Chemical Principles: The Quest for Insight
7th Edition
ISBN: 9781464183959
Author: Peter Atkins, Loretta Jones, Leroy Laverman
Publisher: W. H. Freeman
expand_more
expand_more
format_list_bulleted
Videos
Question
Chapter 5, Problem 5E.1E
(a)
Interpretation Introduction
Interpretation:
The molarity of sodium chloride solution has to be calculated.
Concept introduction:
Number of moles can be calculated by using following formula,
The molality of the solution can be calculated by using following formula
(b)
Interpretation Introduction
Interpretation:
Mass in grams of sodium hydroxide has to be calculated.
Concept introduction:
Refer to part (a)
(c)
Interpretation Introduction
Interpretation:
The molality of urea has to be calculated.
Concept introduction:
Refer to part (a)
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solution
Students have asked these similar questions
13.) Calculate the mass of solid MgSO̟•7H,O that
must be added to precipitate out all of the PO,3- ions in
the form of MgNH¸PO¸ •6H,O from 10.0 mL of a 2.0
M solution of PO,³ (aq)?
The following chemical reaction takes place in aqueous solution:
2 AGNO3(aq)+K,CO;(aq) → Ag,CO;(s)+2 KNO3(aq)
Write the net ionic equation for this reaction.
(a)
A student carries out a titration to find the concentration of some sulfuric acid.
The student finds that 25.00 cm of 0.0880 mol dm aqueous sodium
hydroxide, NaOH, is neutralised by 17.60 cm of dilute sulfuric acid,
H2SO4.
-3
H2SO4(aq) + 2NaOH(aq) → Na2SO,(aq) + 2H2O(1)
(i)
Calculate the amount, in moles, of NaOH used.
answer =
mol
(ii)
Determine the amount, in moles, of H2SO4 used.
answer
mol
(ii) Calculate the concentration, in mol dm3, of the sulfuric acid.
answer =
mol dm3
(b) After carrying out the titration in (a), the student left the resulting solution to
crystallise. White crystals were formed, with a formula of Na2SO4•x H2O
and a molar mass of 322.1 g mol.
What term is given to the x H2O' part of the formula?
Using the molar mass of the crystals, calculate the value of x.
answer =
Chapter 5 Solutions
Chemical Principles: The Quest for Insight
Ch. 5 - Prob. 5A.1ASTCh. 5 - Prob. 5A.1BSTCh. 5 - Prob. 5A.2ASTCh. 5 - Prob. 5A.2BSTCh. 5 - Prob. 5A.3ASTCh. 5 - Prob. 5A.3BSTCh. 5 - Prob. 5A.1ECh. 5 - Prob. 5A.2ECh. 5 - Prob. 5A.3ECh. 5 - Prob. 5A.4E
Ch. 5 - Prob. 5A.5ECh. 5 - Prob. 5A.6ECh. 5 - Prob. 5A.7ECh. 5 - Prob. 5A.8ECh. 5 - Prob. 5A.11ECh. 5 - Prob. 5B.1ASTCh. 5 - Prob. 5B.1BSTCh. 5 - Prob. 5B.2ASTCh. 5 - Prob. 5B.2BSTCh. 5 - Prob. 5B.3ASTCh. 5 - Prob. 5B.3BSTCh. 5 - Prob. 5B.1ECh. 5 - Prob. 5B.2ECh. 5 - Prob. 5B.3ECh. 5 - Prob. 5B.5ECh. 5 - Prob. 5B.7ECh. 5 - Prob. 5C.1ASTCh. 5 - Prob. 5C.1BSTCh. 5 - Prob. 5C.2ASTCh. 5 - Prob. 5C.2BSTCh. 5 - Prob. 5C.3ASTCh. 5 - Prob. 5C.3BSTCh. 5 - Prob. 5C.1ECh. 5 - Prob. 5C.3ECh. 5 - Prob. 5C.4ECh. 5 - Prob. 5C.5ECh. 5 - Prob. 5C.6ECh. 5 - Prob. 5C.7ECh. 5 - Prob. 5C.8ECh. 5 - Prob. 5C.9ECh. 5 - Prob. 5C.10ECh. 5 - Prob. 5C.11ECh. 5 - Prob. 5C.12ECh. 5 - Prob. 5C.15ECh. 5 - Prob. 5C.16ECh. 5 - Prob. 5D.1ASTCh. 5 - Prob. 5D.1BSTCh. 5 - Prob. 5D.1ECh. 5 - Prob. 5D.2ECh. 5 - Prob. 5D.3ECh. 5 - Prob. 5D.4ECh. 5 - Prob. 5D.5ECh. 5 - Prob. 5D.6ECh. 5 - Prob. 5D.7ECh. 5 - Prob. 5D.8ECh. 5 - Prob. 5D.9ECh. 5 - Prob. 5D.10ECh. 5 - Prob. 5D.11ECh. 5 - Prob. 5D.12ECh. 5 - Prob. 5D.13ECh. 5 - Prob. 5D.14ECh. 5 - Prob. 5D.15ECh. 5 - Prob. 5D.16ECh. 5 - Prob. 5D.18ECh. 5 - Prob. 5D.19ECh. 5 - Prob. 5D.20ECh. 5 - Prob. 5E.1ASTCh. 