A balloon filled with helium gas at 9.90 atm occupies 94.9 L. What volume would the balloon occupy in the upper atmosphere, at a pressure of 0.650 atm and a constant temperature? Be sure your answer has the correct number of significant figures.

To determine the volume of the balloon in the upper atmosphere using the initial and final pressure and volume, we can apply Boyle's Law, which states that the product of pressure and volume is constant for a given amount of gas at constant temperature. This can be expressed mathematically as:

\[
P_1 V_1 = P_2 V_2
\]

Where:
- \( P_1 \) = initial pressure (9.90 atm)
- \( V_1 \) = initial volume (94.9 L)
- \( P_2 \) = final pressure (0.650 atm)
- \( V_2 \) = final volume (unknown)

Rearranging the equation to solve for \( V_2 \), we get:

\[
V_2 = \frac{P_1 V_1}{P_2}
\]

Now substituting the known values into the equation:

\[
V_2 = \frac{(9.90 \, \text{atm}) \times (94.9 \, \text{L})}{0.650 \, \text{atm}}
\]

Calculating:

\[
V_2 = \frac{939.51}{0.650} \approx 1445.01 \, \text{L}
\]

Now, we must ensure that the answer is reported with the correct number of significant figures. The initial pressure (9.90 atm) has three significant figures, the volume (94.9 L) has three significant figures, and the final pressure (0.650 atm) has three significant figures. Therefore, the final result should also be expressed with three significant figures:

\[
V_2 \approx 1440 \, \text{L}
\]

Thus, the volume that the balloon would occupy in the upper atmosphere at 0.650 atm, at a constant temperature, is approximately **1440 L**.