CHEMISTRY COMMUNITY

Vy Lu 2B

Yes, n is the number of moles of of e- transferred in the redox reaction.

Vy Lu 2B

I believe that each reaction has a unique frequency factor, but it should be given specifically if required for calculations.

Vy Lu 2B

This is correct; the rate law deriving solely from elementary steps information relies on the coefficients in the steps, since no experimental data is given.

Vy Lu 2B

The rate of reaction in a zero order reaction is not dependent on the concentration of the reactant, whereas the rate of reaction in a first order reaction is directly proportional to the concentration of reactant.

Vy Lu 2B

This might have been an error on the solution manual's part; however, would this reduction be a result of combining two different reduction phrases together? I believe I saw one question that had that similar issue.

Vy Lu 2B

It is true that the rate of oxygen should be negative since it is a reactant, but the negative sign in the formula for unique rate makes sure that no rates are given as negative values.

Vy Lu 2B

order should be anode - liquids/gases - aqueous species // aqueous species - liquids/gases - cathode.

Vy Lu 2B

To calculate the cell potential of the reaction, there are two methods in doing so; one is not switching any signs on any of the standard reduction potentials and simply subtracting them in this manner (cathode - anode) or the other method is flipping the anode potential's sign and then adding the t...

Vy Lu 2B

I believe that it is written this way due to it being written as the standard reduction potential, in which all of the half-reactions are written as reductions rather than oxidations.

Vy Lu 2B

Is the cell potential the same as the potential of the redox reaction?

Vy Lu 2B

The main difference is that the use of porous disk implies that the solutions are not in separate beakers and allows electron flow; overall the function of the porous bridge is same as that of the salt bridge: prevent charge buildup on one electrode.

Vy Lu 2B

K should not have any units since it is a ratio of concentrations/partial pressures.

Vy Lu 2B

Delta S total is equal to zero when the system is under equilibrium or undergoes a reversible process.

Vy Lu 2B

Enthalpy is the amount of heat absorbed or released by a system under constant pressure, which essentially is the heat at constant pressure, qp. Since the change in internal energy is equal to the sum of heat and work, enthalpy can be related to change in internal energy if the system is at constant...

Vy Lu 2B

Delta S (surroundings) is zero under free expansion; delta S (universe) is zero under reversible conditions and equilibrium; delta S (system) should be zero when there is no transfer of heat from and to the system and when the temperature and volume are constant in the system.

Vy Lu 2B

Did the problem specifically indicate that the process was adiabatic for there to be no heat transfer?

Vy Lu 2B

What is the context of this problem's system? Is this process occurring isothermally or reversibly?

Vy Lu 2B

Standard units should be used, so for pressure use atmospheres and for volume use liters; just as long as the units are in accordance with the constants in the equations provided.

Vy Lu 2B

You need to visualize the lewis structures of the individual molecules and deduce which bonds have been added to the original central atom of the molecule.

Vy Lu 2B

I believe you would have to use the ideal gas constant version that cancels out whatever unit is present in the problem.

Vy Lu 2B

When referring to qp, it is the heat quantity that is derived from using the specific heat capacity at constant pressure; professor Lavelle indicated that qp is also equal to enthalpy change.

Vy Lu 2B

K is a constant and units are not regarded in terms of equilibrium constants.

Vy Lu 2B

You would only have to carry out an ICE table for Ka2 if you are focusing on the second deprotonation of that acid; we work on the first and second deprotonations separately.

Vy Lu 2B

When using the equilibrium constant expression, we look at the equilibrium reaction in terms of reactants or products decreasing to increase reactants or products; subtract what will dissociate and add what will be formed. Whether or not the concentrations of H+ or OH- is subtracted depends on the c...

Vy Lu 2B

He avoided using the word "shift" because some students confuse the reaction favoring foward or reverse directions of reaction as actually changing the K constant; sitting just refers to the reaction no longer being at equilibrium and that the reaction will proceed in that specific directi...

Vy Lu 2B

We use the equilibrium constant K to compare it with the current state of the reaction (reaction quotient Q); the reaction will favor the forward direction if Q<K and it will favor the reverse direction if Q>K. Acids do not specifically only favor the forward and bases do not specifically favor the ...

Vy Lu 2B

K does not change when concentration changes, so the ratio remains the same regardless of change in product or reactant concentration occurs.

Vy Lu 2B

The reactant side should have the arrow pointing towards the product side; in terms of equilibrium, regarding the double arrows, the left side is the reactant side and the right side is the product side.

Vy Lu 2B

You are able to use concentration when you are referring to gases as well.

Vy Lu 2B

Which units do we use when evaluating the equilibrium constant with partial pressure? Is atmosphere used or bar used?

Vy Lu 2B

It is based on the chemical formula that you are given; if you are given Zn(NO2)2 and you know that the charge of the nitrite ion is -1, and the compound's overall charge is neutral, then you can derive that the charge on zinc is +2.

Vy Lu 2B

We tell whether a salt is acidic or basic based on whether the pH is changed by the addition of H+ or OH-; a salt is acidic if it forms H30+ when reacting with water in solution and it is basic if it forms OH- when reacting with water in solution.

Vy Lu 2B

In some answers, H2O is written as OH2 because we want to emphasize that the oxygen atom in the water molecule is the partially negative part of the ligand that binds to the metal central atom in the coordination compound.

