Phosphorous-32 (P-32) decays to sulfur-32 (S-32) with a half life of 14 days. What fraction of the original amount of P-32 will remain after 28 days? Group of answer choices

1/4 of the original amount

1/16 of the original amount

1/2 of the original amount

1/8 of the original amount

Because electrons have negligible mass, to account for the mass of the isotope, there must be 16 neutrons, 16 neutrally charged, massive, fundamental particles present in the sulfur nucleus. and thus finally, if we have the 32s2− ion, there are 18 electrons, 16 protons, and 16 neutrons.

257 grams h2o i think

It could possibly be beneficial if the experiment is successful.

Idont really know sorry