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If this energy were used to vaporize water at 100 ∘C, how much water in liters could be vaporized? Assume that the density of water is 1.0 g/mL. The heat of vaporization of water at 100 ∘C is 40.7 kJ/mol. (Hint: Begin by using the enthalpy of vaporization of water to convert between the given number of kilojoules and moles of water.)

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Answer:

0.45L

Explanation:

onyebuchinnaji

The volume of water in liters that the given quantity of heat can vaporize is 0.481 Liter.

Volume of water vaporized

The volume of water vaporized is calculated as follows;

Q = nΔH

where;

  • n is number of moles of the water
  • ΔH is heat of vaporization of water

1088 kJ = n(40.7 kJ/mol)

n = 1088 /40.7

n = 26.73 moles

Mass of water vaporized

m = n x 18g

m = 26.73 x 18

m = 481.14 g

Volume of water vaporized

volume = mass/density

volume = (481.14 g) / (1 g/mL)

volume = 481.14 mL

volume = 0.481 L

Thus, the volume of water in liters that the given quantity of heat can vaporize is 0.481 Liter.

The complete question is below:

The human body obtains 1088 kJ from a candy bar. if this energy were used to vaporize water at 100 ∘C, how much water in liters could be vaporized? Assume that the density of water is 1.0 g/mL. The heat of vaporization of water at 100 ∘C is 40.7 kJ/mol. (Hint: Begin by using the enthalpy of vaporization of water to convert between the given number of kilojoules and moles of water.)

Learn more about heat of vaporization here: https://brainly.com/question/26306578

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