A 20.0–milliliter sample of 0.200–molar K2CO3 so­lution is added to 30.0 milliliters of 0.400–mo­lar Ba(NO3)2 solution. Barium c

Response:

$30.39

Rationale:

a lot of calculations

Atoms have their protons and neutrons situated in the nucleus.

Electrons orbit in the outer shells, also known as orbitals.

Subatomic particles, comprising protons, neutrons, and electrons, reside within the atoms.

Response:

ΔH = -793.6 kJ

Reasoning:

The ΔH for this particular reaction can be calculated through Hess's law, which allows one to combine the ΔH values of the half-reactions to find the overall ΔH:

The half-reactions are as follows:

1) H₂(g) + ¹/₂O₂(g) ⟶ H₂O(g) ΔH₁ = −241.8 kJ

2) X(s) + 2Cl₂(g) ⟶ XCl₄(s) ΔH₂ = +356.9 kJ

3) ¹/₂H₂(g) + ¹/₂Cl₂(g) ⟶ HCl(g) ΔH₃ = −92.3 kJ

4) X(s) + O₂(g) ⟶ XO₂(s) ΔH₄ = −639.1 kJ

5) H₂O(g) ⟶ H₂O(l) ΔH₅ = −44.0 kJ

The calculation involves summing (4) + 4×(3) - (2) - 2×(1) - 2×(5):

(4) X(s) + O₂(g) ⟶ XO₂(s) ΔH = −639.1 kJ

+4×(3) 2H₂(g) + 2Cl₂(g) ⟶ 4HCl(g) ΔH = −369.2 kJ

-(2) XCl₄(s) ⟶ X(s) + 2Cl₂(g) ΔH = -356.9 kJ

-2×(1) 2H₂O(g) ⟶ 2H₂(g) + O₂(g) ΔH = +483.6 kJ

-2×(5) 2H₂O(l) ⟶ 2H₂O(g) ΔH = +88.0 kJ

= XCl₄(s) + 2H₂O(l) ⟶ XO₂(s) + 4HCl(g)

ΔH = -639.1 kJ -369.2 kJ -356.9 kJ +483.6 kJ +88.0 kJ

ΔH = -793.6 kJ

I trust this clarifies things!