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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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If the element A have three protons in the nucleus. find the atomic number of the element A​
sweet-ann [11.9K]

Answer:

ATOMIC number of element A is 3

Explanation:

Atomic number is the same as number of protons of an element.

according to the periodic table, Lithium is the element with 3 protons in its neutral state and it's atomic number is 3.

3 0
3 years ago
The Michaelis constant for pancreatic lipase is 5 mM. At 60 C, lipase is subject to deactivation with a half-life of 8 min. Fat
Dmitriy789 [7]

Explanation:

According to the given data, we will calculate the following.

 Half life of lipase t_{1/2} = 8 min x 60 s/min

                                       = 480 s

Rate constant for first order reaction is as follows.

         k_{d} = \frac{0.6932}{480}

                        = 1.44 \times 10^{-3}s^{-1}


Initial fat concentration S_{o} = 45 mol/m^{3}

                                                = 45 mmol/L

Rate of hydrolysis V_m_{o} = 0.07 mmol/L/s

Conversion X = 0.80

Final concentration (S) = S_{o} (1 - X)

                                      = 45 (1 - 0.80)

                                      = 9 mol/m^{3}

or,                                  = 9 mmol/L

It is given that K_{m} = 5mmol/L

Therefore, time taken will be calculated as follows.

                    t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]

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

            t = -\frac{1}{K_{d}}ln[1 - \frac{K_{d}}{V}{K_{M} ln (\frac{S_{o}}{S}) + (S_{o} - S)]  

             = -\frac{1}{1.44 \times 10^{-3}s^{-1}}ln[1 - \frac{1.44 \times 10^{-3}s^{-1}}{0.07 mmol/L/s
}{K_{M} ln (\frac{45 mmol/L
}{9 mmol/L
}) + (45 mmol/L - 9 mmol/L
)]

              = 1642.83 s \times \frac{1 min}{60 sec}

              = 27.38 min

Therefore, we can conclude that time taken by the enzyme to hydrolyse 80% of the fat present is 27.38 min.

6 0
3 years ago
The most common element in the atmosphere are
Ivanshal [37]
Nitrogen and oxygen are the most prevalent in the atmosphere.
8 0
4 years ago
Many elements that are essential for life,including nitrogen,oxygen,and carbon, are part of what classification?
Bess [88]

Answer:

The answer is emma... C

8 0
4 years ago
True or False? The fatal 1930 incident in Muese Valley, Belgium, was associated with leakage of toxic chemicals from a former du
Westkost [7]

Answer:

FALSE                            

Explanation:

The incident of Muese Valley occured in 1930 due to air pollution.

Muese Valley lies along the river Muese which is situated Huy and Liege, Belgium. This region was crowded with industries including steel manufacturers, glass manufacturers, explosives plants, zinc smelter, etc.

The increase number of industries and population lead to the sources of pollution. Also increase in burning of domestic coal increased pollution surrounding the area.

Air pollution became so severe at this region that people have severe  respiratory problems. Residents suffered from vomiting, retrosternal pain, coughing fits and several experienced nausea. There were fog and smog all over and many people died.

Hence the answer is FALSE.

8 0
3 years ago
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