I believe it is A)fuel is more readily available:)
<u>Answer:</u> The amount remained after 151 seconds are 0.041 moles
<u>Explanation:</u>
All the radioactive reactions follows first order kinetics.
Rate law expression for first order kinetics is given by the equation:
![k=\frac{2.303}{t}\log\frac{[A_o]}{[A]}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7B2.303%7D%7Bt%7D%5Clog%5Cfrac%7B%5BA_o%5D%7D%7B%5BA%5D%7D)
where,
k = rate constant = 
t = time taken for decay process = 151 sec
= initial amount of the reactant = 0.085 moles
[A] = amount left after decay process = ?
Putting values in above equation, we get:
![4.82\times 10^{-3}=\frac{2.303}{151}\log\frac{0.085}{[A]}](https://tex.z-dn.net/?f=4.82%5Ctimes%2010%5E%7B-3%7D%3D%5Cfrac%7B2.303%7D%7B151%7D%5Clog%5Cfrac%7B0.085%7D%7B%5BA%5D%7D)
![[A]=0.041moles](https://tex.z-dn.net/?f=%5BA%5D%3D0.041moles)
Hence, the amount remained after 151 seconds are 0.041 moles
Answer:
148.04 kJ/mol
Explanation:
Let's consider the following thermochemical equation.
NO(g) + 1/2 O₂(g) → NO₂(g) ΔH°rxn = -114.14 kJ/mol
We can find the standard enthalpy of formation (ΔH°f) of NO(g) using the following expression.
ΔH°rxn = 1 mol × ΔH°f(NO₂(g)) - 1 mol × ΔH°f(NO(g)) - 1/2 mol × ΔH°f(O₂(g))
ΔH°f(NO(g)) = 1 mol × ΔH°f(NO₂(g)) - ΔH°rxn - 1/2 mol × ΔH°f(O₂(g)) / 1 mol
ΔH°f(NO(g)) = 1 mol × 33.90 kJ/mol - (-114.14 kJ) - 1/2 mol × 0 kJ/mol / 1 mol
ΔH°f(NO(g)) = 148.04 kJ/mol