A chemist adds 45 ml of a M mercury (II) iodide solution to a reaction flask. Calculate the micromoles of mercury (II) iodide the chemist has added to the flask. Round your answer to significant digits.

4.5 * 10⁴ μmol

Explanation:

The molar concentration for mercury (II) iodide is missing. Let's suppose it is 1.0 M and then you can replace it with your value.

The chemist has 45 mL of a 1.0 M mercury (II) iodide solution. The moles of mercury (II) iodide are:

45 * 10⁻³ L * 1.0 mol/L = 4.5 * 10⁻² mol

1 mole is equal to 10⁶ micromoles. The micromoles corresponding to 4.5 * 10⁻² moles are:

4.5 * 10⁻² mol * (10⁶ μmol/1 mol) = 4.5 * 10⁴ μmol