CHEMISTRY COMMUNITY

Jennie Fox 1D

The graph would be a horizontal line that abruptly drops straight down to zero (once all of the reactants are used up)

Jennie Fox 1D

Used interchangeably

Jennie Fox 1D

E is not a state function, so you can't simply add them together. G is a state function, so that is why we must convert to G.

Jennie Fox 1D

O2 has a charge of 0. For CO2, the O has a charge of -2, and since there are 2 "O"s in CO2 the charge for O is -4. In order to cancel this -4 out, the charge on C must be +4. O2 is reduced since its charge goes from 0 to -4, and thus gained electrons.

Jennie Fox 1D

If you plug it in to the equation, you should get the correct answer. 1/(1/16) is 16.

Jennie Fox 1D

Jennie Fox 1D

It is okay to leave your answer in mmol as long as the question doesn't ask for the answer in mol

Jennie Fox 1D

d[A]/[A] stands for the rate of consumption of [A] during a first order reaction.

Jennie Fox 1D

This is because in second order reactions, half life depends on initial concentration, whereas half life in first order reactions does not depend on concentration.

Jennie Fox 1D

^ They just rearranged the equation. Both versions should yield the same answer

Jennie Fox 1D

Yes, all they did in the book was switch the ln[A]f to the right side, and moved -kt to the left side. ln[A]o-ln[A]f is the same as ln([A]o/[A]f)

Jennie Fox 1D

True, in zero order reactions the rate does not depend on concentration. d[A]=-kdt

Jennie Fox 1D

For second order half life, the half life depends on the initial concentration, t(1/2)=1/(k[A]o) whereas in first order reactions, half life only depends on k

Jennie Fox 1D

For work, I usually show something like 2^a=2 (the concentration doubled, rate doubled), therefore a=1, and the order of [A] is first order. But you need to show work, at least on the quizzes and tests,

Jennie Fox 1D

Yes, once C is determined to be zero order, you can ignore the concentrations for C entirely, since it does not have an effect on the reaction rate.

Jennie Fox 1D

The test will cover 15.1-15.6, which is up to #39 in the homework problems.

Jennie Fox 1D

Up to #39, I checked with a TA!

Jennie Fox 1D

After figuring out the order of each reactant, you just add the orders together to get the overall order

Jennie Fox 1D

Yes! M/s

Jennie Fox 1D

In order to make the rate positive, the negative sign reverses the negative rate of the product being used up in a reaction.

Jennie Fox 1D

Usually you wait until the end to round your answers, but you should definitely ask your TA about it.

Jennie Fox 1D

Rate is equal to k in a zero order reaction, and therefore plotting the points creates a straight line

Jennie Fox 1D

Q and K are constant throughout chemistry, having mainly to do with equilibrium.

Jennie Fox 1D

what kind of dogs do chemists have?

Labratory retrievers

Jennie Fox 1D

No it does not matter, "n" will be the same no matter what.

Jennie Fox 1D

I don't trust atoms....

I heard they make up everything

Jennie Fox 1D

I use Eo(cell) = Eo(cathode) - Eo(anode) so that way I do not need to flip any signs.

Jennie Fox 1D

Nope!

Jennie Fox 1D

Adiabatic only implies that q=0, not delta T

Jennie Fox 1D

For ideal gases when you need to solve for a constant.

Jennie Fox 1D

Where do I find the link?

Jennie Fox 1D

You do. It isn't shown in the solution manual because the reactants in these problems have an Hf of 0 (for example, O2, H2, C(S) all have an Hf=0). Therefore you only need to look at the Hf of the product side (in this case).

Jennie Fox 1D

You won't have to calculate it, it should be given in the problem. They make you look it up in the book for the homework problems, but on the test it will be given.

Jennie Fox 1D

I stated that energy cannot be created nor destroyed, only transferred. I'm pretty sure that is all they would be looking for, because the First Law of Conservation is primarily concerned with the conservation of energy.

Jennie Fox 1D

Joules per Kelvin (J/K)

Jennie Fox 1D

Spontaneous reactions occur without any type of outside intervention/force. Spontaneous reactions favor the formation of products at the current conditions under which the reaction is occurring.

Jennie Fox 1D

For this question, you would use q=mC(delta)T. For part a, where the pressure is held constant (Cp), C would be equal to (5/2)R, where R=8.314. In part b, where the volume is held constant (Cv), C would be equal to (3/2)R, where R is also 8.314. So basically, you would just replace the "C"...

Jennie Fox 1D

The molecules must be on opposite sides and I=equal in quantity to be able to fully cancel them out. For example, if there are 2 modes of CO2 on the reactant side of one equation ands you are adding this equation to one in which there are 2 moles of CO2 on the product side, the CO2's will cancel out.

Jennie Fox 1D

Since enthalpy is a state function, it does not follow a specific pathway, and thus it it okay to use any coefficients as long as you consistently change the enthalpy along with it.

Jennie Fox 1D

717 kJ/mol would have to be given. It can be found in the appendix at the end of the book.

Jennie Fox 1D

Yes, because the cylinder is being supplied with energy, this energy (524) will be positive. In addition, because work is being done on the system by the surroundings, 340 will be positive as well.

