Answer:
704.6 g CO2
Explanation:
MM sucrose = 342.3 g/mol
MM CO2 = 44.01 g/mol
g CO2 = 456.7 g sucrose x (1 mol sucrose/MM sucrose) x (12 moles CO2/1 mol sucrose) x (MM CO2/1mol CO2) = 704.6 g CO2
In order to answer this, we mus know the data for the heat of combustion of propane. This is an empirical data that you can search online. The heat of combustion is -2220 kJ/mol. The molar mass of propane of 44.1 g/mol. The solution is as follows:
ΔH = -2220 kJ/mol (1 mol/44.1 g)(1000g/1kg)(20 kg)
<em>ΔH = -1006802.721 kJ or -1 GJ</em>
Answer: Nucleus
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Answer:
Thus, the joules of heat absorbed when 70.0 grams of water is completely vaporised at its boiling point are 158,200.
Answer:
A
Explanation:
because the object's velocity had no acceleration between time intervals
