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3 years ago

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Isopropyl alcohol is mixed with water to produce a 38.0% (v/v) alcohol solution. How many milliliters of each component are pres

ent in 885 mL of this solution? (Assume that volumes are additive.)

1 answer:

Fed [463]3 years ago

5 0

336.3 ml alcohol 548.7 ml of water

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Three examples of highly corrosive elements.

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3 years ago

The solubility of lead(II) iodide is 0.064 g/100 mL at 20ºC. What is the solubility product for lead(II) iodide?

quester [9]

Answer:

$Ksp=1.07x10^{-8}$

Explanation:

Hello,

In this case, the dissociation reaction is:

$PbI_2(s)\rightleftharpoons Pb^{2+}(aq)+2I^-(aq)$

For which the equilibrium expression is:

$Ksp=[Pb^{2+}][I^-]^2$

Thus, since the saturated solution is 0.064g/100 mL at 20 °C we need to compute the molar solubility by using its molar mass (461.2 g/mol)

$Molar solubility=\frac{0.064g}{100mL}*\frac{1000mL}{1L}*\frac{1mol}{461.2g}=1.39x10^{-3}M$

In such a way, since the mole ratio between lead (II) iodide to lead (II) and iodide ions is 1:1 and 1:2 respectively, the concentration of each ion turns out:

$[Pb^{2+}]=1.39x10^{-3}M$

$[I^-]=1.39x10^{-3}M*2=2.78x10^{-3}M$

Thereby, the solubility product results:

$Ksp=(1.39x10^{-3}M)(2.78x10^{-3}M)^2\\\\Ksp=1.07x10^{-8}$

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Explanation:

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