. The molar heat of vaporization of acetone, C3H6O, is 30.3 kJ/mol at its boiling point. How many kilojoules of heat would be liberated by the condensation of 5.00 g of acetone?

2.61 kilojoules of heat would be liberated by the condensation of 5.00 g of acetone

Explanation:

To convert grams into moles

We have 5.00 g acetone

Heat liberated = moles heat of vapourization

                         =0.0861 mol x 30.3 kJ/mol

                        = 2.61 kJ

Therefore, 2.61 kilojoules of heat would be liberated by the condensation of 5.00 g of acetone

1.consider the following balanced equation and answer parts a through. show calculations for each part.  

3 ag(s) + 4 hno3(aq) → 3 agno3(aq) + no(g) + 2 h2o(l)  

a. calculate the number of moles of agno3 produced by the reaction of 3 moles of ag with excess hno3. if the number to the left of the compound represents the number of moles of the compound which are present, then 3 moles of ag react with 4 moles hno3 to produce 3 moles agno3 (aq).

so the answer to a would be 3 moles.

b. calculate the number of moles of agno3 produced by the reaction of 108 grams of ag with excess hno3.  

i'm not sure how to plug in the grams of silver into an equation using moles to find how many moles are produced.

c. calculate the number of grams of agno3 produced by the reaction of 108 grams of ag with excess hno3  

d. calculate the number of moles of no produced as a byproduct of the reaction of 108 grams of ag with excess hno3.  

e. calculate the number of grams of no produced as a byproduct starting with 108 grams of ag and excess hno3.  

explanation:

a boiling water reactor does not boil the water. a pressurized water reactor boils the water faster. a pressurized water reactor is heated directly by the core. a pressurized water reactor does not use fission.