Limiting reactant : O₂
Mass of N₂O₄ produced = 95.83 g
<h3>Further explanation</h3>
Given
50g nitrous oxide
50g oxygen
Reaction
2N20 + 302 - 2N204
Required
Limiting reactant
mass of N204 produced
Solution
mol N₂O :

mol O₂ :

2N₂O+3O₂⇒ 2N₂O₄
ICE method
1.136 1.5625
1.0416 1.5625 1.0416
0.0944 0 1.0416
Limiting reactant : Oxygen-O₂
Mass N₂O₄(MW=92 g/mol) :

Explanation:
The given data is as follows.
= 0.483,
= 0.173 M,
= 0.433 M,
= 0.306 M,
= 9.0 atm
According to the ideal gas equation, PV = nRT
or, P =
Also, we know that
Density = 
So, P = MRT
and, M = 
= 
= 
= 0.368 mol/L
Now, we will calculate the cell potential as follows.
E = ![E^{o} - \frac{0.0591}{n} log \frac{[Co^{2+}]^{2}[Cl_{2}]}{[Co^{3+}][Cl^{-}]^{2}}](https://tex.z-dn.net/?f=E%5E%7Bo%7D%20-%20%5Cfrac%7B0.0591%7D%7Bn%7D%20log%20%5Cfrac%7B%5BCo%5E%7B2%2B%7D%5D%5E%7B2%7D%5BCl_%7B2%7D%5D%7D%7B%5BCo%5E%7B3%2B%7D%5D%5BCl%5E%7B-%7D%5D%5E%7B2%7D%7D)
= 
= 
= 
= 0.483 - 0.0185
= 0.4645 V
Thus, we can conclude that the cell potential of given cell at
is 0.4645 V.
Answer:
From the graph find the maximum velocity and half it i.e. Vmax/2. Draw a horizontal line from this point till you find the point on the graph that corresponds to it and read off the substrate concentration at that point. This will give the value of Km.
Answer:
kinetic energy
the is because all of them are moving at once
Answer:
14 mol O₂
Explanation:
The reaction between CO and O₂ is the following:
CO + O₂ → CO₂
We balance the equation with a coefficient 2 in CO and CO₂ to obtain the same number of O atoms:
2CO + O₂ → 2CO₂
As we can see from the balanced equation, 1 mol of O₂ is required to react with 2 moles of CO. Thus, the conversion factor is 1 mol of O₂/2 mol CO. We multiply the moles of CO by the conversion factor to calculate the moles of O₂ that are required:
28 mol CO x 1 mol of O₂/2 mol CO = 14 mol O₂