FOUND.OF COLLEGE CHEMISTRY
15th Edition
ISBN: 9781119234555
Author: Hein
Publisher: WILEY
expand_more
expand_more
format_list_bulleted
Concept explainers
Videos
Question
Chapter 10, Problem 28PE
Interpretation Introduction
Interpretation:
The reason why the last electron of potassium goes to the fourth energy level has to be given.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
Describe what happens to the average kinetic energy of ideal gas molecules when the conditions are changed as follows:(a) The pressure of the gas is increased by reducing the volume at constant temperature.(b) The pressure of the gas is increased by increasing the temperature at constant volume.(c) The average velocity of the molecules is increased by a factor of 2.
A sample of 3.42 mol of xenon is confined at low pressure in a volume at a temperature of 86 °C. Describe quantitatively the effects of each of the following changes
on the pressure, the average kinetic energy per molecule in the gas, and the root-mean-square speed.
(a) The temperature is increased to 199 °C
(b) The volume is tripled.
(c) The amount of xenon is decreased to 1.87 mol
Give each answer as a decimal factor of the form: new value factor old value. A factor of 1 means no change.
ChangeP
KEavgms
A sample of 4.72 mol of krypton is confined at low pressure in a volume at a temperature of 68 °C. Describe quantitatively the effects of each of the following changes on the pressure, the average kinetic energy per molecule in the gas, and the root-mean-square speed.
(a) The temperature is decreased to -49 °C.(b) The volume is doubled.(c) The amount of krypton is increased to 6.47 mol.
Give each answer as a decimal factor of the form: new value = factor old value. A factor of 1 means no change.
Chapter 10 Solutions
FOUND.OF COLLEGE CHEMISTRY
Ch. 10.1 - Prob. 10.1PCh. 10.2 - Prob. 10.2PCh. 10.3 - Prob. 10.3PCh. 10.4 - Prob. 10.4PCh. 10.5 - Prob. 10.5PCh. 10.5 - Prob. 10.6PCh. 10.5 - Prob. 10.7PCh. 10 - Prob. 1RQCh. 10 - Prob. 2RQCh. 10 - Prob. 3RQ
Ch. 10 - Prob. 4RQCh. 10 - Prob. 5RQCh. 10 - Prob. 6RQCh. 10 - Prob. 7RQCh. 10 - Prob. 8RQCh. 10 - Prob. 9RQCh. 10 - Prob. 10RQCh. 10 - Prob. 11RQCh. 10 - Prob. 12RQCh. 10 - Prob. 13RQCh. 10 - Prob. 14RQCh. 10 - Prob. 15RQCh. 10 - Prob. 16RQCh. 10 - Prob. 17RQCh. 10 - Prob. 18RQCh. 10 - Prob. 19RQCh. 10 - Prob. 20RQCh. 10 - Prob. 21RQCh. 10 - Prob. 22RQCh. 10 - Prob. 23RQCh. 10 - Prob. 24RQCh. 10 - Prob. 25RQCh. 10 - Prob. 1PECh. 10 - Prob. 2PECh. 10 - Prob. 3PECh. 10 - Prob. 4PECh. 10 - Prob. 5PECh. 10 - Prob. 6PECh. 10 - Prob. 7PECh. 10 - Prob. 8PECh. 10 - Prob. 9PECh. 10 - Prob. 10PECh. 10 - Prob. 11PECh. 10 - Prob. 12PECh. 10 - Prob. 13PECh. 10 - Prob. 14PECh. 10 - Prob. 15PECh. 10 - Prob. 16PECh. 10 - Prob. 17PECh. 10 - Prob. 18PECh. 10 - Prob. 19PECh. 10 - Prob. 20PECh. 10 - Prob. 21PECh. 10 - Prob. 22PECh. 10 - Prob. 23PECh. 10 - Prob. 24PECh. 10 - Prob. 25PECh. 10 - Prob. 26PECh. 10 - Prob. 27PECh. 10 - Prob. 28PECh. 10 - Prob. 29PECh. 10 - Prob. 30PECh. 10 - Prob. 31PECh. 10 - Prob. 32PECh. 10 - Prob. 33PECh. 10 - Prob. 34PECh. 10 - Prob. 35PECh. 10 - Prob. 36PECh. 10 - Prob. 37PECh. 10 - Prob. 38PECh. 10 - Prob. 39PECh. 10 - Prob. 40PECh. 10 - Prob. 41PECh. 10 - Prob. 42PECh. 10 - Prob. 43PECh. 10 - Prob. 44PECh. 10 - Prob. 45PECh. 10 - Prob. 46PECh. 10 - Prob. 47PECh. 10 - Prob. 48PECh. 10 - Prob. 49PECh. 10 - Prob. 50PECh. 10 - Prob. 51AECh. 10 - Prob. 52AECh. 10 - Prob. 53AECh. 10 - Prob. 54AECh. 10 - Prob. 57AECh. 10 - Prob. 58AECh. 10 - Prob. 59AECh. 10 - Prob. 60AECh. 10 - Prob. 61AECh. 10 - Prob. 62AECh. 10 - Prob. 63AECh. 10 - Prob. 64AECh. 10 - Prob. 65AECh. 10 - Prob. 66AECh. 10 - Prob. 67AECh. 10 - Prob. 68AECh. 10 - Prob. 69AECh. 10 - Prob. 70AECh. 10 - Prob. 71AECh. 10 - Prob. 72AECh. 10 - Prob. 73AECh. 10 - Prob. 74AECh. 10 - Prob. 75AECh. 10 - Prob. 76AECh. 10 - Prob. 77AECh. 10 - Prob. 78CECh. 10 - Prob. 79CECh. 10 - Prob. 80CECh. 10 - Prob. 81CECh. 10 - Prob. 82CE
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
- Given that a sample of air is made up of nitrogen, oxygen, and argon in the mole fractions 0.78 N2, 0.21 O2, and 0.010 Ar, what is the density of air at standard temperature and pressure?arrow_forwardYou have two pressure-proof steel cylinders of equal volume, one containing 1.0 kg of CO and the other containing 1.0 kg of acetylene, C2H2. (a) In which cylinder is the pressure greater at 25 C? (b) Which cylinder contains the greater number of molecules?arrow_forwardHow does hydraulic fracturing differ from previously used techniques for the recovery of natural gas from the earth?arrow_forward
