Answer:
k = 100 mol⁻² L² s⁻¹, r= k[A][B]²
Explanation:
A + B + C --> D
[A] [B] [C] IRR
0.20 0.10 0.40 .20
0.40 0.20 0.20 1.60
0.20 0.10 0.20 .20
0.20 0.20 0.20 .80
Comparing the third and fourth reaction, the concentrations of A and C are constant. Doubling the concentration of B causes a change in the rate of the reaction by a factor of 4.
This means the rate of reaction is second order with respect to B.
Comparing reactions 2 and 3, the concentrations of B and C are constant. Halving the concentration of A causes a change in the rate of the reaction by a factor of 2.
This means the rate of reaction is first order with respect to A.
Comparing reactions 1 and 3, the concentrations of A and B are constant. Halving the concentration of A causes no change in the rate of the reaction.
This means the rate of reaction is zero order with respect to C.
The rate expression for this reaction is given as;
r = k [A]¹[B]²[C]⁰
r= k[A][B]²
In order to obtain the value of the rate constant, let's work with the first reaction.
r = 0.20
[A] = 0.20 [B] = 0.10
k = r / [A][B]²
k = 0.20 / (0.20)(0.10)²
k = 100 mol⁻² L² s⁻¹
Answer:
1500 mmol.
Explanation:
Hello,
In this case, for such monovalent potassium species, we can verify that 1 mEq equals 1 mmol, therefore, the required solution is shown below:
Which means that also 1500 mmol of monovalent potassium ions are contained in the liter.
Regards.
Answer:
yes it should react yoyoyoyoyo
I would say chemical but I’m not 100% sure might wanna get a second opinion
