Some chemical changes can be reactive through another chemical change?
        
             
        
        
        
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
 
        
             
        
        
        
It is highly hazardous, it is radioactive
        
             
        
        
        
Answer:
hi I'm sorry I can't I just need points
 
        
             
        
        
        
The empirical formula is the simplest formula attainable while maintaining the ratio so it will be CH2.
Explanation:
 The empirical formula of a chemical compound is the simplistic positive integer ratio of atoms being in a compound. A simple example of this thought is that the empirical formula of sulfur monoxide, or SO, would simply be SO, as is the empirical formula of disulfur dioxide, S2O2.