The second-order reaction 2 Mn(CO)5 → Mn2(CO)10, has a rate constant equal to 3.0 × 109 M-1 s-1 at 25°C. If the initial concentration of Mn(CO)5 is 2.0 × 10-5 M, how long will it take for 90.% of the reactant to disappear?

t = 1.5 E-4 s

Explanation:

∴ a: Mn(CO)5

∴ K = 3.0 E9 M-1 s-1rate constant

∴ T = 25°C

∴ α = 2 second-order

∴ Cao = 2.0 E-5 Minitial concentration

If  Ca = Cao - (Cao×0.90) ⇒ t = ?

⇒ Ca = 2.0 E-5 - ((2.0 E-5)(0.90)) = 2.0 E-6 M

⇒ - δCa/δt = KCa²

⇒ - ∫δCa/Ca² = K∫δt

⇒ [(1/Ca) - (1/Cao)] = K*t

⇒ t = [(1/Ca) - (1/Cao)] / K

⇒ t = [ (1/2 E-6) - (1/2 E-5)] / (3.0 E9)

⇒ t = 1.5 E-4 s

b.false that your answer

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