Considering the definition of pOH and strong base, the pOH of the aqueous solution is 1.14
The pOH (or potential OH) is a measure of the basicity or alkalinity of a solution and indicates the concentration of ion hydroxide (OH-).
pOH is expressed as the logarithm of the concentration of OH⁻ ions, with the sign changed:
pOH= - log [OH⁻]
On the other hand, a strong base is that base that in an aqueous solution completely dissociates between the cation and OH-.
LiOH is a strong base, so the concentration of the hydroxide will be equal to the concentration of OH-. This is:
[LiOH]= [OH-]= 0.073 M
Replacing in the definition of pOH:
pOH= -log (0.073 M)
<u><em>pOH= 1.14 </em></u>
In summary, the pOH of the aqueous solution is 1.14
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1s2 2s2 2p2 i hope this helps
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Answer:
V = 0.63 L
Explanation:
To solve this problem, we need to use the Charle's law which is a law that involves temperature and volume, assuming we have a constant pressure. The problem do not state that the pressure is being altered, so we can safely assume that the pressure is constant (Maybe 1 atm).
Now, as the pressure is constant, the Charle's law is the following:
V₁ / T₁ = V₂ / T₂ (1) V is volume in Liter, and T is temperature in Kelvin.
Using this law with the given data, we solve for V₂:
V₂ = V₁T₂ / T₁
Before we use this expression, let's convert the temperatures to Kelvin:
T₁ = 19 + 273 = 292 K
T₂ = 250 + 273 = 523 K
Now, let's calculate the volume of the balloon:
V₂ = 0.35 * 523 / 292
<h2>
V₂ = 0.63 L</h2>
