CHEMISTRY COMMUNITY

Claudia Luong 4K

Thank you Dr. Lavelle! I was glad to be able to have Dr. Lavelle as a professor for both Chem 14A and 14B. All of the review sessions and lectures were helpful and I enjoyed this class a lot!

Claudia Luong 4K

Typically we are given a fast step and slow step when asked to use this approach

Claudia Luong 4K

Yes, you include both

Claudia Luong 4K

You can use any time unit as long as the same time unit is used for all values in the problem.

Claudia Luong 4K

In isothermal systems, the internal energy does not change so delta U is always equal to zero.

Claudia Luong 4K

I believe the way we've been doing in class is work done by the system is negative and work done on a system is positive.

Claudia Luong 4K

Reaction rate is the actual rate while rate law is the formula (ex. rate=k[A])

Claudia Luong 4K

I believe we can always use platinum.

Claudia Luong 4K

We can use the straight line formula for first-order reactions.

Claudia Luong 4K

In most of the problems we've done, it seems to be explicitly stated in the problem.

Claudia Luong 4K

Solids and liquids do not make much difference in equilibrium. Therefore, they are excluded.

Claudia Luong 4K

Zero-order reactions are not included in the overall rate

Claudia Luong 4K

ICE Table is not needed for this problem, which is why it was probably not used in the solutions.

Claudia Luong 4K

Two lines would indicate a salt bridge instead. A porous disc is indicated by a single line.

Claudia Luong 4K

We can pick any trial and should end up with the correct result

Claudia Luong 4K

I believe Dr. Lavelle may put more of an emphasis on problems based on material learned after the midterm.

Claudia Luong 4K

You create an ICE table subtracting x from the initial concentrations of your reactants and adding x to products. You also multiply x by stoichiometric coefficients.

Claudia Luong 4K

Both should be considered correct

Claudia Luong 4K

Yes. To calculate pH for base, we would need to use Kb and not Ka.

Claudia Luong 4K

We can also use this equation to solve for initial concentrations or rate constant if needed.

Claudia Luong 4K

We will only be tested on the pre-equilibrium approximation

Claudia Luong 4K

Yes, as Dr. Lavelle has stated during lecture.

Claudia Luong 4K

Concentration doesn't have any effect on the rate

Claudia Luong 4K

A lower peak would be seen with a catalyst.

Claudia Luong 4K

I believe that it will be given in the problem whether the reaction is reversible or irreversible.

Claudia Luong 4K

For worksheet 4 #6, I get a different value for the amount of work done. I get 12.4 kJ instead of 10.1kJ. Is there a mistake in the solutions?

Claudia Luong 4K

Longer bond lengths means that the bonds are weaker and the acids are stronger. Strong acids have weak conjugate bases.

Claudia Luong 4K

No, if you take a look at the actual electronegativity values, O is more electronegative than Cl.

Claudia Luong 4K

No, while all amphiprotic substances are amphoteric, not all amphoteric substances are amphiprotic.

Claudia Luong 4K

Yes, the larger the value of Kb, the stronger the base.

Claudia Luong 4K

Strong acids completely dissociate. HCl would completely dissociate into H+ and Cl- ions and therefore won't be able to turn back into HCl while HF is a weaker acid, so it doesn't completely dissociate. Since HF doesn't completely dissociate, F- has the ability to pick up a proton/hydrogen and becom...

Claudia Luong 4K

Size -> larger size -> more polarizable -> weaker bond -> stronger acid
Electronegativity -> higher electronegativity -> pulls electrons very strongly -> more stable -> stronger acid
Resonance -> electrons are more spread out -> more stable -> stronger acid

Claudia Luong 4K

For part c on J.5, for the net ionic equation the solutions manual says it is H+ + OH- -> 2H2O . Shouldn't it be H+ + OH- -> H2O? Or are they maybe using H3O+ instead of H+ in this equation? When do we use H3O+ instead of H+?

Claudia Luong 4K

In larger heavier atoms/molecules, there are more polarizable electrons causing the LDFs to be so strong that they can be even stronger than dipole-dipole forces.

Claudia Luong 4K

This is also due to the fact that fluorine is way more electronegative than hydrogen.

Claudia Luong 4K

NH3 can have Hydrogen bonding (between H and N), dipole-dipole (due to NH3 being a polar molecule), and London dispersion forces. CH4 is nonpolar so it only has London dispersion forces. NH3 has more and also stronger intermolecular forces, so it will have a higher boiling point.

Claudia Luong 4K

The answer above is correct, except I believe when Lyndon went over this in the review session, he said to use ONO instead of NO2.

Claudia Luong 4K

Although H2S has stronger dipole-dipole forces, the LDFs in H2Se are stronger than that of H2S. These forces are strong enough to overcome even the dipole-dipole forces of H2S.

Claudia Luong 4K

Larger and heavier molecules have stronger London dispersion forces. C5H12 is a bigger molecule, therefore C5H12 has stronger LDFs and you will need a higher temperature to break apart these attractive forces so C5H12 has a higher boiling point.

Claudia Luong 4K

All molecules have LDFs. They are forces of attraction that occur when electrons in adjacent atoms occupy positions that make the atoms form temporary dipoles. Molecules have LDFs to hold the molecules together.

