How many molecules are present in 9.19 × 10–4 g H2?
回答 (2)
2017-10-09 3:19 pm
Molar mass of H₂ = 1.008 × 2 g/mol = 2.016 g/mol
No. of moles of H₂ = (9.19 × 10⁻⁴ g) / (2.106 g/mol) = 5.56 × 10⁻⁴ mol
Avagadro constant = 6.022 × 10²³ /mol
No. of H₂ molecules = (5.56 × 10⁻⁴ mol) × (6.022 × 10²³ /mol) = 3.35 × 10²⁰ (H₂ molecules)
No. of moles of H₂ = (9.19 × 10⁻⁴ g) / (2.106 g/mol) = 5.56 × 10⁻⁴ mol
Avagadro constant = 6.022 × 10²³ /mol
No. of H₂ molecules = (5.56 × 10⁻⁴ mol) × (6.022 × 10²³ /mol) = 3.35 × 10²⁰ (H₂ molecules)
2017-10-09 3:22 pm
Molar mass of H2 is 2.015 g/mol, so
9.19 * 10^(-4) g = 4.56 x 10^(-4) moles.
Using Avogadro's number, 6.022 x 10^23 molecules/mole, you get
2.75 x 10^20 molecules.
9.19 * 10^(-4) g = 4.56 x 10^(-4) moles.
Using Avogadro's number, 6.022 x 10^23 molecules/mole, you get
2.75 x 10^20 molecules.
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