# Question #435fb

##### 1 Answer

Here's what I got.

#### Explanation:

The idea here is that you can express the **density** of a substance as the ratio between the **mass** of a given sample of said substance, let's say **volume** it occupies, let's say

#color(blue)(ul(color(black)(rho = m/V)))" " " "color(red)("(*)")#

Now, let's take

You know that

#m_"Au" + m_"Cu" = "668 g"#

If you take

#rho = m/V implies m = rho * V#

This means that you have

#rho_"Au" * V_"Au" + rho_"Cu" * V_"Cu" = "668 g"" " " "color(blue)((1))#

Here

#rho_"Au"# is the density of gold, equal to#"19.3 g cm"^(-3)# #rho_"Cu"# is the density of copper, equal to#"8.9 g cm"^(-3)#

You also know that

#V_"Au" + V_"Cu" = "40 cm"^3 " " " " color(blue)((2))#

You now have two equations with two unknowns. Rewrite equation

#V_"Au" = "40 cm"^3 - V_"Cu"#

and plug in into equation

#19.3 color(red)(cancel(color(black)("g")))"cm"^(-3) * ("40 cm"^3 - V_"Cu") + 8.9 color(red)(cancel(color(black)("g"))) "cm"^(-3) * V_"Cu" = 668 color(red)(cancel(color(black)("g")))#

This is equivalent to

#19.3 color(red)(cancel(color(black)("cm"^(-3)))) * 40 color(red)(cancel(color(black)("cm"^3))) - "19.3 cm"^(-3) * V_"Cu" + "8.9 cm"^(-3) * V_"Cu" = 668#

which gives

#772 - "10.4 cm"^(-3) * V_"Cu" = 668#

and thus

#V_"Cu" = (772 - 668)/"10.4 cm"^(-3) = color(darkgreen)(ul(color(black)("10 cm"^3)))#

Plug this into equation

#V_"Au" = "40 cm"^3 - "10 cm"^3 = color(darkgreen)(ul(color(black)("30 cm"^3)))#

Therefore, you can say that the mixture was made by mixing