<u>Answer:</u> The equilibrium concentration of 
is 0.332 M
<u>Explanation:</u>
We are given:
Initial concentration of = 2.00 M
The given chemical equation follows:
<u>Initial:</u> 2.00
<u>At eqllm:</u> 2.00-2x x x
The expression of 
for above equation follows:
![K_c=\frac{[CO_2][CF_4]}{[COF_2]^2}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5BCF_4%5D%7D%7B%5BCOF_2%5D%5E2%7D)
We are given:
Putting values in above expression, we get:
Neglecting the value of x = 1.25 because equilibrium concentration of the reactant will becomes negative, which is not possible
So, equilibrium concentration of ![COF_2=(2.00-2x)=[2.00-(2\times 0.834)]=0.332M](https://tex.z-dn.net/?f=COF_2%3D%282.00-2x%29%3D%5B2.00-%282%5Ctimes%200.834%29%5D%3D0.332M)
Hence, the equilibrium concentration of is 0.332 M
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Answer:- 1.90 atm
Solution:- It is based on combined gas law equation, PV = nRT
In this equation, P is pressure, V is volume, n is moles of gas, R is universal gas constant and T is kelvin temperature.
If we divide both sides by V then:
We know that, molarity is moles per liter. So, in the above equation we could replace 
by molarity, M of the gas. The equation becomes:
P = MRT
T = 20 + 273 = 293 K
M = 
Let's plug in the values in the equation:
P = 
P = 1.90 atm
So, the pressure of the gas is 1.90 atm.
Answer:
MnO2 M n O 2 is an ionic substance without any oxo-anions, therefore its name has the -ide ending.
Answer:
∆H or Enthalpy of the reaction
Explanation:
If ∆H is +ve
- Reaction is exothermic
- Example:-Combustion, mixing sodium/potassium in water
If ∆H is -ve
- Reaction is endothermic
- Ex:-Melting of ice
