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A chemist determined by measurements that 0.030 moles of barium participated in a chemical reaction. Calculate the mass of barium that participated in the chemical reaction. Round your answer to 2 significant digits.

Sagot :

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 [tex]6.023\times 10^{23}[/tex] of particles.

To calculate the moles, we use the equation:

[tex]\text{Number of moles}=\frac{\text{Given mass}}{\text {Molar Mass}}[/tex]

Given : moles of barium = 0.030

Molar mass of barium = 137 g/mol

[tex]0.030=\frac{x}{137}[/tex]

x= 4.1 g

Thus there are 4.1 g of barium that precipitated.

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