Answer:
a) Chloride dioxide
b) Di nitrogen tetra oxide
c) Potassium phosphide
d) Silver -Ag
e) Aluminium nitride - AlN
f) Silicon dioxide
I) Sulfide
Oxidation is the loss of electrons. Reduction is the gain of electrons. The oxidizing agent is reduced. The reducing agent is oxidized. Cu goes from 0 to +2, it lost electrons S went from +6 to +4, it gained electrons I went from 0 to +5, it lost electrons N went from +5 to +4, it gained electrons.
Answer:
Alkaline is a substance that is over the pH of 7. an acid is a substance that is lower than the pH of 7
When universal indicator is put into an alkaline substance it will turn blue/purple/black. If it is an acid the substance will turn red/orange/yellow
The same will happen if litmus paper is used instaed. If the paper turns blue/purple/black it means that the substance is an alkaline. If the paper turns red/orange/yellow it means that the substance is an acid
If the substance or the litmus paper turns green it means that it is neutral (It has the pH of 7)
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Answer:
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Explanation:
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>
