It's an allele not a chromosome
Answer:
0.01836 M
Explanation:
Again the reaction equation is;
Fe(s) + Mn2+(aq) → Fe2+(aq) + Mn(s)
E°cell= 0.77 V
Ecell= 0.78 V
[Mn2+] = 0.040 M
[Fe2+] = the unknown
n=2
From Nernst's equation;
Ecell= E°cell- 0.0592/n log Q
0.78= 0.77 - 0.0592/2 log [Fe2+] /[0.040]
0.78-0.77= - 0.0592/2 log [Fe2+] /[0.040]
0.01/ -0.0296= log [Fe2+] /[0.040]
-0.3378= log [Fe2+] /[0.040]
Antilog(-0.3378) = [Fe2+] /[0.040]
0.459= [Fe2+] /[0.040]
[Fe2+] = 0.459 × 0.040
[Fe2+] = 0.01836 M
Answer:
K = Ka/Kb
Explanation:
P(s) + (3/2) Cl₂(g) <-------> PCl₃(g) K = ?
P(s) + (5/2) Cl₂(g) <--------> PCl₅(g) Ka
PCl₃(g) + Cl₂(g) <---------> PCl₅(g) Kb
K = [PCl₃]/ ([P] [Cl₂]⁽³'²⁾)
Ka = [PCl₅]/ ([P] [Cl₂]⁽⁵'²⁾)
Kb = [PCl₅]/ ([PCl₃] [Cl₂])
Since [PCl₅] = [PCl₅]
From the Ka equation,
[PCl₅] = Ka ([P] [Cl₂]⁽⁵'²⁾)
From the Kb equation
[PCl₅] = Kb ([PCl₃] [Cl₂])
Equating them
Ka ([P] [Cl₂]⁽⁵'²⁾) = Kb ([PCl₃] [Cl₂])
(Ka/Kb) = ([PCl₃] [Cl₂]) / ([P] [Cl₂]⁽⁵'²⁾)
(Ka/Kb) = [PCl₃] / ([P] [Cl₂]⁽³'²⁾)
Comparing this with the equation for the overall equilibrium constant
K = Ka/Kb
The pH of a solution at 25. 0 °C that contains 2. 95 × 10^-12 m hydronium ions is 13.5.
<h3>What is pH? </h3>
pH is defined as the concentration of the hydrogen bond which is released or gained by the species in the solution which depicts the acidity and basicity of the solution.
<h3>What is pOH? </h3>
pOH is defined as the concentration of the hydronium ion present in solution.
pOH value is inversely proportional to the value of pH.
pH value increases, pOH value decreases and vice versa.
Given,
Total H+ ions = 2.95 ×10^(-12)M
<h3>Calculation of pH</h3>
pH = -log[H+]
By substituting the value of H+ ion in given equation
= log(2.95× 10^(-12) )
= 13.5
Thus we find that the pH of a solution at 25. 0 °C that contains 2. 95 × 10^-12 m hydronium ions is 13.5.
learn more about pH:
brainly.com/question/12942138
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