Answer:
Initial concentration of HI is 5 mol/L.
The concentration of HI after 
is 0.00345 mol/L.
Explanation:
Rate Law: ![k[HI]^2 ](https://tex.z-dn.net/?f=k%5BHI%5D%5E2%0A)
Rate constant of the reaction = k = 
Order of the reaction = 2
Initial rate of reaction = 
Initial concentration of HI =![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
![1.6\times 10^{-7} mol/L s=(6.4\times 10^{-9} L/mol s)[HI]^2](https://tex.z-dn.net/?f=1.6%5Ctimes%2010%5E%7B-7%7D%20mol%2FL%20s%3D%286.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%29%5BHI%5D%5E2)
![[A_o]=5 mol/L](https://tex.z-dn.net/?f=%5BA_o%5D%3D5%20mol%2FL)
Final concentration of HI after t = [A]
t =
Integrated rate law for second order kinetics is given by:
![\frac{1}{[A]}=kt+\frac{1}{[A_o]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3Dkt%2B%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
![\frac{1}{[A]}=6.4\times 10^{-9} L/mol s\times 4.53\times 10^{10} s+\frac{1}{[5 mol/L]}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA%5D%7D%3D6.4%5Ctimes%2010%5E%7B-9%7D%20L%2Fmol%20s%5Ctimes%204.53%5Ctimes%2010%5E%7B10%7D%20s%2B%5Cfrac%7B1%7D%7B%5B5%20mol%2FL%5D%7D)
![[A]=0.00345 mol/L](https://tex.z-dn.net/?f=%5BA%5D%3D0.00345%20mol%2FL)
The concentration of HI after is 0.00345 mol/L.
Answer:can u anser my question
Explanation: On page 17, author Joey Bartolomeo writes “Caleb participated in the investigation by testifying about what happened during the health seminar at his school.” As it is used in that sentence, what does "testify" mean? *
0 points
C. to persuade someone to join a group
B. to disguise or hide from sight
A. to argue so as to make a person agree
D. to talk and answer questions about something after formally promising to tell the truth
Answer:
.08 L or 80 ml
Explanation:
Use the equation V/t = V/t.
.04L / 150K = V / 300K
.04 / 150 * 300 = V
.08 L or 80 ml
The percentage of Chromium in Chromium Oxide is calculated as follow,
Step 1: Calculate Molar mass of Cr₂O₃,
Cr = 51.99 u
O = 16 u
So,
2(51.99) + 3(16) = 103.98 + 48 = 151.98 u
Step 2: Secondly divide molar mass of only chromium with total mass of Cr₂O₃ and multiply with 100.
i.e.
=
× 100
=
68.41 %
So, the %age composition of chromium in chromium oxide is
68.41 %.
Cold blooded animal? it’s a bit vague sorry
