Answer:
Oxygen in hydrogen peroxide oxidizes from -1 to 0.
Explanation:
Oxidation is the loss of electrons. The specie which is oxidized has has elevation in its oxidation state as compared in the reactant and the products.
The given reaction is shown below as:

Manganese in
has oxidation state of +7
Manganese in
has an oxidation state of +2
It reduces from +7 to +2
Oxygen in hydrogen peroxide has an oxidation state of -1.
Oxygen in molecular oxygen has an oxidation of 0.
Thus, oxygen in hydrogen peroxide oxidizes from -1 to 0.
2.392 hectoliters = 239.2 liters. 1 hectoliter = 100 liters.
The volume of the dry gas at stp is calculated as follows
calculate the number on moles by use of PV =nRT where n is the number of moles
n is therefore = Pv/RT
P = 0.930 atm
R(gas contant= 0.0821 L.atm/k.mol
V= 93ml to liters = 93/1000= 0.093L
T= 10 + 273.15 = 283.15k
n= (0.930 x0.093) /(0.0821 x283.15) = 3. 72 x10^-3 moles
At STp 1 mole = 22.4L
what about 3.72 x10^-3 moles
by cross multiplication
volume = (3.72 x10^-3)mole x 22.4L/ 1 moles = 0.083 L or 83.3 Ml
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We have Boltzmann's equation S = k ln W
Boltzmann's constant k = 1.381 x 10^-23 J/K
W = Number of absorption sites
At W = 484, Entropy S1 = 1.381 x 10^-23 ln 484 = 8.537 x 10^-23 J/K
At W = 729, Entropy S2 = 1.381 x 10^-23 ln 729 = 9.103 x 10^-23 J/K
Change of Entropy = S2 - S1 = 0.566 x 10^-23 J/K