Ammonium Iodide
The formula for the ionic compound formed from ammonium and iodide ions is NH4I. Ammonium ion has a charge of +1 and iodide ion has a charge of -1, so they combine in a 1:1 ratio to form a neutral compound.
Oxidation numbers of the elements in NH4I are N: -3 H: +1 (4 times) I: -1 Overall: NH4+ : +1 and I- : -1
The formula for ammonium iodide is NH4I. It consists of one ammonium ion (NH4+) and one iodide ion (I-).
No, NH4I is not a strong base. It is actually a salt formed from ammonium ion (NH4+) and iodide ion (I-) and does not dissociate completely in water to release hydroxide ions, which are characteristic of strong bases.
NH4I + H2O --> NH4OH + H+ + I-HI being a strong acid, remains almost wholly ionised as H+ & I- ionNH4OH being a weak base so you have some more H+ left therefore it is acidic solution. (problem: NH4OH is not an exsisting compound, most of it will react with the strong acid H+ and the remainig will be fully ionized into NH4+)Added:(problem solved: not using NH4OH as it is annon-exsisting compound)NH4I will completely dissolve in ammonium and iodide ions:NH4I --> I- + NH4+of which I- is neutral and NH4+ is quite a weak acid (pKa = 9.25). (The molar standard solution of NH4I will have pH approx. 4.6)
The chemical formula NH4I is for ammonium iodide.
The formula for the ionic compound formed from ammonium and iodide ions is NH4I. Ammonium ion has a charge of +1 and iodide ion has a charge of -1, so they combine in a 1:1 ratio to form a neutral compound.
NH4I is a compound consisting of ammonium cations (NH4+) and iodide anions (I-). It is named ammonium iodide.
The chemical formula for ammonium iodide is NH4I.
yes
Oxidation numbers of the elements in NH4I are N: -3 H: +1 (4 times) I: -1 Overall: NH4+ : +1 and I- : -1
The formula for ammonium iodide is NH4I. It consists of one ammonium ion (NH4+) and one iodide ion (I-).
No, NH4I is not a strong base. It is actually a salt formed from ammonium ion (NH4+) and iodide ion (I-) and does not dissociate completely in water to release hydroxide ions, which are characteristic of strong bases.
Ammonium iodide is NH4I Silver nitrate is AgNO3
When NH4I is dissolved in H2O, it will dissociate into its ions, forming NH4+ and I- ions in the solution. This reaction is considered a dissociation reaction rather than a chemical reaction where new substances are formed.
The name of this compound is iodine heptafluoride.
NH4I + H2O --> NH4OH + H+ + I-HI being a strong acid, remains almost wholly ionised as H+ & I- ionNH4OH being a weak base so you have some more H+ left therefore it is acidic solution. (problem: NH4OH is not an exsisting compound, most of it will react with the strong acid H+ and the remainig will be fully ionized into NH4+)Added:(problem solved: not using NH4OH as it is annon-exsisting compound)NH4I will completely dissolve in ammonium and iodide ions:NH4I --> I- + NH4+of which I- is neutral and NH4+ is quite a weak acid (pKa = 9.25). (The molar standard solution of NH4I will have pH approx. 4.6)