56.279 grams Samarium(Sm) (1 mole Sm/150.36 grams)(6.022 X 10^23/1 mole Sm)
= 2.2540 X 10^23 atoms of Samarium
78.96 g of selenium will have 6.023 x 1023 atoms. So, 30.4 g of selenium will have 2.32 x 1023 atoms.
5 g of sulfur contain 0,94.10e23 atoms.
156,86.1020 atoms of uranium in 6,2 g uranium
The number of atoms is 28,099.10e23.
25.1
Samarium 99 % cost is 0,025 $/g; see the link:http://www.metal-pages.com/metalprices/samarium/.
The number of atoms of lead is 6,68.10e23.
49.1740 g (6.02 x 1023 atoms) / (91.22 g) = 3.25 x 1023 atoms
149 g of calcium contain 22,39.10e23 atoms.
6,687.1023 chlorine atoms
27,30.10e23 atoms
The number of atoms of lead is 6,68.10e23.
78.96 g of selenium will have 6.023 x 1023 atoms. So, 30.4 g of selenium will have 2.32 x 1023 atoms.
5 g of sulfur contain 0,94.10e23 atoms.
156,86.1020 atoms of uranium in 6,2 g uranium
The answer is 3.32*10^23 atoms
6.14x1019 atoms Au