To calculate the molarity of magnesium ions (Mg²⁺) in a solution, first, determine the number of moles of magnesium. The molar mass of magnesium is approximately 24.31 g/mol. Therefore, 48.6 g of Mg corresponds to about 2.00 moles (48.6 g ÷ 24.31 g/mol). Molarity (M) is calculated as moles of solute divided by liters of solution. Thus, the molarity of magnesium ions in 2 liters of water is 2.00 moles ÷ 2 L = 1.00 M.
8.79 grams of magnesium sulfate will remain.
The compound (NH4)2O contains three different elements: nitrogen (N), hydrogen (H), and oxygen (O). There are two ammonium ions (NH4+) in the formula, which contribute nitrogen and hydrogen, along with one oxide ion (O2-). Therefore, the total number of elements present is three.
Ch3coo(-) + h2o ---> ch3cooh + oh(-)
2O2 in a chemical reaction equation means that there are two moles of O2, or oxygen, molecules.
Na+OH-(aq) + H+Cl-(aq) Na+Cl-(aq) + H+2O-(l)
8.79 grams of magnesium sulfate will remain.
The compound is ammonium oxide. You can tell because the NH4+ and O2- ions are present.
There is a compound ammonium hydoxide, NH4OH (cannot be isolated - only in solution) there is no ammonium oxide, if there wrer it would have the forula (NH4)2O
2o mg for 2o gm of soil
2o ml
30...
(cl)2o
a pint is 2o killograhms
about 2o usd
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This formula is for ammonium oxide.
about 44.