INTRODUCTORY CHEMISTRY
8th Edition
ISBN: 2818000163285
Author: CORWIN
Publisher: PEARSON CO
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 10, Problem 14E
Interpretation Introduction
Interpretation:
An explanation regarding the increase in temperature of a gas with increase in its pressure is to be stated.
Concept introduction:
The pressure of the gas is associated with the frequency and energy of collisions of gas molecules.
The three factors that affect the pressure of the gas include:
• An increasing or decreasing volume of the container.
• An increasing or decreasing temperature of the gas.
• An increasing or decreasing number of molecules present in the container.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
For the reaction 2 N2O5(g) → 4 NO2(g) + O2(g), the following mechanism has been proposed:
N2O5 →> NO₂+ NO3_(K1)
NO2 + NO3 → N2O5 (k-1)
NO2 + NO3 →
→
NO2 + O2 + NO (K2)
NO + N2O5-
NO2 + NO2 + NO2 (K3)
d[N₂O5] __2k‚k₂[N2O5]
Indicate whether the following rate expression is acceptable:
dt
k₁₁+ k₂
Given the reaction R + Q → P, indicate the rate law with
respect to R, with respect to P and with respect to P.
Steps and explanations. Also provide, if possible, ways to adress this kind of problems in general.
Chapter 10 Solutions
INTRODUCTORY CHEMISTRY
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
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
- k₁ Given the reaction A B, indicate k-1 d[A] (A). the rate law with respect to A: (B). the rate law with respect to B: d[B] dt dtarrow_forwardk₁ Given the reaction R₂ R + R, indicate k-1 (A). the rate law with respect to R2: (B). the rate law with respect to R: d[R₂] dt d[R] dtarrow_forwardGiven the reaction R+ Q → P, indicate (A). the rate law with respect to P: (B). the rate law with respect to R: (C). the rate law with respect to Q: d[P] dt d[R] dt d[Q] dtarrow_forward
- The reaction for obtaining NO2 from NO and O2 has the rate equation: v = k[NO]2[O2]. Indicate which of the following options is correct.(A). This rate equation is inconsistent with the reaction consisting of a single trimolecular step.(B). Since the overall order is 3, the reaction must necessarily have some trimolecular step in its mechanism.(C). A two-step mechanism: 1) NO + NO ⇄ N2O2 (fast); 2) N2O2 + O2 → NO2 + NO2 (slow).(D). The mechanism must necessarily consist of three unimolecular elementary steps with very similar rate constants.arrow_forwarda. What is the eluent used in the column chromatography here (a “silica plug filtration” is essentially a very short column)? b. The spectroscopy of compound 5b is described in the second half of this excerpt, including 1H-NMR and 13C-NMR (which you will learn about in CHEM 2412L), MS (which you will learn about later in CHEM 2411L) and IR. One of the IR signals is at 3530 cm-1. What functional group does this indicate might be present in compound 5b?arrow_forwardSteps and explanations. Also provide, if possible, ways to adress this kind of problems in general.arrow_forward
- a. The first three lines of this procedure describe the reaction used to make compound 5b. In the fourth line, hexane and sodium bicarbonate are added. What organic lab technique is being used here? b. What is the purpose of the Na2SO4? c. What equipment would you use to “concentrate [a solution] under reduced pressure”?arrow_forwardWhen N,N-dimethylaniline is treated with bromine both the ortho and para products are observed. However when treated with a mixture of nitric acid and sulfuric acid only the meta product is observed. Explain these results and support your answer with the appropriate drawings *Hint amines are bases* N HNO3 H2SO4 N NO2 N Br2 N Br + N 8-8-8 FeBr3 Brarrow_forwardDraw a mechanism that explains the formation of compound OMe SO3H 1. Fuming H2SO4arrow_forward
- Consider the following two acid-base reactions: OH OHI Based on what you know about the compounds and their acidity, which direction would you expect both of these reactions to proceed? Show your reasoning. A pKa table has been provided in case you need it. Functional group Example pka CHA -50 Alkane -35 Amine : NH3 Alkyne RH 25 Water HO-H 169 16 10 Protonated amines NH 10 5 Carboxylic acids OH Hydrochloric acid HCI A chemist intends to run the following reaction on the three substrates shown below: H₂O R-CI product room temp. Cl Cl (1) (2) (3) They find one will react quickly, one slowly, and one will not react at all. Which is which, and why? HINT: What is the reaction they're trying to do? Does that mechanism tell you anything about why something would be favored?arrow_forwardNH3 decomposes through an equilibrium reaction between NH3, H2, and N2. Only one of the options is correct:(A). The mechanism of the NH3 decomposition reaction must necessarily involve the collision of two NH3 molecules to induce a rearrangement of the atoms in this molecule.(B). The molecular weight of the NH3 decomposition reaction is 2 since two NH3 molecules must collide.(C). The rate of the NH3 decomposition reaction must be greater than that of NH3 synthesis, since the former requires two molecules to collide and the latter, four.(D). The NH3 decomposition reaction cannot occur in a single step.arrow_forwardGiven the equilibrium A2 + B2 ⇌ 2 AB where k1 is the rate coefficient of the forward reaction and k-1 is the rate coefficient of the reverse reaction, with the forward reaction being first-order in A2 and B2, and the reverse reaction being second-order in AB. Equilibrium will be reached later if the relative values of the constants are:(A) k1 high and k-1 high(B) k1 high and k-1 low(C) k1 low and k-1 high(D) k1 low and k-1 lowarrow_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
Living By Chemistry: First Edition TextbookChemistryISBN:9781559539418Author:Angelica StacyPublisher:MAC HIGHER
Chemistry for Today: General, Organic, and Bioche...ChemistryISBN:9781305960060Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. HansenPublisher:Cengage Learning
World of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage Learning
Introductory Chemistry: A FoundationChemistryISBN:9781337399425Author:Steven S. Zumdahl, Donald J. DeCostePublisher: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
Living By Chemistry: First Edition Textbook
Chemistry
ISBN:9781559539418
Author:Angelica Stacy
Publisher:MAC HIGHER
Chemistry for Today: General, Organic, and Bioche...
Chemistry
ISBN:9781305960060
Author:Spencer L. Seager, Michael R. Slabaugh, Maren S. Hansen
Publisher:Cengage Learning
World of Chemistry, 3rd edition
Chemistry
ISBN:9781133109655
Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCoste
Publisher:Brooks / Cole / Cengage Learning
Introductory Chemistry: A Foundation
Chemistry
ISBN:9781337399425
Author:Steven S. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning