Answer:
Six electrons are transferred in the formation of Al₂O₃.
Explanation:
Aluminium metal and Oxygen react to form Al₂O₃ as,
2 Al + 3/2 O₂ → Al₂O₃
Oxidation number of Al on left hand side is zero, while than on right hand side in Al₂O₃ is +3. Means it has lost 3 electrons per one atom and six electrons per two atoms. Also, the oxidation number of O at left hand side in O₂ is zero, while that in Al₂O₃ it is -2 per atom and -6 per 3 atoms.
So, two Al atoms have lost 6 electrons and 3 O atoms have gained six electrons.
It is lactic acid the chemical is produced during vigorous exercise when the supply of oxygen is limited or inadequate. <span> It forms when the body breaks down carbohydrates to use for energy during times of low oxygen levels. Hope this answers the question. Have a nice day.</span>
Answer:
The rate of consumption of 
is 2.0 mol/L.s
Explanation:
Applying law of mass action to this reaction-
![-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=-\frac{1}{3}\frac{\Delta [O_{2}]}{\Delta t}=\frac{1}{2}\frac{\Delta [N_{2}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B4%7D%5Cfrac%7B%5CDelta%20%5BNH_%7B3%7D%5D%7D%7B%5CDelta%20t%7D%3D-%5Cfrac%7B1%7D%7B3%7D%5Cfrac%7B%5CDelta%20%5BO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B2%7D%5Cfrac%7B%5CDelta%20%5BN_%7B2%7D%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7B%5CDelta%20%5BH_%7B2%7DO%5D%7D%7B%5CDelta%20t%7D)
where ![-\frac{\Delta [NH_{3}]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BNH_%7B3%7D%5D%7D%7B%5CDelta%20t%7D)
represents rate of consumption of , ![-\frac{\Delta [O_{2}]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BO_%7B2%7D%5D%7D%7B%5CDelta%20t%7D)
represents rate of consumption of 
, ![\frac{\Delta [N_{2}]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BN_%7B2%7D%5D%7D%7B%5CDelta%20t%7D)
represents rate of formation of 
and ![\frac{\Delta [H_{2}O]}{\Delta t}](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BH_%7B2%7DO%5D%7D%7B%5CDelta%20t%7D)
represents rate of formation of 
.
Here rate of formation of is 3.0 mol/(L.s)
From the above equation we can write-
![-\frac{1}{4}\frac{\Delta [NH_{3}]}{\Delta t}=\frac{1}{6}\frac{\Delta [H_{2}O]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B4%7D%5Cfrac%7B%5CDelta%20%5BNH_%7B3%7D%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B1%7D%7B6%7D%5Cfrac%7B%5CDelta%20%5BH_%7B2%7DO%5D%7D%7B%5CDelta%20t%7D)
Here ![\frac{\Delta [H_{2}O]}{\Delta t}=3.0 mol/(L.s))](https://tex.z-dn.net/?f=%5Cfrac%7B%5CDelta%20%5BH_%7B2%7DO%5D%7D%7B%5CDelta%20t%7D%3D3.0%20mol%2F%28L.s%29%29)
So, ![-\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\frac{\Delta [H_{2}O]}{\Delta t}](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BNH_%7B3%7D%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B4%7D%7B6%7D%5Cfrac%7B%5CDelta%20%5BH_%7B2%7DO%5D%7D%7B%5CDelta%20t%7D)
Hence, ![-\frac{\Delta [NH_{3}]}{\Delta t}=\frac{4}{6}\times 3.0 mol/(L.s)=2.0 mol/(L.s)](https://tex.z-dn.net/?f=-%5Cfrac%7B%5CDelta%20%5BNH_%7B3%7D%5D%7D%7B%5CDelta%20t%7D%3D%5Cfrac%7B4%7D%7B6%7D%5Ctimes%203.0%20mol%2F%28L.s%29%3D2.0%20mol%2F%28L.s%29)
You have to balance out those forces and apply the same amount of equal and opposite force to it. I hope I helped ^^
