Assume Boron has (only) two naturally occurring isotopes, B10 and B11. If Boron has an average atomic mass of 10.811amu, B10 has a mass of 10.0 amu and B11 has a mass of 11.0 amu, what are the abundancies of each isotope

B10: 18.9%

B11: 100% - 18.9% = 81.1%

Explanation:

The abundance of the isotope is calculating using;

(M1)(x) + (M2)(1-x) = M(E)

Where;

M1 = Mass of B10 = 10.0 amu

M2 = Mass of B11 = 11 amu

M(E) = Average mass of Boron = 10.811 amu

x = relative abundance

Inserting the values;

10(x)  +  11(1-x) = 10.811

10x + 11 - 11x = 10.811

10x - 11x = 10.811 - 11

-x = -0.189

x = 0.189

Converting to percentage; x = 18.9%

B10: 18.9%

B11: 100% - 18.9% = 81.1%

what is the question?

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.