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MrRa [10]
3 years ago
5

Free radical mono-halogenation of an alkane is typically conducted using bromine versus chlorine because A.the bromine radical i

s more reactive and therefore more selective. B.the chlorine radical is more reactive and therefore more selective. C.the chlorine radical is less reactive and therefore more selective. D.the bromine radical is less reactive and therefore more selective. E.none of these choices.
Chemistry
1 answer:
Nimfa-mama [501]3 years ago
5 0

Answer:Free radical mono-halogenation of an alkane is typically conducted using bromine versus chlorine because the bromine radical is less reactive and therefore more selective.

Explanation: Halogenation occurs when a halogen replaces  one or more hydrogen atoms in an organic compound ie  chlorine or  bromine with the reactivity of the halogens decreasing  in the order of  F2 > Cl2 > Br2 > I2

Since fluorine reacts  explosively making it is difficult to control, and iodine is unreactive. Free radical mono-halogenation of an alkane is typically conducted using bromine versus chlorine  with Chlorination  ie chlorine radical being more reactive and not selective and the Bromination of alkanes ie bromine radical  occurring similarly but slower and less reactive but  more selective which is due to the fact that a bromine atom is less reactive in  the hydrogen abstraction  than a chlorine atom evidence in the  higher bond energy of H-Cl than H-Br.

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2 years ago
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3 years ago
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3 years ago
Realice las siguientes conversiones: a) 72°F a °C, b) 213.8°C a °F, c)180°C a K, d) 315K a °F, e) 1750°F a K, f) 0K a °F.
bonufazy [111]

Answer:

a) 72 °F= 22.22 °C

b)  213.8 °C=  416.84°F

c) 180 °C= 453.15 °K

d) 315 °K=  107.33 °F

e) 1750 °F= 1227.594 °K

f) 0 °K=  -459.67 °F

Explanation:

Para realizar el intercambio de unidades debes tener en cuenta las siguientes conversiones:

  • Fahrenheit a Celsius: C=\frac{F-32}{1.8}
  • Celsius a Fahrenheit: °F= °C*1.8 + 32
  • Celsius a Kelvin: °K= °C + 273.15
  • Kelvin a Fahrenheit: F= (K -273.15)*1.8 + 32
  • Fahrenheit a Kelvin:K=\frac{F-32}{1.8} + 273.15

Entonces se obtiene:

a) 72 °F= \frac{72-32}{1.8}=22.22 °C

b)  213.8 °C= 213.8*1.8 + 32= 416.84°F

c) 180 °C= 180°C + 273.15= 453.15 °K

d) 315 °K= (315 -273.15)*1.8 + 32= 107.33 °F

e) 1750 °F= \frac{1750-32}{1.8} + 273.15= 1227.594 °K

f) 0 °K= (0 -273.15)*1.8 + 32= -459.67 °F

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2 years ago
Question 10 of 25
kap26 [50]

Answer:

Correct option would be A. Mg(s) → Mg2+ + 2e-

Explanation:

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2 years ago
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