Question

Fluoride ion is poisonous in relatively low amounts: 0.2 g of F− per 70 kg of body weight can cause death. Nevertheless, in order to prevent tooth decay, F− ions are added to drinking water at a concentration of 1.00 mg of F− ion per L of water.

How many liters of fluoridated drinking water would a 70−kg person have to consume in one day to reach this toxic level?

How many kilograms of sodium fluoride would be needed to treat a 5.73 × 106−gal reservoir?

Answer 1

a) 200 L of water. b) 3,2210(fifth to the power) kg

Step-by-step explanation:

a) To reach toxic level of a person would have to drink 200 L of water. b) We need 3,2210 to the fifth power kg of fluride for 8,510 to the seventh power gal of water.