5 - Prob. 5E.1BSTCh. 5 - Prob. 5E.2ASTCh. 5 - Prob. 5E.2BSTCh. 5 - Prob. 5E.1ECh. 5 - Prob. 5E.2ECh. 5 - Prob. 5E.11ECh. 5 - Prob. 5E.12ECh. 5 - Prob. 5F.1ASTCh. 5 - Prob. 5F.1BSTCh. 5 - Prob. 5F.2ASTCh. 5 - Prob. 5F.2BSTCh. 5 - Prob. 5F.3ASTCh. 5 - Prob. 5F.3BSTCh. 5 - Prob. 5F.4ASTCh. 5 - Prob. 5F.4BSTCh. 5 - Prob. 5F.5ASTCh. 5 - Prob. 5F.5BSTCh. 5 - Prob. 5F.1ECh. 5 - Prob. 5F.2ECh. 5 - Prob. 5F.3ECh. 5 - Prob. 5F.5ECh. 5 - Prob. 5F.7ECh. 5 - Prob. 5F.9ECh. 5 - Prob. 5F.10ECh. 5 - Prob. 5F.11ECh. 5 - Prob. 5F.12ECh. 5 - Prob. 5F.13ECh. 5 - Prob. 5F.14ECh. 5 - Prob. 5F.15ECh. 5 - Prob. 5F.16ECh. 5 - Prob. 5G.1ASTCh. 5 - Prob. 5G.1BSTCh. 5 - Prob. 5G.2ASTCh. 5 - Prob. 5G.2BSTCh. 5 - Prob. 5G.3ASTCh. 5 - Prob. 5G.3BSTCh. 5 - Prob. 5G.4ASTCh. 5 - Prob. 5G.4BSTCh. 5 - Prob. 5G.5ASTCh. 5 - Prob. 5G.5BSTCh. 5 - Prob. 5G.1ECh. 5 - Prob. 5G.2ECh. 5 - Prob. 5G.3ECh. 5 - Prob. 5G.4ECh. 5 - Prob. 5G.7ECh. 5 - Prob. 5G.8ECh. 5 - Prob. 5G.9ECh. 5 - Prob. 5G.11ECh. 5 - Prob. 5G.12ECh. 5 - Prob. 5G.13ECh. 5 - Prob. 5G.14ECh. 5 - Prob. 5G.15ECh. 5 - Prob. 5G.16ECh. 5 - Prob. 5G.17ECh. 5 - Prob. 5G.19ECh. 5 - Prob. 5G.20ECh. 5 - Prob. 5G.21ECh. 5 - Prob. 5G.22ECh. 5 - Prob. 5H.1ASTCh. 5 - Prob. 5H.1BSTCh. 5 - Prob. 5H.2ASTCh. 5 - Prob. 5H.2BSTCh. 5 - Prob. 5H.1ECh. 5 - Prob. 5H.2ECh. 5 - Prob. 5H.3ECh. 5 - Prob. 5H.4ECh. 5 - Prob. 5H.5ECh. 5 - Prob. 5H.6ECh. 5 - Prob. 5I.1ASTCh. 5 - Prob. 5I.1BSTCh. 5 - Prob. 5I.2ASTCh. 5 - Prob. 5I.2BSTCh. 5 - Prob. 5I.3ASTCh. 5 - Prob. 5I.3BSTCh. 5 - Prob. 5I.4ASTCh. 5 - Prob. 5I.4BSTCh. 5 - Prob. 5I.1ECh. 5 - Prob. 5I.2ECh. 5 - Prob. 5I.3ECh. 5 - Prob. 5I.4ECh. 5 - Prob. 5I.5ECh. 5 - Prob. 5I.6ECh. 5 - Prob. 5I.7ECh. 5 - Prob. 5I.9ECh. 5 - Prob. 5I.10ECh. 5 - Prob. 5I.11ECh. 5 - Prob. 5I.12ECh. 5 - Prob. 5I.13ECh. 5 - Prob. 5I.14ECh. 5 - Prob. 5I.15ECh. 5 - Prob. 5I.16ECh. 5 - Prob. 5I.17ECh. 5 - Prob. 5I.18ECh. 5 - Prob. 5I.19ECh. 5 - Prob. 5I.20ECh. 5 - Prob. 5I.21ECh. 5 - Prob. 5I.22ECh. 5 - Prob. 5I.23ECh. 5 - Prob. 5I.24ECh. 5 - Prob. 5I.25ECh. 5 - Prob. 5I.26ECh. 5 - Prob. 5I.27ECh. 5 - Prob. 5I.28ECh. 5 - Prob. 5I.29ECh. 5 - Prob. 5I.30ECh. 5 - Prob. 5I.32ECh. 5 - Prob. 5I.33ECh. 5 - Prob. 5I.34ECh. 5 - Prob. 5I.35ECh. 5 - Prob. 5I.36ECh. 5 - Prob. 5J.1ASTCh. 5 - Prob. 5J.1BSTCh. 5 - Prob. 5J.3ASTCh. 5 - Prob. 5J.3BSTCh. 5 - Prob. 5J.4ASTCh. 5 - Prob. 5J.4BSTCh. 5 - Prob. 5J.5ASTCh. 5 - Prob. 5J.5BSTCh. 5 - Prob. 5J.1ECh. 5 - Prob. 5J.2ECh. 5 - Prob. 5J.3ECh. 5 - Prob. 5J.4ECh. 5 - Prob. 5J.5ECh. 5 - Prob. 5J.6ECh. 5 - Prob. 5J.9ECh. 5 - Prob. 5J.10ECh. 5 - Prob. 5J.11ECh. 5 - Prob. 5J.12ECh. 5 - Prob. 5J.13ECh. 5 - Prob. 5J.17ECh. 5 - Prob. 5.1ECh. 5 - Prob. 5.2ECh. 5 - Prob. 5.3ECh. 5 - Prob. 5.4ECh. 5 - Prob. 5.5ECh. 5 - Prob. 5.6ECh. 5 - Prob. 5.7ECh. 5 - Prob. 5.8ECh. 5 - Prob. 5.9ECh. 5 - Prob. 5.10ECh. 5 - Prob. 5.11ECh. 5 - Prob. 5.12ECh. 5 - Prob. 5.13ECh. 5 - Prob. 5.14ECh. 5 - Prob. 5.15ECh. 5 - Prob. 5.16ECh. 5 - Prob. 5.17ECh. 5 - Prob. 5.19ECh. 5 - Prob. 5.23ECh. 5 - Prob. 5.24ECh. 5 - Prob. 5.25ECh. 5 - Prob. 5.26ECh. 5 - Prob. 5.27ECh. 5 - Prob. 5.28ECh. 5 - Prob. 5.29ECh. 5 - Prob. 5.30ECh. 5 - Prob. 5.31ECh. 5 - Prob. 5.32ECh. 5 - Prob. 5.33ECh. 5 - Prob. 5.35ECh. 5 - Prob. 5.37ECh. 5 - Prob. 5.38ECh. 5 - Prob. 5.41ECh. 5 - Prob. 5.43ECh. 5 - Prob. 5.44ECh. 5 - Prob. 5.45ECh. 5 - Prob. 5.46ECh. 5 - Prob. 5.47ECh. 5 - Prob. 5.49ECh. 5 - Prob. 5.51ECh. 5 - Prob. 5.53ECh. 5 - Prob. 5.55ECh. 5 - Prob. 5.57ECh. 5 - Prob. 5.58ECh. 5 - Prob. 5.61ECh. 5 - Prob. 5.62E
