Search found 18 matches

by Leon Popa
Fri Dec 08, 2017 10:28 am
Forum: Applying Le Chatelier's Principle to Changes in Chemical & Physical Conditions
Topic: 11.77 textbook problem
Replies: 2
Views: 206

Re: 11.77 textbook problem

Essentially, the problem says the temperature is increasing. This must mean that the reactants need to absorb more energy - an endothermic reaction. Raising the temperature favors the reverse reaction (endothermic) and lowering the temperature favors the forward reaction (exothermic). So in this cas...
by Leon Popa
Sun Nov 26, 2017 11:02 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: CH2Cl2 polarity
Replies: 2
Views: 547

Re: CH2Cl2 polarity

Hi Rain, When I worked on #25, I thought the same thing. My understanding is that because the lewis structure did not have all H atoms or all Cl atoms, around the central carbon atom, it was not truly symmetrical. Its likely that the bond length between a carbon and a hydrogen is not exactly the sam...
by Leon Popa
Sun Nov 26, 2017 10:56 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: delta G
Replies: 2
Views: 192

Re: delta G

I believe delta G refers to Gibbs Free Energy. This is the energy involved in a chemical reaction that can be used to do work. The free energy, or delta G of a reaction is the sum of the reaction's enthalpy (H) and the product of the temperature (Kelvin) and the entropy (S) of the system: essentiall...
by Leon Popa
Fri Nov 17, 2017 6:52 pm
Forum: Naming
Topic: Transition Metal Oxidation States
Replies: 1
Views: 153

Re: Transition Metal Oxidation States

Hi Samantha, From what I understand, elements don't like to lose too many e-. Once they reach a stable configuration, they tend to stay there since the next ionization energy would be too high to remove more e-. For example, Na has 1 valence e-. So Na's starting ionization energy is relatively small...
by Leon Popa
Fri Nov 17, 2017 3:36 pm
Forum: Ionic & Covalent Bonds
Topic: Determining oxidation state
Replies: 3
Views: 491

Re: Determining oxidation state

Hi Jenny, I'm not sure exactly how you would determine the oxidation state by looking at the Lewis Structure (I'm assuming that's what you meant by LS). I do know another way however, which involves using the periodic table. Take MgO as an example of an ionic molecule. Mg is in the second group of t...
by Leon Popa
Thu Nov 09, 2017 3:06 pm
Forum: Dipole Moments
Topic: Dipole Moments
Replies: 1
Views: 200

Re: Dipole Moments

To determine if a molecule has an overall net dipole, you would have to check if all the dipole moments cancel. The dipole moments cancel when they point away from the central atom. BCl3 for example has no net dipole because the dipoles on the Chlorines all point away from the central Boron. Water h...
by Leon Popa
Thu Nov 09, 2017 2:54 pm
Forum: Lewis Structures
Topic: Lewis Structure of SO2
Replies: 1
Views: 806

Re: Lewis Structure of SO2

Hi Ishan, From what I understand, all of those lewis structures are valid. They're essentially resonance structures, but the first structure you mentioned, with two double bonds, is the preferred one. This is because all formal charges are zero. In the structure with one single bond and one double b...
by Leon Popa
Fri Nov 03, 2017 1:30 am
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Lanthanides
Replies: 1
Views: 234

Re: Lanthanides

Hi Leah, To answer this question, I used the diagram that Dr. Lavelle often illustrates when solving problems on the board. This diagram (shown below) indicates that as the principal quantum number (n) increases, the spacing between orbitals becomes smaller. This means that as n gets larger, it beco...
by Leon Popa
Fri Nov 03, 2017 1:12 am
Forum: Trends in The Periodic Table
Topic: Second Ionization Energy
Replies: 2
Views: 244

Re: Second Ionization Energy

Hi Emily, As far as my understanding goes, the second ionization energy of Mg is larger than the first because it always takes more energy to remove an electron from a positively charged ion than from a neutral atom. This means it was easier to remove the first electron when Mg was neutral than to r...
by Leon Popa
Thu Oct 26, 2017 8:58 pm
Forum: Wave Functions and s-, p-, d-, f- Orbitals
Topic: d and s blocks
Replies: 16
Views: 986

