CHEMISTRY COMMUNITY

A De Castro 14B 2H

Mechanism C wouldn't agree with the proposed rate law for the same reason that mechanism A wouldn't: the rate law would include CO. Having a reverse reaction in step 1 means that the reactants used to produce NO 3 can be used in the rate law. Thus, for (c), rate = k[NO 2 ] 2 [CO]. But, again, this d...

A De Castro 14B 2H

The value of the equilibrium constant is only affected by a change in temperature. However, the position of equilibrium itself (i.e. the state when both reactants and products are present in concentrations that tend not to change over time) is affected by many factors. These include change in concen...

A De Castro 14B 2H

Ice melting is a spontaneous process because the order of the water molecules is interrupted, meaning that entropy increases (+∆S). Even though it requires energy (endothermic, +∆H), the reaction usually occurs in high temperatures, resulting in -∆G. The combustion of a natural gas is spontaneous be...

A De Castro 14B 2H

Yes, in the test, K w was K 1 . Once you determine the K 2 value, you can use the rule: K = [H + ][OH - ]. This means you can square root the K 2 value to get [H + ], which you can then plug in pH = -log[H + ]. The pH value you calculate will be the new pH value for neutral solutions. For example, s...

A De Castro 14B 2H

When one substance has a lower molar entropy than another, it means that substance is comparatively more ordered (less disorder). This further indicates that more heat is needed to make it more disordered than the other substance. For example, ice (solid H 2 O) has lower molar entropy than water (li...

A De Castro 14B 2H

Yes, because the rate of that second order reaction is proportional to the squared value of reactant A. However, remember that this applies to reactions that only have one reactant. The overall order of the reaction would be different if there were more than one reactants.

A De Castro 14B 2H

For strong acids, the concentration of H + ions generally equals the concentration of the strong acid. For strong bases, the concentration of OH - ions generally equals the concentration of the strong base. However, for weak acids and bases, their respective concentrations of H + and OH - ions don't...

A De Castro 14B 2H

Reaction mechanisms tell us the order and number of elementary steps that occur to form products. They also tell us whether or not reaction intermediates are formed then consumed. The slowest step in the mechanism, which is usually indicated, determines the rate of the reaction.

A De Castro 14B 2H

In basic solutions, the OH^- ions tend to be on the reactants side in the oxidation half-reaction. Then, the OH^- ions tend to be on the products side in the reduction half-reaction.

A De Castro 14B 2H

The oxidizing agent is the molecule that has the atom whose oxidation state decreases, due to gaining electrons (reduction). Meanwhile, the reducing agent is the molecule that has the atom whose oxidation state increases, due to the loss of electrons (oxidation).

A De Castro 14B 2H

Don’t forget that there is also this equation: ∆G = ∆G° + RTlnQ A lot of homework questions about Gibbs free energy can only be answered with this. It’s generally used when given a temperature value, a K value, and concentations/partial pressures of reactants and products to calculate Q.

A De Castro 14B 2H

E values of half reactions can be used to figure out which one is reduced and oxidized. A more positive reduction potential will tend to be reduced (i.e. gain electrons), while a comparatively more negative reduction potential will tend to be oxidized (i.e. lose electrons).

A De Castro 14B 2H

Reduction potentials can be calculated by comparing the electron transfer of the cell (half reaction) with respect to the Standard Hydrogen Electrode under the conditions of 1M solution, 1 atm, and 25°C (298K). Reduction potentials can also be calculated by using or rearranging this equation: E° cel...

A De Castro 14B 2H

You can use Toolbox K.1 in the 6th edition of the textbook to determine the basic oxidation states. As for metals, it usually depends on the other atoms in the molecule they're in. This is because transition metals can have a range of oxidation states, like the ones in the attached image (the ones i...

A De Castro 14B 2H

If by anode, you mean the electrode where oxidation occurs, and by cathode, you mean the electrode where the reduction takes place, then yes, electrons always flow from anode to cathode. This can be confusing depending on the cell (voltaic or electrolytic) in which the reaction takes place.

A De Castro 14B 2H

What TA's review sessions did y'all find to be the most helpful? Because I've gone to a few where I feel more lost when I leave then helped. From the review sessions for the midterm, I found those by Morris, Nathan, and Ronald the most helpful because they were fun and covered the material in a way...

A De Castro 14B 2H

If by "charge is not given," you mean that the molecule has a neutral charge, then the charge would be 0. For example, H 2 O is a neutral molecule; it has a net charge of 0. Based on the charges from the periodic table, oxygen usually has a 2- charge while hydrogen usually has a + charge. ...

