Answer:
.079 moles of Nirogen gas (N2)
Explanation:
You can see from the equaton that each ONE mole of N2 produces TWO moles of NH3.
Find the number of moles of NH3 produced.
Using Periodic Table : Mole wt of NH3 = 17 gm/mole
2.7 gm / 17 gm/mole = .1588 moles
One half as many moles of N2 are needed = .079 moles
Answer is: both reactions are exothermic.
<span>In exothermic reactions, heat is released and enthalpy of reaction is less than zero (as it show second chemical reaction).
According to Le Chatelier's principle when the reaction
is <span>exothermic heat is included as a
product (as it show first chemical reaction).</span></span>
Answer:
The equilibrium will be shifted to lift with the formation of a brown gelatinous precipitate of Fe(OH)₃.
Explanation:
- Le Chatelier's principle states that <em>"when any system at equilibrium for is subjected to change in concentration, temperature, volume, or pressure, then the system readjusts itself to counteract the effect of the applied change and a new equilibrium is established that is different from the old equilibrium"</em>.
- The addition of NaOH will result in the formation of Fe(OH)₃ precipitate which has a brown gelatinous precipitate.
- The formation of this precipitate cause removal and decrease of Fe³⁺ ions.
- According to Le Chatelier's principle, the system will be shifted to lift to increase Fe³⁺ concentration and reduce the stress of Fe³⁺ removal and readjust the equilibrium again. So, the [Fe(SCN)²⁺] decreases.
- Increasing [Fe³⁺] will produce a yellow color solution that contains a brown gelatinous precipitate of Fe(OH)₃.
Answer: There are 
atoms present in 0.500 mol of 
.
Explanation:
According to the mole concept, there are 
atoms present in 1 mole of a substance.
In a molecule of there is only one carbon atom present. Therefore, number of carbon atoms present in 0.500 mol of are as follows.
Thus, we can conclude that there are atoms present in 0.500 mol of .
Answer : The concentration of NOBr after 95 s is, 0.013 M
Explanation :
The integrated rate law equation for second order reaction follows:
![k=\frac{1}{t}\left (\frac{1}{[A]}-\frac{1}{[A]_o}\right)](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B1%7D%7Bt%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%5BA%5D_o%7D%5Cright%29)
where,
k = rate constant = 
t = time taken = 95 s
[A] = concentration of substance after time 't' = ?
![[A]_o](https://tex.z-dn.net/?f=%5BA%5D_o)
= Initial concentration = 0.86 M
Now put all the given values in above equation, we get:
![0.80=\frac{1}{95}\left (\frac{1}{[A]}-\frac{1}{(0.86)}\right)](https://tex.z-dn.net/?f=0.80%3D%5Cfrac%7B1%7D%7B95%7D%5Cleft%20%28%5Cfrac%7B1%7D%7B%5BA%5D%7D-%5Cfrac%7B1%7D%7B%280.86%29%7D%5Cright%29)
[A] = 0.013 M
Hence, the concentration of NOBr after 95 s is, 0.013 M
