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
Question
Chapter F, Problem L.18E
(a)
Interpretation Introduction
Interpretation:
Concentration of
Concept Introduction:
Acid and base reacts to form salt and water. The concentration of the titrant and volume of the titrant that is required to neutralize the analyte is calculated by volumetric analysis. Stoichiometric point is the one where the quantity of titrant neutralizes the quantity of analyte that is taken.
In order to determine the molarity of analyte, the stoichiometric relationship between titrant and analyte along with molarity of the titrant that is used is considered.
(b)
Interpretation Introduction
Interpretation:
Mass of
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
Sodium 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.
Normal calcium levels in the blood should be between 9.0 and 10.5 mg/dL. Calcium levels in the blood can be determined by adding oxalate ion to precipitate calcium oxalate, CaC2O4, followed by dissolving the precipitate in aqueous acid and titrating the resulting oxalic acid (H2C2O4) with KMnO4.
5 H2C2O4 (aq) + 2 MnO4–(aq) + 6 H+(aq) yields 10 CO2 (g) + 2 Mn2+(aq) + 8 H2O(l)
What is the concentration of Ca2+ (mg/dL) in a 10.0 mL sample of blood if 10.54 mL of 9.88 x 10–4 M KMnO4 solution is needed for the titration.
(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 F Solutions
Chemical Principles: The Quest for Insight
Ch. F - Prob. A.1ASTCh. F - Prob. A.1BSTCh. F - Prob. A.2ASTCh. F - Prob. A.2BSTCh. F - Prob. A.3ASTCh. F - Prob. A.3BSTCh. F - Prob. A.4ASTCh. F - Prob. A.4BSTCh. F - Prob. A.5ASTCh. F - Prob. A.5BST
Ch. F - Prob. A.1ECh. F - Prob. A.2ECh. F - Prob. A.3ECh. F - Prob. A.4ECh. F - Prob. A.5ECh. F - Prob. A.6ECh. F - Prob. A.7ECh. F - Prob. A.8ECh. F - Prob. A.9ECh. F - Prob. A.10ECh. F - Prob. A.11ECh. F - Prob. A.12ECh. F - Prob. A.13ECh. F - Prob. A.14ECh. F - Prob. A.15ECh. F - Prob. A.16ECh. F - Prob. A.17ECh. F - Prob. A.18ECh. F - Prob. A.19ECh. F - Prob. A.20ECh. F - Prob. A.21ECh. F - Prob. A.22ECh. F - Prob. A.23ECh. F - Prob. A.24ECh. F - Prob. A.25ECh. F - Prob. A.26ECh. F - Prob. A.27ECh. F - Prob. A.28ECh. F - Prob. A.29ECh. F - Prob. A.30ECh. F - Prob. A.31ECh. F - Prob. A.32ECh. F - Prob. A.33ECh. F - Prob. A.34ECh. F - Prob. A.35ECh. F - Prob. A.36ECh. F - Prob. A.37ECh. F - Prob. A.38ECh. F - Prob. A.39ECh. F - Prob. A.40ECh. F - Prob. A.41ECh. F - Prob. A.42ECh. F - Prob. B.1ASTCh. F - Prob. B.1BSTCh. F - Prob. B.2ASTCh. F - Prob. B.2BSTCh. F - Prob. B.3ASTCh. F - Prob. B.3BSTCh. F - Prob. B.1ECh. F - Prob. B.2ECh. F - Prob. B.3ECh. F - Prob. B.4ECh. F - Prob. B.5ECh. F - Prob. B.6ECh. F - Prob. B.7ECh. F - Prob. B.8ECh. F - Prob. B.9ECh. F - Prob. B.10ECh. F - Prob. B.11ECh. F - Prob. B.12ECh. F - Prob. B.13ECh. F - Prob. B.14ECh. F - Prob. B.15ECh. F - Prob. B.16ECh. F - Prob. B.17ECh. F - Prob. B.18ECh. F - Prob. B.19ECh. F - Prob. B.20ECh. F - Prob. B.21ECh. F - Prob. B.22ECh. F - Prob. C.1ASTCh. F - Prob. C.1BSTCh. F - Prob. C.2ASTCh. F - Prob. C.2BSTCh. F - Prob. C.1ECh. F - Prob. C.2ECh. F - Prob. C.3ECh. F - Prob. C.4ECh. F - Prob. C.5ECh. F - Prob. C.6ECh. F - Prob. C.7ECh. F - Prob. C.8ECh. F - Prob. C.9ECh. F - Prob. C.10ECh. F - Prob. C.11ECh. F - Prob. C.12ECh. F - Prob. C.13ECh. F - Prob. C.14ECh. F - Prob. C.15ECh. F - Prob. C.16ECh. F - Prob. C.17ECh. F - Prob. C.18ECh. F - Prob. C.19ECh. F - Prob. C.20ECh. F - Prob. C.21ECh. F - Prob. C.22ECh. F - Prob. D.1ASTCh. F - Prob. D.1BSTCh. F - Prob. D.2ASTCh. F - Prob. D.2BSTCh. F - Prob. D.3ASTCh. F - Prob. D.3BSTCh. F - Prob. D.4ASTCh. F - Prob. D.4BSTCh. F - Prob. D.5ASTCh. F - Prob. D.5BSTCh. F - Prob. D.1ECh. F - Prob. D.2ECh. F - Prob. D.3ECh. F - Prob. D.4ECh. F - Prob. D.5ECh. F - Prob. D.6ECh. F - Prob. D.7ECh. F - Prob. D.8ECh. F - Prob. D.9ECh. F - Prob. D.10ECh. F - Prob. D.11ECh. F - Prob. D.12ECh. F - Prob. D.13ECh. F - Prob. D.14ECh. F - Prob. D.15ECh. F - Prob. D.16ECh. F - Prob. D.17ECh. F - Prob. D.18ECh. F - Prob. D.19ECh. F - Prob. D.20ECh. F - Prob. D.21ECh. F - Prob. D.22ECh. F - Prob. D.23ECh. F - Prob. D.24ECh. F - Prob. D.25ECh. F - Prob. D.26ECh. F - Prob. D.27ECh. F - Prob. D.28ECh. F - Prob. D.29ECh. F - Prob. D.30ECh. F - Prob. D.31ECh. F - Prob. D.32ECh. F - Prob. D.33ECh. F - Prob. D.34ECh. F - Prob. D.35ECh. F - Prob. D.36ECh. F - Prob. E.1ASTCh. F - Prob. E.1BSTCh. F - Prob. E.2ASTCh. F - Prob. E.2BSTCh. F - Prob. E.3ASTCh. F - Prob. E.3BSTCh. F - Prob. E.4ASTCh. F - Prob. E.4BSTCh. F - Prob. E.5ASTCh. F - Prob. E.5BSTCh. F - Prob. E.6ASTCh. F - Prob. E.6BSTCh. F - Prob. E.1ECh. F - Prob. E.2ECh. F - Prob. E.3ECh. F - Prob. E.4ECh. F - Prob. E.5ECh. F - Prob. E.6ECh. F - Prob. E.7ECh. F - Prob. E.8ECh. F - Prob. E.9ECh. F - Prob. E.10ECh. F - Prob. E.11ECh. F - Prob. E.12ECh. F - Prob. E.13ECh. F - Prob. E.14ECh. F - Prob. E.15ECh. F - Prob. E.16ECh. F - Prob. E.17ECh. F - Prob. E.18ECh. F - Prob. E.19ECh. F - Prob. E.20ECh. F - Prob. E.21ECh. F - Prob. E.22ECh. F - Prob. E.23ECh. F - Prob. E.24ECh. F - Prob. E.25ECh. F - Prob. E.26ECh. F - Prob. E.27ECh. F - Prob. E.28ECh. F - Prob. E.29ECh. F - Prob. E.30ECh. F - Prob. E.31ECh. F - Prob. E.32ECh. F - Prob. E.33ECh. F - Prob. E.34ECh. F - Prob. F.1ASTCh. F - Prob. F.1BSTCh. F - Prob. F.2ASTCh. F - Prob. F.2BSTCh. F - Prob. F.3ASTCh. F - Prob. F.3BSTCh. F - Prob. F.4ASTCh. F - Prob. F.4BSTCh. F - Prob. F.1ECh. F - Prob. F.2ECh. F - Prob. F.3ECh. F - Prob. F.4ECh. F - Prob. F.5ECh. F - Prob. F.6ECh. F - Prob. F.7ECh. F - Prob. F.8ECh. F - Prob. F.9ECh. F - Prob. F.10ECh. F - Prob. F.11ECh. F - Prob. F.12ECh. F - Prob. F.13ECh. F - Prob. F.14ECh. F - Prob. F.15ECh. F - Prob. F.16ECh. F - Prob. F.17ECh. F - Prob. F.18ECh. F - Prob. F.19ECh. F - Prob. F.20ECh. F - Prob. F.21ECh. F - Prob. F.22ECh. F - Prob. F.23ECh. F - Prob. F.24ECh. F - Prob. F.25ECh. F - Prob. F.26ECh. F - Prob. F.27ECh. F - Prob. F.28ECh. F - Prob. G.1ASTCh. F - Prob. G.1BSTCh. F - Prob. G.2ASTCh. F - Prob. G.2BSTCh. F - Prob. G.3ASTCh. F - Prob. G.3BSTCh. F - Prob. G.4ASTCh. F - Prob. G.4BSTCh. F - Prob. G.1ECh. F - Prob. G.2ECh. F - Prob. G.3ECh. F - Prob. G.4ECh. F - Prob. G.5ECh. F - Prob. G.6ECh. F - Prob. G.7ECh. F - Prob. G.8ECh. F - Prob. G.9ECh. F - Prob. G.10ECh. F - Prob. G.11ECh. F - Prob. G.12ECh. F - Prob. G.13ECh. F - Prob. G.14ECh. F - Prob. G.15ECh. F - Prob. G.16ECh. F - Prob. G.17ECh. F - Prob. G.18ECh. F - Prob. G.19ECh. F - Prob. G.20ECh. F - Prob. G.21ECh. F - Prob. G.22ECh. F - Prob. G.23ECh. F - Prob. G.24ECh. F - Prob. G.25ECh. F - Prob. G.27ECh. F - Prob. G.28ECh. F - Prob. G.30ECh. F - Prob. H.1ASTCh. F - Prob. H.1BSTCh. F - Prob. H.1ECh. F - Prob. H.2ECh. F - Prob. H.3ECh. F - Prob. H.4ECh. F - Prob. H.5ECh. F - Prob. H.6ECh. F - Prob. H.7ECh. F - Prob. H.8ECh. F - Prob. H.9ECh. F - Prob. H.10ECh. F - Prob. H.11ECh. F - Prob. H.12ECh. F - Prob. H.13ECh. F - Prob. H.14ECh. F - Prob. H.15ECh. F - Prob. H.16ECh. F - Prob. H.17ECh. F - Prob. H.18ECh. F - Prob. H.19ECh. F - Prob. H.20ECh. F - Prob. H.21ECh. F - Prob. H.22ECh. F - Prob. H.23ECh. F - Prob. H.24ECh. F - Prob. H.25ECh. F - Prob. H.26ECh. F - Prob. I.1ASTCh. F - Prob. I.1BSTCh. F - Prob. I.2ASTCh. F - Prob. I.2BSTCh. F - Prob. I.3ASTCh. F - Prob. I.3BSTCh. F - Prob. I.1ECh. F - Prob. I.2ECh. F - Prob. I.3ECh. F - Prob. I.4ECh. F - Prob. I.5ECh. F - Prob. I.6ECh. F - Prob. I.7ECh. F - Prob. I.8ECh. F - Prob. I.9ECh. F - Prob. I.10ECh. F - Prob. I.11ECh. F - Prob. I.12ECh. F - Prob. I.13ECh. F - Prob. I.14ECh. F - Prob. I.15ECh. F - Prob. I.16ECh. F - Prob. I.17ECh. F - Prob. I.18ECh. F - Prob. I.19ECh. F - Prob. I.20ECh. F - Prob. I.21ECh. F - Prob. I.22ECh. F - Prob. I.23ECh. F - Prob. I.24ECh. F - Prob. I.25ECh. F - Prob. I.26ECh. F - Prob. J.1ASTCh. F - Prob. J.1BSTCh. F - Prob. J.2ASTCh. F - Prob. J.2BSTCh. F - Prob. J.1ECh. F - Prob. J.2ECh. F - Prob. J.3ECh. F - Prob. J.4ECh. F - Prob. J.5ECh. F - Prob. J.6ECh. F - Prob. J.7ECh. F - Prob. J.8ECh. F - Prob. J.9ECh. F - Prob. J.10ECh. F - Prob. J.11ECh. F - Prob. J.12ECh. F - Prob. J.13ECh. F - Prob. J.14ECh. F - Prob. J.15ECh. F - Prob. J.16ECh. F - Prob. J.17ECh. F - Prob. J.18ECh. F - Prob. J.19ECh. F - Prob. J.20ECh. F - Prob. J.21ECh. F - Prob. J.22ECh. F - Prob. J.23ECh. F - Prob. J.24ECh. F - Prob. K.1ASTCh. F - Prob. K.1BSTCh. F - Prob. K.2ASTCh. F - Prob. K.2BSTCh. F - Prob. K.3ASTCh. F - Prob. K.3BSTCh. F - Prob. K.4ASTCh. F - Prob. K.4BSTCh. F - Prob. K.5ASTCh. F - Prob. K.5BSTCh. F - Prob. K.1ECh. F - Prob. K.2ECh. F - Prob. K.3ECh. F - Prob. K.4ECh. F - Prob. K.5ECh. F - Prob. K.6ECh. F - Prob. K.7ECh. F - Prob. K.8ECh. F - Prob. K.9ECh. F - Prob. K.10ECh. F - Prob. K.11ECh. F - Prob. K.12ECh. F - Prob. K.13ECh. F - Prob. K.14ECh. F - Prob. K.15ECh. F - Prob. K.16ECh. F - Prob. K.17ECh. F - Prob. K.18ECh. F - Prob. K.19ECh. F - Prob. K.20ECh. F - Prob. K.21ECh. F - Prob. K.22ECh. F - Prob. K.23ECh. F - Prob. K.24ECh. F - Prob. K.25ECh. F - Prob. K.26ECh. F - Prob. L.1ASTCh. F - Prob. L.1BSTCh. F - Prob. L.2ASTCh. F - Prob. L.2BSTCh. F - Prob. L.3ASTCh. F - Prob. L.3BSTCh. F - Prob. L.1ECh. F - Prob. L.2ECh. F - Prob. L.3ECh. F - Prob. L.4ECh. F - Prob. L.5ECh. F - Prob. L.6ECh. F - Prob. L.7ECh. F - Prob. L.8ECh. F - Prob. L.9ECh. F - Prob. L.10ECh. F - Prob. L.11ECh. F - Prob. L.12ECh. F - Prob. L.13ECh. F - Prob. L.14ECh. F - Prob. L.15ECh. F - Prob. L.16ECh. F - Prob. L.17ECh. F - Prob. L.18ECh. F - Prob. L.19ECh. F - Prob. L.20ECh. F - Prob. L.21ECh. F - Prob. L.22ECh. F - Prob. L.23ECh. F - Prob. L.24ECh. F - Prob. L.25ECh. F - Prob. L.29ECh. F - Prob. L.30ECh. F - Prob. L.31ECh. F - Prob. L.32ECh. F - Prob. L.33ECh. F - Prob. L.34ECh. F - Prob. L.35ECh. F - Prob. L.37ECh. F - Prob. L.38ECh. F - Prob. L.39ECh. F - Prob. L.40ECh. F - Prob. L.41ECh. F - Prob. L.42ECh. F - Prob. M.1ASTCh. F - Prob. M.1BSTCh. F - Prob. M.2ASTCh. F - Prob. M.2BSTCh. F - Prob. M.3ASTCh. F - Prob. M.3BSTCh. F - Prob. M.4ASTCh. F - Prob. M.4BSTCh. F - Prob. M.1ECh. F - Prob. M.2ECh. F - Prob. M.3ECh. F - Prob. M.4ECh. F - Prob. M.5ECh. F - Prob. M.6ECh. F - Prob. M.7ECh. F - Prob. M.8ECh. F - Prob. M.9ECh. F - Prob. M.10ECh. F - Prob. M.11ECh. F - Prob. M.12ECh. F - Prob. M.13ECh. F - Prob. M.14ECh. F - Prob. M.15ECh. F - Prob. M.16ECh. F - Prob. M.17ECh. F - Prob. M.18ECh. F - Prob. M.19ECh. F - Prob. M.20ECh. F - Prob. M.21ECh. F - Prob. M.22ECh. F - Prob. M.23ECh. F - Prob. M.25ECh. F - Prob. M.26ECh. F - Prob. M.27ECh. F - Prob. M.28E
Knowledge Booster
Similar questions
- The 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_forwardAccording 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_forward
- 1. Sometimes a reaction can fall in more than one category. Into what category (or categories) does the reaction of Ba(OH)2(aq) + H+PO4(aq) fit? acid-base and oxidation-reduction oxidation-reduction acid-base and precipitation precipitationarrow_forwardThe iron content of hemoglobin is determined by destroying the hemoglobin molecule and producing small water-soluble ions and molecules. The iron in the aqueous solution is reduced to iron(II) ion and then titrated against potassium permanganate. In the titration, iron(ll) is oxidized to iron(III) and permanganate is reduced to manganese(II) ion. A 5.00-g sample of hemoglobin requires 32.3 mL of a 0.002100 M solution of potassium permanganate. The reaction with permanganate ion is MnO4(aq)+8H+(aq)+5Fe2+(aq)Mn2+(aq)+5Fe3+(aq)+4H2O What is the mass percent of iron in hemoglobin?arrow_forwardCalculate the molarity of AgNO3 in a solution prepared by dissolving 1.44 g AgNO3 in enough water to form 1.00 L solution.arrow_forward
- Write balanced net ionic equations for the following reactions in acid solution. (a) Liquid hydrazine reacts with an aqueous solution of sodium bromate. Nitrogen gas and bromide ions are formed. (b) Solid phosphorus (P4) reacts with an aqueous solution of nitrate to form nitrogen oxide gas and dihydrogen phosphate (H2PO4-) ions. (c) Aqueous solutions of potassium sulfite and potassium permanganate react. Sulfate and manganese(II) ions are formed.arrow_forwardIf 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_forwardOne method for chemical analysis involves finding a reactant that will precipitate the species of interest, but not the other species. The mass of the precipitate is then used to determine the mass of the species of interest present in the original sample. For example, calcium ions can be precipitated from an aqueous solution by adding of sodium oxalate Na2C2O4. The balanced equation is: Ca2+(aq) + Na2C2O4 (aq) --> CaC2O4 (s) + 2 Na+ (aq) Suppose you had a solution with approximately 15.0 g of calcium ions. Show by calculation whether the addition of 15.0 g of sodium oxalate will precipitate all of the calcium ions present in the sample.arrow_forward
