The rate constant for this second‑order reaction is 0.990 M − 1 ⋅ s − 1 0.990 M−1⋅s−1 at 300 ∘ C. 300 ∘C. A ⟶ products A⟶products How long, in seconds, would it take for the concentration of A A to decrease from 0.650 M 0.650 M to 0.380 M?

Answer : The time taken would be, 1.10 s

Explanation :

The integrated rate law equation for second order reaction follows:

where,

k = rate constant =

t = time taken  = ?

[A] = concentration of substance after time 't' = 0.380 M

= Initial concentration = 0.650 M

Now put all the given values in above equation, we get:

Hence, the time taken would be, 1.10 s

for step 7 table c on the lab "newtons law of action"   the average time it took the car to travel .25 meters   was 1.92 seconds.

the average time it took the car to travel 0.50 meters was 2.61 seconds

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they vary. it used to be meat, grain, bread, candy, beverages, donuts, and fruits/vegetables. now they added exercise in.