The first-order rate constant for the decomposition of N2O5, given below, at 70°C is 6.82 10-3 s-1. Suppose we start with 0.0200 mol of N2O5(g) in a volume of 2.5 L. 2 N2O5(g) → 4 NO2(g) + O2(g) (a) How many moles of N2O5 will remain after 2.5 min? mol (b) How many minutes will it take for the quantity of N2O5 to drop to 0.005 mol? min (c) What is the half-life of N2O5 at 70°C?

Explanation:

For first order reaction the equation is

k is rate constant , t is time , a is initial concentration and a_t is concentration after time t .

Putting the values

= 1.023

.02 / a_t = 2.78

a_t = 7.2 x 10⁻³ moles

b ) Let after t time the mole becomes .005

again putting the values in the equation

409.2 t x 10⁻³ = = 3.689

t = 9.01 min

c ) for half life a_t = a / 2

putting the values in the equation

6.82 x 10⁻³ x 60 x t = ln 2

t = 1.7 min.

in 1783 cavendish was working with his hydrogen gas and found that when he burned it and condensed the vapor it was water. this was very significant since it disproved the greek theory of elements completely since water was not a simple substance but the product of two gases. it was lavoisier who hearing of the experiment gave cavendish's gas the name hydrogen ("water-producer") and added that hydrogen burned by combining with oxygen and that water was a hydrogen-oxygen combination.

explanation:

8 m/s because for him to get the same amount of energy as ray is putting out, bobby, who weighs half as much as ray, needs to run twice as fast.