- One of the chemical controversies of the nineteenth century concerned the element beryllium (Be). Berzelius originally claimed that beryllium was a trivalent element (forming Be3+ ions) and that it gave an oxide with the formula Be2O3. This resulted in a calculated atomic mass of 13.5 for beryllium. In formulating his periodic table, Mendeleev proposed that beryllium was divalent (forming Be2+ ions) and that it gave an oxide with the formula Be2O3. This assumption gives an atomic mass of 9.0. In 1894, A. Combes (Comptes Rendus 1894, p. 1221) reacted beryllium with the anion C5H7O2and measured the density of the gaseous product. Combess data for two different experiments are as follows: I II Mass 0.2022 g 0.2224 g Volume 22.6 cm3 26.0 cm3 Temperature 13C 17C Pressure 765.2 mm Hg 764.6 mm If beryllium is a divalent metal, the molecular formula of the product will be Be(C5H7O2)2; if it is trivalent, the formula will be Be(C5H7O2)3. Show how Combess data help to confirm that beryllium is a divalent metal.arrow_forwardJj.200.arrow_forwardA sample of 3.73 mol of argon is confined at low pressure in a volume at a temperature of 61 C. Describe quantitatively the effects of each of the following changes on the pressure, the average kinetic energy per molecule in the gas, and the root-mean-square speed. (a) The temperature is increased to 177 °C. (b) The volume is tripled. (c) The amount of argon is decreased to 1.96 mol. Give each answer as a decimal factor of the form: new value = factor old value. A factor of 1 means no change. Change KEavg Urms (а) (b) (c)arrow_forward
- In 1897 the Swedish explorer Andree tried to reach the North Pole in a balloon. The balloon was filled with hydrogen gas. The hydrogen gas was prepared from iron splints and diluted sulfuric acid. The reaction is Fe(a) + H,S0, (ag) + PeSO(og) + H;(s) The volume of the balloon was 4800 m and the loss of hydrogen gas during filling was estimated at 20% What mass of iron splints and 98 (by mas) H,S0, were nooded to enre the complete filling of the balloon? Assume a temperature of oC, a presaure of 10 atm during filling nd 100 yield. Mass of Fe Mass of HaS0, -arrow_forwardPhosphorus trichloride is important in the manufacture of insecticides, fuel additives, and flame retardants. Phosphorus has only one naturally occurring isotope, 31P, whereas chlorine has two, 35Cl (75%) and 37Cl (25%). (a) What different molecular masses (in amu) can be found for PCl3? (b) Which is the most abundant? (c) What is the ratio of the effusion rates of the heavi est and the lightest PCl3 molecules?arrow_forwardDecide whether each one of the following statement is correct, not correct, or conditionally correct. Justify your answers in each case. (a) As the pressure increases, a gas molecule travels a shorter distance before it collides with a neighbor. (b) Mean free path of a benzene molecule is larger than that of He.arrow_forward
- (a) An ideal gas occupies a volume of 2.2 cm3 at 20°C and atmospheric pressure. Determine the number of molecules of gas in the container. (b) If the pressure of the 2.2-cm3 volume is reduced to 2.6 ✕ 10−11 Pa (an extremely good vacuum) while the temperature remains constant, how many moles of gas remain in the container?arrow_forward(b) The concentration of SO2 in air was found to be 50. ppb. Calculate its concentration in molecules per uL of air at 1 atm total air pressure and at 27°C.arrow_forwardOn our campus, there are approximately 30.000 students who use 2.5 x 10° kJ of energy per day. Suppose all of that energy comes from the combustion of a carbon- hydrogen-oxygen compound in the presence of excess O2(g). 2.647 g sample of this gaseous carbon-hydrogen-oxygen compound that occupies a volume of 580 mL at 918,6 Torr and 24.00 °C. The products of the combustion of the given amount are 5.059 g CO2(g), 3.106 g H2O(1), and enough heat to raise the temperature of the calorimeter assembly from 24.00 to 38.33 °C. What is the molecular formula of this unknown compound and how many kilograms of this compound are needed to provide enough daily energy to all students on campus? (The heat capacity of the calorimeter is 4.915 kJ/°C.)arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Chemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher:Cengage Learning
Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
Chemistry for Engineering StudentsChemistryISBN:9781337398909Author:Lawrence S. Brown, Tom HolmePublisher:Cengage Learning
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Chemistry for Engineering Students
Chemistry
ISBN:9781337398909
Author:Lawrence S. Brown, Tom Holme
Publisher:Cengage Learning
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
Calorimetry Concept, Examples and Thermochemistry | How to Pass Chemistry; Author: Melissa Maribel;https://www.youtube.com/watch?v=nSh29lUGj00;License: Standard YouTube License, CC-BY