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
Concept explainers
Question
Chapter F, Problem F.1E
(a)
Interpretation Introduction
Interpretation:
Molecular formula of citral has to be written.
(b)
Interpretation Introduction
Interpretation:
Mass percentage of carbon, hydrogen, and oxygen in citral has to be calculated.
Concept Introduction:
Mass percentage composition of the individual elements can be calculated if the formula of the compound containing the elements is known. Mass percentage of an element is calculated using the formula given below;
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
(a) A compound MX2 is formed from the M²+ and X ions. The compound contains 86.8% of X by
mass. In a reaction, a 0.96 g M²+ ion reacts completely with 0.079 mol X¯ ion. Determine the
identity of M²+ and X.
(b) In a reaction, the mixture of 13.4 g of calcium fluoride (CaF₂) and 11.1 g of sulfuric acid
(H₂SO4) was heated to give off gaseous hydrogen fluoride (HF). The reaction is as shown:
CaF2 (s) + H₂SO4 (1) → 2HF (g) + CaSO4 (s)
Determine the maximum amount (grams) of hydrogen fluoride that can be obtained.
(c) A compound containing barium (Ba) and oxygen, weighed 2.018 g, was dissolved to produce
barium ions in solution. The barium ions are then separated from the solution by precipitating
method. The precipitate formed was 3.018 g of BaCrO4. Determine the chemical formula of
the compound.
(d) For the following reaction, determine the mass of sulfur would be needed to obtain 80.0 g of
CS₂. The percent yield of the reaction is consistently 92%.
CH4 (g) + 4S (g) → CS2 (g) + 2H₂S (g)
(a) Calculate the molar mass of ammonium nitrate, NH4NO3. (b) What is the mass percent of nitrogen in ammonium nitrate? (c) How many kilograms of nitrogen are present in a 10.0-lb bag of ammonium nitrate? (d) How many kilograms of ammonium nitrate contain 1.00 lb of nitrogen? (1 lb = 453.6 g)
A compound is composed of carbon, hydrogen, nitrogen and oxygen. When a 1.500-g sample of the compound is completely combusted, it yields 1.476 g of CO2and 0.605 g of H2O. In a separate analysis to determine nitrogen, 1.500 g of the compound is found to produce 0.313 g of N2. (a) Calculate the mass percent of each element in the compound. (b) Determine the empirical formula of the compound. (c) If the compound has a molar mass of 134 g/mol, what is the molecular formula?
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
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
- Chlorine exists mainly as two isotopes, 37Cl and 33Cl. Which is more abundant? How do you know?arrow_forwardPhosphoric acid, H3PO4, can be prepared by the reaction of phosphorus(V) oxide, P4O10, with water. 14P4O10(s)+32H2O(l)H3PO4(aq);H=96.2kJ What is H for the reaction involving 1 mol of P4O10? P4O10(s)+6H2O(l)4H3PO4(aq)arrow_forwardYou take 1.00 g of an aspirin tablet (a compound consisting solely of carbon, hydrogen, and oxygen), burn it in air, and collect 2.20 g CO2 and 0.400 g H2O. You know that the molar mass of aspirin is between 170 and 190 g/mol. Reacting 1 mole of salicylic acid with I mole of acetic anhydride (C4H6O3) gives you 1 mole of aspirin and 1 mole of acetic acid (C2H4O2). Use this information to determine the molecular formula of salicylic acid.arrow_forward
- Tungsten (W) metal, which is used to make incandescent bulb filaments, is produced by the reaction WO3+3H23H2O+W How many grams of H2 are needed to produce 1.00 g of W?arrow_forwardNitrogen fixation in the root nodules of peas and other legumes occurs with a reaction involving a molybdenum-containing enzyme named nitrogenase. This enzyme contains two Mo atoms per molecule and is 0.0872% Mo by mass. Calculate the molar mass of the enzyme.arrow_forwardThe active ingredient in Pepto-Bismo® (an over- the-counter remedy for an upset stomach) is bismuth sub-salicylate, C7H5BiO4. Analysis of a 1.7500-g sample of Pepto-Bismol yields 346 mg of bismuth. What percent by mass is bismuth subsalicylate in the sample? (Assume that there are no other bismuth-containing compounds in Pepto-Bismol.)arrow_forward
