumps the electron shielding. Maybe because when increasing across a period, the electrons added are valence electrons, so I think the electron shielding really isn't increasing much.

Justin Vayakone 1C

No, the ground state of an ion is the ion itself. For example, the ground state of Ca+ is Ca+.

Justin Vayakone 1C

The equation is \textup{c}=\lambda\cdot v . The v actually represents frequency, not velocity. The units of the equation are (\textup{m/s}) = (\textup{m}))(s^{-1}) . As you mentioned, the units don't work if you plug in velocity into that equation. Checking units of each ...

Justin Vayakone 1C

It stands for kilo-electron volt or 1000 electron volts. The SI unit of this is joules. The conversion is $1 \textup{eV}=1.602\times 10^{-19} \textup{J}$

Justin Vayakone 1C

The question is asking for a specific subshell. When we are given an n and l value, we will have a subshell. For part a) since n=5 and l=2, we know the subshell is 5d. Subshell notation is simply just the n value followed by s, p, d, or f subshells. l=0 is the s-subshell, l=1 is the p-subshell, l=2 ...

Justin Vayakone 1C

Electron configuration is just an easy way to see where all of the electrons are in the orbitals. When we are stating all of the orbitals, we start from lowest energy to highest energy. The periodic table is a great way to figure out what orbitals are occupied. Here is a link to see the periodic tab...

Justin Vayakone 1C

Actually, all four of the values would still increase if the electron was in a hydrogen atom. All of the reasoning for the Lithium electron is the same for a hydrogen atom. In both cases, the the electron moves from an s-orbital to a p-orbital, which means the energy and n value is increased. When n...

Justin Vayakone 1C

When elements have z\leq 20 , then the 4s orbital is less in energy than 3d meaning 4s is written first. In other words, elements up to Calcium will have the 4s state come before 3d. But for elements after calcium in the periodic table, the 4s state is actually higher in energy than 3d. For example ...

Justin Vayakone 1C

Each line represents energy released by the movement of a hydrogen electron from one energy level to a lower energy level. Each energy level is represented by a principal quantum number (n). What's common among each series is that their lines all involve the electron being in the same final energy l...

Justin Vayakone 1C

Does it matter if these practice problems are in pen or pencil?

Justin Vayakone 1C

I think they are due every Sunday.

Justin Vayakone 1C

In the seventh edition of the textbook, the 1A problems are right after Topic 1A on pg 9. Regarding turning in homework questions, I believe you can turn in questions just from Review of Chemical Principles if you want, or only from Quantum world, or a mix of the two.

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