N2 + O2 --> 2NO dH° 180.6 kJ
(b) How much heat is absorbed in the oxidation of 5.45 L
of nitrogen measured at 1.00 atm and 273 K? (c) When the oxidation of N2 to NO was completed in a bomb calorimeter, the heat absorbed was measured as 492 J. What mass of nitrogen gas was oxidized?
Can anybody help me out with this question?
#8.85
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Rachel_Harland_3I
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Re: #8.85
b) PV=nRT , V=5.45L P=1atm, R=0.0821, T=273K.
Solve for "n" and you get 0.243 moles.
multiply this by 180.6KJ/1moleN2 to find out the total KJ for 0.243 moles
(43.9KJ)
Solve for "n" and you get 0.243 moles.
multiply this by 180.6KJ/1moleN2 to find out the total KJ for 0.243 moles
(43.9KJ)
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