The volume (in liters) that the gas will occupy if the pressure is increased to 13.5 atm and the temperature is decreased to 15 °C is 15 L
From the question given above, the following data were obtained:
Initial pressure (P₁) = 8.5 atm
Initial volume (V₁) = 24 L
Initial temperature (T₁) = 25 °C = 25 + 273 = 298 K
Final pressure (P₂) = 13.5 atm
Final temperature (T₂) = 15 °C = 15 + 273 = 288 K
<h3>Final volume (V₂) =? </h3>
- The final volume of the gas can be obtained by using the combined gas equation as illustrated below:

Cross multiply
298 × 13.5 × V₂ = 204 × 288
4023 × V₂ = 58752
Divide both side by 4023

<h3>V₂ = 15 L </h3>
Therefore, the final volume of the gas is 15 L
Learn more: brainly.com/question/25547148
Answer:
0.6749 M is the concentration of B after 50 minutes.
Explanation:
A → B
Half life of the reaction = 
Rate constant of the reaction = k
For first order reaction, half life and half life are related by:


Initial concentration of A = ![[A]_o=0.900 M](https://tex.z-dn.net/?f=%5BA%5D_o%3D0.900%20M)
Final concentration of A after 50 minutes = ![[A]=?](https://tex.z-dn.net/?f=%5BA%5D%3D%3F)
t = 50 minute
![[A]=[A]_o\times e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_o%5Ctimes%20e%5E%7B-kt%7D)
![[A]=0.900 M\times e^{-0.02772 min^{-1}\times 50 minutes}](https://tex.z-dn.net/?f=%5BA%5D%3D0.900%20M%5Ctimes%20e%5E%7B-0.02772%20min%5E%7B-1%7D%5Ctimes%2050%20minutes%7D)
[A] = 0.2251 M
The concentration of A after 50 minutes = 0.2251 M
The concentration of B after 50 minutes = 0.900 M - 0.2251 M = 0.6749 M
0.6749 M is the concentration of B after 50 minutes.
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