Is 320 g per 100 g of water.
- 407.4 kJ of heat is released.
<u>Explanation:</u>
We have to write the balanced equation as,
2 C₂H₆(g) + 7O₂ → 4CO₂ + 6H₂O
Here 2 moles of ethane reacts in this reaction.
Now we have to find out the amount of ethane reacted using its given mass and molar mass as,
2 mol C₂H₆ × 30.07 g of C₂H₆ / 1 mol C₂H₆ = 60.14 g of C₂H₆
Heat released = ΔH × given mass / 60.14
= - 1560. 7 kj ×15.7 g / 60. 14 g = -407. 4 kJ
Answer:
fH = - 3,255.7 kJ/mol
Explanation:
Because the bomb calorimeter is adiabatic (q =0), there'is no heat inside or outside it, so the heat flow from the combustion plus the heat flow of the system (bomb, water, and the contents) must be 0.
Qsystem + Qcombustion = 0
Qsystem = heat capacity*ΔT
10000*(25.000 - 20.826) + Qc = 0
Qcombustion = - 41,740 J = - 41.74 kJ
So, the enthaply of formation of benzene (fH) at 298.15 K (25.000 ºC) is the heat of the combustion, divided by the number of moles of it. The molar mass od benzene is: 6x12 g/mol of C + 6x1 g/mol of H = 78 g/mol, and:
n = mass/molar mass = 1/ 78
n = 0.01282 mol
fH = -41.74/0.01282
fH = - 3,255.7 kJ/mol
The answer has to be 22.4L
<span>1.00 atm of each gas, in what direction will the system shift to reach equilibrium</span>
