Aswimming pool is 25.0 ft. long, 18.5 ft. wide, and 9.0 ft. deep. when filled, the water level is 7.0 inches from the top. disinfectant instructions say to add 18 mg of cl per liter of water. how many pounds of chlorine are added to the pool?

1.9841256 kg

Explanation:

Given;

Length of the swimming pool = 25.0 ft = 7.62 m   ( 1 ft = 0.3048 m )

Width of the swimming pool = 18.5 ft =  5.64 m

Depth of the pool = 9.0 ft =

Total depth of the water in the pool when filled = 9 ft - 7 inches = 2.56 m

now,

Volume of the water in the pool = Length × Width × Depth

or

Volume of the water in the pool = 7.62 × 5.64 × 2.56 = 110.2292 m³

also,

1 m³ = 1000 L

thus,

110.2292 m³ = 110229.2 L

also it is given that 18 mg of Cl is added to 1 liter of water

therefore,

In 110229.2 L of water Cl added will be = 110229.2 × 18 = 1984125.6 mg

or

= 1.9841256 kg

the first, second, and fifth options

ok

explanation: