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marusya05 [52]
3 years ago
5

The chemistry lab is 0.3394km from the nurses office. How many feet is the distance?

Chemistry
1 answer:
nlexa [21]3 years ago
6 0
Easy just convert it to feet the answer is <span>1113.51706 Feet

</span>
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Choose one natural resource that is nonrenewable. In three to five sentences, describe two impacts that come from obtaining this
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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}

Regards.

6 0
3 years ago
Need major help. Help me please
jek_recluse [69]
Uhhh, that looks like a game. I don't know how to tell u the answers for that thing -_-
6 0
3 years ago
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