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vaieri [72.5K]
3 years ago
9

If 2.0×10−4 moles of S2O2−8 in 170 mL of solution is consumed in 170 seconds , what is the rate of consumption of S2O2−8?

Chemistry
1 answer:
Ostrovityanka [42]3 years ago
8 0

Answer : The rate of consumption of S_2O_2^{-8} is, 7.0\times 10^{-6}M/s

Explanation : Given,

Moles of S_2O_2^{-8} = 2.0\times 10^{-4}mol

Volume of solution = 170 mL = 0.170 L     (1 L = 1000 mL)

Time = 170 s

First we have to calculate the concentration.

\text{Concentration}=\frac{\text{Moles}}{\text{Volume of solution}}

\text{Concentration}=\frac{2.0\times 10^{-4}mol}{0.170L}

\text{Concentration}=1.2\times 10^{-3}M

Now we have to calculate the rate of consumption.

\text{Rate of consumption}=\frac{\text{Concentration}}{\text{Time}}

\text{Rate of consumption}=\frac{1.2\times 10^{-3}M}{170s}

\text{Rate of consumption}=7.0\times 10^{-6}M/s

Thus, the rate of consumption of S_2O_2^{-8} is, 7.0\times 10^{-6}M/s

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