Fe2O3(s) 2Al(s)Al2O3(s) 2Fe(s) Using standard absolute entropies at 298K, calculate the entropy change for the system when 1.73 moles of Fe2O3(s) react at standard conditions.

Explanation:

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In this case, for the given reaction, we can write the equation to compute the entropy change as shown below:

Letting:

We obtain the entropy change per mole of Fe2O3(s):

Finally, the total entropy change when 1.73 moles of Fe2O3(s) react turns out:

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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.