Answer:
[Ag⁺] = 5.0x10⁻¹⁴M
Explanation:
The product solubility constant, Ksp, of the insoluble salts PbI₂ and AgI is defined as follows:
Ksp(PbI₂) = [Pb²⁺] [I⁻]² = 1.4x10⁻⁸
Ksp(AgI) = [Ag⁺] [I⁻] = 8.3x10⁻¹⁷
The PbI₂ <em>just begin to precipitate when the product [Pb²⁺] [I⁻]² = 1.4x10⁻⁸</em>
<em />
As the initial [Pb²⁺] = 0.0050M:
[Pb²⁺] [I⁻]² = 1.4x10⁻⁸
[0.0050] [I⁻]² = 1.4x10⁻⁸
[I⁻]² = 1.4x10⁻⁸ / 0.0050
[I⁻]² = 2.8x10⁻⁶
<h3>[I⁻] = 1.67x10⁻³</h3><h3 />
So, as the [I⁻] concentration is also in the expression of Ksp of AgI and you know concentration in solution of I⁻ = 1.67x10⁻³M:
[Ag⁺] [I⁻] = 8.3x10⁻¹⁷
[Ag⁺] [1.67x10⁻³] = 8.3x10⁻¹⁷
<h3>[Ag⁺] = 5.0x10⁻¹⁴M</h3>
Answer: ³²P
Explanation:
The radionuclide in question is known as Phosphorus-32. It is an isotope of Phosphorus that is radioactive and has one more neutron than the normal phosphorus does.
Phosphorus does not exist in nature and is created artificially by bombarding the normal stable phosphorus with neutrons. Due to its radioactive nature, it has proven to be useful in the medical industry where it has been used to detect eye tumors as well as in treating some diseases.
Answer:

Explanation:
Half time is period required to desintegrating the half of the initial number of atoms. Then, the total time is:


The answer is B. molality
Answer:
The freezing point of the solution is - 4.39 °C.
Explanation:
We can solve this problem using the relation:
<em>ΔTf = (Kf)(m),</em>
where, ΔTf is the depression in the freezing point.
Kf is the molal freezing point depression constant of water = -1.86 °C/m,
density of water = 1 g/mL.
<em>So, the mass of 575 mL is 575 g = 0.575 kg.</em>
m is the molality of the solution (m = moles of solute / kg of solvent = (465 g / 342.3 g/mol)/(0.575 kg) = 2.36 m.
<em>∴ ΔTf = (Kf)(m</em>) = (-1.86 °C/m)(2.36 m) = <em>- 4.39 °C.</em>
<em>∵ The freezing point if water is 0.0 °C and it is depressed by - 4.39 °C.</em>
<em>∴ The freezing point of the solution is - 4.39 °C.</em>