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EBK CHEMISTRY FOR CHANGING TIMES
14th Edition
ISBN: 8220100663482
Author: MCCREARY
Publisher: PEARSON
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Question
Chapter 1, Problem 3CGP
Interpretation Introduction
Interpretation:
Compare the word in the last column with that in the first column. Discuss any differences in the two definitions. If the word in the last column differs from that in the first column, determine what went wrong in the process.
- Hypothesis
- Theory
- Mixture
- Substance.
Text Entry | Student 1 | Student 2 | Student 3 | Student 4 |
Word | Definition | Word | Definition | Word |
| | | |
Concept Introduction:
A tentative explanation of observed data is known as a hypothesis.
The theory is a set of principles that predict and explain phenomena.
A material system made up of two or more different substances which are mixed but are not combined chemically is said to be a mixture.
Substances are those that can't be broken down into simpler substances.
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Students have asked these similar questions
You have now performed a liquid-liquid extraction protocol in Experiment 4. In doing so, you
manipulated and exploited the acid-base chemistry of one or more of the compounds in your
mixture to facilitate their separation into different phases. The key to understanding how liquid-
liquid extractions work is by knowing which layer a compound is in, and in what protonation state.
The following liquid-liquid extraction is different from the one you performed in Experiment
4, but it uses the same type of logic. Your task is to show how to separate apart Compound
A and Compound B.
. Complete the following flowchart of a liquid-liquid extraction. Handwritten work is
encouraged.
•
Draw by hand (neatly) only the appropriate organic compound(s) in the boxes.
.
Specify the reagent(s)/chemicals (name is fine) and concentration as required in Boxes 4
and 5.
•
Box 7a requires the solvent (name is fine).
•
Box 7b requires one inorganic compound.
• You can neatly complete this assignment by hand and…
b) Elucidate compound D w) mt at 170 nd shows c-1 stretch at 550cm;'
The compound has the ff electronic transitions: 0%o* and no a*
1H NMR Spectrum
(CDCl3, 400 MHz)
3.5
3.0
2.5
2.0
1.5
1.0
0.5 ppm
13C{H} NMR Spectrum
(CDCl3, 100 MHz)
Solvent
80
70
60
50
40
30
20
10
0 ppm
ppm
¹H-13C me-HSQC Spectrum
ppm
(CDCl3, 400 MHz)
5
¹H-¹H COSY Spectrum
(CDCl3, 400 MHz)
0.5
10
3.5
3.0
2.5
2.0
1.5 1.0
10
15
20
20
25
30
30
-35
-1.0
1.5
-2.0
-2.5
3.0
-3.5
0.5
ppm
3.5
3.0
2.5
2.0
1.5
1.0
0.5
ppm
Show work with explanation. don't give Ai generated solution
Chapter 1 Solutions
EBK CHEMISTRY FOR CHANGING TIMES
Ch. 1 - Prob. 1RQCh. 1 - Why do experiments have to be done to support a...Ch. 1 - Why can't scientific methods always be used to...Ch. 1 - How does technology differ from science?Ch. 1 - Prob. 5RQCh. 1 - Prob. 6RQCh. 1 - What is a DQ? What does a large DQ mean? Why is it...Ch. 1 - What derived units of (a) mass and (b) length are...Ch. 1 - What is the Sl-derived unit for volume? What...Ch. 1 - Prefix Symbol Definition tera- T 1012 - M - centi-...
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Similar questions
- Redraw the flowchartarrow_forwardredraw the flowchart with boxes and molecules written in themarrow_forwardPart I. a) Elucidate the structure of compound A using the following information. • mass spectrum: m+ = 102, m/2=57 312=29 • IR spectrum: 1002.5 % TRANSMITTANCE Ngg 50 40 30 20 90 80 70 60 MICRONS 5 8 9 10 12 13 14 15 16 19 1740 cm M 10 0 4000 3600 3200 2800 2400 2000 1800 1600 13 • CNMR 'H -NMR Peak 8 ppm (H) Integration multiplicity a 1.5 (3H) triplet b 1.3 1.5 (3H) triplet C 2.3 1 (2H) quartet d 4.1 1 (2H) quartet & ppm (c) 10 15 28 60 177 (C=0) b) Elucidate the structure of compound B using the following information 13C/DEPT NMR 150.9 MHz IIL 1400 WAVENUMBERS (CM-1) DEPT-90 DEPT-135 85 80 75 70 65 60 55 50 45 40 35 30 25 20 ppm 1200 1000 800 600 400arrow_forward
- • Part II. a) Elucidate The structure of compound c w/ molecular formula C10 11202 and the following data below: • IR spectra % TRANSMITTANCE 1002.5 90 80 70 60 50 40 30 20 10 0 4000 3600 3200 2800 2400 2000 1800 1600 • Information from 'HAMR MICRONS 8 9 10 11 14 15 16 19 25 1400 WAVENUMBERS (CM-1) 1200 1000 800 600 400 peak 8 ppm Integration multiplicity a 2.1 1.5 (3H) Singlet b 3.6 1 (2H) singlet с 3.8 1.5 (3H) Singlet d 6.8 1(2H) doublet 7.1 1(2H) doublet Information from 13C-nmR Normal carbon 29ppm Dept 135 Dept -90 + NO peak NO peak 50 ppm 55 ppm + NO peak 114 ppm t 126 ppm No peak NO peak 130 ppm t + 159 ppm No peak NO peak 207 ppm по реак NO peakarrow_forwardCould you redraw these and also explain how to solve them for me pleasarrow_forwardNonearrow_forward
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