Answer: The standard entropy of vaporization of ethanol is 0.275 J/K
Explanation:
Using Gibbs Helmholtz equation:
For a phase change, the reaction remains in equilibrium, thus 
Given: Temperature = 285.0 K
Putting the values in the equation:
Thus the standard entropy of vaporization of ethanol is 0.275 J/K
B. nuclear to thermal to mechanical to electrical
There are 34 g of oxygen in the container.
We can use the<em> Ideal Gas Law</em> to solve this problem.
But 
, so
and
STP is 0 °C and 1 bar, so
817.567 mm hg the answer for number 2
Answer:
58.9g of SO2 is produced
8g of oxygen remains unconsumed
Explanation:
The burning of Carbon disulfide (CS2) in oxygen. gives the reaction:
CS2 (g) + 3O2 (g) → CO2 (g) + 2SO2 (g)
Molar mass of CS2 = 76.139 g/mol
Molar mass of O2 = 15.99 g/mol
Molar mass of SO2 = 64.066 g/mol
Number of moles of CS2 = 35g/ 76.139 g/mol =0.46 moles
Number of moles of O2 = 30g/15.999 g/mol =1.88 moles
From the chemical reaction
1 mole of CS2 react with 3 moles of O2 to give 2 moles of SO2
Thus 0.46 moles of CS2 reacts to form 2× 0.46 = 0.92 moles of SO2
Mass of SO2 produced = 0.92×64.07 = 58.9g of SO2 is produced
thus 0.46 moles of CS2 reacts with 3 × 0.46 moles of O2 which is =1.38 moles of O2
Thus oxygen is the limiting reactant with 1.88 - 1.38 = 0.496~~0.5 mole remaining
Or 8g of oxygen
58.9g of SO2 is produced
oxygen is the limiting
