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yanalaym [24]
3 years ago
7

The rate constant for this second‑order reaction is 0.760 M−1⋅s−1 at 300 ∘C. A⟶products How long, in seconds, would it take for

the concentration of A to decrease from 0.750 M to 0.330 M?
Physics
1 answer:
Mashcka [7]3 years ago
3 0

Answer:

2.23 s

Explanation:

For a second-order reaction:

1 / [A] = 1 / [A]₀ + kt

Given [A] = 0.330 M, [A]₀ = 0.750 M, and k = 0.760 M⁻¹s⁻¹:

1 / 0.330 = 1 / 0.750 + 0.760t

t = 2.23

It would take 2.23 seconds.

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a=-.78m/s^2

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  • a is the acceleration of the object during this time.
  • t is time

(v_f-v_i)=at is another way to write this equation.

  • The Δ symbol represents "the difference between the initial and final values of a magnitude or vector", so Δv=(v_f-v_i)

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  • if you converted correctly, your answer for v_f will be ≅ 7.6m/s.
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