Search found 23 matches

by Kevin Tam 1J
Mon Mar 13, 2017 8:15 am
Forum: *Alcohols
Topic: Functional groups
Replies: 2
Views: 462

Re: Functional groups

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.
by Kevin Tam 1J
Mon Mar 06, 2017 7:53 am
Forum: *Organic Reaction Mechanisms in General
Topic: SN2 Reactions
Replies: 2
Views: 264

Re: SN2 Reactions

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 ...
by Kevin Tam 1J
Mon Feb 27, 2017 7:22 am
Forum: *Alkanes and Substituted Alkanes (Staggered, Eclipsed, Gauche, Anti, Newman Projections)
Topic: question about nucleophile and electrophile
Replies: 2
Views: 256

Re: question about nucleophile and electrophile

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...
by Kevin Tam 1J
Mon Feb 20, 2017 10:37 am
Forum: General Rate Laws
Topic: Unique rate of reaction
Replies: 2
Views: 261

Re: Unique rate of reaction

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...
by Kevin Tam 1J
Mon Feb 13, 2017 7:52 am
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 2015 midterm
Replies: 1
Views: 211

Re: 2015 midterm

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...
by Kevin Tam 1J
Mon Feb 06, 2017 8:01 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: Example 14.8
Replies: 2
Views: 286

Re: Example 14.8

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.
by Kevin Tam 1J
Mon Jan 30, 2017 8:04 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Hess's Law
Replies: 4
Views: 526

Re: Hess's Law

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 ...
by Kevin Tam 1J
Mon Jan 23, 2017 10:37 am
Forum: Calculating Standard Reaction Entropies (e.g. , Using Standard Molar Entropies)
Topic: Positional/Residual Entropy
Replies: 1
Views: 209

Re: Positional/Residual Entropy

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...
by Kevin Tam 1J
Mon Jan 16, 2017 6:56 pm
Forum: Phase Changes & Related Calculations
Topic: heating curve
Replies: 3
Views: 336

Re: heating curve

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.
by Kevin Tam 1J
Mon Jan 09, 2017 7:24 pm
Forum: Phase Changes & Related Calculations
Topic: Burns Worse from Steam Than Water Example
Replies: 2
Views: 437

Re: Burns Worse from Steam Than Water Example

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...
by Kevin Tam 1J
Mon Nov 28, 2016 10:46 pm
Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
Topic: Heteronuclear Diatomic Molecules
Replies: 2
Views: 331

Re: Heteronuclear Diatomic Molecules

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.
by Kevin Tam 1J
Mon Nov 21, 2016 8:32 am
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: Is a negative pH possible? [ENDORSED]
Replies: 2
Views: 349

Re: Is a negative pH possible? [ENDORSED]

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.
by Kevin Tam 1J
Tue Nov 15, 2016 11:39 am
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Coordination Number for (en) [ENDORSED]
Replies: 6
Views: 873

Re: Coordination Number for (en) [ENDORSED]

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)?
by Kevin Tam 1J
Mon Nov 14, 2016 9:05 am
Forum: Shape, Structure, Coordination Number, Ligands
Topic: Naming Chemical Compounds
Replies: 2
Views: 328

Re: Naming Chemical Compounds

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.
by Kevin Tam 1J
Mon Nov 14, 2016 8:56 am
Forum: Naming
Topic: order of ligand names [ENDORSED]
Replies: 7
Views: 703

Re: order of ligand names [ENDORSED]

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...
by Kevin Tam 1J
Mon Nov 07, 2016 8:15 am
Forum: Naming
Topic: order of ligand names [ENDORSED]
Replies: 7
Views: 703

Re: order of ligand names [ENDORSED]

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.
by Kevin Tam 1J
Mon Oct 31, 2016 7:47 am
Forum: Determining Molecular Shape (VSEPR)
Topic: What is Angular? And what is T-shaped?
Replies: 2
Views: 397

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

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.
by Kevin Tam 1J
Mon Oct 24, 2016 7:59 am
Forum: Lewis Structures
Topic: Homework #3.97
Replies: 2
Views: 279

Re: Homework #3.97

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...
by Kevin Tam 1J
Mon Oct 17, 2016 8:03 am
Forum: Lewis Structures
Topic: Ionization Energy
Replies: 1
Views: 284

Re: Ionization Energy

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...
by Kevin Tam 1J
Mon Oct 10, 2016 6:08 pm
Forum: Limiting Reactant Calculations
Topic: Limiting Reactant [ENDORSED]
Replies: 2
Views: 434

Re: Limiting Reactant [ENDORSED]

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...
by Kevin Tam 1J
Mon Oct 03, 2016 8:48 am
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: 1.57 [ENDORSED]
Replies: 5
Views: 774

Re: 1.57 [ENDORSED]

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...
by Kevin Tam 1J
Mon Sep 26, 2016 12:37 pm
Forum: Molarity, Solutions, Dilutions
Topic: Question on Molarity and Dilution of a Solution: Post-Module Assessment
Replies: 1
Views: 380

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

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...
by Kevin Tam 1J
Sat Sep 24, 2016 9:26 pm
Forum: Empirical & Molecular Formulas
Topic: Assessment Question #9
Replies: 2
Views: 399

Re: Assessment Question #9

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 ...

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