Iodine with one extra electron in its atom
The molar mass of lead(II) iodide (PbI₂) can be calculated by summing the atomic masses of its constituent elements. Lead (Pb) has an atomic mass of approximately 207.2 g/mol, and iodine (I) has an atomic mass of about 126.9 g/mol. Therefore, the molar mass of PbI₂ is calculated as follows: 207.2 g/mol (for Pb) + 2 × 126.9 g/mol (for the two I) = 460.1 g/mol. Thus, the molar mass of lead(II) iodide is approximately 460.1 g/mol.
Calcium iodide is an ionic compound composed of one calcium ion (Ca2+) and two iodide ions (I-). Therefore, there are a total of 3 ions present in calcium iodide.
The chemical equation for Strontium Iodide is SrI2, which represents the compound formed by the elements strontium (Sr) and iodine (I) combining in a 1:2 ratio.
Sodium Iodide is the solute which is created in the reaction and Sodium actetate solution is created. NaI + PbC2H3O2 ---> PbI (Plumbum Iodide) + NaC2H3O2 (Sodium Acetate) solution.
Zinc and Iodide combine to form zinc iodide. The reaction is a combination reaction and also a redox reaction. A combination reaction is when two elements combine to form a single compound. A redox reaction is when one of the element loses an electron (zinc) and the other one gains electrons (iodide). Hope this helps! ~Kim
The elements present in lead phosphate are lead, phosphorus, and oxygen. Lead phosphate has the chemical formula Pb3(PO4)2.
When solutions of lead nitrate and potassium iodide are mixed, a yellow precipitate of lead iodide is formed. This reaction is a double displacement reaction where the lead from lead nitrate reacts with the iodide from potassium iodide to form the insoluble lead iodide.
The compound PbI4 is lead (II) iodide, where the lead ion has a charge of +2 and the iodide ion has a charge of -1.
The reaction between potassium iodide and lead (II) nitrate forms lead iodide, which is a yellow pigment used in paint. Lead iodide provides a vibrant color and good coverage when used in paint formulations.
When reactants lead(II) nitrate and sodium iodide are combined, a double displacement reaction occurs. Lead(II) iodide (insoluble in water) and sodium nitrate are formed, leading to a white precipitate of lead(II) iodide and a solution of sodium nitrate.
The chemical equation for the reaction between lead nitrate (Pb(NO3)2) and potassium iodide (KI) to form lead iodide (PbI2) and potassium nitrate (KNO3) is: Pb(NO3)2 + 2KI → 2KNO3 + PbI2
Potassium Iodide- KI Lead Nitrate- Pb(NO3)2
Lead(II) nitrate and sodium iodide will yield lead(II) iodide and sodium nitrate. This is a double displacement reaction, where the cations and anions switch partners resulting in the formation of two new compounds.
PbI2 has TWO(2) elements. They are ;- Pb ; Lead (Plumbum ; Latin) I ; Iodine . However, there are three (3) atoms in PbI2. Viz: - 1 x Pb(Lead) 2 x I (iodine) 1 + 2 = 3 The name of the sibstance is 'lead iodide'.
A yellow precipitate of lead iodide is formed. This is because potassium iodide reacts with lead nitrate to form lead iodide, which is insoluble in water. The reaction can be represented as: 2KI + Pb(NO3)2 → 2KNO3 + PbI2.
This is a double displacement reaction. 2KI + Pb(NO3)2 --> 2KNO3 + PbI2 Potassium iodide + Lead(II) nitrate --> Potassium nitrate + Lead(II) iodide A bright yellow precipitate will form when these two react.
lead nitrate(Pb(NO3)2 + potassium iodide(KI) = lead iodide(PbI) + potassium nitrate (KNO3)