The reaction of nitrogen dioxide with carbon monoxide

NO2 + CO NO + CO2

is second order in NO2 and zero order in CO.

1.Complete the rate law for this reaction in the box below.
Use the form k[A]m[B]n... , where '1' is understood for m, n ...(don't enter 1) and concentrations taken to the zero power do not appear.

Rate =

2.In an experiment to determine the rate law, the rate constant was determined to be 0.790 M-1s-1. Using this value for the rate constant, the rate of the reaction when [NO2] = 1.23 M and [CO] = 0.221 M would be Ms-1

(1) Rate = k[NO2]^2[CO]

(2) The rate of the reaction is 1.195 Ms^-1

Explanation:

(1) Equation of reaction

NO2 + CO = NO + CO2

The reaction is second-order in NO2 and zero-order in CO

Therefore, rate = k[NO2]^2[CO]^0

(2) Rate = k[NO2]^2[CO]^0

k = 0.790 M^-1s^-1

[NO2] = 1.23 M

[CO] = 0.221 M

Rate = 0.79×1.23^2×0.221^0 = 1.195 Ms^-1

hi my dude or dudette

a. the law of proportion states the combination of 2 atoms in a specific ratio to produce multiple compounds such as the given statement of question ‘1st compound contains 88.8 % cu ’ indicates 88.8 g of cu is present along with 11.2 g oxygen in 100 g of compound whereas the statement ‘2st compound contains 79.9 % cu ’ indicates 79.9 g of cu is present along with 20.1 g oxygen in 100 g of compound

1st compound contains 88.8 g cu

for cu

m = 63 g /mol

atomic % = 88.8 /63 = 1.42

for o

m = 16 g /mol

atomic % = 11.2 /16 = 0.7

b. the emperical formula of 1st compound is

cu1.4o0.7

cu2o