A calibration plot of absorbance vs. concentration (ppm) was obtained with standard known Red 40 dye solutions. The slope of the best-fit straight line of the plot is 0.038 ppm- 1 The absorbance of the dilute unknown sports drink was 0.64 What is the concentration of this dilute unknown sports drink ?

16.84 ppm is the concentration of this dilute unknown sports drink.

Explanation:

Using Beer-Lambert's law :

Formula used :

where,

A = absorbance of solution

C = concentration of solution

l = length of the cell

= molar absorptivity of this solution

The graph between A and C gives straight line with positive slope value. the value of slope corresponds to

Slope =

According to question we have;

The absorbance of the dilute unknown sports drink:

A = 0.64

Slope of the graph =

The concentration of the dilute unknown sports drink = C

16.84 ppm is the concentration of this dilute unknown sports drink.

answer: your mm

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