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>
The answer is B hope this helped
You forgot to attach the text
BTW reduction is the action or fact of making a specified thing smaller or less in amount, degree, or size.
<span>The metal that would more easily lose an electron would be potassium. It is more reactive than sodium. Also, looking on the periodic table, </span><span>from top to bottom for groups 1 and 2, reactivity increases. So, it should be potassium. Hope this answers the question. Have a nice day.</span>
Answer:
87.15%
Explanation:
To find percent yield, we can use this simple equation
Where "Actual" is the amount in grams actually collected from the reaction, and "Theoretical" is, well, the theoretical amount that should have been produced.
They give us these values, so to find the percent yield, just plug the numbers in.
So, the percent yield is 87.15%
An easy trick to remember how to do this is just to divide the smaller number by the bigger number and move the decimal back two places. If you have a percent yield greater than 100%, something is wrong in the reaction.
