The standard enthalpy of vaporization of Ammonia (NH3) is 21.5 kJ/mol at its boiling temperature of -28.2 °C. How much heat in the unit of kJ is released when 55 g of this compound is condensed at -28.2 °C? Please keep your answer to 1 decimal place.
Added by Kevin M.
Step 1
Given: - Mass of NH3 = 55 g - Molar mass of NH3 = 17 g/mol Number of moles (n) = Mass / Molar mass n = 55 g / 17 g/mol n ≈ 3.24 mol Show more…
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