Answer:
Explanation:
For a first order reaction the rate law is:
![v=\frac{-d[A]}{[A]}=k[A]](https://tex.z-dn.net/?f=v%3D%5Cfrac%7B-d%5BA%5D%7D%7B%5BA%5D%7D%3Dk%5BA%5D)
Integranting both sides of the equation we get:
![\int\limits^a_b {\frac{d[A]}{[A]}} \, dx =-k\int\limits^t_0 {} \, dt](https://tex.z-dn.net/?f=%5Cint%5Climits%5Ea_b%20%7B%5Cfrac%7Bd%5BA%5D%7D%7B%5BA%5D%7D%7D%20%5C%2C%20dx%20%3D-k%5Cint%5Climits%5Et_0%20%7B%7D%20%5C%2C%20dt)
where "a" stands for [A] (molar concentration of a given reagent) and "b" is {A]0 (initial molar concentration of a given reagent), "t" is the time in seconds.
From that integral we get the integrated rate law:
![ln\frac{[A]}{[A]_{0} } =-kt](https://tex.z-dn.net/?f=ln%5Cfrac%7B%5BA%5D%7D%7B%5BA%5D_%7B0%7D%20%7D%20%3D-kt)
![[A]=[A]_{0}e^{-kt}](https://tex.z-dn.net/?f=%5BA%5D%3D%5BA%5D_%7B0%7De%5E%7B-kt%7D)
![ln[A]=ln[A]_{0} -kt](https://tex.z-dn.net/?f=ln%5BA%5D%3Dln%5BA%5D_%7B0%7D%20-kt)
![k=\frac{ln[A]_{0}-ln[A]}{t}](https://tex.z-dn.net/?f=k%3D%5Cfrac%7Bln%5BA%5D_%7B0%7D-ln%5BA%5D%7D%7Bt%7D)
therefore k is
Answer:
moles HCl = 101.2 g /36.461 g/mol=2.78
[H+]= 2.78 mol/ 150 L =0.0185
pH = - log 0.0185=1.73
1.73 is the answer
Given equation:
P + O2 → P2O5
In order for the equation to be balanced, the stoichiometry of the atoms of one kind on the reactant side must be equal to that on the product
Reactants Products
P = 1 P = 2
O = 2 O = 5
The balanced equation would be:
4P + 5O2 → 2P2O5
Reactants Products
P = 4 P = 4
O = 10 O = 10
Ans: D)
Answer:
CH3OH + 02 ----> C02 + H20
balanced equation -
CH3OH + 3/202 ----> C02 + 2H20
Use exactly the same process as the one used on another question of yours I answered.
