A 20.0 g lead ball is heated in a Bunsen burner to 705oC. It is then dropped into a 500.0 g water bath. What is the initial temperature of the water if the final temperature is 35 oC? The Cp of lead is 0.13 J/goC. [Remember: CH2O = 4.18 J/g○C]

Solve this in the following order:
Part A: What is the energy lost of the metal? (round to the nearest WHOLE number, type the ABSOLUTE value of energy lost)
Part B: What is the water's temperature change? (round to ONE decimal place!)
Part C: What was the water's initial temperature? (final answer should include ONE decimal place!)

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.

Ansa leaf changing color in autumn;

Remember that rust prevented by placing a sacrificial electrode in electrical contact with the iron. the sacrificial electrode must be composed of a metal that  oxidizes  more easily than iron (that is, it must be below iron in table 18.1).