Answer:
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Answer:
Rate of reaction =
Rate of consumption of A = 
Rate of consumption of B =
Rate of formation of D = 
Explanation:
According to laws of mass action for the given reaction,
![Rate= -\frac{1}{2}\frac{\Delta [A]}{\Delta t}=-\frac{\Delta [B]}{\Delta t}=\frac{1}{2}\frac{\Delta [C]}{\Delta t}=\frac{1}{3}\frac{\Delta [D]}{\Delta t}](https://tex.z-dn.net/?f=Rate%3D%20-%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D%3D-%5Cfrac%7B%5CDelta%20%5BB%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7B%5CDelta%20%5BD%5D%7D%7B%5CDelta%20t%7D)
where, ![-\frac{\Delta [A]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BA%5D%7D%7B%5CDelta%20t%7D)
is rate of consumption of A, ![-\frac{\Delta [B]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BB%5D%7D%7B%5CDelta%20t%7D)
is rate of consumption of B, ![\frac{\Delta [C]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D)
is rate of formation of C and ![\frac{\Delta [D]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BD%5D%7D%7B%5CDelta%20t%7D)
is rate of formation of D
Here ![\frac{\Delta [C]}{\Delta t}=2.7mol.dm^{-3}.s^{-1}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%3D2.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D)
So, Rate of reaction = 
Rate of formation of D = ![(\frac{3}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{3}{2}\times 2.7mol.dm^{-3}.s^{-1})=4.15mol.dm^{-3}.s^{-1}](https://tex.z-dn.net/?f=%28%5Cfrac%7B3%7D%7B2%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%29%3D%28%5Cfrac%7B3%7D%7B2%7D%5Ctimes%202.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D%29%3D4.15mol.dm%5E%7B-3%7D.s%5E%7B-1%7D)
Rate of consumption of A = ![(\frac{2}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{2}{2}\times 2.7mol.dm^{-3}.s^{-1})=2.7mol.dm^{-3}.s^{-1}](https://tex.z-dn.net/?f=%28%5Cfrac%7B2%7D%7B2%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%29%3D%28%5Cfrac%7B2%7D%7B2%7D%5Ctimes%202.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D%29%3D2.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D)
Rate of consumption of B = ![(\frac{1}{2}\times \frac{\Delta [C]}{\Delta t})=(\frac{1}{2}\times 2.7mol.dm^{-3}.s^{-1})=1.35mol.dm^{-3}.s^{-1}](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B2%7D%5Ctimes%20%5Cfrac%7B%5CDelta%20%5BC%5D%7D%7B%5CDelta%20t%7D%29%3D%28%5Cfrac%7B1%7D%7B2%7D%5Ctimes%202.7mol.dm%5E%7B-3%7D.s%5E%7B-1%7D%29%3D1.35mol.dm%5E%7B-3%7D.s%5E%7B-1%7D)
Hey there!:
Given [ OH⁻ ] = 9.00 x 10⁻⁴
We know that : [H⁺ ] [OH⁻ ] = 1 x 10⁻¹⁴
Thus [H⁺ ] = ( 1 x 10⁻¹⁴ ) / ( 9.00 x 10⁻⁴ )
[ H⁺ ] = 1 x 10⁻¹¹ M
therefore:
pH = - log [ H⁺ ]
pH = - log [ 1 x 10⁻¹¹ ]
pH = 10.95
Therfore pH = 10.95
I hope this will help !
Answer:oxidation number H = +1
oxidation number O = -2
let x = oxidation number Cl
in HClO3
+1 + x -6 =0
x = +5
Explanation:
D) Hail. Hail is frozen rain
