The number of iron atoms present in each red blood cell is 1.18 × 10⁹ atoms
From the question, we are to determine the number of iron atoms present in each red blood cell
First, we will determine the mass of iron in one red blood cell
The mass of iron in a red blood cell = [tex]\frac{2.90}{2.64 \times 10^{13} }[/tex]
= 1.098 × 10⁻¹³ g
To determine the number of iron atoms present, we will determine the number of moles of iron atoms present
From the formula
[tex]Number\ of\ moles = \frac{Mass}{Molar\ mass}[/tex]
Number of moles of iron in a red blood cell = [tex]\frac{1.098 \times 10^{-13} }{55.85}[/tex]
Number of moles of iron in a red blood cell = 1.966 × 10⁻¹⁵ mole
Now, for the number of iron atoms present
Number of atoms = Number of moles × Avogadro's constant
Then,
Number of iron atoms present = 1.966 × 10⁻¹⁵ × 6.022 × 10²³
Number of iron atoms present = 1.18 × 10⁹ atoms
Hence, the number of iron atoms present in each red blood cell is 1.18 × 10⁹ atoms
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