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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The correct answer is : a. Salt water
Answer : The enthalpy change for the reaction is, 419.5 kJ
Explanation :
According to Hess’s law of constant heat summation, the heat absorbed or evolved in a given chemical equation is the same whether the process occurs in one step or several steps.
The given chemical reaction is,
Now we have to determine the enthalpy change for the reaction below:
By reversing and then dividing the reaction by 2, we get the enthalpy change for the reaction.
The expression will be:
Therefore, the enthalpy change for the reaction is, 419.5 kJ
As the O3 suggests, the Ozone molecule is made up of 3 Oxygen atoms.
The equation N2+H2= NH3 is balanced. This is because N2 + 3H^2 ➡️ 2NH^3