Vy Lu 2B

If an anion has greater polarizability, then it will more likely be "sharing" its electrons with the cation that is exerting a force on its electrons; hence, molecules with anions having a greater polarizability are not necessarily stronger in bond strength; rather, it is simply more coval...

Vy Lu 2B

It will most likely be that we will have to familiarize ourselves with the common ligands and recognize them in the molecules; naming the coordination compounds will also require us to have knowledge of the naming rules taught in lecture.

Vy Lu 2B

An atomic orbital is a region where the electron will most likely be in around the atom it belongs to. The specific electron lone pair that you are referring to is considered a region of high electron concentration, or density when writing out the shape of the molecule.

Vy Lu 2B

We do work with d orbitals, but I believe that we do not have to know how to draw out complex d orbitals for this class!

Vy Lu 2B

Polarizability is the degree to which a particle's electron clouds can be distorted. Electronegativity is the degree to which a particle wants to pull in electrons. The bigger the polarizability a particle has, the less electronegativity it has; for example, larger atoms have bigger electron clouds ...

Vy Lu 2B

When looking at the shape of molecule being analyzed, if there are different angles between different bonds, how do we know which angles within the shape of the molecule are the bigger angles or smaller angles based on the shape of the molecule?

Vy Lu 2B

The VSEPR model bases off the shape of the molecule by the idea of electron repulsion around the central atom of the molecule; the more lone pairs the central atom has, the more asymmetrical its bounded atoms are, which determines the shape of the molecule itself.

Vy Lu 2B

The difference between electron geometry and molecular shape is whether the lone pairs are taken into account or not; electron geometry demonstrates the complete positions of all shared or lone pair electrons around the central atom, while the molecular shape depicts the physical appearance of the a...

Vy Lu 2B

Each bond determines the properties of the molecule, like whether or not the atoms bonded can rotate freely (sigma bonds) or not (pi bonds).

Vy Lu 2B

Although the repulsion of the bond pairs is greater than that of the repulsion of the lone pairs with one another, lone pairs still repel one another in the atom and hence affect the angle of the bond, possibly increasing the bond that it is near.

Vy Lu 2B

The bonding pairs are what determine the naming of the shape of the molecule; hence, a molecule with four bonding pairs of electrons will always be tetrahedral. If a molecule has two bonding pairs and two lone pairs of electrons (which adds up to four regions of electron density), it is not a tetrah...

Vy Lu 2B

The electron cloud of the atom is not evenly distributed at all times around the atom and may sometimes aggregate instantaneously and temporarily on one side of the atom, therefore leading to a dipole in the atom in which one side of the atom is partially negative and the other side is partially pos...

Vy Lu 2B

When writing out Lewis structures for molecules, I know that we can expand the valence shells of the central atoms of the molecule that belong to period 3 and beyond; is there a limit as to how many extra electrons we can expand on the valence shell of the central atom?

Vy Lu 2B

You should write out the electron configurations of those ions and elements, and from there, you would identify what subshell is not filled completely; apply Pauli exclusion rule and Hund's rule to determine how many unpaired electrons are present in the ions and elements.

Vy Lu 2B

Since we did go over exceptions to the octet rule which may include radicals, it would be best to know and familiarize yourself with radicals!

Vy Lu 2B

The polyatomic ions that you listed have separate lewis structures drawn enclosed by the symbolic representation of its overall charge; to connect the polyatomic ions, you simply list the separate strutures next to one another, as shown in class.

Vy Lu 2B

Degeneracy simply means same energy; it is indicating that the orbitals within that shell have the same energy.

Vy Lu 2B

I find that Khan Academy is a great tool for practicing chem principles and applications!
Obviously, practice problems from the textbook supplemented by office hours and peer learning sessions is helpful as well.

Vy Lu 2B

To derive the de Broglie equation, we rely on these equations: c (speed of light) = (lambda, wavelength)(nu, frequency) E (energy) = (momentum)(velocity) Note that for a photon, E = (momentum)(speed of light, c) because the velocity of a photon is the speed of light and E = (h, planck's constant)(nu...

Vy Lu 2B

I believe that the main gist of wavefunctions is that it is a mathematical representing of the wavelike properties of electrons; an orbital is like a valid wavefunction and it is a region where the electron will most likely be found around the nucleus of the atom.

Vy Lu 2B

I thought that nodal planes were regions where the electron density is zero. The cone is a nodal plane, meaning that the probability density of the atom is zero there?

Vy Lu 2B

We should stick to v (nu) for the symbol representing frequency.

Vy Lu 2B

The unit is supposed to be in kg, as that is the standard SI unit.

Vy Lu 2B

These audio-visual modules are essentially a resource for students to clarify on lecture topics and review old concepts; I believe they do not count towards our grades but it is strongly advised to be used!

Vy Lu 2B

I did the math and I got about 2.957 for the Na subscript and about 5.910 for the F subscript; I think you would be able to round off to the nearest whole number so that Na has a subscript of 3 and F has a subscript of 6? Does anyone agree with this? Unless you multiply everything by 1000 so that th...

Vy Lu 2B

You would first balance out the original chemical equation C4H10(g) + O2(g) → CO2(g) + H2O(g) Since the problem calls for 4 moles of C4H10 (g), we add the coefficient to that molecule: 4C4H10(g) + O2(g) → CO2(g) + H2O(g) After that we simply balance out the rest of the chemical equation following th...

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