armer Brown tests the hypothesis that the more alfalfa sprouts cattle eat, the more milk they produce. He divides his pasture into four fields and puts twenty cattle in each field. Field 1 cows get no alfalfa sprouts, field 2 cows get ten pounds of sprouts per day, field 3 cows get twenty pounds of sprouts, and field 4 cows get thirty pounds. He feeds all groups fifty pounds of regular hay per day, uses the same breed of cattle, and gives them the same amount of water. He measures gallons of milk produced per cow every day What is the independent variable in Farmer Brown's experiment?

Answer is in the photo. I couldn't attach it here, but uploaded it to a file hosting. link below! Good Luck!

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the molecules move in straight line until they collide with each other or with the walls of the container. collisions are perfectly elastic; when two molecules collide, they change their directions and kinetic energies, but the total kinetic energy is conserved.

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