Answer to Question #153930 in Inorganic Chemistry for Morayo

Inorganic Chemistry

Question #153930

What is the amount of gold present in 15.5g of a pure gold wedding ring.(Au=197)

Expert's answer

Solution:

$\nu=\frac{m} {A} =\frac{15.5 g} {197 g/mol} =78.68\times10^{-3}mol$

Answer: $78.68\times10^{-3}mol$

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