Answer:
126.8, Iodine
Explanation:
- mass ×abundance/100
- (126.9045×80.45/100)+(126.0015×17.23/100)+(128.2230×2.23/100)
- 102.1+21.7+3=126.8
<em>IODINE</em><em> </em><em>has</em><em> </em><em>an</em><em> </em><em>atomic</em><em> </em><em>mass</em><em> </em><em>of</em><em> </em>126.8 or 126.9
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:
The answer is the photo attached
Explanation:
1 mole K ------------- 6.02x10²³ atoms
1.83 moles K ------ ?? atoms
1.83 x (6.02x10²³) / 1 =
1.101x10²⁴ atoms of K
hope this helps!
