Answer:
In aqueous solution the pH scale varies from 0 to 14, which indicates this concentration of hydrogen. Solutions with pH less than 7 are acidic (the value of the exponent of the concentration is higher, because there are more ions in the solution) and alkaline (basic) those with a pH higher than 7. If the solvent is pure water, the pH = 7 indicates neutrality of the solution
Explanation:
PH is a measure of how acidic or basic a liquid is. Specifically, from a dissolution. The acidity of a solution is essentially due to the concentration of hydrogen ions dissolved in it. In reality, the ions are not found alone, but are in the form of hydronium ions consisting of one oxygen molecule and three positively charged hydrogen. PH precisely measures this concentration. And to do it, we can use simple and very visual methods.
Answer:
Electrons
Explanation:
In an atom there would be three subatomic particles: Neutrons, electrons, protons. The smallest and lightest in terms of mass is electrons. This is because the nucleus is comprised of the protons and the neutrons, these have a greater mass than electrons as electrons has very little mass that can considered to be 0.
Answer: B,C,D
Explanation: Who ever reads this have a great day and stay safe :}
Answer:

Explanation:
Here, we want to convert J to cal
Mathematically:

Thus, to get our answer in cal, we divide the value given by 4.186
We have that as:
The answer is Ka = 1.00x10^-10.
Solution:
When given the pH value of the solution equal to 11, we can compute for pOH as
pOH = 14 - pH = 14 - 11.00 = 3.00
We solve for the concentration of OH- using the equation
[OH-] = 10^-pOH = 10^-3 = x
Considering the sodium salt NaA in water, we have the equation
NaA → Na+ + A-
hence, [A-] = 0.0100 M
Since HA is a weak acid, then A- must be the conjugate base and we can set up an ICE table for the reaction
A- + H2O ⇌ HA + OH-
Initial 0.0100 0 0
Change -x +x +x
Equilibrium 0.0100-x x x
We can now calculate the Kb for A-:
Kb = [HA][OH-] / [A-]
= x<span>²</span> / 0.0100-x
Approximating that x is negligible compared to 0.0100 simplifies the equation to
Kb = (10^-3)² / 0.0100 = 0.000100 = 1.00x10^-4
We can finally calculate the Ka for HA from the Kb, since we know that Kw = Ka*Kb = 1.0 x 10^-14:
Ka = Kw / Kb
= 1.00x10^-14 / 1.00x10^-4
= 1.00x10^-10