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Answer:

22.7

Explanation:

Initially, calculate the energy released from the sample's mass. The combustion heat represents the energy per mole of the fuel:

ΔHC=qrxnn

We can rearrange this formula to isolate qrxn, remembering to convert the sample's mass into moles:

qrxn=ΔHrxn×n=−3374 kJ/mol×(0.500 g×1 mol213.125 g)=−7.916 kJ=−7916 J

The heat released during the reaction must match the total heat absorbed by both the water and the bomb calorimeter:

qrxn=−(qwater+qbomb)

The heat absorption by the water can be calculated using its specific heat:

qwater=mcΔT

The calorimeter's heat absorption can be derived from its heat capacity:

qbomb=CΔT

Combine both equations into the first equation, substituting the known details, with ΔT=Tfinal−20.0∘C:

−7916 J=−[(4.184 Jg ∘C)(600. g)(Tfinal–20.0∘C)+(420. J∘C)(Tfinal–20.0∘C)]

Distribute each multiplication term and simplify:

−7916 J=−[(2510.4 J∘C×Tfinal)−(2510.4 J∘C×20.0∘C)+(420. J∘C×Tfinal)−(420. J∘C×20.0∘C)]=−[(2510.4 J∘C×Tfinal)−50208 J+(420. J∘C×Tfinal)−8400 J]

Combine the similar terms and simplify:

−7916 J=−2930.4 J∘C×Tfinal+58608 J

Finally, isolate Tfinal:

−66524 J=−2930.4 J∘C×Tfinal

Tfinal=22.701∘C

Round to three significant figures gives the final result as 22.7∘C.

Boyle's law pertaining to ideal gases states that the volume of a gas relates inversely to its pressure when temperature is constant. According to this principle, the relationship between pressure and volume can be expressed as:

This implies:

Utilizing this equation enables us to determine the volume of gas given a specific pressure:

P₁=Initial pressure

V₁=Initial volume

P₂=Final pressure

V₂= Final volume

In this case, the initial pressure P₁ is noted as 85.0 kPa.

The initial volume V₁ is provided as 525 mL.

The final pressure P₂ is recorded as 65.0 kPa.

Thus,

V_{2}=85×525÷65

=686 mL

Consequently, the gas volume will amount to 686 mL.