CHEMISTRY COMMUNITY

Raj_Bains_2C

I believe that it will tell you which way to write it but otherwise either way should be fine.

Raj_Bains_2C

Concentration cells have the same elements on both side but with different concentrations. The difference in concentrations allows the reaction to occur.

Raj_Bains_2C

A salt bridge allows ions to go into solutions and it prevents the build-up of charges.

Raj_Bains_2C

We will most likely have to determine the order of the reaction, but we will be given the necessary information to determine this.

Raj_Bains_2C

Raj_Bains_2C wrote:The half-life for first order dow not depend on the initial concentration.

*does not*

Raj_Bains_2C

The half-life for first order dow not depend on the initial concentration.

Raj_Bains_2C

You would use Kelvin (K).

Raj_Bains_2C

The reaction intermediate does not appear in the reactants or products of the overall reaction. It only appears in the intermediate steps.

Raj_Bains_2C

For first order, the straight line plot is ln[A] vs time. For second order, the straight line plot is 1/[A] vs time. For zero order, the straight line plot is [A] vs time.

Raj_Bains_2C

Q is just relating the products to the reactants when it is not at equilibrium, so you can use concentration of partial pressures to relate them.

Raj_Bains_2C

Thermodynamics only cares about the initial and final states, while kinetics focuses on the pathway of the reaction.

Raj_Bains_2C

I don't think that we will have to derive the actual rate laws, but I think that knowing how to do it will be helpful in answering some questions.

Raj_Bains_2C

This is because the reactant concentration decreases exponentially in a first order reaction.

Raj_Bains_2C

Test 2 will cover everything we have learned in weeks 6 and 7.

Raj_Bains_2C

There is only one way to calculate E that we have learned so far. The method involving the half reactions is a more descriptive way of the method using just the equation.

Raj_Bains_2C

The cathode is the reduction side and the anode is the oxidation side.

Raj_Bains_2C

I think that atm and bar are the two units that can be used to calculate Kp values.

Raj_Bains_2C

This is because entropy is a state function, so the path taken does not matter. Just the final and initial values matter.

Raj_Bains_2C

Steam at 100 C contains more heat than water at 100 C because heat goes into breaking intermolecular bonds.

Raj_Bains_2C

In an open system, the container is left open, so matter (like gasses) and energy (like heat) can escape the system, which is the container. In a closed system, the container has been closed (a cap is placed on a jar, for example), so matter (like gasses) can't escape, but energy (like heat) can sti...

Raj_Bains_2C

This means that in isothermal systems, the heat lost is equal to the work done, but a negative sign is added.

Raj_Bains_2C

U is a state property because the path taken does not matter. Only the initial and final conditions matter.

Raj_Bains_2C

You can use both Kelvin and Celsius because you just need the change in temperature.

Raj_Bains_2C

I don't think that there is any way to prove 100% isolation of any system except for the universe. Even an insulated water bottle or thermos has a very small amount of heat lost, so it is not completely 100% isolated. But, as Lavelle explained in lecture, that heat lost is extremely small compared t...

Raj_Bains_2C

When energy goes into the surroundings in an experiment, it is not lost. It is just going into the universe. The fact that the universe is an isolated system means that it does not lose energy, so energy is conserved overall. The total energy in the universe never changes.

Raj_Bains_2C

Yes. You can use an ICE chart for both partial pressures and concentrations.

Raj_Bains_2C

We did not go over Hess's Law in 14A and I believe that it will be covered later in the quarter in the thermodynamics outline.

Raj_Bains_2C

We are allowed to assume this when K is less than 10^-3.

Raj_Bains_2C

Most of the time you would use ICE tables when solving a problem with weak acids and bases.

Raj_Bains_2C

Decreasing the pressure of one side makes it go toward that side. Increasing the pressure of one side makes it go toward the other side.

Raj_Bains_2C

The x is just a variable that represents the change in concentration, which we are trying to solve for.

Raj_Bains_2C

K is the ratio of the concentrations or partial pressures of the products to reactants at equilibrium. It is a way to tell is a reaction is at equilibrium and it helps us know what needs to happen to take the reaction to equilibrium.

Raj_Bains_2C

We sometimes use KP instead of KC for gases because that is what the problem gives us as the initial information.

Raj_Bains_2C

K is when the reaction is at equilibrium and Q is the current state of the reaction. If Q = K, the reaction is at equilibrium and if Q does not equal K, the reaction is not at equilibrium.

Raj_Bains_2C

A mono-dentate only has one ligand attached to the transition metal, while a poly-dentate has more than one ligand attached to the transit metal. This does affect the naming of the compound.

Raj_Bains_2C

We do not need to calculate the pH of a weak acid solution unless it explicitly gives us the H3O+ concentration.

