Hello!
To determine [H₃O⁺], we need to apply the Henderson-Hasselback equation, since this is a case of an acid and its conjugate base:
![pH=pKa+log( \frac{[A^{-}] }{[HA]} )](https://tex.z-dn.net/?f=pH%3DpKa%2Blog%28%20%5Cfrac%7B%5BA%5E%7B-%7D%5D%20%7D%7B%5BHA%5D%7D%20%29)
Now, we use the definition of pH and clear [H₃O⁺] from there:
![pH=-log[H_3O^{+}]](https://tex.z-dn.net/?f=pH%3D-log%5BH_3O%5E%7B%2B%7D%5D%20)
![[H_3O^{+}] = 10^{-pH} =10^{-3,84}=0,00014 M](https://tex.z-dn.net/?f=%20%5BH_3O%5E%7B%2B%7D%5D%20%3D%2010%5E%7B-pH%7D%20%3D10%5E%7B-3%2C84%7D%3D0%2C00014%20M)
So, the [H₃O⁺] concentration is
0,00014 M
Have a nice day!
<span>Answer:
W must be 5-chloro-2-methylpentane. It can give only 4-methy-1-pentene (Y) upon dehydrohalogenation:
X must be 4-chloro-2-methylpentane. Dehydrohalogenation yields both Y and 4-methyl-2-pentene. (Z)</span>
Answer
is: <span>the
percent ionizationof formic acid is 1,82%.
Chemical reaction: HCOOH(aq) </span>⇄ H⁺(aq)
+ HCOO⁻(aq).<span>
pKa(</span>HCOOH) = 3,77.
Ka(HCOOH) = 1,7·10⁻⁴.
c(HCOOH) = 0,5 M.
<span>
[H</span>⁺]
= [HCOO⁻] = x; equilibrium
concentration.<span>
[HA] = 0,1 M - x.
Ka = [H</span>⁺] · [HCOO⁻] / [HCOOH].<span>
0,00017 = x² / 0,5 M - x.
Solve quadratic equation: x = 0,0091 M.
α = 0,0091 M ÷ 0,5 M · 100% = 1,82%.</span>
Answer is "B - 700,000".<span>
<span>Kinetic energy of a single particle (atom or molecule)<span> is directly proportional to its
temperature according to the following equation.</span></span>
KE = (3kT)/2
<span>Where </span>KE<span> is the
kinetic energy of a single atom/molecule (</span>J<span>), </span>k<span> is the Boltzmann
constant (</span>1.381 × 10</span>⁻²³ J/K<span>) and </span>T<span> is the temperature (</span>K<span>) </span><span>
When temperature increases, then the kinetic
energy increases.
<span>If kinetic
energy of atoms increases, then there will be more motions which create many
collisions.</span></span>
Answer:
Have the same number of electrons in their outer energy levels
Explanation:
Elements in the same group have similar chemical properties because they have the same number of valence electron(s) in their outermost shell.
Chlorine and Iodine have similar chemical properties because they have the same number of valence electron in their outermost shell. This can be seen from their electronic configuration as shown below:
Cl (17) => 1s² 2s²2p⁶ 3s²3p⁵
I (53) => [Kr] 4d¹⁰ 5s²5p⁵
From the above illustration:
Outer shell of Cl (3s²3p⁵) = 2 + 5 = 7 electrons
Outer shell of I (5s²5p⁵) = 2 + 5 = 7 electrons
Since they have the same number of valence electrons, therefore, they will have similar chemical properties.
