The question is incomplete, here is the complete question:
A chemist determines by measurements that 0.0850 moles of fluorine gas participate in a chemical reaction. Calculate the mass of fluorine gas that participates. Be sure your answer has the correct number of significant digits.
Answer: The mass of fluorine gas that is precipitated is 3.23 grams
Explanation:
To calculate the number of moles, we use the equation:
[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text{Molar mass}}[/tex]
We are given:
Moles of fluorine gas = 0.0850 moles
Molar mass of fluorine gas = 38.0 g/mol
Putting values in above equation, we get:
[tex]0.0850mol=\frac{\text{Mass of fluorine gas}}{38.0g/mol}\\\\\text{Mass of fluorine gas}=(0.0850mol\times 38.0g/mol)=3.23g[/tex]
Hence, the mass of fluorine gas that is precipitated is 3.23 grams
