In the average adult human, there are approximately 2.6 *10^13 red blood cells with a total of 2.9 grams of Iron. On average, how many iron atoms are present in each red blood cell? Thanks
2007-05-18 15:02:55 · 3 answers · asked by Giant Donut 2 in Science & Mathematics ➔ Chemistry
2.9 /55.847 = 0.0519 mole Fe
0.0519 / 2.6 x 10^13 = 2.00 x 10 ^-15 mole Fe in each red blood cell
2.00 x 10^-15 x 6.02 x 10^23 = 1.20 x 10^9 iron atoms
2007-05-18 15:24:07 · answer #1 · answered by Anonymous · 1⤊ 0⤋
moles of Fe = 2.9g /55.847
=0.05193 mol
no of Fe atoms in 2.6 x10^13 cells = 0.05193 mol x 6.02 x 10^23 atoms= 3.1260 x 10^22 atoms
thus the no of Fe atoms in 1 red blood cell
=3.1260 x 10^22 atoms/ 2.6 x10^13
=1.202 x 10^9 atoms
2007-05-18 23:49:56 · answer #2 · answered by 1-man-show 3 · 0⤊ 0⤋
Divide total Fe weight by atomic weight of Fe
Multiply the result by Avogadro;s Number
Divide that result by the number of blood cells.
2007-05-18 22:27:42 · answer #3 · answered by cattbarf 7 · 0⤊ 0⤋