Answer: partial pressure of NOBr is 7792 atm
Explanation:
Equilibrium constant is the ratio of the concentration of products to the concentration of reactants each term raised to its stochiometric coefficients.
Equilibrium constant is given as:
![K_{p}=\frac{[p_{NOBr}]^2}{[p_{NO}]^2\times [p_{Br_2}]^1}](https://tex.z-dn.net/?f=K_%7Bp%7D%3D%5Cfrac%7B%5Bp_%7BNOBr%7D%5D%5E2%7D%7B%5Bp_%7BNO%7D%5D%5E2%5Ctimes%20%5Bp_%7BBr_2%7D%5D%5E1%7D)
![28.4=\frac{[p_{NOBr}]^2}{[(119)^2\times (151)^1}](https://tex.z-dn.net/?f=28.4%3D%5Cfrac%7B%5Bp_%7BNOBr%7D%5D%5E2%7D%7B%5B%28119%29%5E2%5Ctimes%20%28151%29%5E1%7D)
![[p_{NOBr}]=7792](https://tex.z-dn.net/?f=%5Bp_%7BNOBr%7D%5D%3D7792)
atm
Partial pressure of NOBr is 7792 atm
The answer to your question is dependent variable.
Answer:
6.44 moles
Explanation:
At STP, 1 mole = 22.4 L
145 L × (1 mole ÷22.4 L) = 6.44 moles
Answer:
Increase
Explanation:
According to Gay-Lussac Law,
The pressure of given amount of a gas is directly proportional to its temperature at constant volume and number of moles.
Mathematical relationship:
P₁/T₁ = P₂/T₂
If the initial temperature and pressure is standard,
Pressure = 1 atm
Temperature = 273.15 K
then we increase the temperature to 400.0 K, The pressure will be,
1 atm / 273.15 K = P₂/400.0K
P₂ = 1 atm × 400.0 K / 273.15 K
P₂ = 400.0 atm. K /273.15 K
P₂ = 1.46 atm
Pressure is also increase from 1 atm to 1.46 atm.
Answer:
That would be hydrogen and helium! :)
