Answer:
Rate = 0.001615 Ms-1
Explanation:
2 NO2 + F2 --> 2 NO2F
The reaction is first order with respect to NO2 and also first order with respect to F2.
The rate law is given as;
Rate = k [NO2] [ F2]
k = 1.58E-4 M-1s-1
[NO2] = 2.84 M
[F2] = 3.60 M
Rate = ?
Inserting the values into the equation, we have;
Rate = 1.58E-4 * 2.84 * 3.60
Rate = 0.001615 Ms-1
I mostly believe in between D and B beacuse K3po4 and caco3 is not an element equation
to have the closest number rounded up
Rate law for the given 2nd order reaction is:
Rate = k[a]2
Given data:
rate constant k = 0.150 m-1s-1
initial concentration, [a] = 0.250 M
reaction time, t = 5.00 min = 5.00 min * 60 s/s = 300 s
To determine:
Concentration at time t = 300 s i.e. ![[a]_{t}](https://tex.z-dn.net/?f=%5Ba%5D_%7Bt%7D)
Calculations:
The second order rate equation is:
![1/[a]_{t} = kt +1/[a]](https://tex.z-dn.net/?f=1%2F%5Ba%5D_%7Bt%7D%20%3D%20kt%20%2B1%2F%5Ba%5D)
substituting for k,t and [a] we get:
1/[a]t = 0.150 M-1s-1 * 300 s + 1/[0.250]M
1/[a]t = 49 M-1
[a]t = 1/49 M-1 = 0.0204 M
Hence the concentration of 'a' after t = 5min is 0.020 M
<span>C. The stratosphere. About 90% of the ozone is in the stratosphere which begins about 8 miles above the Earth's surface.</span>