The Ksp of yttrium iodate, Y(IO3)3 , is 1.12×10−10 . Calculate the molar solubility, s , of this compound.

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

confused

confused             hewdejdnjnjsnd

considering the question as per the attachment

here we can see that the radon nuclide has 222 as the atomic mass number and 86 as the atomic number.

when it undergoes nuclear fission it gives polonium which has atomic mass number as 218 and atomic number will be 84 along with a helium nucleus which has 4 atomic mass number and 2 as its atomic number.

222 = 218 + 4 (atomic mass number)

86 = ? + 2; therefore ? = 86-2 = 84.

the atomic mass number of polonium will be 84.