Chemistry Help!!?
2016-02-26 6:41 pm
How many grams of aluminum hypochlorite are needed to prepare 631 mL of 0.137 M solution?
回答 (1)
2016-02-26 6:48 pm
No. of moles of Al(ClO)₃ = 0.137 × (631/1000) = 0.0865 mol
Molar mass of Al(ClO)₃ = 26.98 + 35.45×3 + 16.00×3 = 181.33 g/mol
Mass of Al(ClO)₃ = 181.33 × 0.0865 = 15.7 g
Molar mass of Al(ClO)₃ = 26.98 + 35.45×3 + 16.00×3 = 181.33 g/mol
Mass of Al(ClO)₃ = 181.33 × 0.0865 = 15.7 g
收錄日期: 2021-04-18 14:33:25
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