Question

a scientist recorded the amount of substance y present in an ongoing chemical reaction. this table gives the amount of substance y over the first 25 minutes. the data can be modeled using an exponential function.

time (min)	5	10	15	20	25

based on the data, which measurement is closest to the amount of substance y that will be present after 60 minutes?

• 0 g
• 1.7 g
• 2.3 g
• 5.2 g

Explanation:

Step1: Define exponential decay model

The general exponential decay function is $Y(t) = Y_0 e^{kt}$, where $Y(t)$ is the amount at time $t$, $Y_0$ is the initial amount, $k$ is the decay rate, and $t$ is time.

Step2: Solve for decay rate $k$

Use $t=5$, $Y(5)=38.7$ and $t=10$, $Y(10)=30.0$.
First, set up the ratio:
$\frac{Y(10)}{Y(5)} = \frac{Y_0 e^{10k}}{Y_0 e^{5k}} = e^{5k}$
Substitute values:
$e^{5k} = \frac{30.0}{38.7} \approx 0.7752$
Take natural log of both sides:
$5k = \ln(0.7752) \approx -0.254$
Solve for $k$:
$k = \frac{-0.254}{5} \approx -0.0508$

Step3: Solve for initial amount $Y_0$

Use $t=5$, $Y(5)=38.7$ and $k\approx-0.0508$:
$38.7 = Y_0 e^{-0.0508 \times 5}$
$38.7 = Y_0 e^{-0.254}$
$Y_0 = \frac{38.7}{e^{-0.254}} \approx \frac{38.7}{0.7752} \approx 50$

Step4: Calculate $Y(60)$

Substitute $t=60$, $Y_0\approx50$, $k\approx-0.0508$ into the model:
$Y(60) = 50 e^{-0.0508 \times 60}$
$Y(60) = 50 e^{-3.048}$
$e^{-3.048} \approx 0.047$
$Y(60) \approx 50 \times 0.047 = 2.35$

Answer:

2.3 g