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kupik [55]
4 years ago
12

Radical chlorination of pentane is a poor way to prepare 1-chloropentane, but radical chlorination of neopentane, (CH3)4C, is a

good way to prepare neopentyl chloride, (CH3)3CCH2Cl. Why?
Chemistry
1 answer:
bija089 [108]4 years ago
4 0
Radical chlorination is not as selective as radical bromination. the rate in which it reacts with primary, secondary or tertiary hydrogens is not that great of a difference. 

So, in pentane, CH3CH2CH2CH2CH3, by the reaction with Cl2, the chlorine group can attach to any of the carbons which would form many isomers and not specifically 1-=chloropentane. In neopentane, (CH3)4C, it is different. each of the H atoms is equivalent. So no matter which one be substituted, only one product, (CH3)CCH2Cl, can be prepared. 
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How many atoms are there in 15.0 grams of Sulfur (S)?
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Answer:

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Explanation:

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The correct answer should be Letter B
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The decomposition of carbon disulfide to carbon monosulfide and sulfur is first order with k=2.8 ×10^-7 at 1000°C .What is the h
RoseWind [281]

Answer:

2.5×10⁶ s

Explanation:

From the question given above, the following data were obtained:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

The half-life of a first order reaction is given by:

Half-life (t½) = 0.693 / Rate constant (K)

t½ = 0.693 / K

With the above formula, we can obtain the half-life of the reaction as follow:

Rate constant (K) = 2.8×10¯⁷ s¯¹

Half-life (t½) =?

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t½ = 2.5×10⁶ s

Therefore, the half-life of the reaction is 2.5×10⁶ s

6 0
3 years ago
A sheet of polyethylene 1.5-mm thick is being used as an oxygen barrier at a temperature of 600K. If the flux is 2.48 x 10-5 kg/
Eddi Din [679]

Complete Question

The complete question is shown on the first uploaded image  

Answer:

The concentration of  high-pressure side is  C_1 =   0.722 \ kg/m^3

Explanation:

From the question we are told that

    The thickness of the polyethylene is  d  =  1.5 \ mm = 0.0015 \ m

     The  temperature is  T  =  600 \   K

      The flux is  JA  =  2.48 *10^{-5} \  kg/m^2\cdot s

      The concentration on the low-pressure side is  C_2 =  0.5 \ kg/m^3

       The initial diffusivity  is  D_o  =  6.2 *10^{-4} \ m^2 /s

       The activation energy for  diffusion is   Q_d  =  41 \ kJ /mol  =  41*10^3 J /mol

Generally the diffusivity  of the oxygen at 600 K can be  mathematically evaluated  as

        D   = D_o * e^{- \frac{Q_d}{R * T  } }  

substituting values  

         D   = 6.2*10^{-4} * e^{- \frac{41 *10^3}{8.314 * 600  } }  

          D   = 1.671 *10^{-7} \ m^2 /s  

Generally the flux is mathematically represented as

          JA  =  D  *  \frac{C_1 -C_2}{d}

Where C_1 is the concentration of oxygen at the higher side

       So  

             C_1 =    d  * \frac{JA}{D}  + C_2

substituting values  

             C_1 =    0.0015   * \frac{2.48*10^{-5}}{1.671*10^{-7}}  + 0.5

              C_1 =   0.722 \ kg/m^3

 

6 0
3 years ago
Given the formula of a compound:
Sindrei [870]

Answer:

C3H6

Explanation:

8 0
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