Answer:
4.83% of acetic acid in the vinegar
Explanation:
Answer:
6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.
Explanation:
![Rate = k[AB]^2](https://tex.z-dn.net/?f=Rate%20%3D%20k%5BAB%5D%5E2)
The order of the reaction is 2.
Integrated rate law for second order kinetic is:
![\frac{1}{[A_t]} = \frac{1}{[A]_0}+kt](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B%5BA_t%5D%7D%20%3D%20%5Cfrac%7B1%7D%7B%5BA%5D_0%7D%2Bkt)
Where, ![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration = 1.50 mol/L
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the final concentration = 1/3 of initial concentration = 
= 0.5 mol/L
Rate constant, k = 0.2 L/mol*s
Applying in the above equation as:-
<u>6.66 s will it take for [AB] to reach 1/3 of its initial concentration 1.50 mol/L.</u>
Specificity. It’s really loose to say that something is fast, since speed can be scalarly linked and relative. I could say that both a car on the highway is fast, but so is the speed of light. The actual speed of something helps to do away with the arbitrary nature of using “fast” and “slow”; however, we’re still at step one of the person who is receiving the information is unfamiliar with the scale that the actual speed is defined in.
Answer:
the heat of formation of isopropyl alcohol is -317.82 kJ/mol
Explanation:
The heat of combustion of isopropyl alcohol is given as follows;
C₃H₇OH (l) +(9/2)O₂ → 3CO₂(g) + 4H₂O (g)
The heat of combustion of CO₂ and H₂O are given as follows
C (s) + O₂ (g) → CO₂(g) = −393.50 kJ
H₂ (g) + 1/2·O₂(g) → H₂O (l) = −285.83 kJ
Therefore we have
3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ which we can write as
3C (s) + 3O₂ (g) → 3CO₂(g) = −393.50 kJ × 3 =
4H₂ (g) + 2·O₂(g) → 4H₂O (l) = −285.83 kJ × 4
3CO₂(g) + 4H₂O (g) → C₃H₇OH (l) +(9/2)O₂ = +2006 kJ/mol
-1180.5 - 1143.32 +2006 = -317.82 kJ/mol
Therefore, the heat of formation of isopropyl alcohol = -317.82 kJ/mol.
