How long will it take for the concentration to reach 0.0500 M?

This question is incomplete, the complete question is;

A general reaction written as A + 2B → C + 2D is studied and yields the following rate law: ∆A/∆t = k A².

The initial concentration is 1.000 M and t½=20.00 s.

How long will it take for the concentration to reach 0.0500 M

it will take 380 seconds for the concentration to reach 0.0500 M

Step-by-step explanation:

Given that;

A + 2B → C + 2D

initial concentration A₀ = 1 M

half life t1/2 = 20sec

concentration after t time [A_t] = 0.0500 M

∆A/∆t = k A²

This is a second order chemical kinetics

for this order; t1/2 = 1 / [A]₀K

A₀ is initial concentration, K is rate constant, t1/2 is half life

so,

k = 1 / [A₀]t1/2

we substitute

K = 1 / 1.000 × 20

K = 0.05 mol⁻¹L/sec

for second reaction order

1/[A_t] = kt + 1/[A₀]

kt = 1/[A_t] - 1/[A₀]

we substitute

kt = 1/0.0500 - 1/1.000

kt = 20 - 1

kt = 19

substitute the value of k

0.05 × t = 19

t = 19 / 0.05

t = 380 sec

Therefore it will take 380 seconds for the concentration to reach 0.0500 M

the answer is 9.8975

explanation:

when you multiply the "39.59" with the 25% it reduces it and shows its final discounted price. hope this !

step-by-step explanation:

step-by-step explanation:

compare