Calculate the number of g of silver nitrate required to prepare 3.25 L of 1.00 M AgNO3. Enter your answer in scientific notation?
回答 (2)
2020-03-16 11:07 am
Molar mass of AgNO₃ = (107.9 + 14.0 + 16.0×3) g/mol = 169.9 g/mol
Moles of AgNO₃ = (1.00 mol/L) × (3.25 L) = 3.25 mol
Mass of AgNO₃ = (3.25 mol) × (169.9 g/mol) = 5.52 × 10² g
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OR:
(3.25 mol AgNO₃ / 1 L solution) × (3.25 L solution) × (169.9 g/mol)
= 5.52 × 10² g
Moles of AgNO₃ = (1.00 mol/L) × (3.25 L) = 3.25 mol
Mass of AgNO₃ = (3.25 mol) × (169.9 g/mol) = 5.52 × 10² g
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OR:
(3.25 mol AgNO₃ / 1 L solution) × (3.25 L solution) × (169.9 g/mol)
= 5.52 × 10² g
2020-03-16 10:55 am
You need 3.25 moles. Multiply that by the molar mass.
收錄日期: 2021-04-18 18:25:48
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