The rate law is first-order in N2O5. At 45 °C the rate constant is 6.08 x 10–4 s–1. What is the rate of the reaction when [N2O5] = 0.100 M, and what happens to the rate when the concentration of N2O5 is doubled to 0.200 M?

Explanation:

Rate constant = 6.08 x 10–4 s–1

What is the rate of the reaction when [N2O5] = 0.100 M..?

A first-order reaction depends on the concentration of one reactant, and the rate law is:

r= −dA / dt = k[A]

Rate of reaction = k [ A ]

where [A] = 0.100 M

Rate of reaction = (6.08 x 10–4) x 0.100

Rate of reaction = 6.08 x 10–5 Ms–1

what happens to the rate when the concentration of N2O5 is doubled to 0.200 M?

Rate of reaction = k [ A ]

[A] = 0.200

Rate of reaction = (6.08 x 10–4) x 0.200

Rate of reaction = 1.216 x 10–4 Ms–1

The rate of the reaction increases to 1.216 x 10–4 Ms–1

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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you gotta add the following molecules so we can give u an answer

explanation: