A) \[0.64\times {{10}^{3}}\]
B) \[0.80\times {{10}^{3}}\]
C) \[1.28\times {{10}^{-1}}\]
D) \[1.60\times {{10}^{-3}}\]
Correct Answer: D
Solution :
[d] \[=-\frac{d[S{{O}_{2}}]}{dt}=1.28\times {{10}^{-3}}g{{L}^{-1}}{{s}^{-1}}\] \[=\frac{1.28\times {{10}^{-3}}}{64}=M{{s}^{-1}}\] \[-\frac{d[S{{O}_{2}}]}{dt}=\frac{d[S{{O}_{3}}]}{dt}\] \[=\frac{1.28\times {{10}^{-3}}}{64}M{{L}^{-1}}{{s}^{-1}}\] \[=\frac{1.28\times {{10}^{-3}}}{64}\times 80g{{L}^{-1}}{{s}^{-1}}\] \[=1.6\times {{10}^{-3}}g\,{{L}^{-1}}{{s}^{-1}}\]You need to login to perform this action.
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