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
Answer:
8 m
Explanation:
3.0 x 10*8 divided by 3.75 x 10*7 = 8 m
Answer:
2H₂ + O₂ → 2H₂O
Explanation:
Chemical equation:
H₂ + O₂ → H₂O
Balance chemical equation:
2H₂ + O₂ → 2H₂O
Step 1:
H₂ + O₂ → H₂O
Left hand side Right hand side
H = 2 H = 2
O = 2 O = 1
Step 2:
H₂ + O₂ → 2H₂O
Left hand side Right hand side
H = 2 H = 4
O = 2 O = 2
Step 3:
2H₂ + O₂ → 2H₂O
Left hand side Right hand side
H = 4 H = 4
O = 2 O = 2
Answer:
The answer is Kr (Krypton).
This is because krypton has an electronic configuration of:
1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6
Taking note of the sequence of electronic configuration:
1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s
It can be seen that Kyrpton's electronic configuration finishes just before the 5s subshell. Therefore, the noble gas notation for an element with valence electrons in the 5s subshell can use [Kr] as a shortcut to denote its electronic configuration. For example:
If an element has 1 valence electron in the 5s subshell, the noble gas notation will be:
[Kr] 5s1
Explanation:
