Assume a quantity decreases by 6.7% in 8.6 hours.
What is the half-life of the substance? (answer should be accurate to two decimal places.)
let the equation be
amount = 1(1/2)^(t/k) , where k is the half-life, and t is number of hours
so (1 - .067) = 1(1/2)^(8.6/k)
logging it ...
log .933 = log (1/2)^(8.6/k)
8.6/k = log .933 / log .5
solve for k, your answer to half-life
To determine the half-life of a substance, we can use the formula:
Half-life = (ln(2) / k)
Where:
- Half-life is the time it takes for the quantity of a substance to decrease by half.
- ln(2) is the natural logarithm of 2, approximately 0.6931.
- k is the decay constant, which represents the rate of decay and can be calculated using the formula k = ln(1 - r), where r is the percentage decrease expressed as a decimal.
In this case, the quantity decreases by 6.7%, so we can calculate r:
r = 6.7% / 100% = 0.067.
Now we can calculate k:
k = ln(1 - 0.067) ≈ -0.0704.
Since the question asks for the answer in hours, we need to convert the decay constant to be in hours as well.
To convert a decay constant from seconds (or any other unit of time) to hours, we divide the constant by the number of seconds in an hour, which is 3600.
k (in hours) = -0.0704 / 3600 ≈ -1.95 × 10^(-5) hours^(-1).
Now we can calculate the half-life using the formula:
Half-life = ln(2) / k ≈ 0.6931 / (1.95 × 10^(-5)) ≈ 35505.13 hours.
Rounded to two decimal places, the half-life of the substance is approximately 35505.13 hours.