The city of Manila held a trial run its newest program which is the Acid and Base Tournament. All of the participants enjoyed the program. But the city mayor noticed that there were traces of spilled liquid on the metal trays where the towels were place. The mayor was anxious because the towels were burnt and the metal trays were damage. The mayor ask for help from the committee and one of the officials suggested to consult you as the chemist. You are going discuss what happened with the damaged metal trays and burnt towels when the liquid was spilled on them. Suggest an alternative material or way to prevent the same incident to happen again.
OUTPUT: You are going to present a cause and effect discussion o the mayor of the city. You will be reporting next week. First, analyze what happened by examining the characteristic of the materials involved. Then, research on the possible solutions that you are proven to be effective. Make a slide presentation of your output.

Not sure, do I need to prepare a presentation?

Explanation:

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