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Ipatiy [6.2K]
3 years ago
5

Which of the following best describes the rock cycle?

Chemistry
2 answers:
DedPeter [7]3 years ago
5 0

Answer:

D. The rock cycle is a series of processes in which one kind of rock is transformed into other kinds.

Explanation:

bekas [8.4K]3 years ago
5 0

Answer:

it is the last one.

<em><u>The </u></em><em><u>rock </u></em><em><u>cycle </u></em><em><u>is </u></em><em><u>a </u></em><em><u>series </u></em><em><u>of </u></em><em><u>processes </u></em><em><u>in </u></em><em><u>which </u></em><em><u>one </u></em><em><u>kind </u></em><em><u>of </u></em><em><u>rock </u></em><em><u>is </u></em><em><u>transformed </u></em><em><u>into </u></em><em><u>other </u></em><em><u>kinds</u></em><em><u>. </u></em>

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Answer:

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

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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

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

Regards.

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