21.42 g of calcium combined with 8.58 g of oxygen in a combustion reaction

Self-ionizes to form an equilibruium system in which [ h_{3} o^{+} ]=[ oh^{-} ]=10^{-7} m

Radon-222 or rn-222 has a half-life of 3.82 days. to determine the fraction of an original sample of rn-222 that remains unchanged after 7.646 days, we calculate the amount of rn-222 remaining after 7.646 days and divide it with the original amount. using 1 gram as a basis and using half-life calculation methods, we can determine that  0.249 or 0.25  grams of rn-222 will remain, and thus the fraction remaining is 0.25 or 1/4.

Elements would be the right answer