Water has 18.0 g/mole and its heat of ovaporization is 40.7 kJ/mole. Determine t

Question

Water has 18.0 g/mole and its heat of ovaporization is 40.7 kJ/mole. Determine the amount of heat that must be removed to condense 99.0 g of water vapor. Show all units. Suppose you remove this heat by vaporizing ammonia (NH3). The heat of vaporization is 23.4 kJ/mole for ammonia. Determine the moles of ammonia that are needed. Ammonia has 17.0 g/mole. Determine the mass of ammonia with units included.
Please help ASAP. Water has 18.0 g/mole and its heat of ovaporization is 40.7 kJ/mole. Determine the amount of heat that must be removed to condense 99.0 g of water vapor. Show all units. Suppose you remove this heat by vaporizing ammonia (NH3). The heat of vaporization is 23.4 kJ/mole for ammonia. Determine the moles of ammonia that are needed. Ammonia has 17.0 g/mole. Determine the mass of ammonia with units included. Water has 18.0 g/mole and its heat of ovaporization is 40.7 kJ/mole. Determine the amount of heat that must be removed to condense 99.0 g of water vapor. Show all units. Suppose you remove this heat by vaporizing ammonia (NH3). The heat of vaporization is 23.4 kJ/mole for ammonia. Determine the moles of ammonia that are needed. Ammonia has 17.0 g/mole. Determine the mass of ammonia with units included.
Please help ASAP.

Explanation / Answer

The heat of condensation is exactly equal to heat of vapourisation, just opposite in sign.

So the heat released when 1 mole is converted to liquid form= -40.7 kJ

heat released when 99 g is converted= -40.7 kJ* (99g/18g) = -223.85 kJ

If heat is removed using NH3, amount of heat to be removed=-223.85 kJ

let mass of ammonia= m

then,

(m/17 g)*(- 23.4 kJ )= -223.85

m= 162.63 g

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