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What is the empirical formula of the oxygen fluoride?
sodium + Na+ and flouride - F- NaF The chemical formula of sodium fluoride is NaF.
Iron III fluoride is FeF3 Note that it is iron III fluoride not iron 111 fluoride.
Formula for aluminum fluoride: AlF3
Sodium Fluoride is made up of one sodium atom and one fluorine atom...therefore, the formula is NaF
40.08 + 2*19.00 = 78.08 g/mol
Manganese IV fluoride: The first thing you should realize right off the bat is that this compound is composed of the elements manganese (Mn) and fluorine (F). My guess is that… it is the IV throwing you off. :) The IV is usually in parentheses and indicates the positive charge on the preceding element. These designations are used if the preceding element can have more than one charge. It is generally only used with the transition metals (iron is the classic example). Therefore, the manganese (Mn) has a charge of +4. Since we also have fluoride, we can create our compound! Fluoride has a -1 charge. So... Mn4+ and F- The combination to give this compound a charge of zero is: MnF4
What is the empirical formula of the compound with a certain fluoride of vanadium is 40.13 percent in vanaium?