a reaction takes place between two gases at 298 k. there are 2.0 moles of gas a and 3.0 moles of gas b. which scenario would  lower  the collision frequency between gases as well as the rate of reaction between gases a and b? a)increase the moles of gas b. b)lower the temperature of the reaction to 250 k. c)compress the gases into a smaller volume. d)introduce a catalyst to the container.

B)Lower the temperature of the reaction to 250 K.

Explanation:

Option B is correct. Lowering the temperature will decrease the kinetic energy of the gas molecules decreasing the frequency of collision and the rate of the reaction.

Option A is incorrect. Increasing the moles or the number of gas molecules of A and B will bring the gas molecules closer to each other, thus increasing the frequency of collision and the rate of the reaction.  

Option C is incorrect. Compressing the gas into a small volume will bring the gas molecules closer to each other, thus increasing the frequency of collision and the rate of the reaction.  

Option D is incorrect. A catalyst works by lowering the activation energy and thus increasing the frequency and orientation of the collision between gas molecules and thus increasing the rate of the reaction.  

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

away from the curb

explanation:

the wheels would then hit the curb if it started to rolling, decreasing the chance of an accident.