Energy change is the sum of heat and work:  ΔE = q + w.  Work is calculated by: w = -PΔV What is the change in energy (in joules) if a reaction absorbs 40.3 J of heat and increases in volume from 0.250L to 0.750L at a constant pressure of 1.056 atm?  Please include the correct sign with your numerical result. (Note:  Make sure your units are consistent when combining energy terms.  The conversion factor you need is:  1 L·atm = 101.3 J)

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