Major species in the solution present in higher concentrations about more than 10% of the other species in the solution. The nature of solute that is dissolved in the solution, or the extent of reaction of different groups with the solvent decide the major species present in the solution.
For weak acids(bases) dissolved in water, the major species will be the undissociated acid(base) and water as the extent of dissociation of the molecular weak acid(weak base) is relatively less. For ionic solutes, the major species after dissolution in water will be the ions that form in the solution in addition to water.
Iron (II) iodide (FeI
)is an ionic solute that is completely soluble in water. The following will be the reaction that represents solubility of FeI2 in water:![FeI_{2}(s) + H_{2}O(l)-->Fe^{2+}(aq)+2I^{-}(aq)](https://tex.z-dn.net/?f=FeI_%7B2%7D%28s%29%20%2B%20H_%7B2%7DO%28l%29--%3EFe%5E%7B2%2B%7D%28aq%29%2B2I%5E%7B-%7D%28aq%29)
So, the major species when iron(II) iodide is dissovled in water will be ![Fe^{2+}, I^{-}, H_{2}O](https://tex.z-dn.net/?f=Fe%5E%7B2%2B%7D%2C%20I%5E%7B-%7D%2C%20H_%7B2%7DO)
Na has a one proton over than electrons.so Na makes a Na+
Cl has a one electron overthan protons.so Cl makes a Cl-
proton has a + affectation
electron has a - affectation ..
Avogadro's number tells you that in one mol of element (sulfur) there are 6.022x10^23 atoms...
So,
1 mol : 6.022x10^23 = x mol : 1.2x10^2
x = 1.99 x 10^-22 mol
*Although this should be correct, you might want to check your question again. You probably misinterpret the question... you want 1.2x10^23 !! :) check it.
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Answer:</h3>
B. It loses an electron and has an octet in its inner-most shell
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Explanation:</h3>
- Rubidium is a metal in the chemical family known as alkali metals.
- Metallic elements form ions known as cation by losing electron(s) from their outermost energy levels to attain an octet configuration.
- Rubidium has one electron in the outermost energy level, and thus, loses the electron to form a cation and forms an octet configuration.
- Like other alkali metals such as potassium and sodium, it forms a cation with a charge of + 1.