What isotope is the product of the alpha (*2He) nuclear decay of Radon-222 (222U)? 222 Rn
86
a +
2

ⁿₐX => ²¹⁸₈₄Po

Explanation:

Let ⁿₐX be the isotope.

Thus, the equation can be written as follow:

²²²₈₆Rn —> ⁴₂α + ⁿₐX

Next, we shall determine the value of 'n' and 'a'. This can be obtained as follow:

222 = 4 + n

Collect like terms

222 – 4 = n

218 = n

Thus,

n = 218

86 = 2 + a

Collect like terms

86 – 2 = a

84 = a

Thus,

a = 84

ⁿₐX => ²¹⁸₈₄Po

²²²₈₆Rn —> ⁴₂α + ⁿₐX

²²²₈₆Rn —> ⁴₂α + ²¹⁸₈₄Po

where are the choice options?

nacl or sodium chloride is an ionic compound. this is because an ionic compound pertains to a compound that consists of positively and negatively charged ions. in this case, na (sodium) has a tendency to lose one electron when it reacts with with other atoms, so the resulting sodium ion contains 11 protons and 10 electrons, thus making it positively charged (na+), when chlorine reacts, it usually gains one electron to form an ion called chloride, thus making it negatively charged ( since the ions are oppositely charged, they attract.