Hey there!:
Molar mass MgCl2 = 95.2110 g/mol
So:
1 mole MgCl2 -------------- 95.2110 g
moles MgCl2 ---------------- 319 g
moles MgCl2 = 319 * 1 / 95.2110
moles MgCl2 = 319 / 95.2110
=> 3.350 moles of MgCl2
Hope that helps!
<span><span>N2</span><span>O3</span><span>(g)</span>→NO<span>(g)</span>+<span>NO2</span><span>(g)</span></span>
<span><span>[<span>N2</span><span>O3</span>]</span> Initial Rate</span>
<span>0.1 M r<span>(t)</span>=0.66</span> M/s
<span>0.2 M r<span>(t)</span>=1.32</span> M/s
<span>0.3 M r<span>(t)</span>=1.98</span> M/s
We can have the relationship:
<span>(<span><span>[<span>N2</span><span>O3</span>]/</span><span><span>[<span>N2</span><span>O3</span>]</span>0</span></span>)^m</span>=<span><span>r<span>(t)/</span></span><span><span>r0</span><span>(t)
However,
</span></span></span>([N2O3]/[N2O3]0) = 2
Also, we assume m=1 which is the order of the reaction.
Thus, the relationship is simplified to,
r(t)/r0(t) = 2
r<span>(t)</span>=k<span>[<span>N2</span><span>O3</span>]</span>
0.66 <span>M/s=k×0.1 M</span>
<span>k=6.6</span> <span>s<span>−<span>1</span></span></span>
Answer:
a. 113 min
Explanation:
Considering the equilibrium:-
2N₂O₅ ⇔ 4NO₂ + O₂
At t = 0 125 kPa
At t = teq 125 - 2x 4x x
Thus, total pressure = 125 - 2x + 4x + x = 125 - 3x
125 - 3x = 176 kPa
x = 17 kPa
Remaining pressure of N₂O₅ = 125 - 2*17 kPa = 91 kPa
Using integrated rate law for first order kinetics as:
![[A_t]=[A_0]e^{-kt}](https://tex.z-dn.net/?f=%5BA_t%5D%3D%5BA_0%5De%5E%7B-kt%7D)
Where,
![[A_t]](https://tex.z-dn.net/?f=%5BA_t%5D)
is the concentration at time t
![[A_0]](https://tex.z-dn.net/?f=%5BA_0%5D)
is the initial concentration
Given that:
The rate constant, k = 
min⁻¹
Initial concentration = 125 kPa
Final concentration = 91 kPa
Time = ?
Applying in the above equation, we get that:-
