Foundations of College Chemistry, Binder Ready Version
Foundations of College Chemistry, Binder Ready Version
15th Edition
ISBN: 9781119083900
Author: Morris Hein, Susan Arena, Cary Willard
Publisher: WILEY
bartleby

Concept explainers

bartleby

Videos

Question
Book Icon
Chapter 15, Problem 17PE

(a)

Interpretation Introduction

Interpretation:

Molarity of each ion found in 1.25 M CuBr2 has to be calculated.

Concept Introduction:

Molarity is quantitatively defined as moles of solute in one liter of solution. For example, 0.070 M AlCl3 indicates that in 1 L volume moles of AlCl3 is 0.070 mol. The formula to evaluate volume from molarity is given as follows:

  M=nV

Here,

V represents volume.

M represents molarity.

n represents number of moles.

(a)

Expert Solution
Check Mark

Explanation of Solution

1.25 M CuBr2 indicates that in 1 L they moles of CuBr2 is 1.25 mol.

CuBr2 can be broken into 1 mol Cu2+ and 2 mol Br as follows:

  CuBr2Cu2++2Br

Since 1 M Cu2+ is furnished by 1 M CuBr2 so molarity of Cu2+ ion due to 1.25 M CuBr2 is calculated as follows:

  Molarity of Cu2+=(1.25 M CuBr2)(1 M Cu2+1 M CuBr2)=1.25 M Cu2+

Since 2 M Br is furnished by 1 M CuBr2 so molarity of Br ions due to 1.25 M CuBr2 is calculated as follows:

  Molarity of Br=(1.25 M CuBr2)(2 M Br1 M CuBr2)=2.5 M Br

Hence, molariry of  Cu2+ and Br is 1.25 M and 2.5 M respectively.

(b)

Interpretation Introduction

Interpretation:

Molarity of each ion found in 3.50 M K3AsO4 has to be calculated.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

K3AsO4 can be broken into 3 mol K+ and  1 mol AsO4 as follows:

  K3AsO43K++AsO4

Since 3 M K+ is furnished by 1 M AlCl3 so molarity of K+ ions due to 3.50 M K3AsO4 is calculated as follows:

  Molarity of K+=(3.50 M K3AsO4)(3 M K+1 M K3AsO4)=10.5 M K+

Since 1 M AsO4 is furnished by 1 M K3AsO4 so molarity of AsO4 ion due to 3.50 M K3AsO4 is calculated as follows:

  Molarity of AsO4=(3.50 M K3AsO4)(1 M AsO41 M K3AsO4)=3.50 M AsO4

Hence, molariry of K+ and AsO4 is 10.5 M and 3.50 M respectively.

(c)

Interpretation Introduction

Interpretation:

Molarity of each ion found in 0.75 M NaHCO3 has to be calculated.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

NaHCO3 can be broken into 1 mol Na+ and  1 mol HCO3 as follows:

  NaHCO3Na++HCO3

Since 1 M Na+ is furnished by 1 M NaHCO3 so molarity of Na+ ions due to 0.75 M NaHCO3 is calculated as follows:

  Molarity of Na+=(0.75 M NaHCO3)(1 M Na+1 M NaHCO3)=0.75 M Na+

Since 1 M HCO3 is furnished by 1 M NaHCO3 so molarity of HCO3 ions due to 0.75 M NaHCO3 is calculated as follows:

  Molarity of HCO3=(0.75 M NaHCO3)(1 M HCO31 M NaHCO3)=0.75 M HCO3

Hence, molariry of Na+ and HCO3 is 0.75 M10.5 M and 0.75 M respectively.

(d)

Interpretation Introduction

Interpretation:

Molarity of each ion found in 0.65 M (NH4)2SO4 has to be calculated.

Concept Introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

(NH4)2SO4 can be broken into 2 mol NH4+ and  1 mol SO42 as follows:

  (NH4)2SO42NH4++SO42

Since 2 M NH4+ is furnished by 1 M (NH4)2SO4 so molarity of NH4+ ions due to 0.65 M (NH4)2SO4 is calculated as follows:

  Molarity of NH4+=(0.65 M (NH4)2SO4)(2 M NH4+1 M (NH4)2SO4)=1.3 M NH4+

Since 1 M SO42 is furnished by 1 M (NH4)2SO4 so molarity of SO42 ion due to 0.65 M (NH4)2SO4 is calculated as follows:

