The chemical formula for copper bromide containing the Cu+ ion is CuBr. This compound consists of a copper cation with a +1 charge (Cu+) and a bromide anion with a -1 charge (Br-), thus requiring one atom of each element to balance the charges.
The formula for iron reacting with copper(II) nitrate is: Fe + Cu(NO3)2 -> Fe(NO3)2 + Cu. This is a single displacement reaction where iron displaces copper from copper nitrate to form iron(II) nitrate and copper metal.
When CuSO4 (copper(II) sulfate) reacts with KBr (potassium bromide), they will exchange ions to form CuBr2 (copper(II) bromide) and K2SO4 (potassium sulfate). This is a double displacement reaction where the cations from each compound switch places.
The balanced chemical equation for this reaction is: 2AlBr3 + 3Cl2 → 2AlCl3 + 3Br2.
Magnesium bromide is the compound formed when magnesium reacts with bromine.
I'm assuming this was meant to be seeking the chemical formula for Iron (III) bromide, also known as Ferric bromide. In this case, the cation is Fe3+ (The iron III part of the name) The anion is Br- (the bromide part of the name) Since it is an ionic compound, you simply have to put the two ions within the compound in a ratio that will cause their net charge to balance out to zero. If we had one Fe3+ ion and three Br- ions, we would cancel out the charge, since (1)(3)+(3)(-1)=0 (1 ion of plus-3 charge + 3 ions of minus-1 charge). The end result is this formula: FeBr3
Copper sulfide. Depending on the valence of copper, you could have CuS or Cu2S
There are two different copper oxide formula's:2Cu + O2 -> 2CuO (black Copper(II) oxide)or4Cu + O2 -> 2Cu2O (red Copper(I) oxide)
The formula for iron reacting with copper(II) nitrate is: Fe + Cu(NO3)2 -> Fe(NO3)2 + Cu. This is a single displacement reaction where iron displaces copper from copper nitrate to form iron(II) nitrate and copper metal.
Cu2+ + I- --> Cu2I The compound created is Copper(I) Iodide
The chemical formula of cupric oxide is CuO.
When CuSO4 (copper(II) sulfate) reacts with KBr (potassium bromide), they will exchange ions to form CuBr2 (copper(II) bromide) and K2SO4 (potassium sulfate). This is a double displacement reaction where the cations from each compound switch places.
lead + bromine ---> lead bromide Pb + Br2 ---> PbBr2 As a chemical compound containing lead, it is categorised as probably carcinogenic to humans. It is a white powder at room temperature.
The balanced chemical equation for this reaction is: 2AlBr3 + 3Cl2 → 2AlCl3 + 3Br2.
Magnesium bromide is the compound formed when magnesium reacts with bromine.
Sodium Bromide
First, it's ClO3 with a lowercase L. There are two possible compounds: copper I chlorate, CuClO3 or copper II chlorate Cu(ClO3)2
I'm assuming this was meant to be seeking the chemical formula for Iron (III) bromide, also known as Ferric bromide. In this case, the cation is Fe3+ (The iron III part of the name) The anion is Br- (the bromide part of the name) Since it is an ionic compound, you simply have to put the two ions within the compound in a ratio that will cause their net charge to balance out to zero. If we had one Fe3+ ion and three Br- ions, we would cancel out the charge, since (1)(3)+(3)(-1)=0 (1 ion of plus-3 charge + 3 ions of minus-1 charge). The end result is this formula: FeBr3