As per the question, the mass of the nitrogen gas m = 22.25 gram.
The latent heat of vaporization of nitrogen = 199.0 j/g
As per the question, the nitrogen gas will condense. During condensation, the nitrogen gas will lose or release heat equal to its latent heat.
Hence, the heat released by nitrogen gas Q = ml = 22.25 × 199.0 J = 4427.75 J.
Hence, the amount of heat released will be 4427.75 J.
<h3>How can you figure out how much heat is in each gram?</h3>
The formula: can be utilized to determine energy. Q = mc ∆T. In the equation, Q stands for energy expressed in joules or calories, m for mass expressed in grams, c for specific heat, and T for temperature change, which is the difference between the final temperature and the initial temperature. Water has a specific heat of 1 calorie/gram °C.
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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
How strong the bonds are between the atoms
