Search found 10 matches

by Caity Colt 2N
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: 393

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 ...
by Caity Colt 2N
Wed Mar 08, 2017 12:33 pm
Forum: Administrative Questions and Class Announcements
Topic: Quiz 3 Winter 2017
Replies: 183
Views: 20004

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...
by Caity Colt 2N
Sun Mar 05, 2017 12:27 pm
Forum: Administrative Questions and Class Announcements
Topic: Quiz 3 Winter 2017
Replies: 183
Views: 20004

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.
by Caity Colt 2N
Tue Feb 21, 2017 10:48 pm
Forum: Administrative Questions and Class Announcements
Topic: Quiz 2 Winter 2017
Replies: 160
Views: 17432

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)
by Caity Colt 2N
Sun Feb 12, 2017 10:28 pm
Forum: Balancing Redox Reactions
Topic: HW 14.3d
Replies: 1
Views: 262

Re: HW 14.3d

The Cl2 would be both the oxidizing and the reducing agent.
by Caity Colt 2N
Sun Feb 05, 2017 10:57 pm
Forum: Balancing Redox Reactions
Topic: Help on problem 14.17
Replies: 4
Views: 685

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)
by Caity Colt 2N
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: 547

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.
by Caity Colt 2N
Sun Jan 22, 2017 3:55 pm
Forum: Phase Changes & Related Calculations
Topic: 8.117 [ENDORSED]
Replies: 1
Views: 313

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
by Caity Colt 2N
Sun Jan 15, 2017 10:39 pm
Forum: Heat Capacities, Calorimeters & Calorimetry Calculations
Topic: Heat capacity
Replies: 1
Views: 259

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...

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