The bond dissociation energy of the Cl - Cl bond is -958 kJ mol^-1.
<h3>What is the dissociation enthalpy?</h3>
Given that;
H-H bond energy = 435 kJ mol^-1
H-Cl bond energy = 431 kJ mol^-1
ΔHfO of HCL(g) = -92kJ mol^-1
Bond dissociation enthalpy of the Cl-Cl bond = x
-92 = 435 + 431 + x
x = -92 - (435 + 431)
x = -958 kJ mol^-1
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I think the answer would be Ionic sodium phosphate (Na3PO4) because it has the greatest boiling point elevation.
Answer:
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I only came up with 2 I couldnt think of a third one
Your answer would be c
hope this helps