Jennie Fox 1D

If the system is doing work on its surroundings, as in this problem, the work is negative.

Jennie Fox 1D

I found this online and found it to be very helpful. "Heat and work, unlike temperature, pressure, and volume, are not intrinsic properties of a system. They have meaning only as they describe the transfer of energy into or out of a system. "Heat & work are the forms of energy in trans...

Jennie Fox 1D

Jennie Fox 1D

Hi, I found this online and found it very helpful. "Heat and work, unlike temperature, pressure, and volume, are not intrinsic properties of a system. They have meaning only as they describe the transfer of energy into or out of a system." "They appear only when there occurs any chang...

Jennie Fox 1D

It depends on what is produced. If H3O+ is produced in the reaction, then you can solve for pH. If OH- is produced, then you can solve for pOH.

Jennie Fox 1D

You use an ICE table when a weak acid or base is being reacted with water. A strong acid or base completely dissociates in water, so you can go straight to the equilibrium constant expression.

Jennie Fox 1D

A coordination compound is the result of a Lewis acid-base reaction in which ligands (anions) bond to the central metal atom, or ion. Ligands are Lewis bases, meaning they have an extra pair to pairs of electrons to donate to the Lewis acid (the central metal atom).

Jennie Fox 1D

All of the bonds are 90 degrees. It says 180 for the two atoms directly across from one another.

Jennie Fox 1D

For this problem, we would need to use a graphing calculator in order to solve for "x". We shouldn't get a problem like this on a test unless we are allowed to use a graphing calculator.

Jennie Fox 1D

The two formulas are the same. You compare Q to K in order to determine which way the reaction will shift (sit). If Q<K, it will shift/sit right (toward products), if Q>K, it will shift/sit left (towards reactants).

Jennie Fox 1D

I believe the length would be closer to a double bond, but in between the lengths of a single and double bond.

Jennie Fox 1D

The shape depends on the number of lone pairs as well as their placement on the molecule.

Jennie Fox 1D

Another thing to note for polarity is in regard to the dipole moments. If the dipole moments are equal and cancel each other out, the molecule is non polar; however, if the dipole moments do not cancel each other out, the molecule is polar.

Jennie Fox 1D

In addition to what has already been said, if the dipole moments around a molecule are equal (cancel each other out), the molecule is non polar. If the dipole moments are not equal (do not cancel each other out), the molecule is polar.

Jennie Fox 1D

@Swetha No, I don't believe so.

Jennie Fox 1D

On the exam and homework schedule, it says that the midterm covers all material up to November 3, so I think it might be. I can ask Lavelle.

Jennie Fox 1D

Since the s-block has lower ionization energies than the p-block, it takes less energy to remove an electron from the s-block than the p-block. Because it can lose electrons easier, the s-block is thus more reactive than the p-block.

Jennie Fox 1D

A half filled D-orbital (d^5) or a filled D-orbital (d^10) is more stable than d^4 or d^9, for example, so the electron from the S-orbital will "jump" to the d-orbital in order to make it more stable. In this case, the electron will jump to the d-orbital in order to completely fill it (d^1...

Jennie Fox 1D

Hey, I've already taken the test and you do not need to know about Heisenberg's uncertainty principle for this test. You are only being tested on Chapter 1 sections 1-5. As for the homework, my TA had us turn in 4 problems that we used to study for the test (so I turned in 4 problems from sections 1...

Jennie Fox 1D

The conversion between KeV and Joules will be given either in the problem or on the conversion/formula sheet. We are not expected to memorize the conversion.

1 eV = 1.6022 x 10^-19 J

Jennie Fox 1D

First, you must convert 140.511 eV to Joules. (1eV = 1.6022 x 10^-19 J).

From there, you use the equation E=hc/lambda and solve for lambda (wavelength).

You should get 8.8237 pm for your answer.

Jennie Fox 1D

In order to solve the problem, you do use the wavelength. In order to solve for n2, you must first find the frequency (v) so that you can plug the frequency into Rydberg's equation. To find the frequency, you use the equation v=c/lambda. Then you use plug this frequency into Rydberg's equation, v=R(...

Jennie Fox 1D

Since you are solving for moles of Cl- ions, you have to convert the 0.0417 mol CuCl2 x 4H2O to moles of Cl-. Since one mole of CuCl2 x 4H2O contains 2 Cl- ions, you take the 0.0417 mol CuCl2 x 4H2O and multiply it by (2 mol Cl- / 1 mol CuCl2 x 4H2O). This is what my work looks like: 0.0417 mol CuCl...

Jennie Fox 1D

Typically, you will get numbers that are very close to a whole number such as 1.97 or 3.08 which you would round to 2 and 3. In many of the problems I have done so far that I've had to multiply in order to get a whole number, I've gotten numbers like 2.67, 1.33 and 3.75 which can easily be multiplie...

Jennie Fox 1D

As long as the significant figures in your answer match the significant figures given in the problem (or whichever number you are using for calculations), you will be fine. You will get a much more accurate answer if you don't round numbers during your calculations. Just make sure that the significa...

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