## Search found 21 matches

Wed Mar 09, 2016 11:11 pm
Forum: *Cycloalkenes
Topic: Winter 2012 Final 6A naming a cycloalkene
Replies: 1
Views: 366

### Winter 2012 Final 6A naming a cycloalkene

For this question, there is a line structure of a cyclohexene. Between the first two carbons is a double bond. Attached to the second carbon (the end of the double bond) is a methyl group. The next carbon along the chain also has a methyl group, and the next carbon has two methyl groups. The solutio...
Wed Mar 09, 2016 10:49 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: 2012 Final Q1B
Replies: 3
Views: 418

### Re: 2012 Final Q1B

The q rxn found represents the DeltaH of that specific reaction. You divide by 4 because you must account for the fact that that q you just solved for represented the reaction of four moles of NO reacting. By dividing by 4 moles, you are left with units of kJ/mol, which are overarching and not speci...
Wed Mar 09, 2016 10:15 am
Forum: Concepts & Calculations Using Second Law of Thermodynamics
Topic: Residual Entropy
Replies: 1
Views: 342

### Re: Residual Entropy

Residual entropy is always taken at 0K. The equation to find this is S=kBlnW, W being the number of possible states it can take. While NO can take two possible states (NO and ON), BF3 can only take one (BF3), giving it a lower residual entropy. Hope this helps!
Tue Mar 01, 2016 9:18 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: determining molecularity 15.85 b
Replies: 2
Views: 369

### determining molecularity 15.85 b

15.85 says that 2I-> I2 is termolecular. does this mean the molecularity is determined by the total reaction, and not the overall reaction?
Sun Feb 21, 2016 8:54 pm
Forum: General Rate Laws
Topic: Pseudo Rate Law
Replies: 2
Views: 1304

### Re: Pseudo Rate Law

An important part of this law is the assumption that the other reactants are in large excess, and then you only measure the isolated, low-concentrate reactant.
Tue Feb 09, 2016 12:25 am
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Winter 2014 1C (Bond Enthalpies)
Replies: 2
Views: 418

### Re: Winter 2014 1C (Bond Enthalpies)

Thank you!
Sun Feb 07, 2016 9:26 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Winter 2014 1C (Bond Enthalpies)
Replies: 2
Views: 418

### Winter 2014 1C (Bond Enthalpies)

Why does the answer not take into account the energy to break the three C-H bonds in CH3OH?
Sun Feb 07, 2016 9:09 pm
Forum: Gibbs Free Energy Concepts and Calculations
Topic: 2011 midterm q4
Replies: 3
Views: 540

### Re: 2011 midterm q4

I'm confused on why 2x is not the top. Wouldn't each reactant have a 1/2 in front of it? And then wouldn't you bring a 4 to the top instead of a two?
Mon Feb 01, 2016 7:05 pm
Forum: Balancing Redox Reactions
Topic: Ch. 14 #14.3a
Replies: 1
Views: 344

### Re: Ch. 14 #14.3a

I believe that part of solving the problem involves you balancing the oxidation half-reaction, so the equation in its given state is not already equal on both sides. You must multiply by 4 in order to get an equal number of electrons on both sides of the equation. However, when I look at the solutio...
Fri Jan 22, 2016 7:09 pm
Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
Topic: Calculating Reaction Enthalpy
Replies: 1
Views: 251

### Calculating Reaction Enthalpy

For problems like 8.75 b, when looking at a molecule like CH3CHCH2 and CH3CH(OH)CH3, do I draw out the lewis structure to see which Carbons have double vs single bonds? Is this something I should intuitively know/is there a trick or select few I should know off of the top of my head without drawing ...
Sun Jan 17, 2016 8:07 am
Forum: Calculating Work of Expansion
Topic: Bond Formation Exothermic/Endothermic
Replies: 2
Views: 1095

### Re: Bond Formation Exothermic/Endothermic

The formation of a bond takes energy that existed in two atoms and, because bonded atoms are more stable and have less energy, releases some of this energy to the outside environment. This would make the formation of a bond exothermic, as this energy is no longer contained and must go somewhere.
Tue Jan 12, 2016 9:55 pm
Forum: Thermodynamic Definitions (isochoric/isometric, isothermal, isobaric)
Topic: When to use C=5/2R or C=3/2R
Replies: 7
Views: 2158

