F. <em>None of the above
</em>
<em>No O atoms are present</em> as reacting substances, only O_2 and H_2O molecules.
O_2 + 2H_2O + 2e^(-) → 4OH^(-)
We must use <em>oxidation numbers</em> to decide whether oxygen or water is the substance reduced.
The oxidation number of O changes from 0 in O_2 to -2 in OH^(-).
A decrease in oxidation number is <em>reduction</em>, so O_2 is the substance reduced.
The oxidation number of O is -2 in both H_2O and OH^(-), so water is <em>neither oxidized nor reduced</em>.
The answer is package c because each cd would cost $1.87, which is the lowest price
Barium carbonate (BaCO₃) <span>will be more soluble in acidic solution than in pure water, because Ksp (solubility constant) in water for this salt is very low.
In acid (for example hydrochloric acid) barium carbonate dissolves more because it forms weak electrolyte carbonic acid:
BaCO</span>₃(s) + 2HCl(aq) → BaCl₂(aq) + H₂CO₃(aq).
Density is defined as mass/volume, and the volume is 36.2-25 mL
so just substitute into the equation to get the answer.
Hope this helps.
