According to this formula:
㏑(K2/K1) = Ea/R(1/T1 - 1/T2)
when K is the rate constant
Ea is the activation energy
R is the universal gas constant
and T is the temperature K
when K is doubled so K2: K1 = 2:1 & R = 8.314 J.K^-1.mol^-1
and T1 = 10 +273 = 283 k & T2 = 21 + 273 = 294 k
So by substitution:
㏑2 =( Ea / 8.314) (1/283 - 1/294 )
∴ Ea = 43588.9 J/mol = 43.6 KJ/mol
Answer:
https://www.clutchprep.com/chemistry/practice-problems/70217/hi-aq-h2o-l-h3o-aq-i-aq-identify-each-as-either-a-bronsted-lowry-acid-bronsted-l
Explanation:
https://www.clutchprep.com/chemistry/practice-problems/70217/hi-aq-h2o-l-h3o-aq-i-aq-identify-each-as-either-a-bronsted-lowry-acid-bronsted-l
Answer:
a list of characteristics that can be used to identify a substance include color, odor, taste, density, melting point, boiling point, conductivity, and hardness.
Let's start with the amount given in percent. Let our basis be 100 grams of compound. So, that means that in this amount, 57.1 g is oxygen and 100-57.1=42.9 g is carbon. Since there is 1:1 atom ratio, it also means that moles oxygen = moles carbon.
Moles = Mass/Relative Mass
Let x be the relative mass of oxygen
57.1/16 = 42.9/x
Solving for x,
<em>x = 12.02 amu</em>