"How many cubic centimeters of ore containing 0.22%..." was answered by our community

Elleanor Mckenzie

Answered question

2021-02-13

How many cubic centimeters of ore containing 0.22% by massgold must be processed to obtain $100.00 worth of gold? The densityof the ore is 8.0

\frac{g}{c} m^{3}

and the price of gold is $418.00 per troyounce. (14.6 troy ounce = 1 pound; 1bpound = 454g)

opsadnojD

Skilled2021-02-14Added 95 answers

Calculate the volume which contains 0.22g mass of gold
volume(

c m^{3}

) = mass(g) /density(g/cm)

= \frac{0.22 g}{8.0 \frac{g}{c} m^{3}}

= 0.0275 c m

calculte the mass of gold which cost $100.00

= \frac{1 troy ounce}{$ 418.00} \cdot \frac{454 g}{14.6 troy ounce} \cdot $ 100.00

=108.6 g
For 0.22g gold it requires

0.0275 c m^{3} = \frac{108.6 g \cdot 0.0275 c m^{3}}{0.22 g}

= 13.57 c m^{3}

is the volume required

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