Answer:
The reactants Cu(SO)₄ and Na₂(NO)₃ are reacting to produce Na₂(SO)₄ and Cu(NO)₃
Explanation:
When we describe a reaction by a chemical reaction, we put the reactants in on the side and the other side the products. A simple arrow (→) represents an irreversible reaction, the double arrow (⇄) represents a reversible reaction (the products are forming the reactants too).
So, for the reaction given, it's irreversible, and the reactants Cu(SO)₄ and Na₂(NO)₃ are reacting to produce Na₂(SO)₄ and Cu(NO)₃.
NaCl, molecules are ionic bonding this is covalent
Br- + 2MnO4- + 2H+ → BrO3- + 2MnO2 + H2O
<h3>Further explanation</h3>
Given
MnO4- + Br- = MnO2 + BrO3-
Required
The half-reaction
Solution
In acidic conditions :
1. Add the coefficient
2. Equalization O atoms (add H₂O) on the O-deficient side.
3. Equalization H atoms (add H⁺ ) on the H -deficient side. .
4. Equalization of charge (add electrons (e) )
5. Equalizing the number of electrons and then adding the two half -reactions together
Oxidation : Br- → BrO3-
Reduction : MnO4- → MnO2
Br- + 3H2O → BrO3-
MnO4- → MnO2 + 2H2O
Br- + 3H2O → BrO3- + 6H+
MnO4- + 4H+ → MnO2 + 2H2O
Br- + 3H2O → BrO3- + 6H+ + 6e- x1
MnO4- + 4H+ + 3e- → MnO2 + 2H2O x2
- Equalizing the number of electrons and then adding the two half -reactions together
Br- + 3H2O → BrO3- + 6H+ + 6e-
2MnO4- + 8H+ + 6e- → 2MnO2 + 4H2O
Br- + 2MnO4- + 3H2O + 8H+ + 6e- → BrO3- + 2MnO2 + 6H+ + 4H2O + 6e-
Br- + 2MnO4- + 2H+ → BrO3- + 2MnO2 + H2O
The answer would be, D. because ive taken the test.