Answer:
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