A De Castro 14B 2H

When writing half reactions for redox reactions in acidic solution, the only acceptable additions are H 2 O, H + , and electrons. For example, balancing this reduction reaction: Cr 2 O 7 2- → Cr 3+ 1. Balance elements in the equation other than oxygen and hydrogen. so, Cr 2 O 7 2- → 2 Cr 3+ 2. Balan...

A De Castro 14B 2H

The salt bridge allows ion transfer between two half reactions by providing an electrical connection within which ions can flow. This keeps both solutions at a neutral state, so electron flow can continue and be measurable through a voltmeter. Salt bridges usually contain an unreactive electrolyte t...

A De Castro 14B 2H

A reaction is spontaneous if it occurs without help or work. A reaction with negative ∆G° is spontaneous, since ∆G° usually results from negative ∆H° and positive ∆S°. The latter means that disorder of the universe increases, which always applies to spontaneous processes. However, remember that all ...

A De Castro 14B 2H

One can tell which ion will be oxidized or reduced based on their standard reduction potentials, E°. A solution with a more positive reduction potential will tend to gain electrons (i.e. to be reduced by oxidizing the new species). Meanwhile, a solution with a more negative reduction potential will ...

A De Castro 14B 2H

Being an inert and unreactive electrode means that it will not give or take any electrons during the reaction. This leaves the reduction and oxidation reactions to the reactants in question.

A De Castro 14B 2H

The change in Gibbs free energy is a state function, i.e. it is additive. Dr. Lavelle showed in lecture that ∆G°_rxn = ΣnG_f°(products) - ΣnG_f°(reactants)

A De Castro 14B 2H

To exhibit the answers above, here's an example: if lnK = 0, then K = e⁰, making K = 1.

A De Castro 14B 2H

One way to remember the left-hand (anode) and right-hand (cathode) electrodes is the acronym AN OIL RIG CAT. This stands for Anode for Oxidation Is Losing elections, while Reduction Is Gaining elections in the Cathode.

A De Castro 14B 2H

In addition to the answers above, Gibbs free energy also allows you to determine whether or not a reaction is likely to occur. Based on the ∆G°= ∆H°- T∆S, negative enthalpy and positive entropy will result to a negative Gibbs free energy, which favors the forward reaction. When enthalpy is positive ...

A De Castro 14B 2H

I answered a bunch of Gibbs Free Energy problems, and a couple of electrochemistry questions because it was taught in lecture yesterday.

A De Castro 14B 2H

K_c is a reaction's equilibrium constant. $Kc = \frac{[prodcuts]}{[reactants]}$
K_w is the ionization constant of water. K_w = [H₃O⁺][OH^-] = 1.0 x 10^-14
pK_w is just the negative log of water's ionization constant. pK_w = -log(pK_w)

A De Castro 14B 2H

I believe both C₅H₅N and NH₃ are in table 12.2. As for the anions, you can find their pK_b values by subtracting the pK_a values of their parent acids (HF and CH₃COOH), found in table 12.1, from 14 (which is pK_w). The lower the pK_b value, the stronger the base is.

A De Castro 14B 2H

ICE tables are often used when you're asked to find equilibrium concentrations, and when you're given initial concentrations and a Kc value. However, you can also be asked to do the other way around: find initial concentrations from equilibrium concentrations and a Kc value. So, no, the concentratio...

A De Castro 14B 2H

Can Q fluctuate around K so that it could be higher at one point and lower at another point? I believe that Q can fluctuate around K, depending on the specific given time you measure Q. For instance, if you measure Q a few seconds after products are added to the system, Q will be greater than K (th...

A De Castro 14B 2H

The delta H values indicate whether the reaction is endothermic or exothermic. When delta H is positive, the reaction is endothermic, simply meaning that heat is in the reactants side. When delta H is negative, the reaction is exothermic, simply meaning that heat is in the products side. Thus, in a)...

A De Castro 14B 2H

Positive delta H is endothermic because the system absorbed heat, which means that there is more energy in the products side. This results to a positive difference in H, when H of reactants is subtracted from H of products. On the other hand, negative delta H is exothermic because the system release...

A De Castro 14B 2H

The more negative the charge of an anion (the greater the number of electrons compared to the number of protons) and bigger its ionic radius, the more polarizable it is. For example P 3- is more polarizable than Cl - . The more positive the charge of a cation (the greater the number of protons compa...

A De Castro 14B 2H

A compound is amphoteric if it is able to react as both a Lewis base (lone pair donor) and as a Lewis acid (lone pair acceptor). Also, metal oxides (Na 2 O, CaO) react with water to form strong bases, while nonmetal oxides (ex. CO 2 , SO 2 ) react with water to form acids. As for As 2 O 3 and Bi 2 O...

