What is the final concentration, in mg/mL, for a solution prepared by adding 12.00 mL of a 3.590 mg/mL Cu2+ stock solution into 100.00 mL volumetric flask and filled with water to the calibration mark?

0.4308 mg/mL

Explanation:

Given data

We can find the final concentration using the dilution rule.

C₁ × V₁ = C₂ × V₂

C₂ = C₁ × V₁ / V₂

C₂ = 3.590 mg/mL × 12.00 mL / 100.00 mL

C₂ = 0.4308 mg/mL

-be a crystalline solid

- be soluble in water

-conduct electricity

hope this : )

explanation:

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