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What is the amount of copper ions present in a sample if 5.25 grams of the sample produces a current of 2.35 A for 2 hours during coulometry analysis? The molar mass of copper is 63.55 g/mol and the sample contains only copper ions.

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Coulometry analysis is based on Faraday's laws of electrolysis, which states that the amount of substance deposited or dissolved at an electrode is directly proportional to the amount of charge passed through the solution. The relationship between the amount of substance and the charge can be expressed as:amount of substance  mol  = charge  Coulombs  /  n * Faraday constant where n is the number of electrons involved in the redox reaction and Faraday constant  F  is approximately 96485 C/mol.For copper ions  Cu , the redox reaction involves the transfer of 2 electrons  n = 2  as shown below:Cu + 2e  CuFirst, we need to calculate the total charge passed through the solution. We are given the current  I  and the time  t  for which the current was applied:I = 2.35 At = 2 hours = 2 * 3600 s = 7200 sCharge  Q  can be calculated using the formula:Q = I * tQ = 2.35 A * 7200 s = 16920 CNow we can calculate the amount of copper ions  in moles  using Faraday's law:amount of Cu  mol  = Q /  n * F amount of Cu  mol  = 16920 C /  2 * 96485 C/mol  = 0.0876 molNow, we can calculate the mass of copper ions in the sample:mass of Cu = amount of Cu  mol  * molar mass of Cumass of Cu = 0.0876 mol * 63.55 g/mol = 5.57 gSo, the amount of copper ions present in the sample is 5.57 grams.

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