Answer:
56.9 mmoles of acetate are required in this buffer
Explanation:
To solve this, we can think in the Henderson Hasselbach equation:
pH = pKa + log ([CH₃COO⁻] / [CH₃COOH])
To make the buffer we know:
CH₃COOH + H₂O ⇄ CH₃COO⁻ + H₃O⁺ Ka
We know that Ka from acetic acid is: 1.8×10⁻⁵
pKa = - log Ka
pKa = 4.74
We replace data:
5.5 = 4.74 + log ([acetate] / 10 mmol)
5.5 - 4.74 = log ([acetate] / 10 mmol)
0.755 = log ([acetate] / 10 mmol)
10⁰'⁷⁵⁵ = ([acetate] / 10 mmol)
5.69 = ([acetate] / 10 mmol)
5.69 . 10 = [acetate] → 56.9 mmoles
Answer:
Explanation:
48.16%
Well, both abundances have to total 100% so is Ag-107 is 51.84%, then Ag-109 must be 100 – 51.84 = 48.16%.
Hope This Helps :)
Burrows, tracks, coprolites, nests and footprints are examples of traces that can be found in a fossil.
Answer:
Azide synthesis is the first method on the table of synthesis of primary amines. The Lewis structure of the azide ion, N3−, is as shown below.
an azide ion
An “imide” is a compound in which an N−−H group is attached to two carbonyl groups; that is,
imide linkage
You should note the commonly used trivial names of the following compounds.
phthalic acid, phthalic anhydride, and phthalimide
The phthalimide alkylation mentioned in the reading is also known as the Gabriel synthesis.
If necessary, review the reduction of nitriles (Section 20.7) and the reduction of amides (Section 21.7).
Before you read the section on reductive amination you may wish to remind yourself of the structure of an imine (see Section 19.8).
The Hofmann rearrangement is usually called the Hofmann degradation. In a true rearrangement reaction, no atoms are lost or gained; however, in this particular reaction one atom of carbon and one atom of oxygen are lost from the amide starting material, thus the term “rearrangement” is not really appropriate. There is a rearrangement step in the overall degradation process, however: this is the step in which the alkyl group of the acyl nitrene migrates from carbon to nitrogen to produce an isocyanate.
Explanation:
