# Question 46628

Sep 30, 2017

$\text{$532,000}$#### Explanation: The first thing that you should know here is that $\textcolor{b l u e}{\underline{\textcolor{b l a c k}{{\text{1 mL" = "1 cm}}^{3}}}}$This means that you can express the volume of the can as ${\text{355 mL" = "355 cm}}^{3}$Now, the density of platinum tells you the mass of exactly ${\text{1 cm}}^{3}$of platinum. In this case, you know that every ${\text{1 cm}}^{3}$of platinum has a mass of $\text{21.4 g}$. You can thus use the density of platinum as a conversion factor to determine the mass of platinum present in the sample. 355 color(red)(cancel(color(black)("cm"^3))) * overbrace("21.4 g"/(1color(red)(cancel(color(black)("cm"^3)))))^(color(blue)("the density of platinum")) = "7,597 g" Finally, you know that you can get $70.00 per gram of platinum, which means that the sample will have a value of

"7,597" color(red)(cancel(color(black)("g"))) * ($70.00)/(1color(red)(cancel(color(black)("g")))) = color(darkgreen)(ul(color(black)("$532,000")))#

The answer is rounded to three sig figs, the number of sig figs you have for the volume of the can.