Answer:
84
Step-by-step explanation:
First, there are 6 pages of work. then multiply 6 by the number of questions (14)
6x14=84
We find the value of N₀ since we are provided with initial conditions.
The condition is that, at time t = 0, the amount of substance contains originally 10 grams.
We substitute:
10 = N₀ (e^(-0.1356)*0)
10 = N₀ (e^0)
N₀ = 10
When the substance is in half-life (meaning, the half of the original amount), it contains 5 grams. We solve t in this case.
5 = 10 e^(-0.1356*t)
0.5 = e^(-0.1356*t)
Multiply natural logarithms on both sides to bring down t.
ln(0.5) = -0.1356*t
Hence,
t = -(ln(0.5))/0.1356
t ≈ 5.11 days (ANSWER)
Half-life = [elapsed time * log (2)] / [log (begng amt / endg amt)]
half-life = (3 days * 0.30102999566) / log (1 / .58)
<span>half-life = 0.903089987 / log </span>
(<span>
<span>
<span>
1.724137931)
</span></span></span>half-life = <span>0.903089987 / 0.23657200643</span>
half-life =
<span>
<span>
<span>
3.8174 days
Source:
http://www.1728.org/halflife.htm
</span></span></span>