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
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