- 4-102 Aspartame, an artificial sweetener used as a sugar substitute in some foods and beverages, has the molecular formula C14H18N2O5. (a) How many mg of aspartame are present in 3.72 × 1026 molecules of aspartame? (b) Imagine you obtain 25.0 mL of aspartame, which is known to have a density of 1.35 g/mL. How many molecules of aspartame are present in this volume? (c) How many hydrogen atoms are present in 1.00 mg of aspartame? (d) Complete the skeletal structure of aspartame, where all the bonded atoms are shown but double bonds, triple bonds, and/or lone pairs are missing. (e) Identify the various types of geometries present in each central atom of aspartame using VSEPR theory. (f) Determine the various relative bond angles associated with each central atom of aspartame using VSEPR theory. (g) What is the most polar bond in aspartame? (h) Would you predict aspartame to be polar or nonpolar? (i) Is aspartame expected to possess resonance? Explain why or why not. (j) Consider the combustion of aspartame, which results in formation of NO2(g) as well as other expected products. Write a balanced chemical equation for this reaction. (k) Calculate the weight of C02(g) that can be prepared from 1.62 g of aspartame mixed with 2.11 g of oxygen gas.arrow_forwardHydrogen cyanide is produced industrially from the reaction of gaseous ammonia, oxygen, and methane: 2NH3(g)+3O2(g)+2CH4(g)2HCN(g)+6H2O(g) If 5.00 103 kg each of NH3, O2, and CH4 are reacted, what mass of HCN and of H2O will be produced, assuming 100% yield?arrow_forwardWhite phosphorus, P4, is prepared by fusing calcium phosphate, Ca3(PO4)2, with carbon, C, and sand, SiO2, in an electric furnace. 2Ca3(PO4)2(s)+6SiO2(s)+10C(s)P4(g)+6CaSiO3(l)+10CO(g) How many grams of calcium phosphate are required to give 30.0 g of phosphorus?arrow_forward
- Chalky, white crystals in mineral collections are often labeled borax, which has the molecular formula Na2B4O7 10H2O, when actually they are partially dehydrated samples with the molecular formula Na2B4O7 5H2O, which is more stable under the storage conditions. Real crystals of borax are colorless and transparent. (a) Calculate the percent mass that the mineral has lost when it partially dehydrates. (b) Is the percent boron by mass the same in both compounds?arrow_forwardSilicon is produced for the chemical and electronics industries by the following reactions. Give the balanced equation for each reaction. a. SiO2(s)+C(s)arefurnaceElectricSi(s)+CO(g) b. Liquid silicon tetrachloride is reacted with very pure solid magnesium, producing solid silicon and solid magnesium chloride. c. Na2SiF6(s) + Na(s) Si(s) + NaF(s)arrow_forwardA mixture of Fe2O3, and FeO was found to contain 72.00% Fe by mass. What is the mass of Fe2O3 in 0.750 g of this mixture?arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- 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 PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage LearningChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistry: An Atoms First ApproachChemistryISBN:9781305079243Author:Steven S. Zumdahl, Susan A. ZumdahlPublisher: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
Chemistry: Principles and Practice
Chemistry
ISBN:9780534420123
Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher:Cengage Learning
Chemistry
Chemistry
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Cengage Learning
Chemistry: An Atoms First Approach
Chemistry
ISBN:9781305079243
Author:Steven S. Zumdahl, Susan A. Zumdahl
Publisher:Cengage Learning
Step by Step Stoichiometry Practice Problems | How to Pass ChemistryMole Conversions Made Easy: How to Convert Between Grams and Moles; Author: Ketzbook;https://www.youtube.com/watch?v=b2raanVWU6c;License: Standard YouTube License, CC-BY