Claudia Luong 4K

We do not need to know these for the final. Just in case, I think it would be good to at least keep in mind that pKa is -log(Ka) and that the bigger the value of Ka is the stronger the acid.

Claudia Luong 4K

The bent shape of O3 is caused by a lone electron pair on O. The lone electron pair on the central atom causes a positive partial charge above and negative partial charge below. These opposing charges cause a net dipole. I3- is linear with the same atom of each side of the central atom; there is no ...

Claudia Luong 4K

Amphiprotic substances can accept and donate protons, amphoteric substances can act as both an acid and a base. All amphiprotic compounds are amphoteric. A clue to determine if a substance is amphoteric is to first check if there are hydrogens/protons to donate in the substance. If not, the substanc...

Claudia Luong 4K

This is because you want the positive charge to be on Cl instead of O (due to Cl being less electronegative than O).

Claudia Luong 4K

s-character of sp: 50%, sp^2: 33%, sp^3: 25%, etc. s-character increases as you go from sp^3 hybridization to sp hybridization. While in sp^3, there are four regions of electron density, sp has two. Less regions of electron density allow for more room between the regions so the bond angles increase ...

Claudia Luong 4K

I know that electronegativity increases as you go right across a period and go up along a group, but between going up a group and going across a period, which causes a larger increase in electronegativity? For example, without being given the exact electronegativity values, can you tell if S and Cl ...

Claudia Luong 4K

For example, for SCl4, I drew the Cl atoms directly top, bottom, left, and right of S, but SCl4 is seesaw-shaped so the solutions manual has it drawn with a different orientation of the Cl atoms (1 directly above S, 1 directly under S, and 2 more Cl atoms in between those 2). Would my Lewis structur...

Claudia Luong 4K

I keep reading that the reason why CH2Cl2 is polar because due to its tetrahedral shape, the dipoles can not cancel each other out but doesn't CH4 also have tetrahedral shape too? I assumed, since the reason for CH2Cl2 being polar is apparently due to the 109.5 degree angles from the tetrahedral sha...

Claudia Luong 4K

When drawing the Lewis structure of the ClO2+ ion, there are different possible structures. One uses double bonds for both the Cl-O bonds while another uses a double bond for one of the Cl-O bonds and a single bond for the other. Both end up with the same overall formal charge of +1. If I am not mis...

Claudia Luong 4K

Some of the problems in Chapter 1 involved knowing concepts such as the Balmer series and the Lyman series, but these were not on the Outline 2 or on the test. Do we need to fully master these concepts for the midterm?

Claudia Luong 4K

I was doing the problems on the worksheet that one of the UAs posted (worksheet #2), and I need help on number 7. The problem is "An excited hydrogen atom emits light with a frequency of 1.14x10 1.14x10^{14} Hz to reach the energy level for which n=4. In what principle quantum level did the ele...

Claudia Luong 4K

I have trouble understanding how to apply Heisenberg's uncertainty principle. Just to confirm, if the problem states, for example, that the speed is known to be +/- 0.55 m/s then the change in velocity that we plug into the equation would be 2(0.55) and not 0.55 right?

Claudia Luong 4K

I am confused on parts b and d on this problem. How does specifying the magnetic quantum number change the number of orbitals and/or the number of electrons that the orbitals can hold? QUESTION 2.29 How many electrons can have the following quantum numbers in an atom: (b) n =4, l= 2, ml= 2; (d) n =...

Claudia Luong 4K

The problem states that in infrared spectroscopy, we express energy in terms of v /c. Therefore, for part (a), v/c = 3600 cm^-1 v = (c)(3600cm^-1) = (2.997 x 10^8 m/s)(3600cm^-1) After this, convert the value for c into terms of cm/s in order to cancel out the units, and you should be left with a va...

Claudia Luong 4K

Hello! I know this is literally the first question for Chapter 2, but I'm having trouble understanding this concept. The solutions manual says that the radius of the atom will increase when the electron in a hydrogen atom undergoes a transition from the 1s-orbital to a 2p-orbital, and I was wonderin...

Claudia Luong 4K

Yes, as others have stated before, you should use kg in order to keep consistency since Planck's constant is in kg. I believe that during the lecture, Dr. Lavelle also said we will use kg due to kg being the standard SI unit for mass!

Claudia Luong 4K

Hello! This may seem like a dumb question, but for all parts of this problem, we are finding the moles of the units specified in each part, so it is unnecessary to use Avogadro's number for any of these parts right? The problem is: Calculate the amount (in moles) of (a) Cu2+ ions in 3.00 g of CuBr2;...

Claudia Luong 4K

This question wants us to write the balanced equation for the combustion of the solid C14H18N2O5 to carbon dioxide gas, liquid water, and nitrogen gas. I understand how to balance the reaction, but I need help on starting the equation!

Claudia Luong 4K

We already know that we have 4.14 grams of phosphorus, so we only need to solve for grams of chlorine. We know we have 27.8 grams of product, so 27.8 g product - 4.14 g P = 23.66 g Cl2 Convert into moles: 4.14g P x 1mol/30.97g = .134 mol P 23.66g Cl2 x 1mol/70.9g = .334 mol Cl2 Now, divide both valu...

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