1/8=(1/2)^3 and the half life of radon is 3.8days
Half life is the time it takes for half of any amount of a radioactive substance to decay into something else.
Therefore for a sample of radon to decay to 1/8 of its original amount, it would take 3 x 3.8days=11.4 days
Answer: 1. 
moles
2. 90 mg
Explanation:
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 
moles of ozone is removed by =
moles of sodium iodide.
Thus moles of sodium iodide are needed to remove 
moles of 
2. 
According to stoichiometry:
1 mole of ozone is removed by 2 moles of sodium iodide.
Thus 0.0003 moles of ozone is removed by =
moles of sodium iodide.
Mass of sodium iodide= 
(1g=1000mg)
Thus 90 mg of sodium iodide are needed to remove 13.31 mg of .
Answer:
1.146 x 10⁴ year.
Explanation:
- The decay of carbon-14 is a first order reaction.
- The rate constant of the reaction (k) in a first order reaction = ln (2)/half-life = 0.693/(5730 year) = 1.21 x 10⁻⁴ year⁻¹.
- The integration law of a first order reaction is:
<em>kt = ln [A₀]/[A]</em>
<em></em>
k is the rate constant = 1.21 x 10⁻⁴ year⁻¹.
t is the time = ??? years.
[A₀] is the initial percentage of carbon-14 = 100.0 %.
[A] is the remaining percentage of carbon-14 = 1/4[A₀] = 25.0 %.
∵ kt = ln [Ao]/[A]
∴ (1.21 x 10⁻⁴ year⁻¹)(t) = ln (100.0%)/[25.0 %]
(1.21 x 10⁻⁴ year⁻¹)(t) = 1.386.
∴ <em>t </em>= 1.386/
(1.21 x 10⁻⁴ year⁻¹) = <em>1.146 x 10⁴ year.</em>
