The reaction 2N2O5 (g) → 2N2O4 (g) + O2 (g) has a reaction rate that is dependent only on the concentration of N2O5 and at a certain temperature has a rate constant k of 0.0168 s-1. If 2.50 moles of N2O5 were placed in a 5.00 liter container at that temperature, how many moles of N2O5 would remain after 1.00 min?

0.910 mol

Explanation:

Let's consider the following reaction.

2 N₂O₅(g) → 2 N₂O₄(g) + O₂(g)

This reaction is of first order with respect to N₂O₅, with a rate constant k = 0.0168 s⁻¹.

We can calculate the concentration at a certain time using the following expression.

where,

The initial concentration of N₂O₅ is:

2.50 mol / 5.00 L = 0.500 M

The concentration of N₂O₅ after 1.00 min (60 s) is:

The moles of N₂O₅ after 1.00 min are:

a: 45.9 meters

b: 4.29 seconds

step-by-step explanation:

a: h(t)=1/2(-9.8)(3)^2+0(3)+90

h(t)=-4.9(3)^2+90

h(t)=-44.1+90

h(t)=45.9 meters

b: 0=1/2(-9.8)(t^2)+0t+90

0=-4.9t^2+90

-90=-4.9t^2

18.37=t^2

t=4.29 seconds

in the second question, the numbers aren't exact, but they're a close approximation.

it's filtration, but this is an assumption so my apologies if i got this