A51.24-g sample of ba(oh)2 is dissolved in enough water to make 1.20 l of solution. how many milliliters of this solution must be diluted with water in order to make 1.00 l of 0.100 m ba(oh)2?

361.5 ml of water will required to make 1.00 L of 0.100 M .

Explanation:

Mass of barium hydroxide = 51.24 g

Volume of solution = 1.20 L

Molarity of the solution =

Let the volume of the solution 0.2766 M to be diluted be

Desired concentration of barium hydroxide solution =

Volume of the 0.1000 M solution =

361.5 ml of water will required to make 1.00 L of 0.100 M .

After 17190 years, Carbon-14 will reach a mass of 15 grams.

Pi is your initial amount (120 grams)
0.5 as your exponential base represents your HALF life hence 0.5
t represents the time elapsed
c represents your given half life value of (5730)
P(t) represents the exponential function of your half life

120/2 =60
60/2 = 30
30 /2 = 15

As you can see above, 3 half lives of 5730 years EACH degraded the carbon-14 to 15 grams. You divide each number by two as it is a HALF life.

P(t)= Pi (0.5) ^t/c
P(17190) = 120 (0.5) ^17190/5730
P(17190) = 120 (0.5) ^3
P(17190) = 15 grams

So your answers should be 3 half lives or after 17190 years have elapsed!