2Na + I2 ==> 2NaI
Sodium Iodide. Here is the balanced reaction eq'm 2Na(s) + I2(s) = 2NaI(s)
The balanced equation for fluorine reacting with sodium iodide is: 2 NaI + F2 → 2 NaF + I2
The ionic compound of sodium iodide is NaI. It is composed of sodium (Na+) ions and iodide (I-) ions held together by ionic bonds.
Solid sodium iodide reacts with water to form sodium ions and iodide ions. No further reactions occur because this compound is a salt of a strong acid and a strong base. Thus, the overall reaction is NaI(s)=> Na+ + I-
When sodium iodide is combined with chlorine, sodium chloride and iodine are produced as the products of the reaction. The balanced equation is: 2NaI + Cl2 → 2NaCl + I2.
Pb(NO3)2(aq) + 2NaI(aq) → PbI2(s) + 2NaNO3(aq) Aqueous lead II nitrate reacts with aqueous sodium iodide to form solid lead II iodide precipitate and aqueous sodium nitrate.
The IUPAC name for sodium iodide is sodium iodide.
Sodium iodide
The chemical equation is: Na+I- (aq) + Ag+[NO3]- (aq) --> AgI (s) + Na+[NO3]- (aq)
The formula for sodium iodide is NaI. It is formed by the combination of sodium (Na) and iodide (I-) ions, with sodium donating an electron to iodine to form a stable compound.
Sodium iodide is a compound and that is its name.
Sodium iodide is an ionic compound. It is composed of sodium cations (Na+) and iodide anions (I-), which are held together by ionic bonds formed through the transfer of electrons from sodium to iodine.