Cliffs AP Chemistry, 3rd Edition

52.(D) The positron is a particle with the same mass as the electron but the opposite charge.
The net effect is to change a proton to a neutron. Begin by writing the nuclear equation.
239
94 Pu →
0
1 e +
A
ZX

Remember that the total of the Aand Zvalues must be the same on both sides of the equation.

Solve for the Zvalue of X: Z+ 1 = 94, so Z= 93.

Solve for the Avalue of X: A+ 0 = 239, so A= 239.

Therefore, you have^23993 X, or^23993 Np.

53.(A)

Ni(s) | Ni2+(aq) || Ag+(aq) | Ag(s)

anode (oxidation) || cathode (reduction)

Ni(s) →Ni2+(aq) + 2e–|| 2 Ag+(aq) + 2e–→2 Ag(s)

By convention, in the representation of the cell, the anode is represented on the left and

the cathode on the right. The anode is the electrode at which oxidation occurs (AN OX),

and the cathode is the electrode at which reduction takes place (RED CAT). The single

vertical lines (|) indicate contact between the electrode and solution. The double vertical

lines (||) represent the porous partition, or salt bridge, between the two solutions in the two

half-cells. The ion concentration or pressures of a gas are enclosed in parentheses.

Take the two equations that decoded the standard cell notation and include the E°reduc-

tion and the E°oxidation voltages:

(Change the sign!)

ox: Ni(s) →Ni2+(aq) + 2e–

red: 2 Ag+(aq) + 2e–→2 Ag (s)

Ecell=+ 105. volts

.

.

E volt

E volt

025

080

ox

red

=+

+

%

%

%

By definition, E°cellvoltages that are positive indicate a spontaneous reaction.

54.(A) Begin by writing as much of an equation as you can:

S 8 (s) + M(s) →MaSb(s)

From the information provided, you can determine that the residue, MaSb(s), weighed

21.00 grams (62.00 −41.00) and that the metal M weighed 14.00 grams (55.00 −41.00).

According to the Law of Conservation of Mass, the sulfur that reacted with the metal must

have weighed 7.00 grams (21.00 −14.00). You can now set up a proportion that relates the

grams of S 8 and M to their respective equivalent weights.

./.

.()

/.

.

grams equiv sulfur

grams sulfur S

x grams equiv

grams M

16 0

700 8 1400

=

Answers and Explanations for the Practice Test