Answer:
Part C: P2 = 0.30 atm
Part D: V1 = 16.22 L.
Explanation:
Part C:
Initial pressure (P1) = 2.67 atm
Initial volume (V1) = 5.54 mL
Final pressure (P2) =.?
Final volume (V2) = 49 mL
The final pressure (P2) can be obtained as follow:
P1V1 = P2V2
2.67 x 5.54 = P2 x 49
Divide both side by 49
P2 = (2.67 x 5.54)/49
P2 = 0.30 atm
Therefore, the final pressure (P2) is 0.30 atm
Part D:
Initial pressure (P1) = 348 Torr
Initial volume (V1) =?
Final pressure (P2) = 684 Torr
Final volume (V2) = 8.25 L
The initial volume (V1) can be obtained as follow:
P1V1 = P2V2
348 x V1 = 684 x 8.25
Divide both side by 348
V1 = (684 x 8.25)/348
V1 = 16.22 L
Therefore, the initial volume (V1) is 16.22 L
Electrons, everything is pretty much based around the likeliness of electrons to be swapped or shared between atoms
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
Learn more:
They are different measurements
