Answer:
3.22×10²³ Cr atoms are in 78.82g of K₂Cr₂O₇
Explanation:
Potassium dichromate → K₂Cr₂O₇
We determine the molar mass → 39.1 g/mol . 2 + 52 g/mol . 2 + 16g/mol . 7 = 294.2 g/mol
1 mol of K₂Cr₂O₇ has 2 moles of K, 2 moles of Cr and 7 moles of O.
1 mol = 294.2 g. Therefore, 294.2 g of compound have 2 moles of Cr
Let's make a rule of three:
294.2 g of K₂Cr₂O₇ have 2 moles of Cr
Then, 78.82 g of K₂Cr₂O₇ may have (78.82 . 2) / 294.2 = 0.536 moles
To calculate the number of atoms, you should know that:
1 mol contains NA atoms (6.02×10²³ particles)
Then 0.536 moles may contain (0.536 . NA) /1 = 3.22×10²³
