Respuesta :
Answer : The rate of consumption of [tex]S_2O_2^{-8}[/tex] is, [tex]7.0\times 10^{-6}M/s[/tex]
Explanation : Given,
Moles of [tex]S_2O_2^{-8}[/tex] = [tex]2.0\times 10^{-4}mol[/tex]
Volume of solution = 170 mL = 0.170 L (1 L = 1000 mL)
Time = 170 s
First we have to calculate the concentration.
[tex]\text{Concentration}=\frac{\text{Moles}}{\text{Volume of solution}}[/tex]
[tex]\text{Concentration}=\frac{2.0\times 10^{-4}mol}{0.170L}[/tex]
[tex]\text{Concentration}=1.2\times 10^{-3}M[/tex]
Now we have to calculate the rate of consumption.
[tex]\text{Rate of consumption}=\frac{\text{Concentration}}{\text{Time}}[/tex]
[tex]\text{Rate of consumption}=\frac{1.2\times 10^{-3}M}{170s}[/tex]
[tex]\text{Rate of consumption}=7.0\times 10^{-6}M/s[/tex]
Thus, the rate of consumption of [tex]S_2O_2^{-8}[/tex] is, [tex]7.0\times 10^{-6}M/s[/tex]
The rate of consumption of S2O2−8 is 7.0*10^-6 m/s.
Calculation of the rate of consumption:
Since
Moles 2.0×10−4
Volume of solution = 170 mL = 0.170 L (1 L = 1000 mL)
Time = 170 s
Now first determine the concentration
So here concentration = moles / volume of solution
= 2.0×10−4 / 0.170L
= 1.2*10^-3
Now the rate of consumption is
= Concentration /time
= 1.2*10^-3 / 170 s
= 7.0*10^-6 m/s.
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