What could you predict about a methane molecule that would most likely be made by using both the quantum model of the atom and john dalton's model?

25 mL of 1 × 10–5 M Co(NO₃)₂ and 75 mL of 5 × 10–4 M Na₂S

500 mL of 7.5 × 10–4 M AlCl₃ and 100 mL of 1.7 × 10–5 M Hg₂(NO₃)₂.

650 mL of 0.0080 M K₂SO₄ and 175 mL of 0.15 M AgNO₃

Explanation:

When 2 compounds that produce an insoluble substance are mixed together, A precipitate will be formed if Q of reaction > Ksp

For the solutions:

1.5 L of 0.025 M BaCl₂ and 1.25L of 0.014 M Pb(NO₃)₂.

Ksp is:

PbCl₂(s) ⇄ Pb²⁺(aq) + 2Cl⁻(aq)

Ksp = 2.4x10⁻⁴ = [Pb²⁺][Cl⁻]²

Molar concentration of each ion is:

[Pb²⁺] =  1.25L ₓ (0.014mol / L) = 0.0175mol / 2.75L = 6.36x10⁻³M

[Cl⁻] = 2 ₓ 1.5L ₓ (0.025mol / L) = 0.075mol / 2.75L = 0.0273M

Replacing in Ksp expression to find Q:

Q = [6.36x10⁻³M][0.0273M]² = 4.73x10⁻⁶

As Q < Ksp, the mixture will not produce a precipitate.

25 mL of 1 × 10–5 M Co(NO₃)₂ and 75 mL of 5 × 10–4 M Na₂S

Ksp is:

CoS(s) ⇄ Co²⁺(aq) + S²⁻(aq)

Ksp = 4.0x10⁻²¹ = [Co²⁺][S²⁻]

Molar concentration of each ion is:

[Co²⁺] =  0.025L ₓ (1x10⁻⁵mol / L) = 2.5x10⁻⁷mol / 0.1L = 2.5x10⁻⁶M

[S²⁻] = 0.075L ₓ (5x10⁻⁴mol / L) = 3.75x10⁻⁵mol / 0.1L = 3.75x10⁻⁴M

Replacing in Ksp expression to find Q:

Q = [2.5x10⁻⁶M][3.75x10⁻⁴M] = 9.38x10⁻⁶

As Q > Ksp, the mixture will produce a precipitate.

500 mL of 7.5 × 10–4 M AlCl₃ and 100 mL of 1.7 × 10–5 M Hg₂(NO₃)₂.

Ksp is:

Hg₂Cl₂(s) ⇄ 2Hg⁺(aq) + 2Cl⁻(aq)

Ksp = 3.5x10⁻¹⁸ = [Hg⁺]²[Cl⁻]²

Molar concentration of each ion is:

[Hg⁺] =  2ₓ0.100L ₓ (1.7x10⁻⁵mol / L) = 3.4x10⁻⁶mol / 0.6L = 5.67x10⁻⁶M

[Cl⁻] = 3 ₓ 0.500L ₓ (7.5x10⁻⁴mol / L) = 1.125x10⁻³mol / 0.6L = 1.88x10⁻³M

Replacing in Ksp expression to find Q:

Q = [5.67x10⁻⁶M]²[1.88x10⁻³M]² = 1.14x10⁻⁶

As Q > Ksp, the reaction will produce a precipitate.

650 mL of 0.0080 M K₂SO₄ and 175 mL of 0.15 M AgNO₃

Ksp is:

Ag₂SO₄(s) ⇄ 2Ag⁺(aq) + SO₄²⁻(aq)

Ksp = 1.5x10⁻⁵ = [Ag⁺]²[SO₄²⁻]

Molar concentration of each ion is:

[Ag⁺] =  0.175L ₓ (0.15mol / L) = 0.02625mol / 0.825L = 0.0318M

[SO₄²⁻] = 0.650L ₓ (0.080mol / L) = 0.052mol / 0.825L = 0.0630M

Replacing in Ksp expression to find Q:

Q = [0.0318M]²[0.0630M] = 6.37x10⁻⁵

As Q > Ksp, the reaction will produce a precipitate.