Answer:
Rate of reaction when concentration of A is doubled=0.018 M/s.
Step-by-step explanation:
We are given that for the multi-step reaction
It is given that the rate-limiting step is unimolecular with A as the sole reactant
It means rate of reaction depend upon the concentration of reactant A only
Rate of reaction =k[A]
[A]=0.110 M
[B]=0.110 M
Rate of reaction =0.0090 M/s
According to rate law
Rate of reaction=k[A]
If concentration of A is doubled then the rate of reaction is given by
Rate of reaction =k[2A]= 2 k[A]=2 initial rate of reaction
Therefore, rate of reaction after substituting values
Rate of reaction ==0.018 M/s
Rate of reaction when concentration of A is doubled=0.018 M/s.
First you need to distribute the 2 into each equation and then add like terms. the answer is 6x + 8
Answer:
x =
x = -1
Step-by-step explanation:
The given equation is,
-2(bx - 5) = 16
Dividing by (-2) on both sides of the equation,
(bx - 5) = -8
By adding 5 on both the sides of the equation,
(bx - 5) + 5 = -8 + 5
bx = -3
Dividing by 'b' on both the sides of the equation,
x =
If b = 3,
x =
x = -1
Answer:
A or d
Step-by-step explanation:
Hope this helps