Answer: Equilibrium constant is 0.70.
Explanation:
Initial moles of 
= 0.35 mole
Volume of container = 1 L
Initial concentration of 
Initial moles of 
= 0.40 mole
Volume of container = 1 L
Initial concentration of 
equilibrium concentration of 
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The given balanced equilibrium reaction is,
Initial conc. 0.35 M 0.40M 0 0
At eqm. conc. (0.35-x) M (0.40-x) M (x) M (x) M
The expression for equilibrium constant for this reaction will be,
![K_c=\frac{[CO_2]\times [H_2O]}{[CO]\times [H_2O]}](https://tex.z-dn.net/?f=K_c%3D%5Cfrac%7B%5BCO_2%5D%5Ctimes%20%5BH_2O%5D%7D%7B%5BCO%5D%5Ctimes%20%5BH_2O%5D%7D)
we are given : (0.35-x)= 0.18
x = 0.17
Now put all the given values in this expression, we get :
Thus the value of the equilibrium constant is 0.70.
A patient receiving an X-ray in a hospital
Answer: 3.01 x 10^24 atoms
Explanation:
Based on Avogadro's law:
1 mole of any substance has 6.02 x 10^23 atoms
So, 1 mole of water = 6.02 x 10^23 atoms
5 moles of water = Z atoms
To get the value of Z, cross multiply
Z x 1 mole = (6.02 x 10^23 atoms x 5 moles)
Z•mole = 30.1 x 10^23 atoms•mole
Divide both sides by 1 mole
Z•mole/1 mole = 30.1 x 10^23 atoms•mole/ 1 mole
Z = 30.1 x 10^23 atoms
[Place the value of Z in standard form]
Z = 3.01 x 10^24 atoms
Thus, there are 3.01 x 10^24 atoms in 5 mole of water
Static Friction, Sliding Friction, Rolling Friction, and finally Fluid Friction.
Answer : The correct option is, 13.7 mole
Solution : Given,
Moles of 
= 27.4 moles
The given balanced chemical reaction is,
From the balanced chemical reaction, we conclude that
As, 2 moles of react with 1 moles of 
So, 27.4 moles of react with 
moles of
Therefore, the number of moles of oxygen required are, 13.7 moles
