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11111nata11111 [884]
3 years ago
15

Give the IUPAC name for each compound.

Chemistry
1 answer:
Kaylis [27]3 years ago
3 0

Answer:

a. 1-fluoro-3,3,4-trimethyl-pentane.

b. 1-iodo-3-ethyl-2-methyl-hexane.

c. 1,3-dichloro-5-dimethyl-hexane.

d. 1-bromo-3-methyl-cyclopentane.

Explanation:

Hello!

In this case, according to the IUPAC rules for the listed alkyl halides, we first need to name the halogens (considering periodic order) then alkyl radicals and finally the parent chain; thus, the names are given below:

a. 1-fluoro-3,3,4-trimethyl-pentane.

b. 1-iodo-3-ethyl-2-methyl-hexane.

c. 1,3-dichloro-5-dimethyl-hexane.

d. 1-bromo-3-methyl-cyclopentane.

e. 5-chloro-1-bromo-1,1,5-trimethyl-pentane (radicals are not clear).

Best regards!

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A sample of an ideal gas in a cylinder of volume 2.67 L at 298 K and 2.81 atm expands to 8.34 L by two different pathways. Path
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Explanation:

  • For path A, the calculation will be as follows.

As, for reversible isothermal expansion the formula is as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

Since, we are not given the number of moles here. Therefore, we assume the number of moles, n = 1 mol.

As the given data is as follows.

              R = 8.314 J/(K mol),          T = 298 K ,

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Now, putting the given values into the above formula as follows.

            W = -2.303 nRT log(\frac{V_2}{V_1})

   = -2.303 \times 1 \times 8.314 J/K mol \times 298 log(\frac{8.34}{2.67})

     = -2.303 \times 1 \times 8.314 J/K mol \times 298 \times 0.494

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Hence, work for path A is -2818.68 J.

  • For path B, the calculation will be as follows.

Step 1: When there is no change in volume then W = 0

Hence, for step 1, W = 0

Step 2: As, the gas is allowed to expand against constant external pressure P_{external} = 1.00 atm.

So,              W = -P_{external} \times \Delta V

Now, putting the given values into the above formula as follows.

               W = -P_{external} \times \Delta V

                   = -1 atm \times (8.34 L - 2.67 L)  

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As we known that, 1 atm L = 101.33 J

Hence, work will be calculated as follows.

       W = -\frac{101.33 J}{1 atm L} \times 5.67 atm L

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Therefore, total work done by path B = 0 + (-574.54 J)

                        W = -574.54 J

Hence, work for path B is -574.54 J.

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