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
<h3><u>Answer</u>;</h3>
≈ 4.95 g/L
<h3><u>Explanation;</u></h3>
The molar mass of KCl = 74.5 g/mole
Therefore; 0.140 moles will be equivalent to ;
= 0.140 moles × 74.5 g/mole
= 10.43 g
Concentration in g/L
= mass in g/volume in L
= 10.43/2.1
= 4.9667
<h3> <u> ≈ 4.95 g/L</u></h3>
The haze is called smog, hope this helps, pls mark brainliest!
Question:
At standard temperature and pressure, the volume of a tire is 3.5L. What is the new pressure if the temperature outside is 296k and its weight causes the volume of the gas is 2.0 L?
Answer:
The new pressure is: 1.896 atm
Explanation:
At standard temperature and pressure, we have:



Outside, we have:


Required
Determine the new pressure
Using combined gas law, we have:

This gives:

Solve for 


