Answer:
5.3 × 10⁻³ kg
Explanation:
There is some info missing. I think this is the original question.
A chemist adds 135.0 mL of a 0.21 M zinc nitrate (Zn(NO₃)₂) solution to a reaction flask. Calculate the mass in kilograms of zinc nitrate the chemist has added to the flask. Be sure your answer has the correct number of significant digits.
We have 135.0 mL of a 0.21 M zinc nitrate (Zn(NO₃)₂) solution. The moles of zinc nitrate are:
0.1350 L × 0.21 mol/L = 2.8 × 10⁻² mol
The molar mass of zinc nitrate is 189.36 g/mol. The mass corresponding to 2.8 × 10⁻² moles is:
2.8 × 10⁻² mol × 189.36 g/mol = 5.3 g
1 kilogram is equal to 1000 grams. Then,
5.3 g × (1 kg/1000 g) = 5.3 × 10⁻³ kg
