A 17.5 mL sample of an acetic acid (CH3CO2H) solution required 29.6 mL of 0.250 M NaOH for neutralization. The concentration of acetic acid was M. A 17.5 mL sample of an acetic acid (CH3CO2H) solution required 29.6 mL of 0.250 M NaOH for neutralization. The concentration of acetic acid was M. 0.591 0.423 129.5 0.296 0.105

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.