Raj_Bains_2C

If the whole molecule has no charge, then it is safe to assume that the two lone pairs do still exist. Therefore, the hybridization is sp3.

Raj_Bains_2C

Yes, this is correct. When an electron drops in energy levels, it gives off energy, so the change in energy is negative.

Raj_Bains_2C

The oxidation number is the overall charge on the complex, while the coordination number is the number of ligands bonded to the central tradition metal.

Raj_Bains_2C

I am not sure what the reason is for this, but I just know that the whole lewis structure is more stable if the H is attached to the O instead of the Se. I thing that most structures like to have the OH groups instead of the H being attached to the central atom.

Raj_Bains_2C

One triple bond should contain one sigma bond and two pi bonds.

Raj_Bains_2C

It will only be non polar if the two outside atoms are identical.

Raj_Bains_2C

1. It is referring to the sigma/pi bonds, but I believe it also wants you to indicate which orbitals are bonding to each other. 2. All three sp^2 hybrid orbitals are identical so you don't have to indicate which one is used. 3. The hybridization of the lone pairs just refers to which hybrid orbitals...

Raj_Bains_2C

P is in the center because it has the lowest ionization energy out of the atoms in the structure.

Raj_Bains_2C

The bonding electrons of the smaller atom would be closer to the nucleus so they would be more attracted to the positive charge from the protons.

Raj_Bains_2C

In high school I learned that greater charges always matter more than a smaller size.

Raj_Bains_2C

I believe that a specific value is preferred but a range should be accepted as a correct answer.

Raj_Bains_2C

This seems correct based on what we have learned so far. As there are more non covalent interactions between molecules, the intermolecular bonds become stronger, causing more stability and higher boiling points.

Raj_Bains_2C

I believe that test 3 will only cover the material from after the midterm to right before the test. It should not cover everything from the beginning of the quarter.

Raj_Bains_2C

Questions on radicals may be on the midterm so it is best to study this topic.

Raj_Bains_2C

I do not know the answer to this but I am just letting you know that we have not covered this in lecture yet and it will not be on the midterm.

Raj_Bains_2C

The only thing you did wrong in this question was that you didn't double the resulting mass of one oxygen atom. Oxygen exists is a diatomic state, so the formula for the oxygen molecule is actually O2.

Raj_Bains_2C

The midterm is going to cover everything we have covered from the beginning of the quarter.

Raj_Bains_2C

Electron affinity is how much an element attracts another electron, so it is directly related to ionization. If an element has a very high or very low electron affinity, it will easily gain or loose an electron, so it will easily ionize.

Raj_Bains_2C

Parallel electrons do exist in the same orbital but they are in different sub-orbitals, as indicated by the m in the n, l, m quantum numbers.

Raj_Bains_2C

I believe they are squared because they are overlapping the p-orbitals, but I am not sure that this is the reason.

Raj_Bains_2C

The midterm time is 6-8 pm and I think that the midterm location will be revealed at a later date.

Raj_Bains_2C

Raj_Bains_3C wrote:I believe that they do always stay in the dame formation, but as Dr, Lavelle said, do not pay too much attention to the shapes because they are only mathematical models.

*same formation, not "dame"

Raj_Bains_2C

I believe that they do always stay in the dame formation, but as Dr, Lavelle said, do not pay too much attention to the shapes because they are only mathematical models.

Raj_Bains_2C

Different metals have different threshold energies because they have a different number of protons, electrons, and energy levels. If the metal has many protons and electrons, there will be more energy levels, resulting in the electrons being further away from the nucleus. These negatively charged e...

Raj_Bains_2C

Different metals have different threshold energies because they have a different number of protons, electrons, and energy levels. If the metal has many protons and electrons, there will be more energy levels, resulting in the electrons being further away from the nucleus. These negatively charged el...

Raj_Bains_2C

I don't think that it matters on the homework problems if we use 3.0 x 10^8 m/s or a more precise value, but for the tests, I would stick to using the value that is on the constants and equations sheet.

Raj_Bains_2C

My TA told me that we were allowed to do any homework problems from previous assignments, so we can turn in some week 1 problems for the week 3 homework. As far as I understand it, we can just do any problems that we feel like we need more practice on.

Raj_Bains_2C

The states of matter should be stated most of the time, but as far as I know it does not affect any calculations made using the balanced chemical equation. I do think that they are just there for analytical purposes and to just show what states of matter the reactants and products are in.

Raj_Bains_2C

It is best to obey the rules of significant figures all the time to retain the accuracy of measurements. It might help you to review the basic rules of significant figures.
Examples: 5.48 x 6.2 = 34 8.2 x 0.6 = 5 9.01 x 2.333 = 21.0

Raj_Bains_2C

You did not need to do any calculations for this problem because it already states that there are equal amounts of atoms. Therefore, none from the pair contains more moles of atoms. They contain equal amounts.

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