EBK FOUNDATIONS OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781119227946
Author: Willard
Publisher: VST
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 13, Problem 10RQ
Interpretation Introduction
Interpretation:
Reason for the increase of vapor pressure with increase in temperature on basis of kinetic theory has to be explained.
Concept Introduction:
Kinetic theory explains the straight and steady motion of the gaseous particles and demonstrates the elastic collisions between the particles. It means that on collision particles do not gain or lose energy. According to this theory number of gaseous particles is directly related to temperature. It also determines the existence of three phases of matter that is solid, liquid and gas.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Please help answer number 2. Thanks in advance.
How do I explain this? Thank you!
When an unknown amine reacts with an unknown acid chloride, an amide with a molecular mass of 163 g/mol (M* = 163 m/z)
is formed. In the infrared spectrum, important absorptions appear at 1661, 750 and 690 cm. The 13C NMR and DEPT spectra
are provided. Draw the structure of the product as the resonance contributor lacking any formal charges.
13C NMR
DEPT 90
200
160
120
80
40
0
200
160
120
80
40
0
DEPT 135
T
200
160
120
80
40
0
Draw the unknown amide.
Select
Dow
Templates
More
Frage
Chapter 13 Solutions
EBK FOUNDATIONS OF COLLEGE CHEMISTRY
Ch. 13.2 - Prob. 13.1PCh. 13.2 - Prob. 13.2PCh. 13.3 - Prob. 13.3PCh. 13.3 - Prob. 13.4PCh. 13.4 - Prob. 13.5PCh. 13.5 - Prob. 13.6PCh. 13.5 - Prob. 13.7PCh. 13.5 - Prob. 13.8PCh. 13.6 - Prob. 13.9PCh. 13.6 - Prob. 13.10P
Ch. 13 - Prob. 1RQCh. 13 - Prob. 2RQCh. 13 - Prob. 3RQCh. 13 - Prob. 4RQCh. 13 - Prob. 5RQCh. 13 - Prob. 6RQCh. 13 - Prob. 7RQCh. 13 - Prob. 8RQCh. 13 - Prob. 9RQCh. 13 - Prob. 10RQCh. 13 - Prob. 11RQCh. 13 - Prob. 12RQCh. 13 - Prob. 13RQCh. 13 - Prob. 14RQCh. 13 - Prob. 15RQCh. 13 - Prob. 16RQCh. 13 - Prob. 17RQCh. 13 - Prob. 19RQCh. 13 - Prob. 20RQCh. 13 - Prob. 21RQCh. 13 - Prob. 22RQCh. 13 - Prob. 23RQCh. 13 - Prob. 24RQCh. 13 - Prob. 25RQCh. 13 - Prob. 26RQCh. 13 - Prob. 27RQCh. 13 - Prob. 28RQCh. 13 - Prob. 29RQCh. 13 - Prob. 30RQCh. 13 - Prob. 31RQCh. 13 - Prob. 32RQCh. 13 - Prob. 33RQCh. 13 - Prob. 34RQCh. 13 - Prob. 35RQCh. 13 - Prob. 36RQCh. 13 - Prob. 37RQCh. 13 - Prob. 38RQCh. 13 - Prob. 39RQCh. 13 - Prob. 40RQCh. 13 - Prob. 41RQCh. 13 - Prob. 42RQCh. 13 - Prob. 43RQCh. 13 - Prob. 1PECh. 13 - Prob. 2PECh. 13 - Prob. 3PECh. 13 - Prob. 4PECh. 13 - Prob. 5PECh. 13 - Prob. 6PECh. 13 - Prob. 7PECh. 13 - Prob. 8PECh. 13 - Prob. 9PECh. 13 - Prob. 10PECh. 13 - Prob. 11PECh. 13 - Prob. 12PECh. 13 - Prob. 13PECh. 13 - Prob. 14PECh. 13 - Prob. 15PECh. 13 - Prob. 16PECh. 13 - Prob. 17PECh. 13 - Prob. 18PECh. 13 - Prob. 19PECh. 13 - Prob. 20PECh. 13 - Prob. 21PECh. 13 - Prob. 22PECh. 13 - Prob. 23PECh. 13 - Prob. 24PECh. 13 - Prob. 25PECh. 13 - Prob. 26PECh. 13 - Prob. 27PECh. 13 - Prob. 28PECh. 13 - Prob. 29PECh. 13 - Prob. 30PECh. 13 - Prob. 31PECh. 13 - Prob. 32PECh. 13 - Prob. 33AECh. 13 - Prob. 34AECh. 13 - Prob. 35AECh. 13 - Prob. 36AECh. 13 - Prob. 38AECh. 13 - Prob. 39AECh. 13 - Prob. 40AECh. 13 - Prob. 41AECh. 13 - Prob. 42AECh. 13 - Prob. 43AECh. 13 - Prob. 44AECh. 13 - Prob. 45AECh. 13 - Prob. 46AECh. 13 - Prob. 47AECh. 13 - Prob. 48AECh. 13 - Prob. 49AECh. 13 - Prob. 50AECh. 13 - Prob. 51AECh. 13 - Prob. 52AECh. 13 - Prob. 53AECh. 13 - Prob. 54AECh. 13 - Prob. 55AECh. 13 - Prob. 56AECh. 13 - Prob. 57AECh. 13 - Prob. 58AECh. 13 - Prob. 59AECh. 13 - Prob. 60AECh. 13 - Prob. 61AECh. 13 - Prob. 62AECh. 13 - Prob. 63AECh. 13 - Prob. 64AECh. 13 - Prob. 65AECh. 13 - Prob. 66AECh. 13 - Prob. 67AECh. 13 - Prob. 69CECh. 13 - Prob. 70CECh. 13 - Prob. 71CECh. 13 - Prob. 72CE
