19 20 21 22 23 24 25 26 27 28 29 Chlorine is used to disinfect swimming pools. The accepted concentration for this purpose is 1 ppm chlorine or 1 g of chlorine per million g of water. Calculate the volume of a chlorine solution (in milliliters) a homeowner should add to their swimming pool to achieve 1.00 ppm chlorine if the solution contains 6.00 percent chlorine by mass and there are \( 4.64 \times 10^{4} \) gallons of water in the pool. Assume that the density of the chlorine solution is the same as that of water, \( 1.00 \frac{\mathrm{~g}}{\mathrm{~mL}} \). Round your answer to 3 significant digits. Note: Reference the Conversion factors for non-Sr units table for additional information. \( \square \) mL. \( \square \) \( \times 10 \)

To calculate the volume of chlorine solution needed to achieve 1 ppm in a pool containing \( 4.64 \times 10^{4} \) gallons of water, we start by finding the total mass of the water. 1 gallon of water weighs approximately 3,785 grams. Therefore, the mass of water in the pool is: \[ 4.64 \times 10^{4} \text{ gallons} \times 3,785 \text{ g/gallon} = 1.754 \times 10^{8} \text{ g} \] To achieve a concentration of 1 ppm, the mass of chlorine required is: \[ 1 \text{ ppm} = \frac{1 \text{ g chlorine}}{1 \text{ million g water}} \times 1.754 \times 10^{8} \text{ g water} = 0.1754 \text{ g chlorine} \] Next, we need to determine how much of the 6.00% chlorine solution is needed to provide 0.1754 g of chlorine. Since the solution is 6.00% chlorine by mass, we can say: \[ 0.06 \times m = 0.1754 \text{ g} \quad \Rightarrow \quad m = \frac{0.1754 \text{ g}}{0.06} \approx 2.9233 \text{ g} \] Since the density of the chlorine solution is \( 1.00 \text{ g/mL} \), this mass translates directly to volume in mL: \[ V = 2.9233 \text{ mL} \] Rounding this to three significant figures gives us: \[ \approx 2.92 \text{ mL} \] Inputting this into the required format: 2.92 mL. 2.92 × 10^0