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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Discuss why the solubility of CaCO33(s) would be greater in distilled water than in tap water that contains 50 mg/L of calcium i

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The greater, the ionic property, the greater will be solubility. water is a covalent compound where calcium carbonate is ionic that's why the solubility of $CaCO_{3}$ (s) would be greater.

<h3>What is solubility?</h3>

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The greater, the ionic property, the greater will be solubility. water is a covalent compound where calcium carbonate is ionic that's why the solubility of $CaCO_{3}$ (s) would be greater in distilled water than in tap water that contains 50 mg/L of calcium ions

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3 years ago

Enter your answer in the provided box. phosphorus pentachloride is used in the industrial preparation of organic phosphorus comp

ankoles [38]

I have a helpful trick for such problems.

Step 1:
Write the equation asked with exact moles given in question,

PCl₃ + Cl₂ → PCl₅ --------(1)
Step 2:
Write equations given in data along with energies,

P₄ + 6Cl₂ → 4PCl₃ ΔH = −1280 kj -----(2)

P₄ + 10Cl₂ → 4PCl₅ ΔH = −1774 kj -----(3)

Step 3:
Compare the equation 1 with eq 2 and 3 and modify eq 2 and 3 according to eq 1 both in terms of reactant and product and number of moles. For example in eq 1 PCl₃ is on left hand side and its number of moles is one, so, modify eq, 2 according to eq 1 both in terms of moles and reactant product sides. i.e.
By inter converting eq 2, the sign of energy will change from negative to positive, and divide eq 2 with 4, to have its ratio's equal to eq 1

PCl₃ → 1/4 P₄ + 1.5 Cl₂ ΔH = +320 KJ

Same steps are done for eq 3, but in this case signs are not changed only moles are changed

1/4P₄ + 2.5Cl₂ → 1/4PCl₅ ΔH = −443.5 kj
Now,
Compare both last equations,

PCl₃ → 1/4 P₄ + 1.5 Cl₂ ΔH = +320 KJ

1/4P₄ + 2.5Cl₂ → 4/4PCl₅ ΔH = −443.5 kj
_____________________________________________

PCl₃ + Cl₂ → PCl₅ ΔH = -123.5 KJ

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3 years ago

Solve for x in the following equation? v=w/x

Aleonysh [2.5K]

Answer:

x = w/v

General Formulas and Concepts:

Pre-Algebra

Equality Properties

Explanation:

Step 1: Define

v = w/x

Step 2: Solve for x

Multiply x on both sides: xv = w
Divide v on both sides: x = w/v

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3 years ago