Answer:
A primary source of energy is a fossil fuel , nuclear fuel , wind or sunlight in an unconverted state . A secondary source of energy is created when a primary source of energy is burned or otherwise converted into a form , like electricity, that can be used for useful work.
Explanation:
What are the option for the questions???
Answer:
Mn^2+(aq) + 4H2O(l) + 5[VO2]^+(aq) + 10H^+(aq) ---------->MnO4^-(aq) + 8H^+(aq) + 5[VO]^2+(aq) + 5H2O(l)
Explanation:
Oxidation half equation:
Mn^2+(aq) + 4H2O(l) ------------> MnO4^-(aq) + 8H^+(aq) + 5e
Reduction half equation:
5[VO2]^+(aq) + 10H^+(aq) + 5e --------> 5[VO]^2+(aq) + 5H2O(l)
Overall redox reaction equation:
Mn^2+(aq) + 4H2O(l) + 5[VO2]^+(aq) + 10H^+(aq) ---------->MnO4^-(aq) + 8H^+(aq) + 5[VO]^2+(aq) + 5H2O(l)
Answer:
6Br⁻ + XeO₃ + 6H⁺ → 3Br₂ + Xe + 3H₂O
Explanation:
First, we need to write the half-reactions:
2Br⁻ → Br₂ + 2e⁻ Oxidation -Balanced yet-
XeO₃ → Xe Reduction
To balance the reduction in acidic aqueous solution we need to add waters in the other side of the reaction as oxygens are present:
XeO₃ → Xe + 3H₂O
And H⁺ as hydrogens from water we have:
XeO₃ + 6H⁺ → Xe + 3H₂O
To balance the charge:
<h3>XeO₃ + 6H⁺ + 6e⁻ → Xe + 3H₂O Reduction -Balanced-</h3><h3 />
To cancel out the electrons of both half-reaction we need to multiply oxidation 3 times:
6Br⁻ → 3Br₂ + 6e⁻
XeO₃ + 6H⁺ + 6e⁻ → Xe + 3H₂O
And the balanced reaction in acidic aqueous solution is the sum of both half-reactions:
<h3>6Br⁻ + XeO₃ + 6H⁺ → 3Br₂ + Xe + 3H₂O </h3>
