Answer:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
Explanation:
First, write the half equations for the reduction of MnO4^- and the oxidation of C2O4^2- respectively. Balance it.
Reduction requires H+ ions and e- and gives out water, vice versa for oxidation.
Reduction:
MnO4^- (aq) + 4H^+ (aq) + 3e- ---> MnO2(s) + 2H2O (l)
Oxidation:
C2O4^2- (aq) + 2H2O (l) ---> 2CO3^2 -(aq) + 4H^+ (aq) + 2e-
Balance the no. of electrons on both equations so that electrons can be eliminated. we can do so by multiplying the reduction eq by 2, and oxidation eq by 3.
2MnO4^- (aq) + 8H^+ (aq) + 6e- ---> 2MnO2(s) + 4H2O (l)
3C2O4^2- (aq) + 6H2O (l) ---> 6CO3^2 -(aq) + 12H^+ (aq) + 6e-
Now combine both equations and eliminate repeating H+ and H2O.
2MnO4^- (aq) + 8H^+ (aq) + 3C2O4^2- (aq) + 6H2O (l) --> 2MnO2(s) + 4H2O (l) +6CO3^2 -(aq) + 12H^+ (aq)
turns into:
2MnO4^- (aq) + 3C2O4^2- (aq) + 2H2O (l) --> 2MnO2(s) +6CO3^2 -(aq) + 4H^+ (aq)
Answer:
Octasulfur is just S8. Eight S atoms in
a sort of crown shape. Sulfur Dioxide is a gas, SO2. Does that help?
Explanation:
The empirical formula : C₆H₂Cl₂O
The molecular formula : C₁₂H₄Cl₄O₂
<h3>Further explanation</h3>
Given
The percentage composition
Required
The empirical formula and the molecular formula
Solution
The mol ratio of the components :
C : H : Cl : O
=44.76/12 : 1.25/1 : 44.05/35.5 : 9.94/16
=3.73 : 1.25 : 1.241 : 0.621 divide by 1.241
= 3 : 1 : : 1 : 0.5 x 2
= 6 : 2 : 2 : 1
The empirical formula : C₆H₂Cl₂O
(Empirical formula)n=molecular formula
(C₆H₂Cl₂O)n=321.97
(160.986)n=321.97
n=2
(C₆H₂Cl₂O)₂=C₁₂H₄Cl₄O₂
The molecular formula : C₁₂H₄Cl₄O₂
Answer:
F⁻(aq) + H⁺(aq) ⇄ HF(aq)
Explanation:
When aqueous solutions of potassium fluoride and hydrochloric acid are mixed, an aqueous solution of potassium chloride and hydrofluoric acid results. The corresponding molecular equation is:
KF(aq) + HCl(aq) ⇄ KCl(aq) + HF(aq)
The full ionic equation includes all the ions and the molecular species. HF is a weak acid so it exists mainly in the molecular form.
K⁺(aq) + F⁻(aq) + H⁺(aq) + Cl⁻(aq) ⇄ K⁺(aq) + Cl⁻(aq) + HF(aq)
The net ionic equation includes only the ions that participate in the reaction (not spectator ions) and the molecular species.
F⁻(aq) + H⁺(aq) ⇄ HF(aq)
