MgF2
The empirical formula of the compound formed between magnesium and fluorine is MgF2. This is because magnesium has a 2+ charge and fluorine has a 1- charge, so one magnesium ion will combine with two fluorine ions to achieve a neutral compound.
The empirical formula for the ionic compound formed by sodium and fluorine is NaF. Sodium is a metal that gives away one electron, becoming Na+, while fluorine is a non-metal that gains one electron, becoming F-. The resulting compound has a 1:1 ratio of sodium to fluorine ions, giving NaF as the empirical formula.
Mg + F2 => MgF2 (Magnesium fluoride, salt, soluble in water)metal + gas .. saltMgF2
The empirical formula for a compound containing 13% magnesium and 87% bromine is MgBr2. This is because the ratio of magnesium to bromine atoms in the compound is 1:2, which corresponds to the formula MgBr2.
The empirical formula of the ionic compound formed by sodium and fluorine is NaF, which is sodium fluoride. Sodium typically forms a +1 cation (Na+) and fluorine typically forms a -1 anion (F-), leading to a one-to-one ratio in the compound.
The empirical formula of the compound formed between magnesium and fluorine is MgF2. This is because magnesium has a 2+ charge and fluorine has a 1- charge, so one magnesium ion will combine with two fluorine ions to achieve a neutral compound.
the empirical formula for fluorine is F. the chemical formula is F2.
The empirical formula for the ionic compound formed by sodium and fluorine is NaF. Sodium is a metal that gives away one electron, becoming Na+, while fluorine is a non-metal that gains one electron, becoming F-. The resulting compound has a 1:1 ratio of sodium to fluorine ions, giving NaF as the empirical formula.
MgF2 is the formula for magnesium fluoride
Mg2F
Mg + F2 => MgF2 (Magnesium fluoride, salt, soluble in water)metal + gas .. saltMgF2
The empirical formula for a compound containing 13% magnesium and 87% bromine is MgBr2. This is because the ratio of magnesium to bromine atoms in the compound is 1:2, which corresponds to the formula MgBr2.
The empirical formula of this compound would be MgO.
The empirical formula of the fluoride of uranium can be determined by converting the percentages to moles. Since uranium has an atomic mass of about 238 g/mol and fluorine 19 g/mol, the ratio of moles of fluorine to uranium is approximately 4:1. Therefore, the empirical formula is UF4.
To find the empirical formula, convert the masses of each element to moles. The molar ratio of carbon to chlorine to fluorine is 1:1:2. Therefore, the empirical formula is CClF2.
The empirical formula for strontium (Sr) is Sr and for fluorine (F) is F.
The empirical formula of the ionic compound formed by sodium and fluorine is NaF, which is sodium fluoride. Sodium typically forms a +1 cation (Na+) and fluorine typically forms a -1 anion (F-), leading to a one-to-one ratio in the compound.