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How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 K to liquify the sample and lower the temperature to 335.0 K? The following physical data may be useful.Hvap = 33.9 kJ/molHfus = 9.8 kJ/molCliq = 1.73 J/g CCgas = 1.06 J/g CCsol = 1.51 J/g CTmelting = 279.0 KTboiling = 353.0 Ka. 95.4 kJ
b. 74.4 kJ
c. 38.9 kJ
d. 67.7 kJ
e. 54.3 kJ

Respuesta :

Answer: d) 67.7KJ

Explanation:

Number of moles of Benzene= 125/78.11=1.60 moles

Q = mc◇T = 125× 1.60×(425-353)= 9540J

Q= ◇Hvap× 1.60=

33.9 × 1.60 = 54.24KJ

Q = mc◇T= 125 × 1.73× (353-335)=3803KJ

Total energy required to remove Carbon= (9540 + 54240+ 3893) J =67,673J =67

7KJ

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