Rate = k * [A]^2 * [B]^1
<span>Use the data from any trial to calculate k. </span>
<span>k = (rate)/([A]^2 * [B]^1) </span>
<span>E.g., for Trial 1, we have </span>
<span>rate = 3.0×10−3 M/s </span>
<span>[A] = 0.50 M </span>
<span>[B] = 0.010 M </span>
<span>Plug those numbers in and crank out the answer. </span>
<span>Now with the calculated value of k, calculate the initial rate for [A] = 0.50 M and [B] = 0.075 M </span>
<span>rate = k * [A]^2 * [B]^1 </span>
<span>k = calculated value </span>
<span>[A] = 0.50 M </span>
<span>[B] = 0.075 M</span>
B and d will work out and a and c will also work out
Answer:80KM is distance. 30KM north is displacement.
Explanation:
Answer:
0.30 M
Explanation:
Using integrated rate law for first order kinetics as:
Where,
is the concentration at time t = ?
is the initial concentration = 1.36 M
k is the rate constant = 0.208 s⁻¹
t = 7.30 seconds
So,