Question 5 of 51 Page 5 of 5 Question 5 (2 points) A pateontologist found a fossil on a recent dig. The fossil contained \( 35 \% \) of the original amount of one of its chemical substance. This substance has a haif-life of 2450 years, A.) The value of the decay constant, \( k \), is \( \square \) (round to 4 deeimal places) B.) The fossil is about \( \qquad \) years old, Blank 1: \( \square \) Btank 2: \( \square \)

To calculate the decay constant \( k \) and the age of the fossil, we can use the exponential decay formula: \[ N(t) = N_0 e^{-kt} \] Where: - \( N(t) \) is the amount of substance remaining, - \( N_0 \) is the original amount of the substance, - \( k \) is the decay constant, - \( t \) is time in years. Given that the half-life is 2450 years, we can find the decay constant \( k \) using the formula: \[ k = \frac{\ln(2)}{t_{1/2}} \] Substituting the half-life: \[ k = \frac{0.6931}{2450} \approx 0.0002835 \] So, Blank 1: \( 0.0003 \). For the age of the fossil, we know that the fossil contains 35% of the original substance: \[ \frac{N(t)}{N_0} = 0.35 \] Set up the decay equation: \[ 0.35 = e^{-kt} \] Taking the natural logarithm: \[ \ln(0.35) = -kt \] Substituting for \( k \): \[ t = -\frac{\ln(0.35)}{0.0002835} \approx 1,377.21 \text{ years} \] So, Blank 2: \( 1377.21 \). So, the answers are: Blank 1: \( 0.0003 \) Blank 2: \( 1377.21 \)