The reaction is second-order in [A].
For the reaction A ⟶ products
The rate law is <em>r</em> = <em>k</em>[A]^<em>m</em>
Consider two Experiments 1 and 2.
<em>r</em>₂/<em>r</em>₁ = {<em>k</em>[A]₂^<em>m</em>}/{k[A]₁^<em>m</em> }
<em>r</em>₂/<em>r</em>₁ = {[A]₂/[A]₁}^<em>m</em>
______________
<em>r</em>₂/<em>r₁</em> = {(⅓[A])/[A]}^m = (⅑<em>r</em>₁)/<em>r</em>₁
(⅓)^<em>m</em>= ⅑
3^<em>m</em> = 9
<em>m </em>= 2
The reaction is second-order in [A].
If <em>dividing the concentration by 3 divides the rate by </em>9, the reaction is <em>second-order</em>.
There is two different types but i’ll just do both meanings just incase.
Graham's Law of Diffusion: the rate of diffusion of one gas through another is inversely proportional to the square root of the density of the gas.
Graham's Law of Effusion: the rate of effusion of a gas is inversely proportional to the square root of the density of the gas.
hopes this helps..!
Answer:
547.7 g of C₆H₁₂O₆
Solution:
The balance chemical equation is as follow,
C₆H₁₂O₆ + 6 O₂ → 6 CO₂ + 6 H₂O
According to equation,
6 moles of O₂ burns = 180.56 g of C₆H₁₂O₆
So,
18.2 moles of O₂ will burn = X g of C₆H₁₂O₆
Solving for X,
X = (18.2 mol × 180.56 g) ÷ 6 mol
X = 547.7 g of C₆H₁₂O₆
The periodic table is a representative of all elements we know today.