The integrated rate law expression for a first order reaction is
![ln\frac{[A_{0}]}{[A_{t}]}=kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA_%7B0%7D%5D%7D%7B%5BA_%7Bt%7D%5D%7D%3Dkt)
where
[A0]=100
[At]=6.25
[6.25% of 100 = 6.25]
k = 9.60X10⁻³s⁻¹
Putting values
taking log of 100/6.25
100/6.25 = 16
ln(16) = 2.7726
Time = 2.7726 / 0.0096 = 288.81 seconds
Answer:
7.5 moles of O₂.
Explanation:
We'll begin by writing the balanced equation for the reaction. This is illustrated below:
2KClO₃ —> 2KCl + 3O₂
From the balanced equation above,
2 moles of KClO₃ decomposed to produce 3 moles of O₂.
Finally, we shall determine the number of mole of O₂ produced by the decomposition of 5 moles of KClO₃. This can be obtained as follow:
From the balanced equation above,
2 moles of KClO₃ decomposed to produce 3 moles of O₂.
Therefore, 5 moles of KClO₃ will decompose to produce = (5 × 3)/ 2 = 7.5 moles of O₂.
Thus, 7.5 moles of O₂ were obtained from the reaction.
False because if the circuit performance
