Respuesta :
Answer:
First problem: 89.07 grams
Second problem: 3910.15 days
Step-by-step explanation:
We can solve these problems with the exponencial equation:
P = Po * (1+r)^t
Where P is the final value, Po is the inicial value, r is the rate and t is the time.
In the first problem, the half-life is 119.77 days and Po = 100, so we have that:
P = 100 * (1-0.5)^(t/119.77)
So after 20 days, we will have:
P = 100 * (1-0.5)^(20/119.77) = 89.07 grams
For the second problem, we have the half-life of 5730 days, and the final value over the inicial value is 62.31%, so:
P/Po = (1-0.5)^(t/5730)
0.6231 = 0.5^(t/5730)
log(0.6231) = log(0.5^(t/5730))
-0.4730 = (t/5730)*log(0.5)
-0.4730 = (t/5730)*(-0.6931)
(t/5730) = 0.6824
t = 0.6824 * 5730 = 3910.15 days
6) The amount of substance left after 20 minutes is; A = 89.07 mg
7) The last time of the volcanic eruption is; 3910.11 years.
6) The formula for radioactive decay is;
A = A_o*2^(-t/h)
Where;
A = Amount remaining after time t
A_o = Initial amount
t = time of decay
h = half-life of substance
We are given;
A_o = 100 mg
h = 119.77 days
t = 20 days
Thus;
A = 100 × 2^(-20/119.77)
A = 89.07 mg
7) We are given;
A/A_o = 62.31% = 0.6231
h = 5730 years
Thus;
A/A_o = 0.6231 = 2^(-t/5730)
Thus;
Log 0.6231 = -(t/5730) log 2
t = (−0.2054/-0.301) × 5730
t ≈ 3910.11 years
Read more at; https://brainly.com/question/4387253
