3.68 grams silver (1 mole Ag/107.9 grams)(6.022 X 10^23/1 mole Ag)
= 2.05 X 10^22 atoms of silver
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2.17*10^22
The number of atoms of lead is 6,68.10e23.
The gram atomic mass of silver is 107.868 and that of gold is 196.967. Equal numbers of gram atoms of different elements contain equal numbers of atoms. Therefore, the mass of gold required to contain twice as many atoms as 2.74 g of gold is (2 X 2.74 X 196.967)/107.868 or 10.0 g of gold, to the justified number of significant digits.
1
149 g of calcium contain 22,39.10e23 atoms.
2.17*10^22
2.17*10^22
It depends on the weight of the silver in the material. 107.89 g of silver will have 6.023 x 1023 atoms of silver
The number of atoms of lead is 6,68.10e23.
The gram atomic mass of silver is 107.868 and that of gold is 196.967. Equal numbers of gram atoms of different elements contain equal numbers of atoms. Therefore, the mass of gold required to contain twice as many atoms as 2.74 g of gold is (2 X 2.74 X 196.967)/107.868 or 10.0 g of gold, to the justified number of significant digits.
1
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.