It is the 4th one that is correct and balanced
Hypothesis
Therefore the last step in the scientific method is proposing a hypothesis or obtaining a conclusion
A skateboard coasting on a flat surface slows down and then comes to a stop because the skateboard runs out of energy, and slows down, also, sense it is on a flat surface and not going downhill, it goes slower.
<span>The molecular formula that describes the problem is
2CH3COOH (aq) + Ca(OH)2 (s) ---> Ca(CH3COO)2 (aq) + 2H2O (l)
The net equation is written as follows:
2CH3COOH- (aq) + 2H+ (aq) + Ca(OH)2 (s) ---> Ca2+ (aq) + 2 CH3COO- (aq) + 2H2O (l)
canceling out spectator ions
2H+ (aq) + Ca(OH)2 (s) ---> Ca2+ (aq) + 2 H2O (l)</span>
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.
