Answer : The number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
Solution : Given,
Volume of solution = 500 ml
Molarity of KOH solution = 0.189 M
Molar mass of KOH = 56 g/mole
Formula used :
Now put all the given values in this formula, we get the mass of solute KOH.
Therefore, the number of grams of solute in 500.0 mL of 0.189 M KOH is, 5.292 grams
The question is incomplete, here is the complete question.
If we start with 1.000 g of strontium-90, 0.805 g will remain after 9.00 yr. This means that the half-life of strontium-90 is ________ yr.
a. 28.8 b. 30.9 c. 35.4 d. 32.2
Answer : The half-life of strontium-90 is 28.8 years.
Explanation :
This is a type of radioactive decay and all radioactive decays follow first order kinetics.
First we have to calculate the rate constant.
Expression for rate law for first order kinetics is given by :
where,
k = rate constant
t = time taken for decay process = 9.00 year
a = initial amount or moles of the reactant = 1.000 g
a - x = amount or moles left after decay process = 0.805 g
Putting values in above equation, we get:
To calculate the half-life, we use the formula :
Therefore, the half-life of strontium-90 is 28.8 years.