For physics the answer is 47500
atoms consist of protons and neutrons in the nucleus , surrounded by electrons that reside in orbitals . Orbitals are classified according to the four quantum number s that represent anyone particular orbitals energy , shape , orientation and the spin of the occupying electron.
hope this helps xx
It really depends on your options.
Since NH3 solution is a weak base, the option that neutralizes a base is an acid. So find an acid among your options, and here are some examples: HNO3, HCl, etc.
I believe it's D! Hope you can help with my question!
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)
