<span>circumstellar is the answer</span>
Answer:
The O is being oxidized, but at the same time, is being reducted.
Explanation:
H₂O₂(l) + ClO₂(aq) → ClO₂(aq) + O₂(g)
In this reaction, we have 4 compounds:
Hydrogen peroxide
Chlorine dioxide (twice)
Oxygen
In both dioxide, the Cl acts with +4 in oxidation state; the oxygen acts with -2.
Oxgen in ground state has 0, as oxidation number.
In peroxide, the H acts with +1 but the oxygen acts with -1.
Peroxide is making the oxidation number from the O in the ClO₂, to decrease (reduction) and to increase in the O, at the ground state.
Hydrogen peroxide is a good reducing and oxidizing agent at the same time.
An aqueous solution in a 55 gallon (208 l drum), characterized by minimal buffering capacity, received 4kg of phenol and 1.5 kg of sodium phenate. What is the ph of the solution. The pka of phenol = 9.98. Mw of phenol and sodium phenate are 94 g/mol and 116 g/mol, respectively.
Volume of solution = 55 gallons = 208.2 L [ 1 gallon = 3.78 L]
moles of phenol = mass / molar mass = 4000 g / 94 = 42.55 moles
moles of sodium phenate = mass / molar mass = 1500 / 116 = 12.93 moles
pKa of phenol = 9.98
We know that the pH of buffer is calculated using Hendersen Hassalbalch's equation
pH = pKa + log [salt] / [acid]
volume is same for both the sodium phenate and phenol has we can directly take the moles of each in the formula
pH = 9.98 + log [12.93 / 42.55] = 9.46
