Answer:
The answer to your question is V = 0.32 L
Explanation:
Data
Volume of NH₃ = ?
P = 3.2 atm
T = 23°C
mass of CaH₂ = 2.65 g
Balanced chemical reaction
6Ca + 2NH₃ ⇒ 3CaH₂ + Ca₃N₂
Process
1.- Convert the mass of CaH₂ to moles
-Calculate the molar mass of CaH₂
CaH₂ = 40 + 2 = 42 g
42 g ------------------ 1 mol
2.65 g -------------- x
x = (2.65 x 1)/42
x = 0.063 moles
2.- Calculate the moles of NH₃
2 moles of NH₃ --------------- 3 moles of CaH₂
x --------------- 0.063 moles
x = (0.063 x 2) / 3
x = 0.042 moles of NH₃
3.- Convert the °C to °K
Temperature = 23°C + 273
= 296°K
4.- Calculate the volume of NH₃
-Use the ideal gas law
PV = nRT
-Solve for V
V = nRT / P
-Substitution
V = (0.042)(0.082)(296) / 3.2
-Simplification
V = 1.019 / 3.2
-Result
V = 0.32 L
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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An electron because that is the only part able to be lost or gained without nuclear action needed