An iron bar is dipped in 250 mL of 0.1mol/L copper (II) sulfate solution. After some time, the mass, m, of the obtained copper deposit is 0.6350 g.1) What spontaneous reaction occurs? 2) Determine the final composition of the solution, in terms of amount and concentration.Molar mass: M(Cu) 63.5 g/mol Somebody Please Explain how to do this question in details!
do you use fertilizer with minerals like iron? if not try a steady diet, plants need nourishment to thrive, not just super thrive. the company claims super thrive is hormones. I would lay off on that and feed the plant.
Fe(s) + Cu+2(aq) ---------------- Cu(s) + Fe+2(aq) The solution contained 0.250 L x 0.1 mol/L or 0.0250 mol of Cu+2 ions The mass of that much Cu+2 is mol x Molar Mass 0.0250 mol x 63.5 g/mol 1.5875 g You took out of solution 0.6350 g so that leaves 1.5875 g - 0.6350 g 0.9525 g of Cu+2 ions left 0.9525 g of Cu is equal to 0.9525 g / 63.5 g/mol 0.015 mol The concentration of CuSO4 now is 0.015 mol / 0.25 L 0.06 mol/L
Overwatering after I left it for dead. miraclegro instead of superthrive.
