Final answer:
To find the number of iodine atoms in 40.7206 grams of pentaselenium heptaiodide, convert the mass to moles of Se5I7, then to moles of iodine atoms, and finally to the number of atoms using Avogadro's number. The calculation reveals there are approximately 1.34 x 10^23 iodine atoms.
Step-by-step explanation:
To determine how many iodine atoms are present in 40.7206 grams of pentaselenium heptaiodide (Se5I7), we start by calculating the molar mass of Se5I7. The molar mass of selenium (Se) is 78.96 g/mol, and the molar mass of iodine (I) is 126.90 g/mol.
Therefore, the molar mass of Se5I7 is:
5(Se) + 7(I) = 5(78.96 g/mol) + 7(126.90 g/mol) = 394.8 g/mol + 887.3 g/mol = 1282.1 g/mol
Next, we use dimensional analysis to convert grams of Se5I7 to moles of Se5I7:
40.7206 g Se5I7 x (1 mol Se5I7 / 1282.1 g Se5I7) = 0.031763 mol Se5I7
Since each molecule of Se5I7 contains 7 iodine atoms, we multiply the moles of Se5I7 by 7 to obtain the moles of iodine atoms:
0.031763 mol Se5I7 x (7 mol I / 1 mol Se5I7) = 0.22234 mol I
Finally, to get the number of iodine atoms, we use Avogadro's number (6.022 x 10^23 atoms/mol):
0.22234 mol I x (6.022 x 10^23 atoms/mol) = 1.339 x 10^23 atoms I
Therefore, there are approximately 1.34 x 10^23 iodine atoms in 40.7206 grams of pentaselenium heptaiodide.