If you had a 0.800 l solution containing 0.0240 m of fe3+(aq), and you wished to add enough 1.38 m naoh(aq) to precipitate all of the metal, what is the minimum amount of the naoh(aq) solution you would need to add? assume that the naoh(aq) solution is the only source of oh−(aq) for the precipitation.

0.041 L = 41.3 mL

Explanation:

This problem we will solve by considering the stoichiometry of the reaction and the definition of molarity.

Number of moles in .800 L solution:

0.800 L x 0.0240 M = 0.800 L x .0240 mol/L = 0.0192  mol Fe³⁺

to form the precipitate Fe(OH)₃ we will need 3 times .0192

mol NaOH required = 0.057

given the concentration of 1.38 mol M NaOH we can calculate how many milliliters of NaOH will contain 0.057 mol:

1.L/1.38 mol NaOH x 0.057 mol NaOH = 0.041 L

0.041 L x 1000 mL/1L = 41.3 mL

it should have an internal chamber and two thermometers sticking out of the top. the internal chamber will also be closed off from the outside.

the correct answer would be b) volume versus temperature, because they are also directly proportional.