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Hemoglobin is a large protein molecule that is responsible for carrying oxygen through the blood. Iron ions are a relatively small component of hemoglobin. There are four Fe2+ions that are part of the much larger hemoglobin structure. In a single red blood cell there are 2.50x108molecules of hemoglobin. If a single Fe2+ion has an atomic radius of 75.1 pm and a redblood cell has a volume of 95 μm3, what percentage of the total red blood cell volume is taken up by Fe2+ions?

User Felix Weis
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Answer:

The percentage of volume taken by iron(II) in the blood cell is 0.0019%.

Step-by-step explanation:

Radius of iron(II) ions ,r= 75.1 pm =
7.51* 10^(-5) \mu m

1 pm = 10^{-6} μm

Volume of sphere =
(4)/(3)\pi r^3

Volume of single iron(II) ion = V


V=(4)/(3)* 3.14* (7.51* 10^(-5) \mu m)^3


V=1.7742* 10^(-12) \mu m^3

Number of iron(II) ions in one hemoglobin structure = 4

Number of hemoglobin structure in blood cell =
2.50* 10^8 molecules

Then number of iron (II) ions in
2.50* 10^8 molecules of hemoglobin:


4* 2.50* 10^8 ions=10^9 ions

Volume of
10^9 ions of iron =
V* 10^9

Volume of the hemoglobin structure,V' =
95 \mu m^3

Percentage volume of iron (II) ions in a single blood cell:


(10^9* V)/(V')* 100


(10^9* 1.7742* 10^(-12) \mu m^3)/(95 \mu m^3)* 100

=
0.0019\%

User Oat Anirut
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