Question

A bag contains a mixture of copper and lead BBs. The average density of the BBs is 9.60g/cm3.Assuming that the copper and lead are pure, determine the relative amounts of each kind of BB.

Express your answers numerically separated by a comma.

Answer 1

To determine the relative amounts of copper and lead BBs in the mixture, we use the concept of density and set up an equation to solve for the relative amounts. By substituting the given values and solving the equation, we can find the relative amounts of each kind of BB.

Step-by-step explanation:

To determine the relative amounts of copper and lead BBs in the mixture, we need to use the concept of density. The average density of the BBs is 9.60g/cm3. Assuming that copper and lead are pure, we can set up the following equation:

Density of Copper * Relative Amount of Copper + Density of Lead * Relative Amount of Lead = Average Density of the Mixture

Since density equals mass divided by volume, we can rewrite the equation as:

Mass of Copper / Volume of Copper * Relative Amount of Copper + Mass of Lead / Volume of Lead * Relative Amount of Lead = Average Density of the Mixture

By substituting the given values and solving the equation, we can find the relative amounts of each kind of BB.

Answer 2

Step-by-step explanation:

Let the percentage of copper by weight present is x. Hence, then the percentage of lead will be (100 - x).

So, total mass = 100 g

Therefore, volume occupied by copper will be as follows.

Volume =
`(mass)/(density)`

=
`(x)/(8.94)`

Volume occupied by lead will be as follows.

Volume =
`(mass)/(density)`

=
`((100 - x))/(11.34)`

Total density =
`\frac{\text{total mass}}{\text{total volume}}`

9.60
`g/cm^(3)` =
`(100)/((x)/(8.94) + (100 - x)/(11.34))`

x = 50.34

Therefore, percentage of lead will be (100 - 50.34) = 49.66.

Thus, we can conclude that percentage of copper is 50.34% and percentage of lead is 49.66%.