Valence electrons is the answer to ur question
A. Removal of a water molecule between each two units
Answer:
BaI₂.5H₂O
Explanation:
Given Data:
Mass of Hydrated BaI₂ = 10.222 g
Mass of dried BaI₂ = 9.520 g
Mass of Water removed = 10.222-9.520 = 0.702 g
M.Mass of BaI₂ = 391.136 g/mol
M.Mass of Water = 18.02 g/mol
Now,
Calculate moles of dried BaI₂ as,
Moles = Mass / M.Mass
Moles = 9.520 g / 391.136 g/mol
Moles = 0.02434 moles
Calculate moles of Water as,
Moles = Mass / M.Mass
Moles = 0.702 g / 18.02 g/mol
Moles = 0.0389 moles
Then,
Calculate Mole ratio of BaI₂ and water as,
= 0.02434 moles BaI₂ / 0.0389 moles Water
= 0.625
Now,
We will convert this mole ratio to a whole number by multiplying it with a nearest integer,
= 0.625 × 8
= 5
Hence, this means for every one mole of BaI there are 5 moles of Water.
Result:
BaI₂.5H₂O
M = m/L
moles of CaO = 2.75g / molar mass
2.75g/ 56.08g = 0.049moles CaO
155mL / 1000 = 0.155 L
M= 0.049moles/0.155L
M=0.316