Question

A chemist adds 435.0mL of a 2.28 M zinc nitrate ZnNO32 solution to a reaction flask. Calculate the millimoles of zinc nitrate the chemist has added to the flask. Round your answer to 3 significant digits.

Answer 1

He added 992 milimoles of zinc nitrate (Zn(NO3)2)

Step-by-step explanation:

Step 1: Data given

Volume of zinc nitrate = 435.0 mL =0.435 L

Molarity of zinc nitrate = 2.28 M

Step 2: Calculate moles zinc nitrate

Moles = molarity* volume

Moles Zn(NO3)2 = 2.28 M * 0.435 L

Moles Zn(NO3)2 = 0.9918 moles

Step 3: Convert moles to milimoles

Moles Zn(NO3)2 = 0.9918 moles

Moles Zn(NO3)2 = 0.9918 * 10^3 milimoles

Moles Zn(NO3)2 = 991.8 milimoles ≈ 992 milimoles

He added 992 milimoles of zinc nitrate (Zn(NO3)2)