When you are tuning an instrument it changes the sound of the instrument
I believe the answer is 50.5 molecules
Molar solubility is number of moles of the solute that can be dissolved per liter of solution before the solution becomes saturated.
The molar solubility of lead(ii) chloride with ksp value of 2.4 × 10e4 can be solve as:
Ksp = s2 = 2.4 × 10e4
s2 = 2.4 × 10e4
s = √(2.4 × 10e4)
s = 154.9 mol/L
Answer:
7.16x10⁻⁸M = [Ag+]
Explanation:
Using the equation:
E(Cell) =E⁰ - 0.0592/2 • log ([Cu2+]/[Ag+]²)
<em>Where E</em>⁰<em>= 0.4249V</em>
<em>E(Cell) = -(-0.0019V) -Measured value-</em>
<em>[Cu2+] = 1M</em>
<em />
Replacing:
0.0019V = 0.4249V - 0.0592/2 • log (1M/[Ag+]²)
-0.423V = - 0.0296 • log (1M/[Ag+]²)
14.29 = log (1M/[Ag+]²)
1.95x10¹⁴ = 1M / [Ag+]²
[Ag+]² = 5.12x10⁻¹⁵M
7.16x10⁻⁸M = [Ag+]
