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Using standard formation enthalpies, calculate the standard reaction enthalpy for this reaction. 6CO2(g) + 6H2O(l)C6H12O6 + 6O2(g) ANSWER: kJ/mol

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batolisis

Answer:

2802.5 KJ/mol

Explanation:

Calculate standard enthalpy of the Reaction :

6CO2(g) + 6H2O(l)C6H12O6 + 6O2(g)

using standard formation enthalpies . hence the standard enthalpy of the reaction = 2802.5 KJ/mol

attached below is the detailed solution

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