If a reaction rate decreases by a factor of one-ninth when a reactant concentration is decreased by one-third, what is the order of the reaction with respect to that reactant?

The reaction is second-order in [A].  

For the reaction A ⟶ products

The rate law is r = k[A]^m

Consider two Experiments 1 and 2.

r₂/r₁ = {k[A]₂^m}/{k[A]₁^m }  

r₂/r₁ = {[A]₂/[A]₁}^m

r₂/r₁ = {(⅓[A])/[A]}^m = (⅑r₁)/r₁

(⅓)^m= ⅑

3^m = 9

m = 2

The reaction is second-order in [A].

If dividing the concentration by 3 divides the rate by 9, the reaction is second-order.

number of molecules / avogadro's number * molecule molecular = mass

explanation:

we define a gram-molecular weight as the mass of one mole of any element or compound. a mole contains avogadro's number of units (molecules) –  

6.022

x

10

23

. so, dividing the number of molecules by avogadro's number gives us the number of moles. multiplying that by the molecular weight gives us the mass. from the periodic table we find the gram-molecular weight of u to be238 and of f to be19. so,  

u

f

6

will have a gram-molecular weight of 352g.