1.28 x 10^21 atoms
PCl3(g) + Cl2(g) PCl5(g)
The number of atoms of lead is 6,68.10e23.
149 g of calcium contain 22,39.10e23 atoms.
27,30.10e23 atoms
6.14x1019 atoms Au
The reaction will proceed to the right. 2 PCl3 will be consumed.
PCl3(g) + Cl2(g) PCl5(g)
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