need chem answer pleaseee

Question

How many milligrams of calcium nitrate are needed to make 0.0629 L of a 2.02 M solution? Supply the answer in scientific notation.

Answer 1

Calcium nitrate = Ca(NO₃)₂

Molar mass = 164 g/mol

2.02 M = 2.02 mol/L

2.02 mol/L x 0.0629 L = 0.127 moles needed

Converting 0.127 mols Ca(NO₃)₂ to mg...

0.127 moles x 164 g/mol x 1000 mg/g = 20,838 mg = 20,800 mg (3 sig.figs.)

In scientific notation, this would be 2.08x10⁴ mg