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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:
the activation energy Ea = 179.176 kJ/mol
it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.
Explanation:
From the given information
Thus;
Because at 113.0°C; the rate is 7 time higher than at 100°C
Hence:
1.9459 =
Ea = 179.176 kJ/mol
Thus; the activation energy Ea = 179.176 kJ/mol
b)
here;
where ;
Now;
t = 7.0245 mins
Therefore; it will take 7.0245 mins for the same food to cook in an open pot of boiling water at an altitude of 10000 feet.