Pls inc. working! If the enthalpy change associated with freezing 3.45 g of water is 1.18kJ, calculate enthalpy change per mole of water.?
回答 (2)
2017-08-19 2:13 pm
Molar mass of H₂O (water) = (1.0×2 + 16.0) g/mol = 18.0 g/mol
No. of moles of H₂O involved in the change = (3.45 g) / (18.0 g/mol) = 0.1917 mol
Enthalpy change per mole of H₂O = (1.18 kJ) / (0.1917 mol) = 6.16 kJ/mol
No. of moles of H₂O involved in the change = (3.45 g) / (18.0 g/mol) = 0.1917 mol
Enthalpy change per mole of H₂O = (1.18 kJ) / (0.1917 mol) = 6.16 kJ/mol
2017-08-19 5:37 pm
Molar mass H2O = 18g/mol
Enthalpy change per mol H2O = 18/3.45*1.18 = 6.16kJ/mol
Enthalpy change per mol H2O = 18/3.45*1.18 = 6.16kJ/mol
收錄日期: 2021-04-18 17:44:42
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