Ionic bonds are formed when one of the two atoms that are reacting has excess electrons and transfer the electrons to the atom that is deficient in electrons. During the formation of the ionic bond, one of the reacting atoms will donate electrons and form positive ion.
First, we must know what happens in the precipitation reaction. This type of reaction is a double replacement reactions. It is consists of two reactant compounds which interchange cations and anions to form two products. One of the products is an insoluble solid called a precipitate. For the precipitation of CaCO₃, there are two consecutive reactions involved:
1. Slaking of quicklime, CaO
CaO + H₂O ⇒ Ca(OH)₂
2. Precipitation
Ca(OH)₂ + CO₂ ⇒ CaCO₃ + H₂O
The ions that make up the H₂O molecule are H⁺ and OH⁻. According to solubility rules, the cation (positively charged ion) is likely to be attracted to an anion (negatively charged ion). Together, they form an ionic bond. This type of bond is when there is a complete transfer of electrons between the two. The Ca²⁺ cation lacks 2 electrons, while the anion OH⁻ has an excess 1 electron. In order to be stable, 1 Ca²⁺ ion and 2 OH⁻ ions must combine.
Therefore, the answer is OH⁻ ion.
Answer:
[ S2- ] = 4.0 E-47 M
Explanation:
- PbS(s) → Pb2+ + S2-
- HgS(s) → Hg2+ + S2-
∴ Ksp PbS = 3.4 E-28 = [Pb2+]*[S2-]
∴ [Pb2+] = 0.181 M
∴ Ksp HgS = 4.0 E-53 = [Hg2+]*[S2-]
∴ [Hg2+] = 0.174 M
∴ Ksp PbS > Ksp HgS ⇒ precipitate first Hg2+:
∴ [ Hg2+ ] = 1.0 E-6 M
⇒ [S2-] = 4.0 E-53 / 1.0 E-6 = 4.0 E-47 M
Answer:
The chiefs
Explanation:
hope my answers is helpful
Answer:
H2SO4 + 2NaOH ---> Na2SO4 + 2H2O
Explanation:
H2SO4 + NaOH ---> Na2SO4 + H2O
Na1 ---> 2Na
H2SO4 + 2NaOH ---> Na2SO4 + H2O
2H + 2H = 4H ----> 2H
H2SO4 + 2NaOH ---> Na2SO4 + 2H2O
4O + 2O -----> 4O + 2O
