Answer:
0.328 atm
Explanation:
Kp is the equilibrium constant calculated based on the pressure, and it depends only on the gas substances. It will be the multiplication of partial pressures of the products raised to their coefficients divided by the multiplication of partial pressures of the reactants raised to their coefficients.
For the equation given, the stoichiometry is 1 mol of NH₃ for 1 mol of H₂S, so they will have the same partial pressure in equilibrium, let's call it p. So:
Kp = pxp
0.108 = p²
p = √0.108
p = 0.328 atm, which is the partial pressure of the ammonia.
We see that is the <em>structure </em>of graphite it possess these attributes.
Graphite
Option A
<h3>Electricity</h3>
Generally, he conduction of electricity is do by elements with free talent electron after bonding.
Therefore, it is save to say that the <em>solid </em>that possess the attributes that allows for conduction will end up conducting.
Hence, we see that is the <em>structure </em>of graphite it possess these attributes.
Graphite
Option A
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Answer:
Explanation:
Hello,
In this case, the law of mass action for the first reaction turns out:
![Kc=\frac{[AsH_3]^2}{[As]^2[H_2] ^3}=1.27](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BAsH_3%5D%5E2%7D%7B%5BAs%5D%5E2%5BH_2%5D%20%5E3%7D%3D1.27)
Now, for the second reaction is:
![Kc=\frac{[As][H_2] ^{3/2}}{[AsH_3]}](https://tex.z-dn.net/?f=Kc%3D%5Cfrac%7B%5BAs%5D%5BH_2%5D%20%5E%7B3%2F2%7D%7D%7B%5BAsH_3%5D%7D)
Therefore, by applying square root for the first reaction, one obtains:
![\sqrt{Kc} =\sqrt{\frac{[AsH_3]^2}{[As]^2[H_2] ^3}} =\sqrt{1.27}](https://tex.z-dn.net/?f=%5Csqrt%7BKc%7D%20%3D%5Csqrt%7B%5Cfrac%7B%5BAsH_3%5D%5E2%7D%7B%5BAs%5D%5E2%5BH_2%5D%20%5E3%7D%7D%20%3D%5Csqrt%7B1.27%7D)
![\frac{\sqrt{[AsH_3]^2} }{\sqrt{[As]^2} \sqrt{[H_2] ^3} } =\sqrt{1.27}](https://tex.z-dn.net/?f=%5Cfrac%7B%5Csqrt%7B%5BAsH_3%5D%5E2%7D%20%7D%7B%5Csqrt%7B%5BAs%5D%5E2%7D%20%5Csqrt%7B%5BH_2%5D%20%5E3%7D%20%7D%20%3D%5Csqrt%7B1.27%7D)
![\sqrt{1.27}=\frac{[AsH_3]}{[As][H_2] ^{3/2}}](https://tex.z-dn.net/?f=%5Csqrt%7B1.27%7D%3D%5Cfrac%7B%5BAsH_3%5D%7D%7B%5BAs%5D%5BH_2%5D%20%5E%7B3%2F2%7D%7D)
Finally, since Kc is asked for the inverse reaction, one modifies the previous equation as:
![Kc'=\frac{1}{\sqrt{1.27} }=\frac{[As][H_2] ^{3/2}}{[AsH_3]}=0.887](https://tex.z-dn.net/?f=Kc%27%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B1.27%7D%20%7D%3D%5Cfrac%7B%5BAs%5D%5BH_2%5D%20%5E%7B3%2F2%7D%7D%7B%5BAsH_3%5D%7D%3D0.887)
Best regards.
Answer:
characteristic properties of an element are the defining properties of that element and it does not change with quantity of the element used.
Explanation:
the amount or the quantity of the element used does not affect the characteristic property of the element. it does not matter is the if the amount or the quantity of the element used in the reaction is large or the small the characteristic properties like boiling point, melting point, density, thermal conductivity, etc remain the same or remain constant.
Well all reactions need energy to start it. The light is the energy that the reaction requires to start it. Sunlight also has no mass in the first place so laws of conservation of mass don't apply to it.
