Benzene has a heat of vaporization of 30.72 kj/mol and a normal boiling point of 80.1°c. at what temperature does benzene boil when the external pressure is 445 torr?

Explanation:

The given data is as follows.

            = = (80 + 273) = 353 K

  Atmospheric pressure = 445 torr

   Heat of vaporization = 30.72 kJ/mol =

                                      = 30720 J/mol

Now, according to Clausius-Clapereryon equation,

Putting the given values into the above equation as follows.

                   -0.5352 =                  

                    = 0.002977

                      = 335.909 K

or,                             =

                                =

Thus, we can conclude that benzene boils at when the external pressure is 445 torr.

2h2o2--> 2h2o + o2

this reaction is of the spontaneous decomposition of hydrogen peroxide down into water and oxygen.

add 2 molecules of hydrogen peroxide and 2 molecules of water. since oxygen is naturally diatomic, the total number of atoms of each element is now the same on both sides of the equation so it is balanced.

this may not be entirely correct but let me know.