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Natali [406]
4 years ago
9

Rank the following from highest to lowest priority according to the Cahn-Ingold-Prelog system: -CH2CH3, -CHCH2, -CCH, -CH3.

Chemistry
1 answer:
mario62 [17]4 years ago
7 0

Answer:

The order will be:

CCH > CHCH₂ > CH₂CH₃> CH₃

Explanation:

According to Cahn-Ingold-Prelog system we rank the groups based on the atomic number of directly attached atom with the chiral carbon.

For example: between C and H, we rank Carbon first.

If the same atoms are attached for different groups then we prioritized based on the second element with highest atomic number.

For example:

Among CH₃ and C₂H₅, the priority will be given to C₂H₅.

If an atom is double or triple bonded to the directly attached atom then each pi bond is considered to be a new atom.

Hence CH=CH₂ means, that there are two carbons attached to CH carbon.

So the order based on above selection rules will be:

CCH > CHCH₂ > CH₂CH₃> CH₃

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Answer:

0.1 mol

Explanation:

Step 1: Given data

  • Volume of the solution of sodium hydroxide (V): 200 cm³
  • Molar concentration of sodium hydroxide (C): 0.5 mol/dm³

Step 2: Convert "V" to dm³

We will use the conversion factor 1 dm³ = 1000 cm³.

200 cm³ × 1 dm³/1000 cm³ = 0.200 dm³

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The molarity of the NaOH solution is equal to the moles of NaOH divided by the volume of solution.

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Answer:

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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
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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

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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

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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})

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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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