Answer: The rate constant at 296 K is
Explanation:
According to the Arrhenius equation,
or,
where,
= rate constant at 266 K =
= rate constant at 296 K = ?
= activation energy for the reaction = 103 kJ/mol = 103000 J/mol
R = gas constant = 8.314 J/mole.K
= initial temperature = 266 K
= final temperature = 296 K
Now put all the given values in this formula, we get
Thus the rate constant at 296 K is
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A first order reaction, with a half-life of 125 s, has 1/16 of the original amount left after 500 seconds.
<h3>What is a first order reaction?</h3>
It is a chemical reaction in which the rate of reaction is directly proportional to the concentration of the reacting substance.
First, we will calculate the rate constant using the following expression.
ln ([A]/[A]₀) = - k × t
ln (1/16[A]₀/[A]₀) = - k × 500 s
k = 5.55 × 10⁻³ s⁻¹
where,
- [A] is the final concentration of the reactant.
- [A]₀ is the initial concentration of the reactant.
- k is the rate constant.
- t is the elapsed time.
Next, we can calculate the half-life (th) using the following expression.
th = ln 2 / k = ln 2 / (5.55 × 10⁻³ s⁻¹) = 125 s
A first order reaction, with a half-life of 125 s, has 1/16 of the original amount left after 500 seconds.
Learn more about first order reactions here: brainly.com/question/518682
Answer:
Now u have 48 g of O2. There. Fore mole=weight/M. W. Of oxygen. Therefor 3mole.
After that if we to multiply the avogadro number with it. So 3 *NA
Now u want only atom calculation then we have 2 molecule of oxygen then multiply it with 2 too.
So final claculation is =3*2*NA.
Explanation:
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