### Re: When to use C=5/2R or C=3/2R

What does the R stand for in this equation?
Sun Jan 10, 2016 6:24 pm
Forum: Phase Changes & Related Calculations
Topic: Question 8.9
Replies: 3
Views: 1764

### Re: Question 8.9

In 8.9, regarding the answer in the solutions, I follow that 1.48 L*atm comes from converting torr to atm and then multiplying by the change in volume. However, the solutions then suggest to convert to J using 8.314 J*K/mol over .08206 L*atm/K*mol. How do we know to use these? and how does the K can...
Sat Jan 09, 2016 11:33 pm
Forum: Thermodynamic Systems (Open, Closed, Isolated)
Topic: Which can be exchanged: matter or energy?
Replies: 4
Views: 746

### Re: Which can be exchanged: matter or energy?

Matter in a closed system can not be exchanged with its surroundings, however the energy is still transferred between surroundings and system. Much like a saucepan with a lid on it on a stove, even though all the matter stays in the saucepan, it still is heated by its surroundings.
Fri Dec 04, 2015 9:15 pm
Forum: Photoelectric Effect
Topic: 2 on Fall 2008 Final
Replies: 1
Views: 404

### 2 on Fall 2008 Final

Q2A of theFall 2008 Final Exam asks: "In one photoelectric experiment the wavelength of an electron emitted from a metallic surface is 4.00 x 10^(-10) m. What is the velocity of the ejected electron in this experiment? The answer uses the equation v=h/m*lamba What is this and how is this equati...
Sun Nov 29, 2015 11:28 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: balanced reactions
Replies: 2
Views: 498

### balanced reactions

Should we assume that questions and reactions given in questions are balanced? I ask because there was a question on quiz three (number 7 on my quiz) in which the given reaction was not balanced. My TA later explained to me that this was because an enzyme was left out of the given equation. Should w...
Sun Nov 15, 2015 8:49 pm
Forum: Bond Lengths & Energies
Topic: Stronger Ionic/Covalent Bonds
Replies: 3
Views: 1055

### Re: Stronger Ionic/Covalent Bonds

One way to determine which is stronger would be to calculate the dissociation energy, or how much energy it would take to break a bond. The greater this energy is, the stronger a bond is. Ionic bonds have a much higher dissociation energy than covalent bonds.
Sun Nov 08, 2015 11:26 pm
Forum: Bond Lengths & Energies
Topic: Bonding Orbitals versus Anti-Bonding Orbitals
Replies: 2
Views: 619

### Bonding Orbitals versus Anti-Bonding Orbitals

Why is it that Anti-Bonding Orbitals are higher in energy than bonding orbitals?
Mon Nov 02, 2015 12:00 am
Forum: Photoelectric Effect
Topic: Wave-particle duality
Replies: 1
Views: 334

### Re: Wave-particle duality

Emag radiation can be modeled as a wave in that it can be modeled with a variety of wavelengths and frequencies, and these models proved that light has properties of waves. however, in order to have a more thorough understanding of light, it is necessary to understand light as a particle, too. The d...
Sun Oct 25, 2015 10:49 pm
Forum: Ionic & Covalent Bonds
Topic: Ionic and Covalent Character
Replies: 2
Views: 482

### Re: Ionic and Covalent Character

Even though all bonds between atoms have some covalent character (because the electron density is shared between the atoms), how ionic versus covalent a bond is gets determined by the difference in electronegativity between the atoms, not really by an atom's size. A larger difference points to a mor...
Sat Oct 10, 2015 12:07 pm
Forum: Electron Configurations for Multi-Electron Atoms
Topic: Electron Removal
Replies: 1
Views: 393

### Electron Removal

For Chapter 2, problem 47, part b, the question asks for the type of orbital from which an electron would be removed to form the +1 ion of Mn. I know that the Electron configuration is [Ar]3d5 4s2. Why does the solutions manual say the electron will come from 4s orbital, not 3d orbital?