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)
0.25 moles of CO2 is present in 11 grams of CO2.
Explanation:
A mole represents the number of chemical entities in an element or molecule.
Number of moles of an element or molecule is determined by the formula:
The Number of moles (n) = weight of the atom given ÷ atomic or molecular weight of the one mole of the element or molecule.
Themolar mass of one mole of carbon dioxide is:
12+ ( 16×2)
= 44 gram/mole
The given weight is 44 grams of carbon dioxide.
Putting the values in the equation,
n= 11 gms÷44 gms/ mole
n = 0.25 mole
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Equate the gravitational force to the electrostatic force:
<span>KC²/D² = Gm²/D² → C = m√[G/K] = 7.6√[6.67E-11/9E9] = 6.54E-10 coulombs </span>
<span>Number of electrons N = 6.24E18*C = 4.083E9 electrons</span>
<h2>
Answer:</h2>
The magnesium ribbon, <u>D. It forms a material to cast the tool mark</u>.
<h2>
Explanation:</h2>
When a magnesium ribbon is burnt in the presence of oxygen it gives out strong light and heat is produced. Apart from it, it leads to the production of substance called as magnesium oxide which is formed as the product due to the reaction of magnesium with the oxygen present in the air.
Tool marks are the mark which is created by tools while using them. In order to identify or locate them castes made up of magnesium oxide is utilized. When this is pasted on the suspected area, the tool mark of the suspected tool gets pasted on it.