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.Similar questions
- Identify the unknown compound from its IR and proton NMR spectra. C4H6O: 'H NMR: 82.43 (1H, t, J = 2 Hz); 8 3.41 (3H, s); 8 4.10 (2H, d, J = 2 Hz) IR: 2125, 3300 cm¹ The C4H6O compound liberates a gas when treated with C2H5 MgBr. Draw the unknown compound. Select Draw с H Templates Morearrow_forwardPlease help with number 6 I got a negative number could that be right?arrow_forward1,4-Dimethyl-1,3-cyclohexadiene can undergo 1,2- or 1,4-addition with hydrogen halides. (a) 1,2-Addition i. Draw the carbocation intermediate(s) formed during the 1,2-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,2-addition product formed during the reaction in (i)? (b) 1,4-Addition i. Draw the carbocation intermediate(s) formed during the 1,4-addition of hydrobromic acid to 1,4-dimethyl-1,3-cyclohexadiene. ii. What is the major 1,4-addition product formed from the reaction in (i)? (c) What is the kinetic product from the reaction of one mole of hydrobromic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (d) What is the thermodynamic product from the reaction of one mole of hydrobro-mic acid with 1,4-dimethyl-1,3-cyclohexadiene? Explain your reasoning. (e) What major product will result when 1,4-dimethyl-1,3-cyclohexadiene is treated with one mole of hydrobromic acid at - 78 deg * C ? Explain your reasoning.arrow_forward
- Give the product of the bimolecular elimination from each of the isomeric halogenated compounds. Reaction A Reaction B. КОВ CH₂ HotBu +B+ ко HOIBU +Br+ Templates More QQQ Select Cv Templates More Cras QQQ One of these compounds undergoes elimination 50x faster than the other. Which one and why? Reaction A because the conformation needed for elimination places the phenyl groups and to each other Reaction A because the conformation needed for elimination places the phenyl groups gauche to each other. ◇ Reaction B because the conformation needed for elimination places the phenyl groups gach to each other. Reaction B because the conformation needed for elimination places the phenyl groups anti to each other.arrow_forwardFive isomeric alkenes. A through each undergo catalytic hydrogenation to give 2-methylpentane The IR spectra of these five alkenes have the key absorptions (in cm Compound Compound A –912. (§), 994 (5), 1643 (%), 3077 (1) Compound B 833 (3), 1667 (W), 3050 (weak shoulder on C-Habsorption) Compound C Compound D) –714 (5), 1665 (w), 3010 (m) 885 (3), 1650 (m), 3086 (m) 967 (5), no aharption 1600 to 1700, 3040 (m) Compound K Match each compound to the data presented. Compound A Compound B Compound C Compound D Compoundarrow_forward7. The three sets of replicate results below were accumulated for the analysis of the same sample. Pool these data to obtain the most efficient estimate of the mean analyte content and the standard deviation. Lead content/ppm: Set 1 Set 2 Set 3 1. 9.76 9.87 9.85 2. 9.42 9.64 9.91 3. 9.53 9.71 9.42 9.81 9.49arrow_forward
