The task of physical chemistry 27

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Uploaded: 16.10.2017
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When 298,16° standard enthalpy of combustion of cigicigijingle (CIS-С9Н16) is equal to - 1351,6 kcal*mol-1. The combustion process can be represented by the equation: CIS-С9Н16 (g) + 13 O2 (g) -> 9СО2 (g) +8Н2О (W). a) Calculate the standard enthalpy of formation of 1 mole of tishetshkin from simple substances at 298,16° To b) What will be the answer if the standard state of carbon is adopted diamond instead of graphite? With (diamond) -> C (graphite) H298 = - 0,45 kcal.

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