What is the correct formula for Tin (IV) chromate nonahydrate?
1 answer:
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Through manipulation of equations, we are able to obtain the equation:
![-pOH= log [ OH^{-}]](https://tex.z-dn.net/?f=-pOH%3D%20log%20%5B%20OH%5E%7B-%7D%5D%20)
Then we can transform the equation into:
![[ OH^{-}]= 10^{-pOH}](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%3D%2010%5E%7B-pOH%7D%20%20)
Then we are able to plug in the pOH and directly get [OH-]:
![[ OH^{-}] = 10^{-6.48}](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%20%3D%2010%5E%7B-6.48%7D%20)
![[ OH^{-}]=3.31* 10^{-7} M](https://tex.z-dn.net/?f=%5B%20OH%5E%7B-%7D%5D%3D3.31%2A%2010%5E%7B-7%7D%20M%20%20)
Then we can transform the equation into:
Then we are able to plug in the pOH and directly get [OH-]:
We need to know the relationship between atmospheric pressure and the density of gas particles in an area of increasing pressure.
The relationship is: As air pressure in an area increases, the density of the gas particles in that area increases.
For any gaseous substance, density of gas is directly proportional to pressure of gas.
This can be explained from idial gas edquation:
PV=nRT
PV=RT [where, w= mass of substance, M=molar mass of substance]
PM=RT
PM=dRT [where, d=density of thesubstance]
So, for a particular gaseous substance (whose molar mass is known), at particular temperature, pressure is directly related to density of gaseous substance.
Therefore, as air pressure in an area increases, the density of the gas particles in that area increases.