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
- Actually, the carbon in CO2(g) is thermodynamically unstable with respect to the carbon in calcium carbonate(limestone). Verify this by determining the standardGibbs free energy change for the reaction of lime,CaO(s), with CO2(g) to make CaCO3(s).arrow_forwardThe carbon dioxide exhaled in the breath of astronauts is often removed from the spacecraft by reaction with lithium hydroxide 2LiOH(s)+CO2(g)Li2CO3(s)+H2O(l) Estimate the grams of lithium hydroxide required per astronaut per day. Assume that each astronaut requires 2.50 103 kcal of energy per day. Further assume that this energy can be equated to the heat of combustion of a quantity of glucose, C6H12O6, to CO2(g) and H2O(l). From the amount of glucose required to give 2.50 103 kcal of heat, calculate the amount of CO2 produced and hence the amount of LiOH required. The H for glucose(s) is 1273 kJ/mol.arrow_forwardUse the appropriate tables to calculate H for (a) the reaction between MgC03(s) and a strong acid to give Mg2+(aq), CO2(g), and water. (b) the precipitation of iron(III) hydroxide from the reaction between iron(III) and hydroxide ions.arrow_forward
- According to the Resource Conservation and Recovery Act (RCRA), waste material is classified as toxic and must be handled as hazardous if the lead concentration exceeds 5 mg/L. By adding chloride ion, the lead ion will precipitate as PbCl2, which can be separated from the liquid portion. Once the lead has been removed, the rest of the waste can be sent to a conventional waste treatment facility. How many grams of sodium chloride must be added to 500 L of a waste solution to reduce the concentration of the Pb2+ ion from 10 to 5 mg/L?arrow_forwardSimple acids such as formic acid, HCOOH, and acetic acid, CH3COOH, are very soluble in water; however, fatty acids such as stearic acid, CH3(CH2)16COOH, and palmitic acid, CH3(CH2)14COOH, are water-insoluble. Based on what you know about the solubility of alcohols, explain the solubility of these organic acids.arrow_forwardA volume of 125.0 mL of 0.565 M KOH (aq) is diluted with 425.0 mL of pure water. What is the final concentration (in molarity) of KOH? Report to three significant figures.arrow_forward
- If weak acids ionize only a few percent in aqueous solution, why is it possible to fully neutralize a weak acid by reacting it with the stoichiometric equivalent of sodium hydroxide solution, NaOH(aq)?arrow_forwardSodium hydrogen carbonate (NaHCO,, also known as sodium bicarbonate or "baking soda", can be used to relieve acid indigestion. Acid indigestion is the burning sensation you get in your stomach when it contains too much hydrochloric acid (HCI) , which the stomach secretes to help digest food. Drinking a glass of water containing dissolved NaHCO, neutralizes excess HCl through this reaction: 圖 HCl(aq) + NaHCO3(aq) → NaCl(aq) + H,O(1) + CO,(g) The CO, gas produced is what makes you burp after drinking the solution. Suppose the fluid in the stomach of a man suffering from indigestion can be considered to be 50. mL of a 0.089 M HCI solution. What mass of NaHCO, would he need to ingest to neutralize this much HCl ? Be sure your answer has the correct number of significant digits.arrow_forwardWhat volume of 0.250 M HCl(aq) will completely react with 50.0 mL of 0.115 M NaOH(aq)? Write the balanced reaction and show the stoichiometry. Write the reaction of acetate ion in water. Predict the approximate pH of a solution with acetate ion present.arrow_forward
