(a) Interpretation: The molarity of a solution of HClO 4 of amount 25.00mL is to be determined by the student. The student is also using 0.731M KOH. He forgets to add the indicator in solution after the addition of KOH of volume 42.35 mL into the solution. At this condition, he takes the pH of the solution which is 12.39. It has to be determined if the student go beyond equivalence point or not. Concept introduction: The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration. It can also be known as volumetric titration.
(a) Interpretation: The molarity of a solution of HClO 4 of amount 25.00mL is to be determined by the student. The student is also using 0.731M KOH. He forgets to add the indicator in solution after the addition of KOH of volume 42.35 mL into the solution. At this condition, he takes the pH of the solution which is 12.39. It has to be determined if the student go beyond equivalence point or not. Concept introduction: The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration. It can also be known as volumetric titration.
Solution Summary: The author explains how the molarity of HClO 4 is to be determined by the student, who forgets to add the indicator in the solution.
The molarity of a solution of HClO4 of amount 25.00mL is to be determined by the student. The student is also using 0.731M KOH. He forgets to add the indicator in solution after the addition of KOH of volume 42.35 mL into the solution. At this condition, he takes the pH of the solution which is 12.39. It has to be determined if the student go beyond equivalence point or not.
Concept introduction:
The titration is described as the process by which the concentration of the dissolved substance is to be determined in terms of the smallest amount of reagent of a given concentration. It can also be known as volumetric titration.
Interpretation Introduction
(b)
Interpretation:
The molarity of strong acid is to be determined.
Concept introduction:
The molarity can be calculated as follows:
molarity=number of moles of solute volume of solution(L)
The relation between pH and pOH is as follows:
pH+pOH=14
The pOH can be calculated as follows:
pOH=−log10[OH−]
Interpretation Introduction
(c)
Interpretation:
The volume of KOH is to be determined which is added in excess if he added. Or the volume of KOH is to be determined which is required to add if he didn’t add.
Rank the labeled protons (Ha-Hd) in order of increasing acidity, starting with the least acidic.
НОН НЬ
OHd
Онс
Can the target compound at right be efficiently synthesized in good yield from the unsubstituted benzene at left?
?
starting
material
target
If so, draw a synthesis below. If no synthesis using reagents ALEKS recognizes is possible, check the box under the drawing area.
Be sure you follow the standard ALEKS rules for submitting syntheses.
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Note for advanced students: you may assume that you are using a large excess of benzene as your starting material.
C
:0
T
Add/Remove step
G
The following equations represent the formation of compound MX. What is the AH for the
electron affinity of X (g)?
X₂ (g) → 2X (g)
M (s) → M (g)
M (g)
M (g) + e-
AH = 60 kJ/mol
AH = 22 kJ/mol
X (g) + e-X (g)
M* (g) +X (g) → MX (s)
AH = 118 kJ/mol
AH = ?
AH = -190 kJ/mol
AH = -100 kJ/mol
a)
-80 kJ
b)
-30 kJ
c)
-20 kJ
d)
20 kJ
e)
156 kJ
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