- Draw the Zaitsev product famed when 2,3-dimethylpentan-3-of undergoes an El dehydration. CH₂ E1 OH H₁PO₁ Select Draw Templates More QQQ +H₂Oarrow_forwardComplete the clean-pushing mechanism for the given ether synthesia from propanol in concentrated sulfurica140°C by adding any mining aloms, bands, charges, nonbonding electron pairs, and curved arrows. Draw hydrogen bonded to cayan, when applicable. ore 11,0 HPC Step 1: Draw curved arrows Step 2: Complete the intend carved Q2Q 56 QQQ Step 3: Complete the intermediate and add curved Step 4: Modify the structures to draw the QQQ QQQarrow_forward6. In an experiment the following replicate set of volume measurements (cm3) was recorded: (25.35, 25.80, 25.28, 25.50, 25.45, 25.43) A. Calculate the mean of the raw data. B. Using the rejection quotient (Q-test) reject any questionable results. C. Recalculate the mean and compare it with the value obtained in 2(a).arrow_forward
- A student proposes the transformation below in one step of an organic synthesis. There may be one or more reactants missing from the left-hand side, but there are no products missing from the right-hand side. There may also be catalysts, small inorganic reagents, and other important reaction conditions missing from the arrow. • Is the student's transformation possible? If not, check the box under the drawing area. • If the student's transformation is possible, then complete the reaction by adding any missing reactants to the left-hand side, and adding required catalysts, inorganic reagents, or other important reaction conditions above and below the arrow. • You do not need to balance the reaction, but be sure every important organic reactant or product is shown. + T G OH де OH This transformation can't be done in one step.arrow_forwardMacmillan Leaming Draw the major organic product of the reaction. 1. CH3CH2MgBr 2. H+ - G Select Draw Templates More H о QQarrow_forwardDraw the condensed structure of 3-hydroxy-2-butanone. Click anywhere to draw the first atom of your structure.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Chemistry: Matter and ChangeChemistryISBN:9780078746376Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl WistromPublisher:Glencoe/McGraw-Hill School Pub Co
World of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage Learning
General, Organic, and Biological ChemistryChemistryISBN:9781285853918Author:H. Stephen StokerPublisher:Cengage Learning
Principles of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage Learning
Introductory Chemistry: An Active Learning Approa...ChemistryISBN:9781305079250Author:Mark S. Cracolice, Ed PetersPublisher:Cengage Learning
Chemistry: Matter and Change
Chemistry
ISBN:9780078746376
Author:Dinah Zike, Laurel Dingrando, Nicholas Hainen, Cheryl Wistrom
Publisher:Glencoe/McGraw-Hill School Pub Co
World of Chemistry, 3rd edition
Chemistry
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Brooks / Cole / Cengage Learning
General, Organic, and Biological Chemistry
Chemistry
ISBN:9781285853918
Author:H. Stephen Stoker
Publisher:Cengage Learning
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
Introductory Chemistry: An Active Learning Approa...
Chemistry
ISBN:9781305079250
Author:Mark S. Cracolice, Ed Peters
Publisher:Cengage Learning
Types of Matter: Elements, Compounds and Mixtures; Author: Professor Dave Explains;https://www.youtube.com/watch?v=dggHWvFJ8Xs;License: Standard YouTube License, CC-BY