A certain substance in a laboratory is accidentally exposed to a reagent and starts to decay. However, this goes on for some time before it is noticed. Once the lab scientists notice what is happening, they decide to measure the remainder of this phenomenon until the substance has completely decayed. 

**Hours**

**Mass of**
**substance (g)**

4

34.5

5

21.7

6

13.7

7

8.6

8

5.4

This type of decay is known to follow an exponential curve. About how many grams of the substance were initially exposed to the reagent?
​​​​

To solve this problem, we need to model the decay using an exponential function. An exponential decay model can be described by the equation:

\[
M(t) = M_0 e^{-kt}
\]

where:
- \( M(t) \) is the mass of the substance at time \( t \).
- \( M_0 \) is the initial mass of the substance (what we are trying to find).
- \( k \) is the decay constant.
- \( t \) is the time in hours.

From the given data, we have:

| Hours (\( t \)) | Mass (\( M(t) \)) |
|-----------------|-------------------|
| 4               | 34.5              |
| 5               | 21.7              |
| 6               | 13.7              |
| 7               | 8.6               |
| 8               | 5.4               |

### Step 1: Establish two equations using the data points
We can use two data points to create a system of equations to solve for \( M_0 \) and \( k \). Let's use the data for \( t = 4 \) and \( t = 5 \).

1. At \( t = 4 \), \( M(4) = 34.5 \):
   \[
   34.5 = M_0 e^{-4k}
   \]
   
2. At \( t = 5 \), \( M(5) = 21.7 \):
   \[
   21.7 = M_0 e^{-5k}
   \]

### Step 2: Solve for \( k \)
Divide the second equation by the first to eliminate \( M_0 \):

\[
\frac{21.7}{34.5} = \frac{M_0 e^{-5k}}{M_0 e^{-4k}}
\]

\[
\frac{21.7}{34.5} = e^{-5k + 4k}
\]

\[
\frac{21.7}{34.5} = e^{-k}
\]

Now, solve for \( k \):

\[
e^{-k} = \frac{21.7}{34.5}
\]

\[
-k = \ln\left(\frac{21.7}{34.5}\right)
\]

\[
k = -\ln\left(\frac{21.7}{34.5}\right)
\]

Let's calculate \( k \):

\[
k \approx -\ln\left(\frac{21.7}{34.5}\right) \approx -\ln(0.629) \approx 0.462
\]

### Step 3: Solve for \( M_0 \)
Use the value of \( k \) and one of the original equations to find \( M_0 \). We'll use \( t = 4 \) and \( M(4) = 34.5 \):

\[
34.5 = M_0 e^{-4 \cdot 0.462}
\]

\[
34.5 = M_0 e^{-1.848}
\]

\[
34.5 = M_0 \times 0.157
\]

\[
M_0 = \frac{34.5}{0.157}
\]

\[
M_0 \approx 219.7
\]

### Conclusion
The initial mass of the substance, \( M_0 \), that was exposed to the reagent was approximately **219.7 grams**.

Would you like more details, or do you have any questions?

Here are 5 related questions you might consider:

1. How can we determine the half-life of the substance using the decay constant \( k \)?
2. What would the mass of the substance be after 10 hours?
3. How would the model change if the decay was not exponential but followed a different pattern, such as linear decay?
4. Can you determine the time at which only 1 gram of the substance remains?
5. How does the decay constant \( k \) affect the rate of decay?

**Tip:** When dealing with exponential growth or decay, always try to express the relationships in terms of natural logarithms (base \( e \)) to simplify the calculations.

Learn how to find the initial mass of a substance exposed to a reagent using an exponential decay model. Step-by-step solution using data points and exponential function analysis provided. Suitable for Grades 10-12.

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