Search found 5 matches

by 904676178
Sat Mar 17, 2018 6:16 pm
Forum: Arrhenius Equation, Activation Energies, Catalysts
Topic: chapter 15 #109
Replies: 1
Views: 198

chapter 15 #109

for question 109 I am confused as to why exactly they used k', we are given two different temperatures and I used the equation ln k2/k1=Ea/R (1/T1-1/T2), I am wondering why they used k' for k2 in the equation? is it because the temperature is lower for the second temperature given (the first reactio...
by 904676178
Fri Feb 23, 2018 1:22 am
Forum: Galvanic/Voltaic Cells, Calculating Standard Cell Potentials, Cell Diagrams
Topic: 14.15a [ENDORSED]
Replies: 3
Views: 154

Re: 14.15a [ENDORSED]

First, identify which component is oxidized and which is reduced by assigning oxidation numbers. Then, look up the relevant half-reactions in Appendix 2B that match the half-reactions of oxidation and reduction you just identified. If the cell is to be galvanic (AKA voltaic, spontaneous) then E ° c...
by 904676178
Thu Feb 22, 2018 8:54 pm
Forum: Balancing Redox Reactions
Topic: 14.11
Replies: 2
Views: 174

Re: 14.11

Yes, but the catch with this problem is that the battery does not have a positive voltage, as the Eº of the anode is greater than that of the cathode. Pt(s) | O 2 (g) | H + (aq) || OH - (aq) | O 2 (g) | Pt(s) Overall reaction: 4H 2 O(l) → 4H + (aq) + 4OH - (aq) Anode: O 2 (g) + 4H + (aq) → 2H 2 O(l...
by 904676178
Sat Feb 10, 2018 12:07 am
Forum: Administrative Questions and Class Announcements
Topic: Lecture Slides
Replies: 6
Views: 310

Re: Lecture Slides

while the lectures are not online the search section on chemistry community has been very helpful. Searching a topic can bring up posts from years before and allows people to view old powerpoint's that were made, pdf files (such as practice review questions), and lecture questions that people in the...
by 904676178
Tue Jan 30, 2018 12:56 pm
Forum: Entropy Changes Due to Changes in Volume and Temperature
Topic: 9.11
Replies: 2
Views: 117

Re: 9.11

I believe yes we do use that r value (8.314J/K*MOL) and the atm should just cancel each other out when you divide (p2/p1). so you are left with JK-1 in answer.

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