how do i solve the following problem?

Nitrogen dioxide decomposes to nitric oxide and oxygen via the reaction:

2NO2 → 2NO + O2

In a particular experiment at 300°C, [NO2] drops from 0.0100 to 0.00650 M in 100 s. The rate of
Disappearance of NO2 for this period is __________ M/s.
3.5x10^-5 is supposedly the answer, but how do i get to that answer?

(0.0100-0.00650)/100 =

(.1M-.0065M)/(100sec)=3.5e-5

To find the rate of disappearance of NO2, you need to calculate the change in concentration of NO2 per unit of time. This can be done using the formula:

Rate = (change in concentration of NO2) / (change in time)

Given:
Initial concentration of NO2, [NO2]initial = 0.0100 M
Final concentration of NO2, [NO2]final = 0.00650 M
Change in time, Δt = 100 s

Now, we can substitute these values into the formula:

Rate = ([NO2]final - [NO2]initial) / Δt

Rate = (0.00650 M - 0.0100 M) / 100 s

Simplifying the equation:

Rate = -0.0035 M / 100 s

Rate = -3.5x10^-5 M/s (Note the negative sign indicates the rate of disappearance)

Therefore, the rate of disappearance of NO2 for this period is 3.5x10^-5 M/s.

To find the rate of disappearance of NO2, you can use the formula:

Rate = ∆[NO2] / ∆t

Where ∆[NO2] is the change in concentration of NO2 and ∆t is the change in time.

Given:
Initial concentration, [NO2]initial = 0.0100 M
Final concentration, [NO2]final = 0.00650 M
Time interval, ∆t = 100 s

First, find the change in concentration (∆[NO2]):

∆[NO2] = [NO2]final - [NO2]initial
= 0.00650 M - 0.0100 M
= -0.0035 M

Note the negative sign, indicating a decrease in concentration.

Next, substitute these values into the rate formula:

Rate = ∆[NO2] / ∆t
= -0.0035 M / 100 s
= -3.5 × 10^-5 M/s

It seems there was an error in the answer you provided. The correct rate of disappearance of NO2 for this period is -3.5 × 10^-5 M/s.