Sodium-25 was to be used in an experiment, but it took 3.0 minutes to get the sodium from the reactor to the laboratory. If 5.0 mg of sodium-25 was removed from the reactor, how many mg of sodium-25 were placed in the reaction vessel 3.0 minutes later, given that the half-life of sodium-25 is 60 seconds?

Let's solve the problem about the decay of sodium-25 together!

### Step-by-Step Solution

1. Understanding the Half-life:
- Sodium-25 has a half-life of 60 seconds. This means that every 60 seconds, half of the sodium-25 will have decayed.

2. Time Calculation:
- You need to find out how much sodium-25 remains after 3.0 minutes. First, we convert this time into seconds since the half-life is given in seconds:
[tex]\[
3.0 \text{ minutes} = 3.0 \times 60 = 180 \text{ seconds}
\][/tex]

3. Decay Formula:
- The decay of a radioactive isotope like sodium-25 can be calculated using the formula:
[tex]\[
N(t) = N_0 \times \left(\frac{1}{2}\right)^{\frac{t}{t_{1/2}}}
\][/tex]

- Where:
- [tex]\(N(t)\)[/tex] is the remaining amount after time [tex]\(t\)[/tex]
- [tex]\(N_0\)[/tex] is the initial amount (5.0 mg in this case)
- [tex]\(t\)[/tex] is the total time elapsed (180 seconds)
- [tex]\(t_{1/2}\)[/tex] is the half-life of the isotope (60 seconds)

4. Calculating Remaining Sodium-25:
- Substitute the known values into the formula:
[tex]\[
N(180) = 5.0 \times \left(\frac{1}{2}\right)^{\frac{180}{60}}
\][/tex]

- Now, calculate the exponent:
[tex]\[
\frac{180}{60} = 3
\][/tex]

- Therefore, the formula becomes:
[tex]\[
N(180) = 5.0 \times \left(\frac{1}{2}\right)^3 = 5.0 \times \frac{1}{8}
\][/tex]

- Calculating this gives:
[tex]\[
N(180) = 5.0 \times 0.125 = 0.625 \text{ mg}
\][/tex]

So, after 3.0 minutes, 0.625 mg of sodium-25 would remain and be placed in the reaction vessel.