Answer: Rate law=
, order with respect to A is 1, order with respect to B is 2 and total order is 3. Rate law constant is 
Explanation: Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.
![Rate=k[A]^x[B]^y](https://tex.z-dn.net/?f=Rate%3Dk%5BA%5D%5Ex%5BB%5D%5Ey)
k= rate constant
x = order with respect to A
y = order with respect to A
n = x+y = Total order
a) From trial 1:
(1)
From trial 2:
(2)
Dividing 2 by 1 :![\frac{4.8\times 10^{-2}}{1.2\times 10^{-2}}=\frac{k[0.10]^x[0.40]^y}{k[0.10]^x[0.20]^y}](https://tex.z-dn.net/?f=%5Cfrac%7B4.8%5Ctimes%2010%5E%7B-2%7D%7D%7B1.2%5Ctimes%2010%5E%7B-2%7D%7D%3D%5Cfrac%7Bk%5B0.10%5D%5Ex%5B0.40%5D%5Ey%7D%7Bk%5B0.10%5D%5Ex%5B0.20%5D%5Ey%7D)
therefore y=2.
b) From trial 2:
(3)
From trial 3:
(4)
Dividing 4 by 3:![\frac{9.6\times 10^{-2}}{4.8\times 10^{-2}}=\frac{k[0.20]^x[0.40]^y}{k[0.10]^x[0.40]^y}](https://tex.z-dn.net/?f=%5Cfrac%7B9.6%5Ctimes%2010%5E%7B-2%7D%7D%7B4.8%5Ctimes%2010%5E%7B-2%7D%7D%3D%5Cfrac%7Bk%5B0.20%5D%5Ex%5B0.40%5D%5Ey%7D%7Bk%5B0.10%5D%5Ex%5B0.40%5D%5Ey%7D)
, x=1
Thus rate law is ![Rate=k[A]^1[B]^2](https://tex.z-dn.net/?f=Rate%3Dk%5BA%5D%5E1%5BB%5D%5E2)
Thus order with respect to A is 1 , order with respect to B is 2 and total order is 1+2=3.
c) For calculating k:
Using trial 1: ![1.2\times 10^{-2}=k[0.10]^1[0.20]^2](https://tex.z-dn.net/?f=1.2%5Ctimes%2010%5E%7B-2%7D%3Dk%5B0.10%5D%5E1%5B0.20%5D%5E2)
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Answer:
It depends on their melting and/or their boiling points, because the heat provides the particles with kinetic energy to break the electrosatic bonds in the substances, which can differ in strength
Explanation:
Answer:
Explanation:
<em>0.5 i go to k12 i jus took the test</em>
The correct answer is (A) from 2nd to 3rd shell.
The explanation :
when a gain of energy is the shift of the electrons from a shell of low energy to the shell of high energy
and we have here 2nd shell is the shell of low energy, and 3rd shell is the shell of high energy.
∴ (A) from 2nd to 3rd shell is the correct answer.