Answer:
pH = 4.27. Porcentaje de disociación: 0.03%
Explanation:
El pH de un ácido débil, HX, se obtiene haciendo uso de su equilibrio:
HX(aq) ⇄ H⁺(aq) + X⁻(aq)
Donde la constante de equilibrio, Ka, es
Ka = 1.65x10⁻⁸ = [H⁺] [X⁻] / [HX]
Como los iones H⁺ y X⁻ vienen del mismo equilibrio podemos decir:
[H⁺] = [X⁻]
[HX] es:
20g * (1mol/55g) = 0.3636moles / 2.100L = 0.1732M
Reemplazando es Ka:
1.65x10⁻⁸ = [H⁺] [H⁺] / [0.1732M]
2.858x10⁻⁹ = [H⁺]²
5.35x10⁻⁵M = [H⁺]
pH = -log[H⁺]
<h3>pH = 4.27</h3>
El porcentaje de disociacion es [X⁻] / [HX] inicial * 100
Reemplazando
5.35x10⁻⁵M / 0.1732M * 100
<h3>0.03%</h3>
Answer:
d.have a mass of 1 amu..
Explanation:
Does not match a as the answer for a is electrons B.electron circle around the nucleus because neutrons are inside the nucleus.C.proton and neutrons make the nucleus so the only answer left is .d .which is correct
Answer:
2,4,4-trimethyl-2-pentene yields mixture of
and ![(CH_{3})_{3}CHO](https://tex.z-dn.net/?f=%28CH_%7B3%7D%29_%7B3%7DCHO)
Explanation:
In ozonolysis (hydrolysis step involve a reducing agent such as Zn,
etc.), a pi bond is broken to form ketone/aldehyde.
Ketone is formed from di-substituted side of double bond and aldehyde is formed from mono-substituted side of double bond.
Ozoznolysis involves two consecutive steps : (1) formation of ozonide, (2) hydrolysis of ozonide.
Hydrolysis can be done with/without using reducing agent. Carboxylic acid/carbon dioxide/ketone is produced when hydrolysis is done without using reducing agent.
Here, 2,4,4-trimethyl-2-pentene yields mixture of
and ![(CH_{3})_{3}CHO](https://tex.z-dn.net/?f=%28CH_%7B3%7D%29_%7B3%7DCHO)
Reaction steps are shown below.
Answer:
I think its 1.2 cause I divided 15.5 with 12 and got 1.2 as an answer
Answer:
<em>no</em><em> </em>
Explanation:
our heart muscles never get tired, because it has to pump blood in our body 72 times a minute, it is made of special cardiac muscles which helps it to perform it's function without getting tired ....
<em>i</em><em> </em><em>hope</em><em> </em><em>it</em><em> </em><em>helped</em><em>.</em><em>.</em><em>.</em><em>.</em><em>.</em>