Ammonia, nh3, is used as a refrigerant. at its boiling point of –33 oc, the standard enthalpy of vaporization of ammonia is 23.3 kj/mol. how much heat is released when 50.0 g of ammonia is condensed at –33 oc? –0.466 kj–7.94 kj–36.6 kj–68.4 kj–1.17 x 103 kj

-68.4 kJ

Explanation:

The standard enthalpy of vaporization = 23.3 kJ/mol

which means the energy required to vaporize 1 mole of ammonia at its boiling point (-33 °C).

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.

Thus,  Q = -23.3 kJ/mol

Where negative sign signifies release of heat

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

Thus, amount of heat released when 50 g of ammonia condensed at -33 °C= -68.4 kJ, where negative sign signifies release of heat.

the electron configuration that you gave me has 16 electrons ergo it has 16 protons ergo it is the element sulfur.