CHEMISTRY COMMUNITY

Kevin Tam 1J

I believe only the functional groups listed in the course reader will need to be known. Dr. Lavelle didn't emphasize the other ones (like thioether and phenol). Hope this helps.

Kevin Tam 1J

Yeah, SN2 reactions stand for bimolecular nucleophilic substitution reactions. A nucleophile from somewhere else will replace the nucleophile that is already on the molecule. I think that the pattern usually involves the "original" nucleophile breaking off because of its negative polarity ...

Kevin Tam 1J

To tentatively answer your first question, I think that the size of the compound does affect how readily electrons are donated/accepted? Take, for example, F- and Br-. If you were to compare which one of these elements is the stronger nucleophile, you would see that Br- is larger and thus has higher...

Kevin Tam 1J

I've been trying to decipher the distinction between the average reaction rate and the unique average rate. On pg. 612-613, the book says that the average reaction rate is defined as average rate of consumption of R = - Δ[R]/Δt On the other hand, the book says that the unique average rate of reactio...

Kevin Tam 1J

In this question, you are essentially separating the entropy into two calculations: one involving only volume change at constant temperature, and the other involving only temperature change at constant volume. You have to always hold one of the variables constant in order to solve for the other vari...

Kevin Tam 1J

I think that, for the most part, you just reverse the reaction that would, when you add up the E naught values, give you a positive E naught value. You would want the E naught to be positive, which would indicate that the overall reaction is spontaneous.

Kevin Tam 1J

To add on, I would say that you would want to always refer back to your "goal" equation. Compare your "goal" equation to the equations that are given to you, and, as aforementioned in the previous comments, multiply the equations by a coefficient that will ensure that you end up ...

Kevin Tam 1J

To my understanding, positional/residual entropy is calculated as the absolute entropy (the equation, I think, was derived through molecular/statistical analysis). The equation for absolute entropy is S=Kb*lnW. On the other hand, I think that the "regular entropy" that you are talking abou...

Kevin Tam 1J

It is because all of the heat during the phase change (i.e. from solid to liquid) is going into converting the solid to a liquid. In the case of water, the heat is being used to break the hydrogen bonds (to be more specific on which bonds are being broken). That is why temperature remains constant.

Kevin Tam 1J

Yes. The heat released from steam is the sum of the heat released during the phase change from gas to liquid (as represented by the second horizontal line on a heating curve of water) and the heat released during the cooling of liquid water. This sum is much larger than merely the heat released duri...

Kevin Tam 1J

I think the rule is that for Z<8, the pi orbitals are lower in energy than the sigma orbitals, but for Z greater than or equal to 8, the pi orbitals are higher in energy than the sigma orbitals.

Kevin Tam 1J

No, the pH scale only goes from 0 to 14. When you take the negative log of a number between 10^-14 and 1 (which I think are the range of values you are going to be dealing with), you will always get a positive pH.

Kevin Tam 1J

Do molecules/anions outside the coordination sphere contribute to the compound's coordination number? For example, for [Cu(NH3)4]SO4.H2O, is the coordination number 4 (since each NH3 is monodentate and there are four of them), or is the coordination number 6 (including the SO4 and H2O)?

Kevin Tam 1J

I think it is also important to note that a polydentate ligand can have multiple bonds with the central cation. Take for example (en) and (dien). I think one will have to count (en) as 2 bonds and (dien) as 3 bonds, since (en) is bidentate and (dien) is tridentate.

Kevin Tam 1J

I think that for H2O, it is already implied/understood that the oxygen (not the hydrogen) is bonded to the central cation, so writing it either way (as H2O or OH2) is fine, although I think it is better to write it as OH2. In other words, there is only one possible way for the water to bond to the c...

Kevin Tam 1J

The NC ligand suggests that the nitrogen (N) is bonded to the central cation. However, the CN ligand suggests that the carbon (C) is bonded to the central cation.

Kevin Tam 1J

I'd like to add on to the definition of an angular/bent molecule.

The central atom of a bent molecule can have either 1 or 2 lone pairs.

The trick is when there are 3 lone pairs. In the case of 3 lone pairs, the molecule becomes linear.

Kevin Tam 1J

I think that drawing diagonal bonds is the only way to satisfy the condition that 1 P atom is connected to 3 other P atoms. But I think that you could also draw it like NH3. Put 1 P atom in the center and the 3 P atoms symmetrically around it. The only added step would be to connect all of the outer...

Kevin Tam 1J

As for determining the exact ionization energy level of each element, I don't know if such a task is within or beyond the scope of this course. These are the general trends for ionization energy. As you go across a period , the ionization energy generally increases . The exceptions for this periodic...

Kevin Tam 1J

Another way of doing this problem is to take one of the reactants and calculate the required amount of the second reactant using a mole ratio. (22.4g NH3/17.03g mol^-1 NH3) = 1.32 mol NH3 40.5g O2/32.00g mol^-1 O2) = 1.27 mol O2 Let's take 1.32 mol NH3 and multiply it by the mole ratio, 5 mol O2/4 m...

Kevin Tam 1J

For the Balmer series, the final energy level is always n=2. So, the wavelengths 653.6, 486.1, 434.0, and 410.2 nm correspond to n=3, n=4, n=5, and n=6 respectively. Since the last wavelength, 410.2 nm, corresponds to n=6, the next wavelength should logically correspond to n=7. To solve for the wave...

Kevin Tam 1J

For this solution, I am not going to use M1V1=M2V2. I am just going to be using dimensional analysis, which should generate the same answer. First, convert grams to moles. Preferably in the same step, divide the moles by liters to get M (molarity). (5.00g KMnO4 x (1 mol KMnO4/158.04g KMnO4))/0.15000...

Kevin Tam 1J

First Step: Divide all the gram values by 8.00 g and multiply them by 100. 3.27g/8.00g x 100 = 40.9% C 0.366g/8.00g x 100 = 4.58% H 4.37g/8.00g x 100 = 54.6% O Second Step: Change the percentages to grams. Convert the new values in grams to moles. 40.9g C x (1 mol C/12.01g C) = 3.41 mol C 4.58g H x ...

CHEMISTRY COMMUNITY

Search found 23 matches

Re: Functional groups

Re: SN2 Reactions

Re: question about nucleophile and electrophile

Re: Unique rate of reaction

Re: 2015 midterm

Re: Example 14.8

Re: Hess's Law

Re: Positional/Residual Entropy

Re: heating curve

Re: Burns Worse from Steam Than Water Example

Re: Heteronuclear Diatomic Molecules

Re: Is a negative pH possible? [ENDORSED]

Re: Coordination Number for (en) [ENDORSED]

Re: Naming Chemical Compounds

Re: order of ligand names [ENDORSED]

Re: order of ligand names [ENDORSED]

Re: What is Angular? And what is T-shaped?

Re: Homework #3.97

Re: Ionization Energy

Re: Limiting Reactant [ENDORSED]

Re: 1.57 [ENDORSED]

Re: Question on Molarity and Dilution of a Solution: Post-Module Assessment

Re: Assessment Question #9