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How many moles of ionic species are present in 1.0 L of a solution marked 1.0 M mercury(I) nitrate?

according to stoichiometry of reaction, 1 mole of $\mathrm{HgCl}_{2}$ , will produce 1 mole of $\mathrm{Hg}^{2+}$ and 2 moles of $\mathrm{Cl}^{-}$ . Or, One mole of $\mathrm{HgCl}_{2}$ will produce 3 mole of ionic species.

Chemistry 101

Chapter 18

Representative Metals, Metalloids, and Nonmetals

Nonmetals Chemistry

University of Central Florida

Rice University

University of Kentucky

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to figure out how many moles of my own expiate seas are in one leader of this one water solution, uh, first made formula of mercury nitrate. Mercury one is a little funny. His H G to two plus in nitrate is another three minus. So the Ionic compound farm from these two ions is HT two and 32 Now, when it dissolves, it forms the mercury moron I on which is a treat to plus and clearly two moles. Event of three minus. So there is a total of three onyx species in the solution. Now let's go back and recall that one point. Oh Moeller means one point of malls for every warm leader. Now we're told that we have one leader a solution. So I have 1.11 point of moles of rectory Want nitrate. So I have one of these and three on a species. So my one mole of mercury one nitrate, as we said, makes three Janek species of That's a total of three times as many molds of ionic species. So I have 3.0 moles. Bionic species, one mole of mercury, two moles of my treat

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