Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
8th Edition
ISBN: 9780134421377
Author: Charles H Corwin
Publisher: PEARSON
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Question
Chapter 10, Problem 19ST
Interpretation Introduction
Interpretation:
The reason as to why a half-full plastic water bottle in a refrigerator is crushed after an hour is to be stated.
Concept introduction:
The temperature at which the atmospheric pressure becomes equal to the vapor pressure is known as the boiling point. Colligative properties are those properties of a material that depends on the number of a particle of the substance present in the material.
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יווי
10
20
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70
3.5
3
2.5
2
1.5
1
[ppm]
3.5
3
2.5
2
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6 [ppm]
1
1.5
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3.5
2H2S(g)+3O2(g)→2SO2(g)+2H2O(g)
A 1.2mol sample of H2S(g) is combined with excess O2(g), and the reaction goes to completion.
Question
Which of the following predicts the theoretical yield of SO2(g) from the reaction?
Responses
1.2 g
Answer A: 1.2 grams
A
41 g
Answer B: 41 grams
B
77 g
Answer C: 77 grams
C
154 g
Answer D: 154 grams
D
Part VII. Below are the 'HNMR, 13 C-NMR, COSY 2D- NMR, and HSQC 2D-NMR (similar with HETCOR but axes are reversed) spectra of an
organic compound with molecular formula C6H1003 - Assign chemical shift values to the H and c atoms of the
compound. Find the structure. Show complete solutions.
Predicted 1H NMR Spectrum
4.7 4.6 4.5 4.4 4.3 4.2 4.1 4.0 3.9 3.8 3.7 3.6 3.5 3.4 3.3 3.2 3.1 3.0 2.9 2.8 2.7 2.6 2.5 2.4 2.3 2.2 2.1 2.0 1.9 1.8 1.7 1.6 1.5 1.4 1.3 1.2 1.1
f1 (ppm)
Predicted 13C NMR Spectrum
100
f1 (ppm)
30
220 210 200 190 180
170
160 150 140 130 120
110
90
80
70
-26
60
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46
30
20
115
10
1.0 0.9 0.8
0
-10
Chapter 10 Solutions
Introductory Chemistry: Concepts and Critical Thinking (8th Edition)
Ch. 10 - Prob. 1CECh. 10 - Prob. 2CECh. 10 - Prob. 3CECh. 10 - Prob. 4CECh. 10 - Prob. 5CECh. 10 - Prob. 6CECh. 10 - Prob. 7CECh. 10 - Prob. 8CECh. 10 - Prob. 9CECh. 10 - Prob. 10CE
Ch. 10 - Prob. 1KTCh. 10 - Prob. 2KTCh. 10 - Prob. 3KTCh. 10 - Prob. 4KTCh. 10 - Prob. 5KTCh. 10 - Prob. 6KTCh. 10 - Prob. 7KTCh. 10 - Prob. 8KTCh. 10 - Prob. 9KTCh. 10 - Prob. 10KTCh. 10 - Prob. 11KTCh. 10 - Prob. 12KTCh. 10 - Prob. 13KTCh. 10 - Prob. 14KTCh. 10 - Prob. 15KTCh. 10 - Prob. 16KTCh. 10 - Prob. 17KTCh. 10 - Prob. 18KTCh. 10 - Prob. 19KTCh. 10 - Prob. 20KTCh. 10 - Prob. 21KTCh. 10 - Prob. 22KTCh. 10 - Prob. 23KTCh. 10 - Prob. 24KTCh. 10 - Prob. 1ECh. 10 - Prob. 2ECh. 10 - Prob. 3ECh. 10 - Prob. 4ECh. 10 - Prob. 5ECh. 10 - Prob. 6ECh. 10 - Prob. 7ECh. 10 - Prob. 8ECh. 10 - Prob. 9ECh. 10 - Prob. 10ECh. 10 - Prob. 11ECh. 10 - Prob. 12ECh. 10 - Prob. 13ECh. 