A) 490 atm
B) 250 atm
C) 480 atm
D) 420 atm
Correct Answer: A
Solution :
\[{{p}_{0}}=500\,atm\] \[K=\frac{2.303}{t}{{\log }_{10}}\frac{{{p}_{0}}}{{{p}_{t}}}\] \[3.38\times {{10}^{-5}}=\frac{2.303}{10\times 60}\log \frac{500}{{{p}_{t}}}\] or \[0.00880=\log \frac{500}{{{p}_{t}}}\Rightarrow \frac{500}{1.02}=490\,atm\]You need to login to perform this action.
You will be redirected in
3 sec