4. Static, sliding,rolling,and fluid friction
Answer is: Cl and Na.
sodium and chlorine are in third period and they have very different properties. Sodium is solid metal and chlorine is gaseous nonmetal.
They form compound NaCl (Sodium chloride), because sodium lost one valence electron and form cation Na⁺, chlorine gain one electron and form anion Cl⁻.
Electron configuration of sodium atom: ₁₁Na 1s² 2s² 2p⁶ 3s¹.
Electron configuration of chlorine atom: ₁₇Cl 1s² 2s² 2p⁶ 3s² 3p⁵.
Other examples are metal-metal pairs and they do not form cation and anion.
Is there any answer choices before i get started on working out the problem
Answer:
ΔH₁₂ = -867.2 Kj
Explanation:
Find enthalpy for 3H₂ + O₃ => 3H₂O given ...
2H₂ + O₂ => 2H₂O ΔH₁ = -483.6 Kj
3O₂ => 2O₃ ΔH₂ = + 284.6 Kj
_____________________________
3(2H₂ + O₂ => 2H₂O) => 6H₂ + 3O₂ => 6H₂O (multiply by 3 to cancel O₂)
6H₂ + 3O₂ => 6H₂O ΔH₁ = 3(-483.6 Kj) = -1450.6Kj
2O₃ => 3O₂ ΔH₂ = -284.6Kj (reverse rxn to cancel O₂)
_______________________________
6H₂ + 2O₃ => 6H₂O ΔH₁₂ = -1735.2 Kj (Net Reaction - not reduced)
________________________________
divide by 2 => target equation (Net Reaction - reduced)
3H₂ + O₃ => 3H₂O ΔH₁₂ = (-1735.2/2) Kj = -867.2 Kj