- Someone is attempting to measure the H3PO4(aq) content of an unknown solution by both titration with NaOH(aq) and calorimetry. The person measures 100 mL of stock NaOH(aq) solution into a buret but forgets to write down its concentration. They then measure out 50.0 mL of unknown H3PO4(aq) solution (which has an indicator) and places this solution in a constant-pressure calorimeter for the titration experiment. The initial temperature of all the solutions is 22.9 Celsius. The student adds 46.84 mL of NaOH(aq) from the buret to the H3PO4(aq) solution in the calorimeter to reach the equivalence point (color change for indicator). The final solution temperature is 33.5 Celsius. Assume density and specific heat of all solutions are the same as that of pure water and that volumes are additive. a) what is the balanced molecular equation for the acid-base reaction that takes place in the calorimeter? b) Calculate delta HRXN (enthralpy) for the above acid-base reaction c) what is the…arrow_forwardA student uses 7.500 grams of oxalic acid (H₂C₂O₄) to make 250.0 cm³ of solution in a volumetric flask. She then transfers 20.0 cm³ of the solution to a conical flask with a pipette. What is the concentration in mol·dm⁻³ of the oxalic in the conical flask? [Round the answer to three decimals. Do not type in the percentage sign. Use a decimal point.] *arrow_forwardComplete the balanced chemical reaction for the following weak base with a strong acid. In this case, write the resulting acid and base as its own species in the reaction. CH:NH2(aq) + HCIO:(aq)- |3- 2- 4- 2+ |3+ 4+ 1 2 3 4 5 6 7 8 O2 O3 D4 Os |1 6. 10 (s) (1) (g) (aq) + H2O H3O+ H CI OH- C Reset x H2O Deletearrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
Chemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning
Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning
Principles of Modern ChemistryChemistryISBN:9781305079113Author:David W. Oxtoby, H. Pat Gillis, Laurie J. ButlerPublisher:Cengage Learning
General Chemistry - Standalone book (MindTap Cour...ChemistryISBN:9781305580343Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; DarrellPublisher:Cengage Learning
Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage Learning
Chemistry: Principles and Reactions
Chemistry
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:Cengage Learning
Chemistry & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
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
Principles of Modern Chemistry
Chemistry
ISBN:9781305079113
Author:David W. Oxtoby, H. Pat Gillis, Laurie J. Butler
Publisher:Cengage Learning
General Chemistry - Standalone book (MindTap Cour...
Chemistry
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Cengage Learning
Chemistry: The Molecular Science
Chemistry
ISBN:9781285199047
Author:John W. Moore, Conrad L. Stanitski
Publisher:Cengage Learning
The Laws of Thermodynamics, Entropy, and Gibbs Free Energy; Author: Professor Dave Explains;https://www.youtube.com/watch?v=8N1BxHgsoOw;License: Standard YouTube License, CC-BY