How much energy must be removed from a 125 g sample of benzene (molar mass= 78.11 g/mol) at 425.0 k to liquify the sample and lower the temperature to 335.0 k? the following physical data may be useful.δhvap = 33.9 kj/molδhfus = 9.8 kj/molcliq = 1.73 j/g°ccgas = 1.06 j/g°ccsol = 1.51 j/g°ctmelting = 279.0 ktboiling = 353.0 k

Explanation:

First, we will calculate the number of moles of benzene as follows.

      No. of moles =

                            =

                            = 1.60 moles

We assume that the heat energy is removed from the benzene in three separate steps:

1) In cooling the gaseous sample from 425K to 353K (its boiling point)

2) In liquefying the sample at 353 K

3) In cooling the liquid from 353 K to 335 K

So, we will first calculate the energy removed from the benzene at each step, then add them together to find the total amount of energy removed from the benzene.

1)  Q =

       = 125 x 1.06 x (425 - 353)

       = 9540J

or,    = 9.540 kJ        (as 1 kJ = 1000 J)

2) Also,  Q =

                  = 33.9 x 1.6

                 = 54.24 kJ

3)  Q =

       = 125 x 1.73 x (353 - 335)

       = 3893J

       = 3.893kJ

Hence, total = 9.540 + 54.240 + 3.893 = 67.673 kJ

Thus, we can conclude that 67.673 kJ energy must be removed from a 125 g given sample of benzene under the given conditions.

1. a)c-14 has two more protons

2. d)they have different numbers of protons, neutrons, and valence electrons; they are all gases at room temperature.

3. c) iron and nickel.