Why is a chemical equilibrium described as dynamic?

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01-06-2017
Chemistry

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Why is a chemical equilibrium described as dynamic?

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All of the following expressions have a value between 0 and 1, excepta. \frac{(-3^{7})}{(-3)^9}b. 4^{-10}× 4^{6}c. \frac{(8^3)^3}{8^{-4}} d. ( \frac{2}{5} ^{8})

find the circumference of a circle whose radius is 15.4cm

All of the following expressions have a value between 0 and 1, excepta. \frac{(-3^{7})}{(-3)^9}b. 4^{-10}× 4^{6}c. \frac{(8^3)^3}{8^{-4}} d. ( \frac{2}{5} ^{8})

find the circumference of a circle whose radius is 15.4cm

What value of x makes the equation true ? 4(2x-4)=16

TrinityWheeler72 TrinityWheeler72 · Answer 1 · 2017-06-12T15:28:17+02:00

TrinityWheeler72 TrinityWheeler72

12-06-2017

Because when equilibrium is reached, the reaction is still occurring in both directions, its just that rate (foreward)=rate(reverse) so there is no net change in the concentrations of the reactions or products.

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aprilmartinezz2 aprilmartinezz2 · Answer 2 · 2018-11-27T06:59:23+01:00

aprilmartinezz2 aprilmartinezz2

27-11-2018

Because both reactants and products continue to form. Option (c) is the answer.

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