Question Solved1 Answer The human body obtains \( 915 \mathrm{~kJ} \) of energy from a candy bar. If this energy were used to vaporize water at \( 100.0{ }^{\circ} \mathrm{C} \), how much water (in litres) could be vaporized? (Assume the density of water is \( 1.00 \mathrm{~g} \mathrm{~mL}^{-1} \).)

JQU7LD The Asker · Chemistry
The human body obtains \( 915 \mathrm{~kJ} \) of energy from a candy bar. If this energy were used to vaporize water at \( 100.0{ }^{\circ} \mathrm{C} \), how much water (in litres) could be vaporized? (Assume the density of water is \( 1.00 \mathrm{~g} \mathrm{~mL}^{-1} \).)
Transcribed Image Text: The human body obtains \( 915 \mathrm{~kJ} \) of energy from a candy bar. If this energy were used to vaporize water at \( 100.0{ }^{\circ} \mathrm{C} \), how much water (in litres) could be vaporized? (Assume the density of water is \( 1.00 \mathrm{~g} \mathrm{~mL}^{-1} \).)
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Transcribed Image Text: The human body obtains \( 915 \mathrm{~kJ} \) of energy from a candy bar. If this energy were used to vaporize water at \( 100.0{ }^{\circ} \mathrm{C} \), how much water (in litres) could be vaporized? (Assume the density of water is \( 1.00 \mathrm{~g} \mathrm{~mL}^{-1} \).)
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Given: 915kJ from candy bar, water d=1.00gmL^(-1) Find: L(H_(2)O) vaporized at 100.0^(@)COther: Delta_("vap ")H^(@)=40.7kJmol^(-1)Conceptual Plan: q rarrmolH_(2)OrarrgH_(2)OrarrmLH_(2)OrarrLH_(2)O(1(mol))/(40.7(kJ))quad(18.01(g))/(1(mol))quad(1.00(mL))/(1.00(g))quad( ... See the full answer