How many molecules are there in 85.4 g of PF3?
回答 (1)
2016-06-12 4:56 pm
Molar mass of PF₃ = (30.97 + 19.00×3) g/mol = 87.97 g/mol
No. of moles of PF₃ = (85.4 g) / (87.97 g/mol)
1 mole of PF₃ contains 6.022 × 10²³ PF₃ molecules.
No. of PF₃ molecules = [(85.4 g) / (87.97 g/mol)] × (6.022 × 10²³ /mol) = 5.85 × 10²³
No. of moles of PF₃ = (85.4 g) / (87.97 g/mol)
1 mole of PF₃ contains 6.022 × 10²³ PF₃ molecules.
No. of PF₃ molecules = [(85.4 g) / (87.97 g/mol)] × (6.022 × 10²³ /mol) = 5.85 × 10²³
收錄日期: 2021-04-18 14:58:09
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