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Helen [10]
3 years ago
8

The density of gold is 19.3 g/cm3. what volume would 131.2 g of gold have?

Chemistry
1 answer:
Luda [366]3 years ago
5 0
Volume would be 6.79792764 cm³. Just divide 131.2 by 19.3

ρ=Mass/Volume
Volume= mass/density
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Qsp > Ksp, BaCO3 will precipitate

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The equation of the reaction is;

Na2CO3 + BaBr2 -------> 2NaBr + BaCO3

Since BaCO3 may form a precipitate we can determine the Qsp of the system.

Number of moles of Na2CO3 = 0.96g/106 g/mol = 9.1 * 10^-3 moles

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