Why cannot pure bromine be used in free-radical reactions with alkenes?

The correct answer is due to its reactivity with alkanes.

Explanation:

The reactivity of the different halogens varies considerably. The relative reactivity rate of bromine is 7 * 10^-11. Due to this bromine behavior, the reaction of brominated alkanes is slow and requires high levels of UV irradiation, therefore it is not possible to use pure bromine in free radical reactions with alkanes.

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the ideal gas law equation is as follows

pv = nrt

where p - pressure

v - volume

n - number of moles

r - universal gas constant

t - temperature

so if the pressure, volume and temperature are already known

we are left with n and r

since r is the universal gas constant that has a known fixed value then r too is known

so we are left with 'n'

once we know temperature volume and pressure

we can find the number of moles of gas present using the ideal gas law equation

this is not my own answer, this is an answer for the same question by the user blahblahmali. i do not claim ownership of this account.

c, liquids evaporate, creating a gas