Answer:
-68.4 kJ
Explanation:
<u>The standard enthalpy of vaporization = 23.3 kJ/mol</u>
<u>which means the energy required to vaporize 1 mole of ammonia at its boiling point (-33 °C).</u>
To calculate heat released when 50.0 g of ammonia is condensed at -33 °C.
This is the opposite of enthalpy of vaporization which means that same magnitude of heat is released.
<u>Thus, Q = -23.3 kJ/mol</u>
<u>Where negative sign signifies release of heat</u>
Given: mass of 50.0 g
Molar mass of ammonia = 17.034 g/mol
Moles of ammonia = 50.0 /17.034 moles = 2.9353 moles
Also,
1 mole of ammonia when condenses at -33 °C releases 23.3 kJ
2.9412 moles of ammonia when condenses at -33 °C releases 23.3×2.9353 kJ
<u>Thus, amount of heat released when 50 g of ammonia condensed at -33 °C= -68.4 kJ, where negative sign signifies release of heat.</u>
Mention four reasons why the poll ordinance failed
Answer:
Upwelling is the natural process which brings cold, nutrient-rich water to the surface. A huge upwelling regularly occurs off the coast of Peru, which enjoys a large fishing industry as a result. Upwelling is a process in which currents bring deep, cold water to the surface of the ocean.
Explanation:
good luck
Answer:
hope this helps
Explanation:
The atoms of hydrogen have smaller mass than oxygen. Thus their speeds have to higher in order to produce the same average kinetic energies.
