A farmer uses triazine herbicide to control pigweed in his field. For the first few years, the triazine works well and almost all the pigweed dies; but after several years, the farmer sees more and more pigweed. Which of these explanations best explains what happened?

A. The herbicide company lost its triazine formula and started selling poor-quality triazine.
B. Triazine-resistant weeds were more likely to survive and reproduce
C. Natural selection caused the pigweed to mutate, creating a new triazine-resistant species
D. Triazine-resistant pigweed has less efficient photosynthesis metabolism.

B. Triazine-resistant weeds were more likely to survive and reproduce

Explanation:

According to darwin's theory of evolution, variation is already present in some members of a population and this variation lets them survive in the adverse condition and those who do not have that variation which helps in survival are lost with time.

So as before using triazine herbicide the major population of the weed were not resistant to this herbicide so in the first few years the nonresistant weeds were lost and only resistant weed which was very less in number survived.

So after several years these resistant weeds reproduced and transferred their gene to their offsprings and became predominant in the field. Therefore the correct answer is B.

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 answer is na (sodium)