A voltaic cell prepared using zinc andiodine has the following cell notation. Zn(s) | Zn2+(aq) ||I-(aq) | I2(s) |C(graphite)

Which of the following equations correctly represents the balanced, spontaneous, cell reaction?

a. 2I-(aq) +Zn2+(aq) --->I2(s) + Zn(s)
b. I2(s) + Zn(s)---> 2I-(aq) +Zn2+(aq)
c. 2I-(aq) + Zn(s)---> I2(s) +Zn2+(aq)
d. I2(s) +Zn2+(aq) --->2I-(aq) + Zn(s)
e. None of the above, since graphite must be in the equation.

b. I2(s) + Zn(s)---> 2I-(aq) +Zn2+(aq)

Explanation:

Firstly, you can rule out any equation that doesn't have a balanceof charges outright. Therefore, the third and fourthquestions are silly. Graphite is generally regarded as an'inert' electrode, meaning that it will not react in a voltaiccell. This leaves only the first or the second as a viableanswer.

Logically, iodide and zinc in solution will not spontaneously formtheir elements.

This leaves only the second choice as a viable answer.

I2(s) + Zn(s)---> 2I-(aq) +Zn2+(aq)

is the correct choice.

fire could start, buildings collapsing, and ash falling from the sky.

explanation:

this is because the buildings or plants could catch on fire and the buildings would most likely collapse. lastly ash would fall because all the burning would cause ash to fall from the sky.

hope this .

part 1: the half-life of the element is 22 years.

part 2: it will take 132.02 year for the sample to decay from 308.0 g to 4.8125 g.

explanation:

part 1: there are two ways to determine the half life time.

first way: the half life time can be determined directly from its definition as it is the time required for the sample to decay to its half concentration.

from the data the initial concentration was 45.0 g and decayed to its half value (22.5 g) in 22.0 years.

second way: the radio active decay is considered first order reaction so we can use the laws of first order reactions.

k = (1/t) ln (a/a-x), where k is the rate constant of the reaction,

t is the time of the reaction

a is the initial concentration of the reactant

a-x is the remaining concentration at time (t)

a = 45.0 g the initial concentration at t = 0 year.

at t = 22 year, a-x = 22.5 g

then k = (1/22 year) ln (45.0 g/ 22.5 g) = 0.0315 year-1.

the half-life time (t1/2) = ln 2/ k = 0.693/ k =0.693/ 0.0315 = 22 years.

part 2:

using the same law k = (1/t) ln (a/a-x),

using the given data; a = 308.0 g and a-x = 4.8125 g and the calculated k from the part 1 (k = 0.0315 year-1)

t = (1/k) ln (a/a-x) = (1/0.0315) ln (308.0/ 4.8125) = 132.02 years