To solve this problem, we need to use the formula for exponential decay, which is:

N = N0 * e^(-λt)

where:
N is the final amount of the substance,
N0 is the initial amount of the substance,
λ is the decay constant,
t is the time elapsed.

We are given that N = 800 μg, N0 = 1.30 mg = 1300 μg, and t = 7.57 s. We can plug these values into the formula and solve for λ:

800 = 1300 * e^(-λ * 7.57)

Divide both sides by 1300:

0.6154 = e^(-λ * 7.57)

Take the natural logarithm of both sides:

ln(0.6154) = -λ * 7.57

Solve for λ:

λ = -ln(0.6154) / 7.57 ≈ 0.071 s^-1

The half-life (T) of a substance is related to the decay constant by the formula:

T = ln(2) / λ

Plug in the value of λ we just found:

T = ln(2) / 0.071 ≈ 9.77 s

So, the half-life of the substance is approximately 9.77 seconds, rounded to two significant digits.

Question

To solve this problem, we need to use the formula for exponential decay, which is:

N = N0 * e^(-λt)

where:
N is the final amount of the substance,
N0 is the initial amount of the substance,
λ is the decay constant,
t is the time elapsed.

We are given that N = 800 μg, N0 = 1.30 mg = 1300 μg, and t = 7.57 s. We can plug these values into the formula and solve for λ:

800 = 1300 * e^(-λ * 7.57)

Divide both sides by 1300:

0.6154 = e^(-λ * 7.57)

Take the natural logarithm of both sides:

ln(0.6154) = -λ * 7.57

Solve for λ:

λ = -ln(0.6154) / 7.57 ≈ 0.071 s^-1

The half-life (T) of a substance is related to the decay constant by the formula:

T = ln(2) / λ

Plug in the value of λ we just found:

T = ln(2) / 0.071 ≈ 9.77 s

So, the half-life of the substance is approximately 9.77 seconds, rounded to two significant digits.

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