J.R. S. answered 03/22/19
Ph.D. University Professor with 10+ years Tutoring Experience
O2(g) + O2(g) + O2(g) ==> 2O3(g)
KO2(s) ==> K(s) + O2(g) ∆H = 283 kJ
K(s) + O2(g) ==> KO2(s) ∆H = -283 kJ
Madison L.
asked 03/22/19a) Write the balanced chemical equation that represents the standard heat of formation of O3(g) at 298 K.
Use the pull-down boxes to specify states. Write fractions with a slash, such as 1/2 for one half. If a box is not needed leave it blank.
_____ (state) + ____ (state) + ______ (state) --> _____ (state)
(b) The standard enthalpy change for the following reaction is 283 kJ at 298 K.
KO2(s)--> K(s) + O2(g)
What is the standard heat of formation of KO2(s)? ____ kJ/mol
J.R. S. answered 03/22/19
Ph.D. University Professor with 10+ years Tutoring Experience
O2(g) + O2(g) + O2(g) ==> 2O3(g)
KO2(s) ==> K(s) + O2(g) ∆H = 283 kJ
K(s) + O2(g) ==> KO2(s) ∆H = -283 kJ
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