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Chemistry, 03.02.2021 03:10 Sariyahhall1

P4O10(s) = -2984 kJ/mol H2O(l) = -285.9 kJ/mol H3PO4(s) = -1279 kJ/mol Calculate the change in enthalpy for the following process: P4O10(s) + 6H2O(l) β†’ 4H3PO4(s)

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P4O10(s) = -2984 kJ/mol H2O(l) = -285.9 kJ/mol H3PO4(s) = -1279 kJ/mol Calculate the change in entha...
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