#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"#.