To solve the problem, we substitute $t = 140$ into the formula:

$y = 24 \left(\frac{1}{2}\right)^{\frac{t}{45}}$

Substituting $t = 140$:

$y = 24 \left(\frac{1}{2}\right)^{\frac{140}{45}}$

Simplify $\frac{140}{45}$:

$\frac{140}{45} \approx 3.1111$

Thus: $y = 24 \left(\frac{1}{2}\right)^{3.1111}$

Calculate $\left(\frac{1}{2}\right)^{3.1111}$:

$\left(\frac{1}{2}\right)^{3.1111} \approx 0.116$

Now multiply by 24: $y \approx 24 \times 0.116 = 2.784 \, \text{grams.}$

Final Answer:

After 140 years, approximately 2.784 grams of the radioactive element remains.

Would you like a step-by-step explanation or details on how the exponential decay works?

Here are 5 follow-up questions you might explore:

1. What happens to the mass after two half-lives?
2. How does the half-life affect the decay rate?
3. Can this formula be used for other substances with different half-lives?
4. How would the graph of this equation look over 200 years?
5. How does the initial amount affect the decay process?

Tip: Always double-check your exponentiation when dealing with exponential decay problems. Misplacing decimals can lead to significant errors.