Question

Question #435fb

Answer 1

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 `m`, and the volume it occupies, let's say `V`.

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

Now, let's take `m_"Au"` to be the mass of gold and `m_"Cu"` the mass of copper used to make the mixture.

You know that

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

If you take `V_"Au"` to be the volume of gold and `V_"Cu"` to be the volume of copper in the mixture, you can rewrite the above equation by using equation `color(red)("(*)")`

`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 `color(blue)((2))` as

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

and plug in into equation `color(blue)((1))` to find

`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 `color(blue)((2))` to find the value of `V_"Au"`

`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 `"30 cm"^3` of gold and `"10 cm"^3` of copper.