To calculate the mass of barium that participated in the chemical reaction, you need to use the molar mass of barium. The molar mass of barium (Ba) is approximately 137.33 g/mol.
Given:
- Moles of barium (\( n \)) = 0.075 mol
- Molar mass of barium (\( M \)) = 137.33 g/mol
The mass (\( m \)) of barium can be calculated using the formula:
\[ m = n \times M \]
Now, substituting the values:
\[ m = 0.075 \, \text{mol} \times 137.33 \, \text{g/mol} \]
\[ m = 10.29975 \, \text{g} \]
Since the given number of moles (0.075) has two significant digits, the mass should also be reported with two significant digits. Hence, we round 10.29975 g to two significant digits:
\[ m \approx 10 \, \text{g} \]
Therefore, the mass of barium that participated in the chemical reaction is approximately **10 grams**.
A chemist determined by measurements that 0.075 moles of barium participated in a chemical reaction. Calculate the mass of barium that participated in the chemical reaction.
Be sure your answer has the correct number of significant digits.
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