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Levart [38]
3 years ago
9

18. Which two particle diagrams each represent a sample of one substance? Key O = atom of oxygen = atom of hydrogen 0 = atom of

chlorine 8 OO 1 நம் மும் 20 2000 (1) I and III (3) II and IV (2) II and III (4) I and II​
Chemistry
1 answer:
Mars2501 [29]3 years ago
3 0

Answer:

1 and 11

Explanation:

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Given that a chlorine-oxygen bond has an enthalpy of 243 kJ/mol , an oxygen-oxygen bond has an enthalpy of 498 kJ/mol , and the
Alecsey [184]

Explanation:

The chemical equation is as follows.

      \frac{1}{2}Cl_{2}(g) + \frac{1}{2}O_{2}(g) \rightarrow ClO(g)

And, the given enthalpy is as follows.

    \frac{1}{2}Cl_{2}(g) + O_{2}(g) \rightarrow ClO_{2}(g);  \Delta H = 102.5 kJ

    Cl-Cl = 243 kJ/mol,      O=O = 498 kJ/mol

Since, the bond enthalpy of Cl-Cl is not given so at first, we will calculate the value of Cl-Cl as follows.

   \Delta H = \sum \text{bond broken in reactants} - \sum \text{bond broken in products}

    102.5 = [(\frac{1}{2})x + 498] - [(2)(243)]

    102.5 = (\frac{1}{2})x + 498 - 486

     102.5 - 12 = \frac{x}{2}

           x = 181 kJ

Now, total bond enthalpy of per mole of ClO is calculated as follows.

       \Delta H = \sum E(\text{bond broken in reactants}) - \sum (\text{bond broken in products})

              x = [(\frac{1}{2})181 + (\frac{1}{2})498] - 243

                 = 339.5 - 243

                 = 96.5 kJ

Thus, we can conclude that the value for the enthalpy of formation per mole of ClO(g) is 96.5 kJ.

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Answer:

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Explanation:

Have a great day :)

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