CHEMISTRY COMMUNITY

MadelynNguyen1F

free expansion is the unrestrained expansion of a gas into a volume

MadelynNguyen1F

The work done on the system will be positive if a system is compressed and delta-V is negative

MadelynNguyen1F

The professor also stated on a previous chem community post that " a monatomic ideal has 3 degrees of freedom and each contributes (1/2)nRT towards the internal energy. So the total internal energy would be (3/2)nR. And since equating both we get Cv = 3/2R. And you know Cp-Cv = R so Cp = Cv + R...

MadelynNguyen1F

An example of an insulated system would be hot liquid inside of a thermos.

MadelynNguyen1F

A closed system has a fixed amount of matter but can exchange energy with its surroundings. An open system can exchange both matter and energy with its surroundings. An isolated system has no contact with its surroundings.

MadelynNguyen1F

Yes, this varies because diatomic gases have a greater number of degrees of freedom compared to monoatomic gases.

MadelynNguyen1F

Yes, negative work is when work leaves the system.

MadelynNguyen1F

Delta H is enthalpy while q is heat flow.

MadelynNguyen1F

I believe Dr.Lavelle drew out the structure of each reactant and product in order to help us visualize which bonds were being broken.

MadelynNguyen1F

Yes, Dr.Lavelle mentioned that method 2, the bond enthalpy method, is the least accurate because it uses averages from many different molecules.

MadelynNguyen1F

The modules are an extremely helpful review for subjects you might be a little rusty on, and I recommend doing the assessments to complement the lectures and homework.

MadelynNguyen1F

Yes, if there is the same chemical on both the reactant side and product side, then it cancels out when you add the equations.

MadelynNguyen1F

Yes Kp is the equilibrium constant expressed in terms of partial pressures of gases

MadelynNguyen1F

K doesn't change when concentrations change. However, a change of temperature can change an equilibrium constant.

MadelynNguyen1F

Since the number of moles is equal on both sides, there is no net change in order to restore equilibrium.

MadelynNguyen1F

Equilibrium constants are calculated using activity, and activity is unitless, so K is also unitless.

MadelynNguyen1F

Only the position of equilibrium is changed if you add a liquid. The reaction shifts to ensure that the expression for K remains the same.

MadelynNguyen1F

Brackets represent concentration, so I've seen them used in Kc problems, while I've seen both parentheses and nothing used with equilibrium problems concerning partial pressure

MadelynNguyen1F

I believe you can just draw the lines, and that the triangles and dashes were just used to help us fully understand the structure.

MadelynNguyen1F

Any elements with an atomic number larger than 10 have access to the d orbital and can have an expanded octet.

MadelynNguyen1F

If there are resonance structures for one of the molecules, then the double bond for that molecule will be longer than the double bond for the other.

MadelynNguyen1F

You have to account for lone pairs, as they count as areas of electron density.

MadelynNguyen1F

We should know the main bond angles for each structure.

MadelynNguyen1F

Cations have polarizing power while anions have polarizability.

MadelynNguyen1F

You would use the normal equation to calculate the formal charge but when you add up all of the formal charge, it should equal the charge of the ion.

MadelynNguyen1F

Yes! The equation [valence electrons - non-bonding valence electrons - (bonding electrons/2) ] would also work

MadelynNguyen1F

Yes, it would be 10 electrons because the d subshell has 5 orbitals with two electrons each.

MadelynNguyen1F

Polarizing power refers to cations because the electrons around the nucleus of the anion are drawn to the cation.

MadelynNguyen1F

I believe we just need to know that single bonds would have a longer bond length compared to double and triple bonds. The bond length also corresponds to bond strength.

MadelynNguyen1F

Two double bonds, because of the formal charges, but the actual structure is a hybrid of all three.

MadelynNguyen1F

The expanded octet applies to atoms where n is greater than or equal to 3 because they have access to the d orbital.

MadelynNguyen1F

Someone asked about this in class and the answer was to plug in the diameter.

MadelynNguyen1F

I believe Dr. Lavelle said that we should just know the concepts behind it.

MadelynNguyen1F

I believe that atoms with a full octet are an exception though.

MadelynNguyen1F

Electron affinity going from left to right of the periodic table increases because the amount of protons an element has also increases. There is an attractive charge between the positively charged protons in the nucleus and the negatively charged electrons. So, the outer electrons are more attracted...

MadelynNguyen1F

It is a one to one ratio so 1000 electrons.

MadelynNguyen1F

I think we also need to know which axis each lobe lays on.

MadelynNguyen1F

I think 4s might fill up first because of electron repulsion. I was looking online and it said that was related to the fact that 4s fills up before 3d.

MadelynNguyen1F

Yes, you would use the equation lambda=h/mv, and make sure to convert the given values into SI units.

MadelynNguyen1F

1 MHz is equal to 10^6 Hz, so you would multiply by 10^6.

MadelynNguyen1F

For part A, you can use the equation lambda=hc/Energy of the electron or (1/2mv^2. Then, you plug in Planck's constant and the speed of light on the top and the mass of an electron and the given velocity(3.6x10^6 when converted from km/s to m/s) on the bottom. The answer would be 33.7 nm, so the wav...

MadelynNguyen1F

Yes, but the question should specify the units you need.

MadelynNguyen1F

I have the sixth edition and I haven't had a problem. I'm pretty sure the sixth edition is the one we're supposed to have.

MadelynNguyen1F

No, but two reactants can be used up in equal amounts. Neither of them limits the other.

MadelynNguyen1F

I think I remember Dr. Lavelle saying that we would be given unit conversions if it was a unit we wouldn't normally know.

MadelynNguyen1F

I remember that question from the module assessment and the answer is no.

MadelynNguyen1F

Yeah I was also a little rusty but the video modules really helped and I reviewed my old AP chem notes so I'm feeling more confident for the first test. The solutions manual is also extremely helpful.

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