Calculate the work done during a reaction in which the internal volume is compressed from 83L to 13L against an outside pressure of 4.4 atm. OLatm; No work is done 424 L.atm OB. 65 Latm 308 L.atm OD. -424 L.atm OE.

How much work in L*atm is done by 0.440 moles of a gas that has an initial volume of 6.5 liters and expands under the following conditions: 303K and 1.05 atm?

A 2 mole sample of gas expands from a volume of 4 L to 10 L under a constant pressure of 12.5 atm before then dropping to a pressure of 5 atm (while remaining at 10 L). 2 moles of gas expands from a volume of 4L to 10 L under a constant temperature (300K). Which situation requires more work? Why?

For the chemical reaction below, which statement is true about the reaction? N2H4 (1) + O2 (g) → N2 (g) + 2 H20 (g) AH° = -543 kJ-mol-1 A. Work is leaving the system. B. There is no work done in the reaction. C. Work is entering the system. D. The process is endothermic. Reset Selection

Explanation please

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A 104.4mL sample of 1.8M HNO3 is mixed with a 800mL sample of 0.5M KOH in a Styrofoam cup. If both solutions were initially at 25.2\deg C, and the enthalpy of the neutralization reaction is -57 kJ/mole of H2O formed, what is the final temperature of the mixture? The volume of the formed H2O can be ignored. Assume that the solution has a density of 1.00 g/mL and a specific heat of 4.184 J/g\deg C, and that the Styrofoam cup has an insignificant heat capacity. (enter one decimal place)