In this case, according to the given information about the oxidation numbers anf the compounds given, it turns out possible to figure out the oxidation number of manganese in both MnI2, manganese (II) iodide and MnO2, manganese (IV) oxide, by using the concept of charge balance.
Thus, we can define the oxidation state of iodine and oxygen as -1 and -2, respectively, since the former needs one electron to complete the octet and the latter, two of them.
Next, we can write the following 
, since manganese has five oxidation states, and it is necessary to calculate the appropriate ones:
Next, we multiply each anion's oxidation number by the subscript, to obtain the following:
Thus, the correct choice is Manganese has an oxidation number of +2 in Mnl2 and +4 in MnO2.
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Answer:
Was the metal oxidized?
Cu = No | No | No
Fe = Yes | No | No
Zn = Yes | Yes | No
Oxidizing Agent:
Cu = None | None | None
Fe = Cu, NO3 | None | None
Zn = Cu, NO3 | Fe, NO3 | None
Answer:
They are called insulators
Its going to be 2.81 x 1023 atoms
Answer:
The four coefficients in order, separated by commas are 1, 8, 5, 6
Explanation:
We count the atoms in order to balance this combustion reaction. In combustion reactions, the products are always water and carbon dioxide.
C₅H₁₂ + ?O₂→ ?CO₂ + ?H₂O
We have 12 hydrogen in right side and we can balance with 6 in the left side. But the number of oxygen is odd. We add 2 in the right side, so we have 24 H, and in the product side we add a 12.
As we add 2 in the C₅H₁₂, we have 10 C, so we must add 10 to the CO₂ in the product side.
Let's count the oxygens: 20 from the CO₂ + 12 from the water = 32.
We add 16 in the reactant side. Balanced equation is:
2C₅H₁₂ + 16O₂→ 10CO₂ + 12H₂O
We also can divide by /2 in order to have the lowest stoichiometry
C₅H₁₂ + 8O₂→ 5CO₂ + 6H₂O
