What is the formula for the theoretical extraction efficiency of a liquid-liquid extraction, in terms of the fraction of the analyte left in the original phase? if nobody has ever made you calculate this before, the chemistry libretexts page on liquid-liquid extractions will be .

Explanation:

Hello,

The requested formula is:

Whereas is the cumulative fraction remaining after the extraction, the distribution coefficient and the the volume for the aqueous and organic phases

It is suitable when we assume that the analyte is originally at the aqueous phase and the aforesaid fraction stands for the efficiency as well (https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Book%3A_Analytical_Chemistry_2.0_(Harvey)/07%3A_Collecting_and_Preparing_Samples/7.7%3A_Liquid%E2%80%93Liquid_Extractions)

Best regards.

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.