To determine the amount of ozone in air, scientists pass the air sample through a solution containing potassium iodide, which
removes the ozone according to the following equation:
O3 (g) + 2 KI(aq) + H2O (l) O2 (g) + I2 (s) + 2 KOH(aq)
a. How many grams of potassium iodide are needed to
react with 52.7 mg of O3?
b. According to the above reaction, how many grams of potassium hydroxide is formed together with
24.8 grams of iodine?

It would b be

Explanation:

I think sorry if u get it wrong

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

- "the choice that is not an indicator of a chemical change is "state of matter changes". more common than not, chemical reactions produce energy in the form of light or heat. along with energy, they also produce a new substance called the product that could be in any state of matter (solid, gas, or liquid)."