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a_sh-v [17]
3 years ago
12

A chemist determined by measurements that 0.030 moles of barium participated in a chemical reaction. Calculate the mass of bariu

m that participated in the chemical reaction. Round your answer to 2 significant digits.
Chemistry
1 answer:
schepotkina [342]3 years ago
3 0

Answer: 4.1 g of barium precipitated.

Explanation:

According to avogadro's law, 1 mole of every substance occupies 22.4 L at STP and contains avogadro's number 6.023\times 10^{23} of particles.

To calculate the moles, we use the equation:

\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}

Given : moles of barium = 0.030

Molar mass of barium = 137 g/mol

0.030=\frac{x}{137}

x= 4.1 g

Thus there are 4.1 g of barium that precipitated.

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