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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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Such reactants determines the extent of chemical reaction.

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Explanation:

The kinetic theory of gases has two significant law which forms the backdrop of motion of gases. They are Charle's law and Boyle's law. As per Charle's law, the volume of any gas molecule at constant pressure is directly proportional to the temperature of the molecule.

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