.250 mole
7.02
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28.09 = .250
You can't have 7.610 atoms of silicon - number of atoms has to be a whole number, because it's a count; a tally. Perhaps you mean how many grams would be in 7.610 moles of silicon. That would be 7.61moles x 28.1g/mol = approx. 214g Si.
For this problem, the Atomic Mass is required. Take the mass in grams and divide it by the atomic mass. Then multiply it by Avogadro's constant, 6.02 × 1023.
7.82 grams Si / (28.1 grams) × (6.02 × 1023 atoms) = 1.68 × 1023 atoms
.250 mole
7.02
------
28.09 = .250
There are 1.34×10²³ atoms. You look for the number of moles in the 6.26 g of Si, then you multiply by the Avogadro constant.
6.34 grams Si ( 1 mole Si/28.09 grams)(6.022 X 1023/1 mole Si)
= 1.36 X 1023 atoms of silicon
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237g / 28 gmol-1 = 8.46mol. 8.46 x (6.02x1023) = 5.09x1024 silicon atoms.
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
11 g x 1 mol Si/28.0855 g Si= 0.39 mol Si 205g He x 1mol He/4.002602 g He=51.2 mol He
237g / 28 gmol-1 = 8.46mol. 8.46 x (6.02x1023) = 5.09x1024 silicon atoms.
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