Le chatelier's principle describes the effect of changing conditions on a chemical reaction that was previously at equilibrium. choose the statement that correctly describes the possible effects. a. adding more carbon dioxide to a combustion reaction will increase the rate of the reaction. b.when oxygen is removed from a combustion reaction, the reaction slows down or stops. c.when water is added to a combustion reaction, it will increase the rate of the reaction.

.B.When oxygen is removed from a combustion reaction, the reaction slows down or stops.

Explanation:

Le Chatelier's principle -

The direction of the reaction by changing the concentration can be determined by Le Chatelier's principle,

It states that ,

When a reaction is at equilibrium , Changing the concentration , pressure,  temperature disturbs the equilibrium , and the reaction again tries to attain equilibrium by counteracting the change.

The combustion reaction of carbon dioxide is as follows -

C + O₂ → CO₂

Hence ,

Removing O₂ from the system , i.e. decreasing the concentration of O₂ ,  according to Le Chatelier , the reaction will move in backward direction , to increase the amount of reduced O₂ .

Hence, reaction will go in backward direction.

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