Suppose a layer of oil is on the top of a beaker of water. water in many oils is slightly soluble, so its concentration is so low that we can treat it as an ideal solute. 1) suppose that at 283 k, the equilibrium concentration is 9 × 10-4 water molecules per oil molecule, and it takes 2.208 × 10-20 j to transfer one water molecule into the oil. what is the equilibrium concentration at 293 k, assuming that nothing else changes

Explanation:

It is given that energy to transfer one water molecule is J/molecule

As it is known that in 1 mole there are atoms.

So, energy in 1 mole = J/mol

                                  = 13.3 kJ/mol

As,    

Putting the given values in the above formula as follows.

                  = 0.08377

                        = 1.213 =

                  =

                                             =

                                             = water molecules per oil molecule

Thus, we can conclude that the equilibrium concentration at 293 K, assuming that nothing else changes is .

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.