Answer: pH = 3.15
Explanation: Solved in the attached picture.
The integrated rate law for a second-order reaction is given by:
![\frac{1}{[A]t} = \frac{1}{[A]0} + kt](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BA%5Dt%7D%20%3D%20%20%20%5Cfrac%7B1%7D%7B%5BA%5D0%7D%20%2B%20kt%20)
where, [A]t= the concentration of A at time t,
[A]0= the concentration of A at time t=0
<span>k =</span> the rate constant for the reaction
<u>Given</u>: [A]0= 4 M, k = 0.0265 m–1min–1 and t = 180.0 min
Hence, ![\frac{1}{[A]t} = \frac{1}{4} + (0.0265 X 180)](https://tex.z-dn.net/?f=%20%5Cfrac%7B1%7D%7B%5BA%5Dt%7D%20%3D%20%5Cfrac%7B1%7D%7B4%7D%20%2B%20%280.0265%20X%20180%29%20)
<span> = 4.858</span>
<span><span><span>Therefore, [A]</span>t</span>= 0.2058 M.</span>
<span>
</span>
<span>Answer: C</span>oncentration of A, after 180 min, is 0.2058 M
Answer:
Metallic bonding
Explanation:
Metals have low ionization energies. Therefore, their valence electrons are easily delocalized (attracted to the neighbouring metal atoms). These delocalized electrons are then not associated with a specific metal atom. Since the electrons are “free”, the metal atoms have become cations, and the electrons are free to move throughout the whole crystalline structure.
We say that a metal consists of an array of cations immersed in a sea of electrons
.
The electrons act as a “glue” holding the cations together.
Metallic bonds are the attractive forces between the metal cations and the sea of electrons.
In an NaK alloy, for example, the Na and K atoms contribute their valence electrons to the "sea". The atoms aren’t bonded to each other, but they are held in place by the metallic bonding.