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3 years ago

Calculate the molar solubility of PbCO3 when the Ksp of PbCO3 is 1.5*10^-15 at 25°C.

1 answer:

7 0

First, write the dissociation equation for PbCO3 which is PbCO3 = Pb^(2+) + CO3^(2-). Then, write the Ksp equation which is Ksp = [Pb^(2+)][CO3^(2-)]. The equation suggests that there is 1:1 molar ratio between the Pb^(2+), CO3^(2-) and the dissolved PbCO3. Thus in equation form, we can represent them as x. The Ksp equation is then: 1.5*10^-15 = x^2. The molar solubility of PbCO3, x, is then equal to 3.87*10^-8 moles per liter.

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Answer: option D) energy was absorbed and entropy increased.

Explanation:

1) Given balanced equation:

2H₂O (l) + 571.6 kJ → 2 H₂ (g) + O₂(g).

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So, the first part is that energy was absorbed.

3) As for the entropy, it is a measure of the disorder or radomness of the system.

Since, two molecules of liquid water were transformed into three molecules of gas, i.e. more molecules and more kinetic energy, therefore the new state has a greater degree of radomness, is more disordered, and you conclude that the entropy increased.

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3 years ago

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erastovalidia [21]

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Explanation:

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2 years ago

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agasfer [191]

<h2>Answer:</h2><h2>The percentage of the family’s total annual electricity that is used to run the two air conditioners for the three summer months = 19.4 %</h2>

Explanation:

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The percentage of the family’s total annual electricity that is used to run the two air conditioners for the three summer months = $\frac{2100}{10800}100$ = 19.4 %

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