10 - Prob. 14ECh. 10 - Prob. 15ECh. 10 - Prob. 16ECh. 10 - Prob. 17ECh. 10 - Prob. 18ECh. 10 - Prob. 19ECh. 10 - Prob. 20ECh. 10 - Prob. 21ECh. 10 - Prob. 22ECh. 10 - Prob. 23ECh. 10 - Prob. 24ECh. 10 - Prob. 25ECh. 10 - Prob. 26ECh. 10 - Prob. 27ECh. 10 - Prob. 28ECh. 10 - Prob. 29ECh. 10 - Prob. 30ECh. 10 - Prob. 31ECh. 10 - Prob. 32ECh. 10 - Prob. 33ECh. 10 - Prob. 34ECh. 10 - Prob. 35ECh. 10 - Prob. 36ECh. 10 - Prob. 37ECh. 10 - Prob. 38ECh. 10 - Prob. 39ECh. 10 - Prob. 40ECh. 10 - Prob. 41ECh. 10 - Prob. 42ECh. 10 - Prob. 43ECh. 10 - Prob. 44ECh. 10 - Prob. 45ECh. 10 - Prob. 46ECh. 10 - Prob. 47ECh. 10 - Prob. 48ECh. 10 - Prob. 49ECh. 10 - Prob. 50ECh. 10 - Prob. 51ECh. 10 - Prob. 52ECh. 10 - Prob. 53ECh. 10 - Prob. 54ECh. 10 - Prob. 55ECh. 10 - Prob. 56ECh. 10 - Prob. 57ECh. 10 - Prob. 58ECh. 10 - Prob. 59ECh. 10 - Prob. 60ECh. 10 - Prob. 61ECh. 10 - Prob. 62ECh. 10 - Prob. 63ECh. 10 - Prob. 64ECh. 10 - Prob. 65ECh. 10 - Prob. 66ECh. 10 - Prob. 67ECh. 10 - Prob. 68ECh. 10 - Prob. 69ECh. 10 - Prob. 70ECh. 10 - Prob. 71ECh. 10 - Prob. 72ECh. 10 - Prob. 73ECh. 10 - Prob. 74ECh. 10 - Prob. 75ECh. 10 - Prob. 76ECh. 10 - Prob. 77ECh. 10 - Prob. 78ECh. 10 - Prob. 79ECh. 10 - Prob. 80ECh. 10 - Prob. 81ECh. 10 - Prob. 82ECh. 10 - Prob. 83ECh. 10 - Prob. 84ECh. 10 - Prob. 1STCh. 10 - Prob. 2STCh. 10 - Prob. 3STCh. 10 - Prob. 4STCh. 10 - Prob. 5STCh. 10 - Prob. 6STCh. 10 - Prob. 7STCh. 10 - Prob. 8STCh. 10 - Prob. 9STCh. 10 - Prob. 10STCh. 10 - Prob. 11STCh. 10 - Prob. 12STCh. 10 - Prob. 13STCh. 10 - Prob. 14STCh. 10 - Prob. 15STCh. 10 - Prob. 16STCh. 10 - Prob. 17STCh. 10 - Prob. 18STCh. 10 - Prob. 19ST
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- Nonearrow_forward4. Draw and label all possible isomers for [M(py)3(DMSO)2(CI)] (py = pyridine, DMSO dimethylsulfoxide).arrow_forwardThe emission data in cps displayed in Table 1 is reported to two decimal places by the chemist. However, the instrument output is shown in Table 2. Table 2. Iron emission from ICP-AES Sample Blank Standard Emission, cps 579.503252562 9308340.13122 Unknown Sample 343.232365741 Did the chemist make the correct choice in how they choose to display the data up in Table 1? Choose the best explanation from the choices below. No. Since the instrument calculates 12 digits for all values, they should all be kept and not truncated. Doing so would eliminate significant information. No. Since the instrument calculates 5 decimal places for the standard, all of the values should be limited to the same number. The other decimal places are not significant for the blank and unknown sample. Yes. The way Saman made the standards was limited by the 250-mL volumetric flask. This glassware can report values to 2 decimal places, and this establishes our number of significant figures. Yes. Instrumental data…arrow_forward
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