The percent of pyridine ( C 5 H 5 N ) that leads to the formation of pyridium ion ( C 5 H 5 NH + ) in a 0.10 M aqueous solution of pyridine is to be calculated. Concept introduction: The pH of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution. The pH of a solution is calculated by the formula, pH = − log [ H + ] At equilibrium, the equilibrium constant expression is expressed by the formula, K b = Concentration of products Concentration of reactants The percent ionization of an acid is calculated by the formula, Percent ionization = Equilibrium concentration of [ OH − ] Initial concentration of the base × 100
The percent of pyridine ( C 5 H 5 N ) that leads to the formation of pyridium ion ( C 5 H 5 NH + ) in a 0.10 M aqueous solution of pyridine is to be calculated. Concept introduction: The pH of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution. The pH of a solution is calculated by the formula, pH = − log [ H + ] At equilibrium, the equilibrium constant expression is expressed by the formula, K b = Concentration of products Concentration of reactants The percent ionization of an acid is calculated by the formula, Percent ionization = Equilibrium concentration of [ OH − ] Initial concentration of the base × 100
Interpretation: The percent of pyridine
(C5H5N) that leads to the formation of pyridium ion
(C5H5NH+) in a
0.10M aqueous solution of pyridine is to be calculated.
Concept introduction: The
pH of a solution is defined as a figure that expresses the acidity of the alkalinity of a given solution.
The
pH of a solution is calculated by the formula,
pH=−log[H+]
At equilibrium, the equilibrium constant expression is expressed by the formula,
Predict the organic products that form in the reaction below:
OH
H+
H+
+
☑
Y
Note: You may assume you have an excess of either reactant if the reaction requires more than one of those molecules to form the products.
In the drawing area below, draw the skeletal ("line") structures of the missing organic products X and Y. You may draw the structures in any arrangement that
you like, so long as they aren't touching.
Click and drag to start drawing a
structure.
✓
m
Determine the structures of the missing organic molecules in the following reaction:
+ H₂O
+H
H+
Y
Z
☑
☑
Note: Molecules that share the same letter have the exact same structure.
In the drawing area below, draw the skeletal ("line") structures of the missing organic molecules X, Y, and Z. You may draw the structures in any arrangement
that you like, so long as they aren't touching. Molecule X shows up in multiple steps, but you only have to draw its structure once.
Click and drag to start drawing a structure.
AP
+
Please help, this is all the calculations
i got!!! I will rate!!!Approx mass of
KMnO in vial: 3.464
4
Moss of beaker 3×~0. z Nax200:
= 29.9219
Massof weacerv after remosimgain
N2C2O4. Need to fill in all the
missing blanks.
ง
ง
Approx mass of KMnO4 in vials 3.464
Mass of beaker + 3x ~0-304: 29.9219
2~0.20
Miss of beaker + 2x-
29.7239
Mass of beaker + 1x~0.2g Naz (204
29-5249
Mass of beaver after removing as
qa Na₂ C₂O
T1
T2
T3
Final Buiet reading
Initial butet reading (int))
Hass of NaOr used for Titration
-reading (mL)
calculation Results:
8.5ml
17mL
27.4mL
Oml
Om
Oml
T1
T2
T3
Moles of No CO
Moles of KMO used
LOF KM. O used
Molenty of KMNO
Averagem Of KMOWL
Chapter 13 Solutions
Bundle: Chemistry: An Atoms First Approach, Loose-leaf Version, 2nd + OWLv2 with Student Solutions Manual, 4 terms (24 months) Printed Access Card
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