<u>Answer:</u> The hydroxide ion concentration and pOH of the solution is
and 2.88 respectively
<u>Explanation:</u>
We are given:
Concentration of barium hydroxide = 0.00066 M
The chemical equation for the dissociation of barium hydroxide follows:

1 mole of barium hydroxide produces 1 mole of barium ions and 2 moles of hydroxide ions
pOH is defined as the negative logarithm of hydroxide ion concentration present in the solution
To calculate pOH of the solution, we use the equation:
![pOH=-\log[OH^-]](https://tex.z-dn.net/?f=pOH%3D-%5Clog%5BOH%5E-%5D)
We are given:
![[OH^-]=(2\times 0.00066)=1.32\times 10^{-3}M](https://tex.z-dn.net/?f=%5BOH%5E-%5D%3D%282%5Ctimes%200.00066%29%3D1.32%5Ctimes%2010%5E%7B-3%7DM)
Putting values in above equation, we get:

Hence, the hydroxide ion concentration and pOH of the solution is
and 2.88 respectively
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Empirical formula of ionic compound is FeO. In which the composition of atoms is 1 : 1.
Empirical formula of an ionic compound is defined as the formula which gives whole number ratio of atoms of various elements present in molecule of compund.
mass of iron in compound = 34.95 g
molar mass of iron = 55.8 g
mass of oxygen in compound = 15.05 g
molar mass of oxygen = 32 g
number of moles of iron present in the compound are ratio of mass of iron in compound/ molar mass of iron
number of moles of iron in compound= 34.95 / 55.8 = 0.6263 ~ 1
number of moles oxygen in compound= 15.05/ 32 = 0.473 ~ 0.5
the ratio of the number of oxygen atoms to number of iron atoms present in one formula unit of iron compund is 2×0.5 / 1 = 1 : 1
Hence , the required empirical formula of iron compound is FeO.
To learn more about Emiprical formula, refer:
brainly.com/question/1439914
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Answer:
n₂ = 2.55 mol
Explanation:
Given data:
Initial number of moles = 0.758 mol
Initial volume = 80.6 L
Final volume = 270.9 L
Final number of moles = ?
Solution:
Formula:
V₁/n₁ = V₂/n₂
V₁ = Initial volume
n₁ = initial number of moles
V₂ = Final volume
n₂ = Final number of moles
now we will put the values in formula.
V₁/n₁ = V₂/n₂
80.6 L / 0.758 mol = 270.9 L/ n₂
n₂ = 270.9 L× 0.758 mol / 80.6 L
n₂ = 205.34 L.mol /80.6 L
n₂ = 2.55 mol