The heat of the reaction, in kJ, when 4.18 g of the hydrocarbon are combusted 775.70 kJ.
The heat energy is given as :
q = m c ΔT + Ccal ΔT
q = ( 974 g× 4.184 ×6.9) + 624 ×6.9
q = 32424.59 J
moles of hydrocarbon = 0.0418 mol
heat of combustion = 32424.59 J / 0.0418 mol
= 775707.89 J
= 775.70 kJ
Thus, A 4.18 g sample of a hydrocarbon is combusted in a bomb calorimeter that contains 974 g of water. the temperature of the water increases by 6.9 °C when the hydrocarbon is combusted. the calorimeter constant for the calorimeter was determined to be 624 J/°C. what is the heat of the reaction is 775.70 kJ.
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Answer:
The answer to the 2nd question is C.giants.
Explanation:
Answer:
a) 0 J
b) -2.67x10² J
c) -2.09x10³ J
Explanation:
For an isothermic expansion (with constant temperature) the work (W) is :
W = -pΔV, where p is the pressure and ΔV the volume variation. The minus signal is used because the compression is positive and ΔV is negative (Vf < Vi).
a) In vacuum, the relative pressure is 0 atm, so the work:
W = -0x(5.7 - 1.3)
W = 0 J
b) For a constant pressure of 0.60 atm
W = -0.6atmx(5.7 - 1.3)L = -2.64 L.atm
1 L.atm = 101.3 J
W = -2.64x101.3 = -2.67x10² J
c) For a pressure of 4.7 atm
W = -4.7atmx(5.7 - 1.3) L = - 20.68 atm.L
1 atm.L = 101.3 J
W = -20.68x101.3 = -2.09x10³ J
