Oxidation-Reduction Reactions Suggested Reading Thus the oxidation number for oxygen in calcium oxide is -2. ... In effect, each calcium atom loses two electrons to form Ca2+ ions, and each O atom in O2 gains two electrons to form O2- ions. The net result is a transfer of electrons from calcium to oxygen, so this reaction is an oxidation-reduction reaction.
+<u>O²</u><u>(</u><u>g</u><u>)</u><u>=</u><u>2</u><u>CaO</u><u>(</u><u>s</u><u>)</u>
Explanation:
we can conclude that in the reaction there is both reduction and oxidation.
Manufacturing A Drill That Can Be Used To Drill For Energy Sources At The Bottom Of The Ocean.
Explanation:
I may be correct but I give good luck to you- ^w
Answer: Electrons are organized into shells and subshells around nuclei. The electron configuration states the arrangement of electrons in shells and subshells. Valence electrons are in the highest numbered shell.
Answer:
At anode - 
At cathode - 
Explanation:
Electrolysis of NaBr:
Water will exist as:

The salt, NaBr will dissociate as:

At the anode, oxidation takes place, as shown below.

At the cathode, reduction takes place, as shown below.

Answer:
9.80g of O2.
Explanation:
Step 1:
The balanced equation for the reaction. This is given below:
2KClO3 —> 2KCl + 3O2
Step 2:
Determination of the mass of KClO3 that decomposed and the mass of O2 produced from the balanced equation. This is illustrated below:
Molar Mass of KClO3 = 39 + 35.5 + (16x3) = 122.5g/mol
Mass of KClO3 from the balanced equation = 2 x 122.5 = 245g
Molar Mass of O2 = 16x2 = 32g/mol
Mass of O2 from the balanced equation = 3 x 32 = 96g.
From the balanced equation above, 245g of KClO3 decomposed to produce 96g of O2.
Step 3:
Determination of the mass of O2 produced in the decomposition of 25.0g of KClO3. This is illustrated below:
From the balanced equation above, 245g of KClO3 decomposed to produce 96g of O2.
Therefore, 25g of KClO3 will decompose to produce = (25x96)/245 =
9.80g of O2.
Therefore, 9.80g of O2 is produced from the decomposition of 25g of KClO3.