Explain what happens when energy is increased
on the three phases of matter.

If we begin with a cube of ice composed water molecules in a crystalized form we are dealing with water molecules with their lowest amount of energy. These molecules have such low amounts of energy that their movement is almost imperceptible. This form of matter for water is at and below 0 C.

If we add heat to the ice thereby increasing the energy in the molecules of water these molecules begin to vibrate at much higher velocities. These vibrations allow for the molecules of water to escape their positions in the crystal and begin to flow into a liquid. This range of temperature is from 0 C to 100 C.

If we continue to add energy and drive the temperature up to 100 C the molecules of water will vibrate so rapidly that they will break away from the liquid form and move into a gaseous state as steam escaping into the the air. This vaporization actually takes place constantly at the surface of the water creating a thin vapor layer at the surface of any body of water.

This is a simplification of the chemical process but basically the amount of energy in the molecules will determine the phase of matter.

Explanation:

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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

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