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Answer 1 · 2017-05-19T19:25:23

$1.$ $11.34$ $"mg"$

$2.$ $256.00$ $"mg"$

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

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 $0.6$ $"kg"$ :

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

$Rightarrow 11.34$ $"mg" = x$

$therefore x = 11.34$ $"mg"$

Therefore, a chick of mass $0.6$ $"kg"$ would contain $11.34$ $"mg"$ of $"PCB"$ s.

$2.$ The average concentration of $"PCB"$ s in the body tissue of a human is $4.00$ $"ppm"$ .

Let's convert the units of the concentration from $"ppm"$ to $"mg"$ / $"kg"$ :

$Rightarrow 1$ $"ppm"$ $= 1$ $"mg"$ / ""kg""

$Rightarrow 4.00$ $"ppm"$ $= 4.00$ $"mg"$ / $"kg"$

We need to find the mass of $"PCB"$ s found in a $64$ $"kg"$ 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 $64$ $"kg"$ :

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

$Rightarrow 256.00$ $"mg" = x$

$therefore x = 256.00$ $"mg"$

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