Re: d and s blocks

As far as my understanding goes, The s, p, d and f orbitals are ordered on the basis of how many electrons can fit in each orbital. Because s orbitals can only fit 2 electrons, s is written. p-orbitals can fit 6 electrons total, d-orbitals can fit 10 electrons total, and f-orbitals can fit 14 electr...
by Leon Popa
Thu Oct 26, 2017 8:53 pm
Forum: Quantum Numbers and The H-Atom
Topic: Problem 2.29 parts, b and c
Replies: 2
Views: 225

Re: Problem 2.29 parts, b and c

Hi Kaitlin, The way I understand 2.29, you only have to find the number of electrons based on how specific the directions are. For example, in part a, we're given n=2 and l=1. When l=0, we know we're looking at an s-orbital. When l=1, we're dealing with a p-orbital. For l=2, d-orbital and l=3, p-orb...
by Leon Popa
Fri Oct 20, 2017 12:01 am
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Chapter 1 #43 [ENDORSED]
Replies: 7
Views: 593

Re: Chapter 1 #43 [ENDORSED]

Hi Michelle, I did some research into the one-dimensional box concept - here's what I understood. Apparently, the problem is asking us to the imagine the atom itself as a box or container in which a single particle is confined. In this case that particle is the electron. Ultimately we are finding th...
by Leon Popa
Thu Oct 19, 2017 11:49 pm
Forum: Bohr Frequency Condition, H-Atom , Atomic Spectroscopy
Topic: Chapter 1 #55
Replies: 1
Views: 266

Re: Chapter 1 #55

Hi Michelle, For part b, I used the formula E = h*V, where h is planck's constant (6.626 * 10^-34 Js) and V is the frequency you calculated in part a. If you got V = 1.1 * 10^14 Hz as I did, then multiply that by planck's constant to get the energy in joules. Hz or Hertz, are the same thing as "...
by Leon Popa
Fri Oct 13, 2017 10:36 am
Forum: DeBroglie Equation
Topic: De Brogile Equation
Replies: 1
Views: 223

Re: De Brogile Equation

Hey Alex, Well, to start, Joules can also be written as (kg * m^2) / s^2. So just looking at the units in λ = h/mv, we have: λ = h / m * v λ = J*s / kg * m/s λ = (kg * m^2/s^2)*s / kg * m/s λ = (kg * m^2 / s) / kg * m/s λ = (kg * m^2) / (kg * m) λ = m^2 / m λ = m That's how I cancel things out to ge...
by Leon Popa
Fri Oct 13, 2017 10:14 am
Forum: Properties of Light
Topic: Converting Units [ENDORSED]
Replies: 11
Views: 809

Re: Converting Units [ENDORSED]

Hey Hammad, From what I've observed, in such a case I would convert the 2.90 * 10^-9 m to nanometers, considering that the problem is asking for the wavelength. If however, the your answer to a certain problem was 2.90 * 10^-38 m, it is best left in that form since the conversion is now severely com...
by Leon Popa
Thu Oct 12, 2017 11:06 am
Forum: Properties of Light
Topic: When is the Date of Test #2?
Replies: 1
Views: 200

When is the Date of Test #2?

To whom it may also concern, Dr. Lavelle posted on his site: "Test 2 (in Discussion Section) covers Chapter 1 up to the end of section 1.5 in the textbook. For the assigned homework it is Chapter 1 questions 3-41." I'm not clear on when exactly this is. Does he mean discussion section of t...
by Leon Popa
Thu Oct 05, 2017 9:30 pm
Forum: Limiting Reactant Calculations
Topic: L39 .. I don't understand it. May someone please explain it to me?! Thank You.
Replies: 2
Views: 512

Re: L39 .. I don't understand it. May someone please explain it to me?! Thank You.

Hey Jason, Here's how I approach the problem: We have 1.50 grams of Tin, which is abbreviated "Sn" on the periodic table. They tell us the 1.50 g of Sn is placed in a crucible (a metal container) that has capacity of 26.45 grams. We also know that the Tin reacts in this container with oxyg...
by Leon Popa
Mon Oct 02, 2017 7:17 pm
Forum: Balancing Chemical Reactions
Topic: Fundamental H.13
Replies: 4
Views: 432

Re: Fundamental H.13

Hey William, Here's my take on how to balance NO + O2 -> NO2: I placed a coefficient of 2 in front of NO on the left and NO2 on the right. Doing this means there are now a total of 2 Nitrogen atoms and 4 oxygen atoms on each side of the equation. NO + O2 -> 2NO2 2NO + O2 -> 2NO2 Looks balanced. Hope...

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