A 100 gram sample of naturally occurring boron contains 19.78 % of B-10 and 80.22 % of B-11. Which numerical setup can be used to determine the atomic mass of naturally occurring B

The required numerical set up is;

RAM = (19.78% * 10) + (80.22% * 11)

Explanation:

The relative atomic mass is the sum of the products of the mass of each isotope and its relative abundance.

Hence the relative atomic mass of Boron is obtained form;

RAM = (19.78% * 10) + (80.22% * 11)

RAM = (1.978) + (8.8242)

RAM =10.8022

n has four pairs of electrons in the valence shell, electrons repel one another ,n has 7 protons in its nucleus, and since n has 7 protons, it must fill the second shell, giving it 4 pairs of electrons. the electrons form 3 bonds and 1 lone pair of electrons. each pair of electrons repels the other pairs, so they are equally far apart.

so all of the above

- "the choice that is not an indicator of a chemical change is "state of matter changes". more common than not, chemical reactions produce energy in the form of light or heat. along with energy, they also produce a new substance called the product that could be in any state of matter (solid, gas, or liquid)."