Using ideal gas equation, PV = nRT, and since there is no volume change and amount change, the equation is now P = kT, where k =nR/V. Temperature must be in kelvin
From the given, k = (0.82)/ (21 + 273) = 2.78 x 10^-3
Substituting T = -3.5+273, P = 0.75 atm
In 0.1 M KOH, thymol blue indicator would appear blue.
Answer:
The initial concentration of ethanal was 0.1590 mol/L.
Explanation:
Integrated rate law for second order kinetic:
![k=\frac{1}{t}(\frac{1}{[A]}-\frac{1}{[A]_o})](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%29)
k = Rate constant =![6.73\times 10^{-3} L mol s](https://tex.z-dn.net/?f=6.73%5Ctimes%2010%5E%7B-3%7D%20L%20mol%20s)
t = Time elapsed = 50.0 s
=initial concentration of ethanal
[A] = Concentration of ethanal left after time t = 0.151 mol/L
On substituting the value:
![6.73\times 10^{-3} L mol s=\frac{1}{50.0 s}(\frac{1}{0.151 mol/L}-\frac{1}{[A_o]})](https://tex.z-dn.net/?f=6.73%5Ctimes%2010%5E%7B-3%7D%20L%20mol%20s%3D%5Cfrac%7B1%7D%7B50.0%20s%7D%28%5Cfrac%7B1%7D%7B0.151%20mol%2FL%7D-%5Cfrac%7B1%7D%7B%5BA_o%5D%7D%29)
![[A]_o=0.1590 mol/L](https://tex.z-dn.net/?f=%5BA%5D_o%3D0.1590%20mol%2FL)
The initial concentration of ethanal was 0.1590 mol/L.
Potential or mechanical energy