To solve this problem, we need to find the amount of copper ions in the 0.1 M CuSO solution after the electrolysis process.First, let's find the number of moles of copper in the 35-mg copper wire:Copper has a molar mass of 63.5 g/mol.35 mg = 0.035 gNumber of moles of copper = mass / molar massNumber of moles of copper = 0.035 g / 63.5 g/mol = 5.51 x 10^-4 molDuring the electrolysis process, the copper wire is oxidized, and the copper ions Cu are released into the solution. The number of moles of Cu ions released is equal to the number of moles of copper in the wire.Now, let's find the charge passed during the electrolysis process:Charge Q = Current I x Time t Q = 0.15 A x 10 min x 60 s/min = 90 CNext, we need to find the number of moles of electrons transferred during the electrolysis process:1 mole of electrons = 96,485 C/mol Faraday's constant Number of moles of electrons = Charge / Faraday's constantNumber of moles of electrons = 90 C / 96,485 C/mol = 9.33 x 10^-4 molSince the copper ions have a charge of +2 Cu , the number of moles of Cu ions released is half the number of moles of electrons transferred:Number of moles of Cu ions released = 9.33 x 10^-4 mol / 2 = 4.665 x 10^-4 molNow, let's find the total number of moles of Cu ions in the 0.1 M CuSO solution after the electrolysis process:Initial moles of Cu ions in the solution = 0.1 M assuming 1 L of solution Initial moles of Cu ions = 0.1 molTotal moles of Cu ions after electrolysis = Initial moles + Moles released from the wireTotal moles of Cu ions = 0.1 mol + 4.665 x 10^-4 mol = 0.1004665 molTherefore, the amount of copper ions in the 0.1 M CuSO solution after the electrolysis process is 0.1004665 mol.