<h2>
Answer:True</h2>
Explanation:
Heterogeneous mixture is a mixture with non-uniform composition.
The properties of the mixture like concentration may change for different parts of the mixture.
Colloids contain solute particles of size 
.The presence of these particles makes the mixture heterogeneous.
Suspensions contain solute particles of size 
.These particles settle to the bottom of the mixture which makes the composition of the bottom different from the top.
So,colloids and suspensions are two types of heterogeneous mixtures.
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
The answer is they follow a patter for valence electrons.
The synthetic compound Zn(C2H3O2)2 is zinc acetic acid derivation. The molecular composition demonstrates that the mixes incorporate zinc, spoke to by Zn; carbon, spoke to by C; hydrogen, spoke to by H; and oxygen, spoke to by O. Below is the <span>an equation that shows how the anion acts as a base:
</span>
C2H3O2−(aq)+H2O(l)⇌<span>HC2H3O2(aq)+OH−(aq)</span>
