It is b i think bc it looks like it would be it and my tudor helped me
Answer:
pH = 4.34
Explanation:
pH= -1/2(logKa) -1/2(log C)
= -1/2( log 5.98*10^-8) -1/2(log 0.0353)
=-1/2(-7.22)-1/2(-1.45)
=3.61+0.725= 4.34
Answer:
RbF
mgo
nh4cl
because electrons are lost by and element forming a cation and gained by the other element forming an anion and held together by electrostatic forces
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
This is what i got The KB expression for aniline c6h5nh2 is: For C6H5NH2 + H2O >< C6H5NH3+ <span>OH-Kb = 4.3 x (10 ^ -10) = [C6H5NH3+][OH-] / [C6H5NH2]
hope this helps:)
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