Sean plated an unknown metal onto his silver earring which initially weighed 1.41 g. He constructs an electrolytic cell using his ring as one of the electrodes. After running the cell, 0.022 moles of the unknown metal was plated onto his ring and the mass of the ring increased to 3.952 g. What is the atomic weight of the unknown metal in g/mol

115 g/mol

Explanation:

Step 1: Given data

Initial mass of the earring (mi): 1.41 gFinal mass of the earring (mf): 3.952 gMoles of the unknown metal deposited (n): 0.022 mol

Step 2: Calculate the mass of the unknown metal deposited

An earring is plated in an electrolytic cell. The mass of the unknown metal deposited can be calculated using the following expression.

m = mf - mi

m = 3.953 g - 1.41 g

m = 2.54 g

Step 3: Calculate the molar mass (M) of the unknown metal

0.022 moles of the metal have a mass of 2.54 g. The molar mass of the metal is:

M = m/n

M = 2.54 g / 0.022 mol

M = 115 g/mol

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.

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