Answer : The time taken for the reaction is, 28 s.
Explanation :
Expression for rate law for first order kinetics is given by :
![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 = 0.0632
t = time taken for the process = ?
![[A_o]](https://tex.z-dn.net/?f=%5BA_o%5D)
= initial amount or concentration of the reactant = 1.28 M
![[A]](https://tex.z-dn.net/?f=%5BA%5D)
= amount or concentration left time 't' = 
Now put all the given values in above equation, we get:
Therefore, the time taken for the reaction is, 28 s.
Answer:
The pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
Explanation:
Given data:
[OH⁻] = 9.50 ×10⁻⁹M
pOH = ?
Solution:
pOH = -log[OH⁻]
Now we will put the value of OH⁻ concentration.
pOH = -log[9.50 ×10⁻⁹M]
pOH = 8
Thus the pOH of HNO₃ solution that ha OH⁻ concentration 9.50 ×10⁻⁹M is 8.
It’s soft which makes It low energy
D) All three bulbs would go out because the circuit would no longer have a ground thereby deeming it "open".
