Search found 13 matches
- Sat Mar 18, 2017 4:48 pm
- Forum: Method of Initial Rates (To Determine n and k)
- Topic: Course Reader Final Practice "2016 Final" Q4 part B
- Replies: 5
- Views: 1186
Re: Course Reader Final Practice "2016 Final" Q4 part B
Also, can you clarify how to determine which is the slow step in addition to clarifying what the actual answer is.
- Wed Mar 15, 2017 5:23 pm
- Forum: Concepts & Calculations Using First Law of Thermodynamics
- Topic: 5/2R vs 3/2R
- Replies: 2
- Views: 536
Re: 5/2R vs 3/2R
Also, I believe you use 3/2R under constant volume and 5/2R under constant pressure.
- Wed Mar 08, 2017 8:49 pm
- Forum: *Alkanes
- Topic: Iso, Neo, and Tert
- Replies: 3
- Views: 1016
Re: Iso, Neo, and Tert
Can someone explain when to use these?
- Sun Mar 05, 2017 2:59 pm
- Forum: Arrhenius Equation, Activation Energies, Catalysts
- Topic: Homework 15.61
- Replies: 1
- Views: 523
Homework 15.61
The rate constant of the first-order reaction 2N2O -->2N2 +O2 is 0.76 s^-1 at 1000K and 0.82 s^-1 at 1030K. Calculate the activation energy of the reaction. How are you supposed to solve this without the "A" in the Arrhenius equation?
- Wed Feb 22, 2017 11:58 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Quiz 2 Winter 2017
- Replies: 160
- Views: 25551
Re: Quiz 2 Winter 2017
I'm not sure why I keep getting this wrong but for number 6 I keep getting 1.66 for the value of k. Could somebody please help? For #6, you have to use the 2nd order integrated rate law and just plug everything in: 1/[0.44M] = k(11 min) + 1/[.95M]. Then subtract 1/[.95M] on both sides to get 1.22= ...
- Wed Feb 22, 2017 11:52 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Quiz 2 Winter 2017
- Replies: 160
- Views: 25551
Re: Quiz 2 Winter 2017
SarahTian3k wrote:Could someone please explain number 7? I keep getting 7.46x10-3 min-1
I think you accidentally plugged in 0.733 M instead of 0.773 M.
- Mon Feb 13, 2017 10:47 pm
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Homework Question 14.41b
- Replies: 1
- Views: 378
Homework Question 14.41b
Can someone explain how to solve this problem? I know I have to use the Nernst Equation but what do I do with the pH?
|Pt(s) |H2(g,1 bar)|H+(aq,pH=4.0)||H+(aq,pH=3.0)|H2(g,1 bar)|Pt(s)
|Pt(s) |H2(g,1 bar)|H+(aq,pH=4.0)||H+(aq,pH=3.0)|H2(g,1 bar)|Pt(s)
- Sun Feb 12, 2017 11:34 am
- Forum: Appications of the Nernst Equation (e.g., Concentration Cells, Non-Standard Cell Potentials, Calculating Equilibrium Constants and pH)
- Topic: Question 14.35b
- Replies: 1
- Views: 370
Question 14.35b
Can someone explain why the solution manual says lnk=(1)(.12V)/(0.02569v)? I don't understand where that number on the denominator came from. I did part b the same way as I did part a but got part b wrong.
- Sat Feb 04, 2017 11:37 pm
- Forum: Balancing Redox Reactions
- Topic: Pneumonic Device
- Replies: 2
- Views: 867
Re: Pneumonic Device
The mnemonic device I always used was OIL RIG
OIL = Oxidation is loss [of electrons]
RIG = Reduction is gain [of electrons]
OIL = Oxidation is loss [of electrons]
RIG = Reduction is gain [of electrons]
- Tue Jan 31, 2017 11:43 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Quiz 1 Preparation Answers
- Replies: 130
- Views: 25842
Re: Quiz 1 Preparation Answers
So for number 11, do we just assume that the deltaH given is per mole?
- Sat Jan 28, 2017 2:31 pm
- Forum: Entropy Changes Due to Changes in Volume and Temperature
- Topic: Homework Problem 9.25
- Replies: 1
- Views: 469
Re: Homework Problem 9.25
You can use log rules to move the 6.02x10^23 to the front and the calculation should be fine.
- Sat Jan 21, 2017 7:12 pm
- Forum: Student Social/Study Group
- Topic: Post All Chemistry Jokes Here
- Replies: 9651
- Views: 3930443
Re: Post All Chemistry Jokes Here
What do you do to a dead chemist?
You Barium
You Barium
- Sat Jan 14, 2017 4:46 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Problem 8.65
- Replies: 3
- Views: 663
Re: Problem 8.65
2 NO(g) + O2 (g) -> 2NO2(g) H°= -114.1 kJ 4 NO2(g) + O2(g) -> 2 N2O5(g) H°= -110,2 First, I divided the second equation by 2 to get 2NO2(g) + 1/2O2(g) --> N2O5(g) with a dH= -55.2. Then I added the equations together to get 2NO(g) + 3/2O2(g) --> N205(g) with a dH= -169.2. The equation for NO is N2(g...