Answer:
D
Explanation:
John is not a very good businessman.
:D
Explanation:
Half life of zero order and second order depends on the initial concentration. But as the given reaction slows down as the reaction proceeds, therefore, it must be second order reaction. This is because rate of reaction does not depend upon the initial concentration of the reactant.
a. As it is a second order reaction, therefore, doubling reactant concentration, will increase the rate of reaction 4 times. Therefore, the statement a is wrong.
b. Expression for second order reaction is as follows:
![\frac{1}{[A]} =\frac{1}{[A]_0} +kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%20%3D%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%20%2Bkt)
the above equation can be written in the form of Y = mx + C
so, the plot between 1/[A] and t is linear. So the statement b is true.
c.
Expression for half life is as follows:
![t_{1/2}=\frac{1}{k[A]_0}](https://tex.z-dn.net/?f=t_%7B1%2F2%7D%3D%5Cfrac%7B1%7D%7Bk%5BA%5D_0%7D)
As half-life is inversely proportional to initial concentration, therefore, increase in concentration will decrease the half life. Therefore statement c is wrong.
d.
Plot between A and t is exponential, therefore there is no constant slope. Therefore, the statement d is wrong
M1 = 17.45 M
M2 = 0.83 M
V2 = 250 ml
M1. V1= M2. V2
V1 = (M2. V2)/M1 = (0.83× 250)/ 17.45= 11.89 ml
Answer:
The answer is C. The partial pressure of hydrogen will be unchanged.
Explanation:
⇒ 
Argon with electronic configuration
(that is atomic number 18) is an inert gas making it unreactive and it's addition to the reaction has no effect on the partial pressure of either the reactant or production or the state of the system.
The partial pressure of hydrogen will remain unchanged on the addition of Argon.