# Question ab0ea

May 19, 2017

$1.$ $11.34$ $\text{mg}$

$2.$ $256.00$ $\text{mg}$

#### Explanation:

$1.$ The average concentration of the $\text{PCB}$s found in the chicks is $18.9$ $\text{mg}$ / $\text{kg}$. Also, the mass of a single chick is $0.6$ $\text{kg}$.

Let's set this information up as a ratio:

Rightarrow frac(18.9 " mg")(1 " kg")= frac(x)(0.6 " kg")

Multiply both sides by $0.6$ $\text{kg}$:

Rightarrow frac(18.9 " mg" times 0.6 " kg")(1 " kg") = frac(x times 0.6 " kg")(0.6 " kg")

$R i g h t a r r o w 11.34$ $\text{mg} = x$

$\therefore x = 11.34$ $\text{mg}$

Therefore, a chick of mass $0.6$ $\text{kg}$ would contain $11.34$ $\text{mg}$ of $\text{PCB}$s.

$2.$ The average concentration of $\text{PCB}$s in the body tissue of a human is $4.00$ $\text{ppm}$.

Let's convert the units of the concentration from $\text{ppm}$ to $\text{mg}$ / $\text{kg}$:

$R i g h t a r r o w 1$ $\text{ppm}$ $= 1$ $\text{mg}$ / ""kg""

$R i g h t a r r o w 4.00$ $\text{ppm}$ $= 4.00$ $\text{mg}$ / $\text{kg}$

We need to find the mass of $\text{PCB}$s found in a $64$ $\text{kg}$ human.

So let's set up another ratio using this information:

Rightarrow frac(4.00 " mg")(1 " kg") = frac(x)(64 " kg")

Multiply both sides by $64$ $\text{kg}$:

Rightarrow frac(4.00 " mg" times 64 " kg")(1 " kg") = frac(x times 64 " kg")(64 " kg")#

$R i g h t a r r o w 256.00$ $\text{mg} = x$

$\therefore x = 256.00$ $\text{mg}$

Therefore, the mass of $\text{PCB}$s present in a $64$ $\text{kg}$ person's body is $256.00$ $\text{mg}$.