Yes
When sodium atoms react with iodine atoms, they undergo a redox reaction to form sodium iodide. Sodium atoms lose an electron to form Na+ ions, while iodine atoms gain an electron to form I- ions. These ions then combine to form sodium iodide crystals.
Yes. Rubidium is an alkali metal in the sodium group. It will react with iodine to form rubidium iodide:- 2Rb+ I2 -> 2RbI
The compound is actually called Sodium Iodide. 1 molecule of Sodium Iodide contains 1 atom of Sodium and 1 atom of Iodine. Chemical formula = NaI
Sodium Iodide. Here is the balanced reaction eq'm 2Na(s) + I2(s) = 2NaI(s)
The word equation for the reaction between sodium and iodine is: sodium + iodine → sodium iodide.
The correct formula for the product of lead(II) nitrate reacting with sodium iodide is lead(II) iodide, which has the formula PbI2. This is because lead(II) ions from lead(II) nitrate react with iodide ions from sodium iodide to form lead(II) iodide as the product.
You can separate a mixture of iodine solid and sodium iodide by using the difference in solubility of the two compounds. Since sodium iodide is soluble in water while iodine is not, you can dissolve the mixture in water to dissolve the sodium iodide, leaving the solid iodine behind. The two can then be separated by filtration.
Sodium iodide has the chemical formula NaI - it is composed from sodium and iodine.
Sodium iodide is composed of sodium (Na) atoms and iodine (I) atoms. Sodium contributes the positive charge, while iodine provides the negative charge.
Yes, potassium iodide would react with sodium carbonate to form potassium carbonate and sodium iodide. This reaction is a double displacement reaction where the cations and anions are exchanged between the compounds.
The ionic compound of sodium iodide is NaI. It is composed of sodium (Na+) ions and iodide (I-) ions held together by ionic bonds.
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.