Answer:
different isotopes of na element
Chemical Reactions and Moles of Reactants and Products
That is, it requires 2 moles of magnesium and 1 mole of oxygen to produce 2 moles of magnesium oxide. If only 1 mole of magnesium was present, it would require 1 ÷ 2 = ½ mole of oxygen gas to produce 2 ÷ 2 = 1 mole magnesium oxide.
This is true. Water is the solvent in aqueous solutions
LiBr.
<h3>Explanation</h3>
Note that the group number in this answer refers to the new IUPAC group number, which ranges from 1 to 18. Counts from the left. Start with the first two column (group 1 and 2), go on to the transition elements (Sc, Ti, etc. in group 3 through 12), and continue with the nonmetals (group 13 through 18).
Li is a group 1 metal. As a metal, it tends to form positive ions ("cations"). Metals in group 1 and 2 are <em>main group</em> metals. The charge on main group metal ions tends to be the same as the group number of the metal. Li is in group 1. The charge on an Li ion will be +1. Formula of the Li ion will be
.
Br is a group 17 nonmetal. As a nonmetal, it tends to form negative ions ("anions"). The charge on nonmetal ions excepting for H tends to equal the group number of the nonmetal minus 18. Br is in group 17. The charge on a Br ion will be 17 - 18 = -1. Formula of the Br ion will be 
All the ions in an ionic compound carry charge. However, some of the ions like
are positive. Others ions like
are negative. Charge on the two types of ions balance each other. As a result, the compound is <em>overall</em> neutral.
1 × (+1) + 1 × (-1) = 0. The positive charge on one
ion balances the negative charge on one
ion. The two ions would pair up at a 1:1 ratio.
The empirical formula for an ionic compound shows all the ions in the compound. Positive ions are written in front of negative ions.
is positive and
is negative. The formula shall also show the simplest ratio between the ions. For the compound between Li and Br, a 1:1 ratio will be the simplest. The "1" subscript in an empirical formula can be omitted. Hence the formula: LiBr.
The first thing we need to do here is to recognize the unit of molarity and the units of the given percentage of nitric acid.
Molarity is mol HNO3 / L of solution. This is our aim
The given percentage is 0.68 g HNO3/ g solution
multiplying this with density to convert g solution into mL solution and dividing with the molecular weight of HNO3 (63 g/mol) to convert g HNO3 to mol. Therefore we obtain
0.016 mol/ mL or 16.23 mol/ L (M)