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

Chemistry

1 answer:

torisob [31]3 years ago

3 0

Answer:

1.427x10^-3mol per L

Explanation:

$Y(IO_{3} )_{3} ---- Y^{3+} +IO_{3} ^{3-}$

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

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A chemist wants to extract copper metal from copper chloride solution. The chemist places 0.50 grams of aluminum foil in a solut

Irina-Kira [14]

Answer:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

Explanation:

It is a stichiometry problem.

We should write the balance equation of the mentioned chemical reaction:

2Al + 3CuCl₂ → 3Cu + 2AlCl₃.

It is clear that 2.0 moles of Al foil reacts with 3.0 moles of CuCl₂ to produce 3.0 moles of Cu metal and 2.0 moles of AlCl₃.

Also, we need to calculate the number of moles of the reported masses of Al foil (0.50 g) and CuCl₂ (0.75 g) using the relation:

n = mass / molar mass

The no. of moles of Al foil = mass / atomic mass = (0.50 g) / (26.98 g/mol) = 0.0185 mol.

The no. of moles of CuCl₂ = mass / molar mass = (0.75 g) / (134.45 g/mol) = 5.578 x 10⁻³ mol.

From the stichiometry Al foil reacts with CuCl₂ with a ratio of 2:3.

∴ 3.85 x 10⁻³ mol of Al foil reacts completely with 5.578 x 10⁻³ mol of CuCl₂ with (2:3) ratio and CuCl₂ is the limiting reactant while Al foil is in excess.

From the stichiometry 3.0 moles of CuCl₂ will produce the same no. of moles of copper metal (3.0 moles).
So, this reaction will produce 5.578 x 10⁻³ mol of copper metal.
Finally, we can calculate the mass of copper produced using:

mass of Cu = no. of moles x Atomic mass of Cu = (5.578 x 10⁻³ mol)(63.546 g/mol) = 0.354459 g ≅ 0.36 g.

So, the answer is:

Approximately 0.36 grams, because copper (II) chloride acts as a limiting reactant.

5 0

3 years ago

Give the nuclear symbol for the isotope of silicon for which a=29? Enter the nuclear symbol for the isotope (e.G., 42he).

quester [9]

The mass number is the summation of number of proton and neutron present in a nucleus of an atom. For the neutral atom the number of positive charge (number of proton) must be equal to the number of electrons. The number of electrons present in an atom is the atomic number of the atom. The standard way to express the mass number (a) and atomic number (m) of a atom (say X) is $x^{a}_{m}$ . Now for silicon number of electron or atomic number is 14. And the mass number (a) given 29. Thus the expression nucleus of silicon will be $Si^{29}_{14}$