The third beaker: since it has the most amount of salt and the same amount of water as all the other beakers . So it has the highest concentrated amount :)
Answer is: 0.0213M.
Reaction: N₂O₄ (g) ↔ 2NO₂ (g).
Kc = 0,211 at <span>100°C.
Kc - </span><span>equilibrium constant.
</span>c (N₂O₄) = [N₂O₄] = 0,00251M.
c (NO₂) = [NO₂] = ?
Kc = [NO₂]² / [N₂O₄]
[NO₂]² = Kc · [N₂O₄]
[NO₂] = √(Kc · [N₂O₄<span>]) </span>
[NO₂] = √(0.211 · <span>0.00215M) </span>
[NO₂] = 0.0213M.
Answer:
2H^+ +2e -> h2(g)
Explanation:
Since hydrogen is reduced, look for a decrease in oxidation number and the gain of electrons.
Since according to ideal gas law, PV=nRT and you have 1.2atm, plug in the value for volume but make sure to convert to L instead of mL and convert temperature to Kelvin