The ratio of the areas of the signals in the h NMR spectrum of pentan-3-ol is 6: 4: 1: 1. The correct option is A.
<h3>What is a NMR spectrum?</h3>
Nuclear magnetic resonance spectroscopy is a spectroscopy that shows the detailed structure and chemical environment of a chemical element.
Pentan-3-ol contain 12 hydrogen atoms. In H-NMR spectra, hydrogen atoms have same environment gives a signal.
There are 4 different kinds of signals due of the 4 different environment experienced by these 12 hydrogens.
Thus, the ratio of the areas of the signals in the h NMR spectrum of pentan-3-ol is 6: 4: 1: 1. The correct option is A.
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Answer:
0.52 L.
Explanation:
Let P be the initial pressure.
From the question given above, the following data were obtained:
Initial pressure (P1) = P
Initial volume (V1) = 1.04 L
Final pressure (P2) = double the initial pressure = 2P
Final volume (V2) =?
The new volume (V2) of the gas can be obtained by using the the Boyle's law equation as shown below:
P1V1 = P2V2
P × 1.04 = 2P × V2
1.04P = 2P × V2
Divide both side by 2P
V2 = 1.04P /2P
V2 = 0.52 L
Thus, the new volume of the gas is 0.52 L.
Answer:
S = 21.92 %
F = 78.08 %
Explanation:
To find the percent composition of each element in SF6, we must find the molar mass of SF6 first.
Molar mass of SF6 = 32 + 19(6)
= 32 + 114
= 146g/mol
mass of Sulphur (S) in SF6 = 32g
mass of Fluorine (F) in SF6 = 114g
Percent composition = mass of element/molar mass of compound × 100
- % composition of S = 32/146 × 100 = 21.92%.
- % composition of F = 114/146 × 100 = 78.08%.
Answer:There is colder and drier air coming
Explanation:
