Search found 21 matches

Mon Mar 13, 2017 4:58 pm
Forum: Calculating Work of Expansion
Topic: Latm --> J
Replies: 1
Views: 356

Re: Latm --> J

Yes I believe you should although it does depend on what units the question asks for.
Mon Mar 06, 2017 3:49 pm
Forum: Reaction Mechanisms, Reaction Profiles
Topic: Fast/Slow Steps in Reaction Mechanisms
Replies: 1
Views: 347

Re: Fast/Slow Steps in Reaction Mechanisms

I believe that is correct.
Mon Feb 27, 2017 2:39 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Arrhenius [ENDORSED]
Replies: 4
Views: 739

Re: Arrhenius[ENDORSED]

I don't think so. The third quiz will probably focus on organic chemistry.
Fri Feb 24, 2017 12:04 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: Catalysts and equilibrium constant
Replies: 4
Views: 501

Re: Catalysts and equilibrium constant

I'm not too sure what you mean by equilibrium, but a catalyst speeds up a reaction by lowering activation energy. If the reaction is at equilibrium, I think the equilibrium constants don't change if you add a catalyst.
Mon Feb 13, 2017 12:37 am
Forum: Gibbs Free Energy Concepts and Calculations
Topic: How to treat deltaE when flipping equations?
Replies: 3
Views: 602

Re: How to treat deltaE when flipping equations?

The sign changes when we flip the half reactions. However, we do not multiply it by the constant.
Mon Feb 06, 2017 12:03 am
Forum: Phase Changes & Related Calculations
Topic: Quiz #1 Return
Replies: 1
Views: 281

Re: Quiz #1 Return

I'm pretty sure we'll get them back in our discussions this week.
Mon Jan 30, 2017 11:29 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 9.25
Replies: 1
Views: 357

Re: 9.25

When you draw the various lewis structures of SO2F2 you can see that the total number of ways of arranging the molecules in the crystal is 6. Since the question wants the residual molar entropy, W=6^6.022 x 10^23. Using the equation S = klnW, you can find the residual molar entropy with a little log...
Mon Jan 23, 2017 12:25 am
Forum: Gibbs Free Energy Concepts and Calculations
Topic: Spontaneity
Replies: 1
Views: 317

Re: Spontaneity

A system is spontaneous when delta G is negative. There's a bit more to it and I think pages 36 and 37 in the course reader does a great job of outlining the specifics. Hope this helps!
Wed Jan 18, 2017 8:53 pm
Forum: Concepts & Calculations Using First Law of Thermodynamics
Topic: Chapter 8 Question #3 (The First Law) [ENDORSED]
Replies: 1
Views: 303

Re: Chapter 8 Question #3 (The First Law)[ENDORSED]

Since w = -P(change in volume), we first have to calculate the change in volume. Since we know the radius (from the diameter) and the distance the pump is compressed, we can find the change in volume through -(pi)(r^2)(d). We then convert this to liters using (1L / 1000 cm^3). Then we can plug every...
Wed Jan 11, 2017 7:10 pm
Forum: Administrative Questions and Class Announcements
Topic: Quiz Practice
Replies: 5
Views: 582

Re: Quiz Practice

There are practice quizzes in the course reader. I don't think they replace the lowest quiz score anymore.
Tue Nov 29, 2016 1:36 am
Forum: Calculating pH or pOH for Strong & Weak Acids & Bases
Topic: 12.61
Replies: 4
Views: 795

Re: 12.61

I don't think you have to worry too much about the nominal part. In the problem you are using an ICE table. The nominal concentration (your initial concentration) and the equilibrium is listed in the answers, but I think that the book skipped the change part. You can add it in if you want. Hope this...
Thu Nov 24, 2016 7:12 am
Forum: Administrative Questions and Class Announcements
Topic: Quiz 3
Replies: 2
Views: 681

Re: Quiz 3

You could probably pick up the quiz next week. I would check with your TA during their office hours.
Sat Nov 19, 2016 6:18 pm
Forum: Equilibrium Constants & Calculating Concentrations
Topic: ICE table question
Replies: 2
Views: 391

Re: ICE table question

I don't think you would. Multiplying by the moles applies to the change in molar concentration.
Sun Nov 13, 2016 8:28 pm
Forum: Student Social/Study Group
Topic: Help with Coordination Compounds?
Replies: 6
Views: 1096

Re: Help with Coordination Compounds?

I found this video to be helpful: https://youtu.be/RKXYxTOMpuA
Hopes this helps!
Fri Nov 04, 2016 5:42 pm
Forum: *Molecular Orbital Theory (Bond Order, Diamagnetism, Paramagnetism)
Topic: Molecular orbital theory question
Replies: 2
Views: 317

Re: Molecular orbital theory question

Bonding molecular orbitals are in phase meaning that they are constructive while anti bonding molecular orbitals are out of phase meaning that they are destructive. The bonding molecular orbital has a lower energy than the anti bonding molecular orbital so the electrons fill the bonding molecular or...
Sat Oct 29, 2016 12:56 pm
Forum: Determining Molecular Shape (VSEPR)
Topic: Can you present VSEPR theory without having to deal with Lewis Structures?
Replies: 1
Views: 300

Re: Can you present VSEPR theory without having to deal with Lewis Structures?

Depending on what the problem gives you I would still always draw the lewis structures just to be safe.
Sat Oct 22, 2016 12:32 am
Forum: Hybridization
Topic: textbook problem #19e
Replies: 1
Views: 338

Re: textbook problem #19e

I believe it has something to do with stability. When transition metals like Ni ionizes, they lose the s electrons first due to the different ionization levels of s and p.
Thu Oct 13, 2016 8:58 pm
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Ch.1 Problem 43
Replies: 2
Views: 551

Re: Ch.1 Problem 43

I'm not sure if I can explain this, but I found this link really helpful: http://chem.libretexts.org/Core/Physica ... sional_box
Sat Oct 08, 2016 11:09 am
Forum: Heisenberg Indeterminacy (Uncertainty) Equation
Topic: Question on #7 on Heisenberg Uncertainty module (pre/post assessment question) [ENDORSED]
Replies: 1
Views: 598

Re: Question on #7 on Heisenberg Uncertainty module (pre/post assessment question)[ENDORSED]

I believe what you did was correct. For further clarification try this link: viewtopic.php?t=2438
Sat Oct 01, 2016 12:30 pm
Forum: Molarity, Solutions, Dilutions
Topic: Clarification for G.7
Replies: 2
Views: 427

Re: Clarification for G.7

510 g refers to the aqueous solution so you would start the problem by finding the amount of KNO3 (in grams) in the solution. Since 5.45% of the solution is KNO3 you can multiply 510 and 5.45 to find out how much KNO3 you need. Hope this helps!
Sun Sep 25, 2016 4:54 pm
Forum: Administrative Questions and Class Announcements
Topic: Textbooks
Replies: 3
Views: 505

Re: Textbooks

For the most part, the material covered in each book is close to being the same, but if you plan on doing the homework problems, the 5th and 6th edition problems do not align meaning that a problem in the 6th edition might be another problem in the 5th edition (E.32 in the 6th edition is E.28 in the...