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.
The grams of oxygen that are required to produce 1 mole of H₂O is 16 g ( answer B)
<u><em> calculation</em></u>
2 CH₄ + 2NH₃ +3 O₂ → 2HCN + 6H₂O
step 1: use the mole ratio to find moles of O₂
from equation above the mole ratio of O₂: H₂O is 3:6 therefore the moles of O₂ = 1 mole x3/6 =0.5 moles
step 2: find mass of O₂
mass= moles x molar mass
from periodic table the molar mass of O₂ = 16 x2= 32 g/mol
mass O₂ = 0.5 moles x 32 g/mol = 16 g (answer B)
Answer:
What is the net ionic equation for a reaction between HCl and NaOH?
Explanation:
A salt is a neutral ionic compound. Let's see how a neutralization reaction produces both water and a salt, using as an example the reaction between solutions of hydrochloric acid and sodium hydroxide. The overall equation for this reaction is: NaOH + HCl → H2O and NaCl
Hope that helped.
