How long would it take for 1.50 mol of water at 100.0 ? c to be converted completely into steam if heat were added at a constant rate of 25.0 j/s ?

The molar latent heat of vaporization of a substance is the amount of energy necessary to transform one mole of a substance from a liquid to a gas at constant temperature. In the case of water, this is 40,650 J/mol. Assuming no energy is lost, the total amount of energy required will be:

Q = 1.5 * 40,650
Q = 60,975 J

The time taken will be:
60,975/25 s

The vaporization will take 2,439 seconds.

Answerthis solution will not form a precipitateexplanationsolubility equilbrium:         pbf₂(s)  ⇄ pb²⁺(aq) + 2 f⁻(aq)the solubility product expression is    (we exclude solids from the equilibrium expression).now suppose we have the trial ion product q, which we can define as    then, we can interpret  q as the product of the concentration of the ions that we have and ksp as the product of the ion concentrations that are needed to form a saturated solution.if q > ksp, then that implies that there will be a surplus of ions that exceed the amount that is required to form a saturated solution. a precipitate will then form (the precipitate continually  forms until the surplus ions are removed and q is decreased to ksp's value).if q = ksp, the saturated solution that forms is barely saturated and the precipitate will be minimal.if q < ksp, then the ion concentration will not be enough to form a saturated solution (so no precipitate).using our information, we calculate the trial ion product:     since q = 2.7  × 10⁻⁸ < ksp =  4  ×  10⁻⁸, a precipitate will not form.

Ibelieve the answer is b.2200 watts, because p=vi; p=220(10); then p=2200hope my answer has you : )