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The rate for the reaction 2N2O(5(g)) to ...

The rate for the reaction `2N_2O_(5(g)) to 4NO_(2(g)) + O_(2(g))` is `2.4 xx 10^(-5) mol "lit"^(-1) "sec"^(-1)`. If the rate is `3.0 xx 10^(-5) "sec"^(-1)` then the concentration of `N_2O_5` in `mol "lit"^(-1)` is:
(Fill your anwer by multiply with 10)

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Rate = `k [N2_O_5] `
Hence `2.4 xx 10^(-5) = 3.0 xx 10^(-5) IN_2 O_5]`
or `[N_2O_5] = 0.8 molL^(-1)`
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