86. Solving a chemistry problem
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21. The heat of combustion of sucrose under standard conditions is -5645 kJ / mol. Change in the isobaric potential of the reaction for the formation of sucrose
12C(t) 11H2(g) 5.5O2(g)=C12H22O11(t) equals -1555.5 kJ.
Determine the change in the entropy of the reaction.
12C(t) 11H2(g) 5.5O2(g)=C12H22O11(t) equals -1555.5 kJ.
Determine the change in the entropy of the reaction.
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