Consider the following equilibrium reaction:            

The reaction A → D has the energy profile shown in the graph. Based on this energy profile, choose the false statement.

Consider the voltaic cell diagram:  Al (s) | Al3⁺(aq) || Fe2⁺ (aq) | Fe (s) What is the standard cell potential, E°cell, (in V) for this voltaic cell? Hint: Refer to the table of standard reduction potentials provided. Half Reaction E° red (V) Ag+ (aq)  + e–  →  Ag (s)  +0.80 Fe3+ (aq)  +   e–  → Fe2+ (aq)   +0.77 Cu2+ (aq)  + 2 e–  → Cu (s)    +0.34 Sn4+ (aq) + 2 e–  → Sn2+ (aq)   +0.15 2H+ (aq) + 2 e–  →  H2 (g)  0.00 Pb2+ (aq) + 2 e –  → Pb (s)  –0.13 Sn2+ (aq)  +  2 e–  → Sn (s)  –0.14 Ni2+ (aq) + 2 e –  → Ni (s)   –0.28 Cd2+ (aq) + 2 e –  → Cd (s)   –0.40 Fe2+ (aq) + 2 e –  → Fe (s) –0.44 Cr3+ (aq) + 3 e– →  Cr (s)      –0.74 Zn2+ (aq) + 2 e –  → Zn (s)  –0.76 Al3+ (aq) + 3 e– →  Al (s) –1.66  

The diagrams here represent solutions at various stages in the titration of a weak base B (such as NH3) with HCl. Which image best represents the solution in the beyond the equivalence point? 

Phophorous-32 decays by beta emission to form sulfur-32.   NOTE: In case the picture does not show: The picture indicates that 24 P-32 atoms decay to 3 P-32  atoms in 9 min. What is the decay constant, k, (in min-1) for P-32?

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!  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!

The following figure shows a red-black tree (RBT) T in which a circle denotes a red node, a square denotes a black node, and the NIL nodes are not shown (to save space). The number inside a circle/square is the key value of the corresponding node. The label (upper-case letter) next to a node is a pointer pointing to the memory location of the corresponding node. You should use the label when referring to a node.   (a) Suppose that we want to insert key 26 into the tree in the figure. We first allocate memory for a tree node P and set its color to red and its key to 26. Then we insert it into tree T as if inserting into a binary search tree. After BST insertion (before RBT insertion fixup), the parent of P is [a]   (b) Suppose that we want to insert key 26 into the tree in the figure. We first allocate memory for a tree node P and set its color to red and its key to 26. Then we insert it into tree T as if inserting into a binary search tree. After BST insertion (before RBT insertion fixup), which property of RBT is violated? [b]   (c) Suppose that we want to insert key 26 into the tree in the figure. We first allocate memory for a tree node P and set its color to red and its key to 26. Then we insert it into tree T as if inserting into a binary search tree. In the resulting RBT,  (after RBT insertion fixup), the color of node E is [c]   (d) Suppose that we want to insert key 26 into the tree in the figure. We first allocate memory for a tree node P and set its color to red and its key to 26. Then we insert it into tree T as if inserting into a binary search tree. In the resulting RBT,  (after RBT insertion fixup), the left child of node P is [d]   (e) Suppose that we want to insert key 26 into the tree in the figure. We first allocate memory for a tree node P and set its color to red and its key to 26. Then we insert it into tree T as if inserting into a binary search tree. In the resulting RBT,  (after RBT insertion fixup), the parent of node P is [e]   (f) Suppose that we want to delete the node D with key 10 from the RBT. After the BST deletion, we need to perform deletion fixup. What case is this deletion fixup?  [f]   (g) Suppose that we want to delete the node D with key 10 from the RBT. In the resulting RBT (after the deletion fixup), what is the left child of node A? [g]   (h) Suppose that we want to delete the node D with key 10 from the RBT. In the resulting RBT (after the deletion fixup), what is the color of node H? [h]   (i) Suppose that we want to delete the node D with key 10 from the RBT. In the resulting RBT (after the deletion fixup), what is the left child of node H? [i]   (j) Suppose that we want to delete the node D with key 10 from the RBT. In the resulting RBT (after the deletion fixup), what is the right child of node H? [j]

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

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 first cell (index to the table) probed by Hash-Search(T, 18) is