  Molarity of SO42=(0.65 M (NH4)2SO4)(1 M SO421 M (NH4)2SO4)=0.65 M SO42

Hence, molariry of NH4+ and SO42 is 0.75 M1.3 M and 0.65 M respectively.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!
Students have asked these similar questions
2. Predict the NMR spectra for each of these two compounds by listing, in the NMR tables below, the chemical shift, the splitting, and the number of hydrogens associated with each predicted peak. Sort the peaks from largest chemical shift to lowest. **Not all slots must be filled** Peak Chemical Shift (d) 5.7 1 Multiplicity multiplate .......... 5.04 double of doublet 2 4.98 double of doublet 3 4.05 doublet of quartet 4 5 LO 3.80 quartet 1.3 doublet 6 Peak Chemical Shift (d) Multiplicity
Interpreting NMR spectra is a skill that often requires some amount of practice, which, in turn, necessitates access to a collection of NMR spectra. Beyond Labz Organic Synthesis and Organic Qualitative Analysis have spectral libraries containing over 700 1H NMR spectra. In this assignment, you will take advantage of this by first predicting the NMR spectra for two closely related compounds and then checking your predictions by looking up the actual spectra in the spectra library. After completing this assignment, you may wish to select other compounds for additional practice. 1. Write the IUPAC names for the following two structures: Question 2 Question 3 2. Predict the NMR spectra for each of these two compounds by listing, in the NMR tables below, the chemical shift, the splitting, and the number of hydrogens associated with each predicted peak. Sort the peaks from largest chemical shift to lowest. **Not all slots must be filled**
11:14 ... worksheets.beyondlabz.com 3. To check your predictions, click this link for Interpreting NMR Spectra 1. You will see a list of all the - compounds in the spectra library in alphabetical order by IUPAC name. Hovering over a name in the list will show the structure on the chalkboard. The four buttons on the top of the Spectra tab in the tray are used to select the different spectroscopic techniques for the selected compound. Make sure the NMR button has been selected. 4. Scroll through the list of names to find the names for the two compounds you have been given and click on the name to display the NMR spectrum for each. In the NMR tables below, list the chemical shift, the splitting, and the number of hydrogens associated with each peak for each compound. Compare your answers to your predictions. **Not all slots must be filled** Peak Chemical Shift (d) Multiplicity 1 2 3 4 5

Chapter 15 Solutions

Foundations of College Chemistry, Binder Ready Version

Ch. 15 - Prob. 3RQCh. 15 - Prob. 4RQCh. 15 - Prob. 5RQCh. 15 - Prob. 6RQCh. 15 - Prob. 7RQCh. 15 - Prob. 8RQCh. 15 - Prob. 9RQCh. 15 - Prob. 10RQCh. 15 - Prob. 11RQCh. 15 - Prob. 12RQCh. 15 - Prob. 13RQCh. 15 - Prob. 14RQCh. 15 - Prob. 15RQCh. 15 - Prob. 16RQCh. 15 - Prob. 17RQCh. 15 - Prob. 18RQCh. 15 - Prob. 19RQCh. 15 - Prob. 20RQCh. 15 - Prob. 21RQCh. 15 - Prob. 22RQCh. 15 - Prob. 23RQCh. 15 - Prob. 24RQCh. 15 - Prob. 25RQCh. 15 - Prob. 26RQCh. 15 - Prob. 27RQCh. 15 - Prob. 28RQCh. 15 - Prob. 1PECh. 15 - Prob. 2PECh. 15 - Prob. 3PECh. 15 - Prob. 4PECh. 15 - Prob. 5PECh. 15 - Prob. 6PECh. 15 - Prob. 7PECh. 15 - Prob. 8PECh. 15 - Prob. 9PECh. 15 - Prob. 10PECh. 15 - Prob. 11PECh. 15 - Prob. 12PECh. 15 - Prob. 13PECh. 15 - Prob. 14PECh. 15 - Prob. 15PECh. 15 - Prob. 16PECh. 15 - Prob. 17PECh. 15 - Prob. 18PECh. 15 - Prob. 19PECh. 15 - Prob. 20PECh. 15 - Prob. 21PECh. 15 - Prob. 22PECh. 15 - Prob. 23PECh. 15 - Prob. 24PECh. 15 - Prob. 25PECh. 15 - Prob. 26PECh. 15 - Prob. 27PECh. 15 - Prob. 28PECh. 15 - Prob. 29PECh. 15 - Prob. 30PECh. 15 - Prob. 31PECh. 15 - Prob. 32PECh. 15 - Prob. 33PECh. 15 - Prob. 34PECh. 15 - Prob. 35PECh. 15 - Prob. 36PECh. 15 - Prob. 37PECh. 15 - Prob. 38PECh. 15 - Prob. 39PECh. 15 - Prob. 40PECh. 15 - Prob. 41PECh. 15 - Prob. 42PECh. 15 - Prob. 43PECh. 15 - Prob. 44PECh. 15 - Prob. 45AECh. 15 - Prob. 46AECh. 15 - Prob. 47AECh. 15 - Prob. 48AECh. 15 - Prob. 49AECh. 15 - Prob. 50AECh. 15 - Prob. 51AECh. 15 - Prob. 52AECh. 15 - Prob. 53AECh. 15 - Prob. 54AECh. 15 - Prob. 55AECh. 15 - Prob. 56AECh. 15 - Prob. 57AECh. 15 - Prob. 58AECh. 15 - Prob. 59AECh. 15 - Prob. 60AECh. 15 - Prob. 61AECh. 15 - Prob. 62AECh. 15 - Prob. 63AECh. 15 - Prob. 64AECh. 15 - Prob. 65AECh. 15 - Prob. 66AECh. 15 - Prob. 67AECh. 15 - Prob. 68AECh. 15 - Prob. 69AECh. 15 - Prob. 70AECh. 15 - Prob. 71AECh. 15 - Prob. 72AECh. 15 - Prob. 73CECh. 15 - Prob. 74CE
Knowledge Booster
Background pattern image
Chemistry
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
SEE MORE QUESTIONS
Recommended textbooks for you
Text book image
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
General Chemistry | Acids & Bases; Author: Ninja Nerd;https://www.youtube.com/watch?v=AOr_5tbgfQ0;License: Standard YouTube License, CC-BY