Two chemist are trying to determine if their water sample is a mixture or pure substance. in their investigation they first filter the sample of dirty water with gravel. the water is then clearer then before. see image. what does this change tell the chemist about the chemical make up of the sample the dirty water is only a mixture because it was not completely filtered by the gravel. dirty water is a mixture of solid particles and liquid. it is both a mixture and pure substance. more information is needed to answer the question. the dirty water is only a pure substance now that it has been filtered by the gravel

The correct answer is option B. Dirty water is a mixture of solid particles and liquid. It is both a mixture and pure substance.

The dirty water sample has both gravel and liquid water in it. After filtration the gravel is removed so the water sample looks clearer than before filtration. Liquid water is a pure substance because it is a compound that is made up of elements hydrogen and oxygen. Now the gravel is only physically combined with the liquid water, thus giving the water sample properties of a mixture. And like any mixture, gravel is physically separated through filtration from the liquid water.

Thus the water sample of the chemists is both a mixture and pure substance.

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

catya that's not even a b or c