Answer:
e) pH is independent of concentration.
Explanation:
a) It is a mixture of a weak acid and its conjugate base. <em>TRUE. </em>A buffer is defined as a mixture of a weak acid and its conjugate base or a weak base and its conjugate acid.
b) Resists pH changes because it reacts with added acid or base. <em>TRUE. </em>Thermodynamically, the reaction of added acid or base is faster with the buffer mixture than with H⁺ or OH⁻ ions of the solutions.
c) The maximum buffer capacity is at pH = pKa. <em>TRUE. </em>The buffer capacity is pka±1. For this, buffer capacity is maximum in pka.
d) pH is dependent on the solution ionic strength and temperature. <em>TRUE.</em> Ionic strength and temperature are factors that influence concentrations of ions in solutions as the H⁺ ion that is the responsible
e) pH is independent of concentration. <em>FALSE. </em>pH in a buffer depends completely of concentrations of the acid and its conjugate base or vice versa.
I hope it helps!
Answer:
Contains DNA, Contains Ribosomes, Lacks a nucleus
Explanation:
Answer:
Oxygen
Explanation:
First, list out all the ions in the aqueous solution:
Ag+, NO3-
H+, OH-
In the anode, the substance lose electrons to undergo oxidation.
From the 4 ions, only OH- can lose electrons to form water and oxygen,
4OH- ---> O2 + 2H2O + 4e-
While others tend to gain electrons to form new substances instead (they undergo reduction).
Oxygen is the gas produced.
I believe the answer would be a.
Hello!
The half-life is the time of half-disintegration, it is the time in which half of the atoms of an isotope disintegrate.
We have the following data:
mo (initial mass) = 43 g
m (final mass after time T) = ? (in g)
x (number of periods elapsed) = ?
P (Half-life) = 20 minutes
T (Elapsed time for sample reduction) = 80 minutes
Let's find the number of periods elapsed (x), let us see:






Now, let's find the final mass (m) of this isotope after the elapsed time, let's see:




I Hope this helps, greetings ... DexteR! =)