Yes, chromium(II) oxide (CrO2) is an ionic compound. It is composed of chromium ions (Cr2+) and oxide ions (O2-) held together by ionic bonds.
CrO2 is a covalent compound. It consists of two nonmetals (chromium and oxygen) which share electrons to form covalent bonds within the molecule.
Chromium oxide (Cr2O3) is an ionic compound, since it is made up of a metal and a nonmetal.
H3(CrO2)4 is the chemical formula of the peroxychromic acid.
No.MgO has a giant ionic lattice structure with strong covalent bonds.for ionic compounds,they can only conduct electricity in aqueous or molten state because the giant ionic lattice structure had collapsed and ions are free to move about to conduct electricity.but in solid state,the ions will be held in fixed position and thus is unable to conduct electricity:)
ionic bond
CrO2 is a covalent compound. It consists of two nonmetals (chromium and oxygen) which share electrons to form covalent bonds within the molecule.
Chromium oxide (Cr2O3) is an ionic compound, since it is made up of a metal and a nonmetal.
CrO2-4
Chromium Dioxide. :)
I can't find any references to a "CRO2" anywhere, nor does it appear to be a military nomenclature. Please check again and repost with the correct nomenclature.
CrO2this is the incorrect formula. For Chromium II oxide the formula is actually CrO without the 2. The reason for this is simple. the II after chromium indicates that it has a charge of +2, and the oxygen, we know is in group 16, has a charge of -2. The formula CrO2 is actually the formula for chromium IV oxide>
Balance the oxidation states on the atoms in the molecule. An oxide always contains oxygen in -2 oxidation state , in ionic compounds this is O2-. As the chromium is Cr(IV) in +4 oxidation state, ( shown as an ion Cr4+ ) you can balance the states or charges which ever way to get CrO2 (which you can write as Cr4+ (O2-)2 )
Because chromium is a plurivalent chemical element; four oxides are known today: CrO2, CrO3, Cr2O3, Cr3O4.
H3(CrO2)4 is the chemical formula of the peroxychromic acid.
To find the volume of the Fe2+ solution needed for titration, we can use the equation M1V1 = M2V2 where M1 is the concentration of Fe2+ (3.85 M), V1 is the volume of Fe2+ solution needed, M2 is the concentration of CrO2-4 (0.125 M), and V2 is the volume of CrO2-4 solution (250.0 mL). Rearranging the equation gives us V1 = (M2*V2) / M1 = (0.125 M * 250.0 mL) / 3.85 M ≈ 8.11 mL. So, approximately 8.11 milliliters of the Fe2+ solution are needed to titrate 250.0 mL of the CrO2-4 solution.
PtO2 is ionic
Potassium iodide is ionic.