10.0 ml of a 0.100 mol l–1 solution of a metal ion m2+ is mixed with 10.0 ml of a 0.100 mol l–1 solution of a substance l. the following equilibrium is established:
m2+(aq) + 2l(aq) picture ml22+(aq)
at equilibrium the concentration of l is found to be 0.0100 mol l–1. what is the equilibrium concentration of ml22+, in mol l–1?
someone me

The concentration of at equilibrium is 0.045 M.

Explanation:

We are given:

For the given chemical equation:

Initially:             0.100M        0.100M

At eqllm:          0.100 - x      0.100 - 2x          x

We are also given:

Equating the two values:

Hence, the concentration of at equilibrium is 0.045 M.

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

- decrease the number of plants and animals living in the water.