- Sodium hydrogen carbonate NaHCO3 , 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 HCl , which the stomach secretes to help digest food. Drinking a glass of water containing dissolved NaHCO3 neutralizes excess HCl through this reaction: HCl (aq) + NaHCO3 (aq) → NaCl (aq) + H2O (l) + CO2 (g)The CO2 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 0.081 M HCl solution. What mass of NaHCO3 would he need to ingest to neutralize this much HCl ? Be sure your answer has the correct number of significant digits.arrow_forwardSodium hydrogen carbonate NaHCO3 , 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 HCl , which the stomach secretes to help digest food. Drinking a glass of water containing dissolved NaHCO3 neutralizes excess HCl through this reaction: HCl (aq) + NaHCO3 (aq) → NaCl (aq) + H2O (l) + CO2 (g)The CO2 gas produced is what makes you burp after drinking the solution. Suppose the fluid in the stomach of a woman suffering from indigestion can be considered to be 150.mL of a 0.023 M HCl solution. What mass of NaHCO3 would she need to ingest to neutralize this much HCl ? Be sure your answer has the correct number of significant digits.arrow_forwardSodium hydrogen carbonate NaHCO3, 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 HCl, which the stomach secretes to help digest food. Drinking a glass of water containing dissolved NaHCO3 neutralizes excess HCl through this reaction: HCl(aq)+NaHCO3(aq)→NaCl(aq)+H2O(l)+CO2(g) The CO2 gas produced is what makes you burp after drinking the solution. Suppose the fluid in the stomach of a woman suffering from indigestion can be considered to be 200.mL of a 0.089M HCl solution. What mass of NaHCO3 would she need to ingest to neutralize this much HCl? Be sure your answer has the correct number of significant digits.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Chemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistry & Chemical ReactivityChemistryISBN:9781133949640Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage LearningChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage Learning
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningGeneral 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 LearningChemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher: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 & Chemical Reactivity
Chemistry
ISBN:9781133949640
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
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
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 & Chemical Reactivity
Chemistry
ISBN:9781337399074
Author:John C. Kotz, Paul M. Treichel, John Townsend, David Treichel
Publisher:Cengage Learning