A De Castro 14B 2H

OH 2 is often used to make it clear that it is the oxygen (not the hydrogen) that is bonded to the central atom. People can overlook that and make mistakes, especially when drawing Lewis structures. Other than that, as long as you know that the bond constitutes electrons that belong to oxygen, I bel...

A De Castro 14B 2H

Yes, there is a difference between concentration and molarity. As you indicated with "concentration as g/L of Kg/L," concentration is the ratio of the amount of solute per amount of solution. Meanwhile, molarity is a unit of concentration that relates the amount of moles of a solute per li...

A De Castro 14B 2H

In terms of electron geometry, sp 3 d generally corresponds with the trigonal bipyramidal shape, which has a VSEPR formula of AX 5 . It can definitely differ with regards to molecular geometry though, as stated by the prior answers. Nonetheless, sp 3 d hybridization indicates that five hybrid orbita...

A De Castro 14B 2H

One equation include sulfur: S + O 2 → SO 2 2SO 2 + O 2 → 2SO 3 SO 3 + H 2 O → H 2 SO 4 Sulphur, which comes from burning of coal, reacts with oxygen in the air to form sulfur dioxide (an acidic oxide). Sulfur dioxide is then oxidized, forming sulfur trioxide. The sulfur trioxide reacts with either ...

A De Castro 14B 2H

For SeF₃⁺, I believe the molecular geometry is trigonal pyramid, AX₃E. The central atom is Se, each F is single bonded to Se, then Se would have a lone pair as well. You can calculate the formal charge of each atom to check if this is correct. Hope this helps.

A De Castro 14B 2H

I believe that it would be good to know those shapes even if they weren't covered in lecture, since some VSEPR shapes (like T-shaped) are in the homework questions on the outline.

A De Castro 14B 2H

A VSEPR model shape is not always polar, even if the central atom has lone pairs. An example of a shape where lone pairs are symmetrical and cancel out each other's charges is square planar, AX 4 E 2 . This is because the lone pairs are in axial position away from each other (i.e. 180° away, on oppo...

A De Castro 14B 2H

A single bond is always a sigma bond.
A double bond is always one sigma bond and one pi bond.
A triple bond is always one sigma bond and two pi bonds.

A De Castro 14B 2H

I've always thought of sigma bonds as "head on collisions" between two orbitals, while pi bonds are just sideways overlaps between two orbitals. Once orbitals have "collided," they can't collide anymore, i.e. once a sigma bond has been formed between two atoms/molecules, another ...

A De Castro 14B 2H

I was confused about that question and s-character as well since we haven't really directly discussed it in depth, but a chem mod explains it pretty well here: viewtopic.php?t=1255

A De Castro 14B 2H

I believe that VSEPR notation will be on the test because many of the homework problems ask us to write the VSEPR formulas of various species. VSEPR formulas are relatively easy to determine, so knowing both ways (the one taught in class and your own method) definitely won't hurt you!

A De Castro 14B 2H

Yes, I also believe that's correct, even if one of the electron densities isn't involved in a bond with another atom or molecule.

A De Castro 14B 2H

Is the energy of a hybrid orbital equal to the average of the energies of the orbitals that make it up? For example, is the energy of an sp3 hybrid orbital equal the average of the energies of one s orbital and three p orbitals?

A De Castro 14B 2H

Based on Monday’s lesson (11/19) on hybridization, is it safe to assume that at least one hybrid orbital constitutes a sigma bond?

A De Castro 14B 2H

London Van Der Waals Forces are fluctuating dipoles that result from fluctuating electron distributions, due to electrostatic interactions between molecules. They are always present and attractive. Yes, according to Dr. Lavelle's lecture, they are synonymous to dispersion and induced dipole-induced ...

A De Castro 14B 2H

Do rod-shaped molecules induce dipole moments in spherical molecules?

A De Castro 14B 2H

Will we use the potential energy equation in test 3 and/or the final exam?

A De Castro 14B 2H

Hi, I attached an image that shows how the Rydberg equation is derived. Hope it helps. Also, I do think that such questions will be on the exam because it shows understanding of the concepts, but prof said that he doesn't like using that equation. He's rather us calculate frequency (nu) from ΔE dire...

A De Castro 14B 2H

Yes, I'm sure. Check this out too if you're still uncertain: viewtopic.php?f=19&t=34622&p=113448

A De Castro 14B 2H

According to prof Lavelle's "Wave Properties of Electrons and the DeBroglie Equation" module, that experiment supports the hypothesis because the only time a diffraction pattern can be observed is when whatever is being used is acting like a wave. In this case, light diffracts as it become...

