Answer:
The amount of energy required to remove an electron from an isolated gaseous atom in its gaseous state.
Explanation:
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<u>Answer:</u> The value of
for the reaction at 690 K is 0.05
<u>Explanation:</u>
We are given:
Initial pressure of
= 1.0 atm
Total pressure at equilibrium = 1.2 atm
The chemical equation for the decomposition of phosgene follows:
![COCl_2(g)\rightleftharpoons CO(g)+Cl_2(g)](https://tex.z-dn.net/?f=COCl_2%28g%29%5Crightleftharpoons%20CO%28g%29%2BCl_2%28g%29)
Initial: 1 - -
At eqllm: 1-x x x
We are given:
Total pressure at equilibrium = [(1 - x) + x+ x]
So, the equation becomes:
![[(1 - x) + x+ x]=1.2\\\\x=0.2atm](https://tex.z-dn.net/?f=%5B%281%20-%20x%29%20%2B%20x%2B%20x%5D%3D1.2%5C%5C%5C%5Cx%3D0.2atm)
The expression for
for above equation follows:
![K_p=\frac{p_{CO}\times p_{Cl_2}}{p_{COCl_2}}](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7Bp_%7BCO%7D%5Ctimes%20p_%7BCl_2%7D%7D%7Bp_%7BCOCl_2%7D%7D)
![p_{CO}=0.2atm\\p_{Cl_2}=0.2atm\\p_{COCl_2}=(1-0.2)=0.8atm](https://tex.z-dn.net/?f=p_%7BCO%7D%3D0.2atm%5C%5Cp_%7BCl_2%7D%3D0.2atm%5C%5Cp_%7BCOCl_2%7D%3D%281-0.2%29%3D0.8atm)
Putting values in above equation, we get:
![K_p=\frac{0.2\times 0.2}{0.8}\\\\K_p=0.05](https://tex.z-dn.net/?f=K_p%3D%5Cfrac%7B0.2%5Ctimes%200.2%7D%7B0.8%7D%5C%5C%5C%5CK_p%3D0.05)
Hence, the value of
for the reaction at 690 K is 0.05
Answer: The pressure after the tire is heated to 17.3°C is 167 kPa
Explanation:
To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.
Mathematically,
![\frac{P_1}{T_1}=\frac{P_2}{T_2}](https://tex.z-dn.net/?f=%5Cfrac%7BP_1%7D%7BT_1%7D%3D%5Cfrac%7BP_2%7D%7BT_2%7D)
where,
are the initial pressure and temperature of the gas.
are the final pressure and temperature of the gas.
We are given:
![P_1=149kPa\\T_1=-14^0C=(273-14)=259K\\P_2=?=27.5psi\\T_2=17.3^0C=(273+17.3)=290.3K](https://tex.z-dn.net/?f=P_1%3D149kPa%5C%5CT_1%3D-14%5E0C%3D%28273-14%29%3D259K%5C%5CP_2%3D%3F%3D27.5psi%5C%5CT_2%3D17.3%5E0C%3D%28273%2B17.3%29%3D290.3K)
Putting values in above equation, we get:
![\frac{149}{259}=\frac{P_2}{290.3}\\\\P_2=167kPa](https://tex.z-dn.net/?f=%5Cfrac%7B149%7D%7B259%7D%3D%5Cfrac%7BP_2%7D%7B290.3%7D%5C%5C%5C%5CP_2%3D167kPa)
Hence, the pressure after the tire is heated to 17.3°C is 167 kPa
Radioactive is the most penetrating nuclear radiation
Ok thanks for the valuble info.