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Missing question: The first-order rate constant for the decomposition of N2O5, 2N2O5(g)→4NO2(g) + O2(g) at 70°C is 6.82×10−3 s−1. Suppose we start with 2.70×10−2 mol of N2O5(g) in a volume of 1.8 L .
c₀(N₂O₅) = 0,027 mol ÷ 1,8 L.
c₀(N₂O₅) = 0,015 mol/L.
c(N₂O₅) = 0,019 mol/ 1,8 L = 0,01055 mol/L.
k = 6,82·10⁻³ s⁻¹.
ln c(N₂O₅) = ln c₀(N₂O₅) - k·t.
t = (ln c₀(N₂O₅) - ln c(N₂O₅)) ÷ k.
t = 0,35 ÷ 0,00682 s⁻¹.
t = 51 s = 0,86 min.

batolisis

The number of minutes it will take is : 0.86 min

Determine the number of minutes for N₂O₅ to drop to 1.9*10⁻² mol

first step :

C₀(N₂O₅) = 0.027 mol / 1.8 L.

               = 0.015 mol/l

Also

C ( N₂O₅ ) = 0.019 / 1.8 l

                 = 0.01055 mol/l

Given that K = 6.82 * 10⁻³ s⁻¹

Final step : Calculate the time required

we will apply the formula below

In C₀(N₂O₅) - ln C(N₂O₅) = kt

therefore : t ( time ) = In C₀(N₂O₅) - ln C(N₂O₅) / k

                                = ( 0.015 - 0.01055 ) / 6.82 * 10⁻³

                                = 51 secs = 0.86 min

Hence we can conclude that The number of minutes it will take is : 0.86 min

Learn more about N₂O₅ : https://brainly.com/question/18672473

Attached below is the complete question

The first-order rate constant for the decomposition of N₂O₅.

2N₂O₅(g) → 4NO₂(g) + O₂(g) at 70°C is 6.82×10⁻³ s−1.

Suppose we start with 2.70×10−2 mol of N2O5(g) in a volume of 1.8 L.

How many minutes will it take for the quantity of n2o5 to drop to   1.9×10−2 mol

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