Given the following reduction potentials:                  …

Given the following reduction potentials:                                                               Eored Cr3+(aq)  + 3e- →  Cr(s)                      –0.74 V Fe3+(aq)  +  e- →  Fe2+(aq)                 +0.77 V Calculate the equilibrium constant, K, for the following reaction.               Cr(s)    + 3    Fe3+(aq)   →   Cr3+(aq) + 3 Fe2+(aq) Hint:  Remember that a voltaic cell runs on a spontaneous redox reaction.   Hint: Make sure to determine the total moles of electrons transferred for the cell.  

Copper metal, Cu (s), is produced by running a current throu…

Copper metal, Cu (s), is produced by running a current through a solution of Cu2+ in an electrolytic cell. How many grams of copper metal can be deposited from Cu²⁺(aq) when a current of 1.50 A is run for 75 min? Hint:  F = 96,500 C/mol e− Hint:  1 A = 1 C/s

Congratulations! You have completed (and survived) CHM 116! …

Congratulations! You have completed (and survived) CHM 116!  It was a pleasure to share the semester with you, and I hope you have a truly fantastic summer!  This is a free question – Pick YES if you would like to get the free points!   Grades will be calculated as quickly as possible after the Final Exam closes on Tuesday, May 6 at 11:59 pm MST.  Announcements will go out to let you know that point totals have been added to Canvas.  Please don’t email with questions about your grades – I won’t have any answers for you if you haven’t seen an announcement.  And don’t worry, you will get a chance to review your scores before grades are submitted to the University.   Congratulations on crossing the finish line!

This question is concerned with hashing with open addressing…

This question is concerned with hashing with open addressing, where the table size is 13 (indexed from 0 to 2) and the (linear) probing sequence is defined by h'(k) = k mod 13 and h(k, i) = (h'(k) + i) mod 13. Assume that the content of the hash table T is as follows: T[0] = 13 T[1] = 14 T[2] = DELETED T[3] = 15 T[4] = NIL T[5] = 5 T[6] = DELETED T[7] = 19 T[8] = NIL T[9] = 9 T[10] = 23 T[11] = 24 T[12] = 25   The last cell (index to the table) probed by Hash-Search(T, 18) is