Search found 10 matches
- Sat Mar 18, 2017 10:20 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: Number 5B from 2015 Final
- Replies: 2
- Views: 592
Re: Number 5B from 2015 Final
You want to find the maximum cell potential, you want the largest, positive E˚cell possible. Since: Cd2+ + 2e- --> Cd(s) E˚= -0.40V Ag+ + 1e= --> Ag(s) E˚= +0.80V Flipping the Cd reaction would give the largest E˚cell since you would reverse the sign to E˚= +0.40, it would be the anode since the Cd ...
- Wed Mar 08, 2017 12:33 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Quiz 3 Winter 2017
- Replies: 183
- Views: 29642
Re: Quiz 3 Winter 2017
For number 5, how do we know the first step is the slow one? Also, could someone explain when and how to use iso- in naming? Or just how to do #1? For 5, the first step is the slow one because it involves breaking two bonds (Carbon-carbon double bond and the H-Br bond). Step two only involves the b...
- Sun Mar 05, 2017 12:27 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Quiz 3 Winter 2017
- Replies: 183
- Views: 29642
Re: Quiz 3 Winter 2017
Blake_Katsev_2E wrote:Shouldnt number 4 be 5-butyl instead of 4-butyl?
If you start counting at the top H3C then go around counterclockwise, you should get 4-butyl.
- Tue Feb 21, 2017 10:48 pm
- Forum: Administrative Questions and Class Announcements
- Topic: Quiz 2 Winter 2017
- Replies: 160
- Views: 24721
Re: Quiz 2 Winter 2017
Hi, can someone show me how they did #7, I know you use the first order integrated rate law, but when I work it out I am not getting the answer that is posted. ln[A] = -kt+ln[A](initial) ln(0.622M) = -k(22min) + ln(0.773M) ln(0.622M/0.773M) = -k(22min) -0.00988min^(-1) = -k k= 9.88 *10^(-3) min^(-1)
- Tue Feb 14, 2017 5:18 pm
- Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
- Topic: What's the difference between a Galvanic and Voltaic Cell?
- Replies: 1
- Views: 680
Re: What's the difference between a Galvanic and Voltaic Cell?
Galvanic and Voltaic Cell are just two names for the same thing
- Sun Feb 12, 2017 10:28 pm
- Forum: Balancing Redox Reactions
- Topic: HW 14.3d
- Replies: 1
- Views: 391
Re: HW 14.3d
The Cl2 would be both the oxidizing and the reducing agent.
- Sun Feb 05, 2017 10:57 pm
- Forum: Balancing Redox Reactions
- Topic: Help on problem 14.17
- Replies: 4
- Views: 1045
Re: Help on problem 14.17
a.) MnO4-(aq) + H+(aq) + 5e- Mn2+(aq) + 4H2O(l) (cathode)
5[Fe2+(aq) Fe3+(aq) + e-] (anode)
b.) MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) Mn2+(aq) +5Fe3+(aq) +4H2O(l)
5[Fe2+(aq) Fe3+(aq) + e-] (anode)
b.) MnO4-(aq) + 5Fe2+(aq) + 8H+(aq) Mn2+(aq) +5Fe3+(aq) +4H2O(l)
- Sun Jan 29, 2017 9:26 pm
- Forum: Reaction Enthalpies (e.g., Using Hess’s Law, Bond Enthalpies, Standard Enthalpies of Formation)
- Topic: Standard enthalpies of formation
- Replies: 3
- Views: 823
Re: Standard enthalpies of formation
You can use either equation, but because the equation 2C + O2 --> 2CO is for two moles of CO, you would need to multiply the ∆H°by 2. Then you should end up getting the same answer.
- Sun Jan 22, 2017 3:55 pm
- Forum: Phase Changes & Related Calculations
- Topic: 8.117 [ENDORSED]
- Replies: 1
- Views: 463
Re: 8.117 [ENDORSED]
∆U = ∆H - P∆V
-P∆V = -∆RT
=-(2/3mol)(8.314 J*K^(-1)*mol^(-1))(298K)
=-1651.7 J
=-1.65 kJ
∆H= (1/3)(-318kJ)
= -106kJ
∆U = -106kJ – 1.65kJ
=-108kJ
-P∆V = -∆RT
=-(2/3mol)(8.314 J*K^(-1)*mol^(-1))(298K)
=-1651.7 J
=-1.65 kJ
∆H= (1/3)(-318kJ)
= -106kJ
∆U = -106kJ – 1.65kJ
=-108kJ
- Sun Jan 15, 2017 10:39 pm
- Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
- Topic: Heat capacity
- Replies: 1
- Views: 360
Re: Heat capacity
Monatomic molecules only have translational thermal energy, but more complex molecules also have rotation and vibration motions which absorb energy. Therefore, the energy is divided up between these motions. Since only translational energy contributes to temperature increase, the temperature increas...