A De Castro 14B 2H

Delta v would be 0.2 because 23 +/- 0.1 indicates that possible values lie within 22.9 and 23.1
Since 23.1 - 22.9 = 0.2, delta v is 0.2

A De Castro 14B 2H

The f subshell corresponds to l=3, so possible m values would be -3, -2, -1, 0, 1, 2, and 3. Not sure why the solutions manual would say otherwise though.

A De Castro 14B 2H

I took the test today and that conversion was not in it. Nonetheless, it's an important conversion to keep in mind.

A De Castro 14B 2H

Brevin Hensley 3F wrote:Since Test 2 is suppose to be all quantum material up to and including quantum numbers, does this mean electron material such as configurations and electron affinity will not be on the test?

Yes, configurations and electron affinity will not be on the test.

A De Castro 14B 2H

You must know about the quantum numbers and their possible values. So yes, test 2 covers material about orbitals, but I don't think that should be your biggest concern.

A De Castro 14B 2H

Abby De La Merced 3F wrote:Does anyone know if the test is strictly problem based or will there be conceptual questions as well?

In the test I took at section today, both types of questions were present, so I suggest studying/practicing both.

A De Castro 14B 2H

With this problem, use the deBroglie Equation: λ = h/p, where p is momentum = mass_electron * velocity. You can rearrange the equation to calculate the velocity: v = h/(λ * mass_electron)

A De Castro 14B 2H

Considering that it's from the ultraviolet spectrum, it's evident that the line is from the Lyman series. The lines in this series result from hydrogen's electron going from n≥2 to n=1. Thus, in the problem, n₂ in the Rydberg equation would be 1. Hope this helps you get started!

A De Castro 14B 2H

The "l=n-1" indicates that the shell (principal quantum number, n) limits the number of subshells (angular momentum quantum number, l) that can be inside it. Say n=2, thus the only possible l values are 0 and 1, which further means that the only subshells within the n=2 shell are s and p r...

A De Castro 14B 2H

I'd say review/practice equations that are related to velocity and kinetic energy.

A De Castro 14B 2H

In addition to what everyone else has stated, the half filled d subshell in chromium and completely filled d subshell in copper are more stable configurations because they have comparatively lower energies than [Ar] 3d^5 4s^1 and [Ar] 3d^9 4s^2. I believe this is affected by Hund's rule as well; &qu...

A De Castro 14B 2H

Hi, there are slight differences between the definitions of "work function" and "threshold energy." You can refer to this viewtopic.php?t=15646 for explanations.

A De Castro 14B 2H

Hi! For this equation, you use the M_initial x V_initial = M_final x V_final formula. Essentially, what you're asked to calculate is M_final. To do that, just rearrange the formula so it becomes M_final = (M_initial x V_initial) / V_final.

A De Castro 14B 2H

Exciting an electron (such as that in hydrogen) to a higher energy level means moving it from say n=1 to n=2, given that the frequency of the incoming light that the electron absorbs matches the energy difference between n=1 and n=2. That electron will stay in its "excited state" at n=2 fo...

A De Castro 14B 2H

If this question is the one about calculating the wavelength of radiation generated by the electronic transition from n=4 to n=2, I suggest watching the video module entitled Atomic Spectra posted on the chemistry site. You can go to the "Worked Example" part (54:10); professor breaks down...

A De Castro 14B 2H

Hi, I prepared for the test using the Fundamentals exercises in the textbook, and I believe that was enough for me to not fail. So, studying using the homework and modules would be sufficient! I feel like the thing you should prepare for is timing, since you're given 40 minutes for a test that's wor...

A De Castro 14B 2H

Extensive properties depend on size and the amount of matter in the sample that you have. An example of an extensive property is mass. The more you have of something in one sample, the heavier it gets, thus the mass changes. On the other hand, intensive properties depend only on the type of matter i...

A De Castro 14B 2H

From part a of the question, you should have calculated the amount of CuCl2 4H2O in moles. Then, you can multiply that mole value by the moles of H2O in CuCl2 4H2O, which is 4. Afterwards, you can multiply that product by Avogadro's number to get the number of H2O molecules present. I've attached a ...

A De Castro 14B 2H

I believe that your process is correct, Andrea. However, I rounded my answer to 40.08 gmol^-1 because the question started with 4 significant figures. Also, since the molar mass is 40.08 gmol^-1, the metal would be calcium.

A De Castro 14B 2H

Both are correct in a way, since 1 mol = 6.022x10²³ molecules/atoms/formula units/things. But I believe it is more correct to say "6.022x10²³ molecules of CH4" instead of just "1 molecule of CH4".

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