Respuesta :
Answer: The mass of chlorine needed by the plant per day is [tex]9.4625\times 10^8mg[/tex]
Explanation:
We are given:
Volume o water treated per day = 25,000,000 gallons
Converting this volume from gallons to liters, we use the conversion factor:
1 gallon = 3.785 L
So, [tex]\frac{3.785L}{1\text{ gallon}}\times 25,000,000\text{ gallons}=9.4625\times 10^7L[/tex]
Amount of chlorine applied for disinfection = 10 mg/L
Applying unitary method:
For 1 L of water, the amount of chlorine applied is 10 mg
So, for [tex]9.4625\times 10^7L[/tex] of water, the amount of chlorine applied will be [tex]\frac{10mg}{1L\times 9.4625\times 10^7L}=9.4625\times 10^8mg[/tex]
Hence, the mass of chlorine needed by the plant per day is [tex]9.4625\times 10^8mg[/tex]
