Answer:
1. C4H8 + 6O2 -----> 4CO2 + 4H20
2. 3836.77 kcal
Explanation:
1. Balanced equation for the complete combustion of cyclobutane:
C4H8 + 6O2 -----> 4CO2 + 4H20
2. Heat of combustion of cyclobutane = 650.3 kcal/mol
Molecular weight of cyclobutane, C4H8 = 56.1 g/mol
Mole of C4H8 : mass of cyclobutane/Molecular weight of cyclobutane
Mole of C4H8 = 331/56.1 = 5.9 mol
Energy released during combustion = 5.9 mol × 650.3 kcal/mol = 3836.77kcal
Therefore the energythat is released during the complete combustion of 331 grams of cyclobutane is 3836.77kcal
Answer : The solubility of a gas in water at 1 atm pressure is, 0.4436 g/L
Solution : Given,
Solubility of a gas in water = 1.22 g/L (at 2.75 atm pressure)
At pressure = 1 atm, Solubility of a gas = ?
Now we have to calculate the solubility of a gas.
At 2.75 atm pressure, the solubility of a gas in water = 1.22 g/L
At 1 atm pressure, the solubility of a gas in water = 
Therefore, the solubility of a gas in water at 1 atm pressure is, 0.4436 g/L
Density because a substance always has the same density, but mass can be different and volume depends on mass