A 1.0 mol sample of He(g) at 25°C is mixed with a 1.0 mol sample of Xe(g) at 50°C. Which of the following correctly predicts the changes in average kinetic energy and the average speed of the Xe(g) atoms that will occur as the mixture approaches thermal equilibrium? Average Speed of Xe Atoms Average Kinetic Energy of Xe Atoms
(A) Will increase Will increase
(B) Will increase Will decrease
(C) Will decrease Will increase
(D) Will decrease Will decrease

option D.

Both the average kinetic energy and the average speed of the Xe(g) atoms that will occur as the mixture approaches thermal equilibrium will decrease.

Explanation:

Kinetic Energy is simply said to be the energy of motion, that is when an object moved from one position to another the energy that it gains is called the kinetic energy. It is represented mathematically by 1/2mv². Where m is the mass of the object and v is the velocity of the object.

The average Kinetic energy is the average of all kinetic energies of a particle. The average Kinetic energy is represented mathematically by 3/2RT.  

The temperature of Xe which is at 50°C is greater that the temperature of He which is at  at 25°C will decrease when it reaches equilibrium after mixing. Hence,  the average kinetic energy and the average speed of the Xe(g) atoms that will occur as the mixture approaches thermal equilibrium will decrease.

THE GUILDING PRINCIPLE: We have that the average speed and KE decreases with temperature and vice versa for real gases.

answer: a chemical reaction

explanation:

im not wrong

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