A chemist adds 490.0 mL of a 4.65•10^-5 mol/L zinc oxalate (ZnC2O4) solution to a reaction flask. Calculate the mass in milligrams of zinc oxalate the chemist has added to the flask. Round your answer to three significant digits

3.50 mg ZnC2O4 . -three significant figures

Explanation:

moles ZnC2O4 = 4.65 x 10^-5 moles/L x .490 L = 2.2785 x 10^-5 moles

mass ZnC2O4 = moles ZnC2O4 x MM ZnC2O4 = 2.2785 x 10^-5 x 153.41 = 349.545 x 10^-5 = 3.5 x 10^-3 g = 3.50 mg ZnC2O4

answer; i believe the correct answer is (solid gold);