Answer:
1.427x10^-3mol per L
Explanation:
I could use ⇌ in the math editor so I used ----
from the question each mole of Y(IO3)3 is dissolved and this is giving us a mole of Y3+ and a mole of IO3^3-
Ksp = [Y^3+][IO3-]^3
So that,
1.12x10^-10 = [S][3S]^3
such that
1.12x10^-10 = 27S^4
the value of s is 0.001427mol per L
= 1.427x10^-3mol per L
so in conclusion
the molar solubility is therefore 1.427x10^-3mol per L
Answer:
113.8g
Explanation:
Statement of problem: mass of 1.946mole of NaCl
Given parameters:
Number of moles of NaCl = 1.946mole
Unknown: mass of NaCl
Solution
To find the mass of NaCl, we apply the concept of moles which expresses the relationship between number of moles and mass according to the equation below:
Number of moles =
To find the molar mass of NaCl:
the atomic mass of Na = 23g
atomic mass of Cl = 35.5g
Molar mass of NaCl = (23 + 35.5) = 58.5gmol⁻¹
Mass of NaCl = Number of moles x molar mass of NaCl
Mass of NaCl = 1.946 x 58.5 = 113.8g