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what is the percent yield a precipitate formed when 15.4 mL of 1.25 molar copper two...

what is the percent yield a precipitate formed when 15.4 mL of 1.25 molar copper two nitrate is mixed with 25.3 mL of 0.975 molar sodium hydroxide if the theoretical yield is 0.752 g

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Answer #1

Given in Volume of Cul NO₂ = 15.4 mL or 0.0154L. molarity of Cultos) = 1,25 m Hence number of moles of Cu(N/2= na molasiti xCu(10), + 2 MOOH - ) Cu(OH), t 2 MANO3 0.02466 Hven 0.01425 melh 0.01233 mole 0.02464 Quluneet 0.01233 0.0246 2.024.66 . o m

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