Complete the table below for calculatingthe molar mass of the ionic compoundchromium(II) sulfide.FormulaMolar mass of ionNumber ofMass of ion in one mole ofionschromium(II) sulfideCationCr2+g/molmolAnionS2.g/molmolgMolar mass chromium(II) sulfide =g/mol

Complete the table below for calculating the molar mass of the ionic compound chromium(II) sulfide. Number of Mass of ion in one mole of Formula Molar mass of ion ions chromium(II) sulfide Cation Cr2+ g g/mol mol Anion s2- g g/mol mol Molar mass chromium(II) sulfide = g/mol

Complete the table below for calculating the molar mass of the ionic compound chromium(II) sulfide. Number of Mass of ion in one mole of Formula Molar mass of ion ions chromium(II) sulfide Cation Cr2+ g g/mol mol Anion s2- g g/mol mol Molar mass chromium(II) sulfide = g/mol

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter19: The Chemistry Of The Main-group Elements

Section: Chapter Questions

Problem 72QRT

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Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) A. What is the %m/m of the solution? B. What is the molarity of the solution? C. What is the mole fraction of the solvent?
A well is punched into a slab of gelatin and a solution ofequimolar amounts of NaCl and KCl is placed in it. After anhour the concentration of Na+ and K+ present in the well ismeasured. Which ion, Na+ or K+, would be present ingreater abundance in the well? Explain your answer.
Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) 1. What is the % m/m of the solution? 2. What is the mole fraction of the solvent? 3. What is the molarity of the solution? Based from the calculated molarity of the solution in #3, you are to prepare 250 mL of the said molar concentration. If you have an available 2M stock solution of Palmitic acid, How many mL of the 2M palmitic acid stock solution will you need? and; 4. How many mL of solvent do you need to add to the amount gotten from the 2M Palmitic acid stock solution in order to prepare 250 mL total volume of your desired solution?
Calculate the percentage mass composition of the following substancea. Ethanol ( C2H6O )b. Gypsum (CaSO4 . 2H2Oc. Pyrite (FeS2)With solution.
What is the mass percentage of C in codeine, C₁₈H₂₁NO₃? Provide an answer to two decimal places.
Acording to the folowing chemical equation, what mass of zinc phosphate (MM= 386.11) will be obtained when 9.05 x 1022 atoms of zinc react with 3.74mL phosphoric acid (specific gravity= 1.88, MM=97.99). 3 Zn(s) + 2 H3PO4(l) -> 3 H2(g) + Zn3(PO4)2 (s)
The phosphorus in a 0.2091-g sample was precipitated as the slightly soluble (NH4)3PO4 • 12 MoO3. This precipitate was filtered, washed, and then redissolved in acid. Treatment of the resulting solution with an excess of Pb2+ resulted in the formation of 0.2922 g of PbMoO4. Express the results of this analysis in terms of percent P2O5.
Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) If the calculated molarity is 0.51M, you are to prepare 250 mL of the said molar concentration. If you have an available 2M stock solution of Palmitic acid, How many mL of the 2M palmitic acid stock solution will you need? and; How many mL of solvent do you need to add to the amount gotten from the 2M Palmitic acid stock solution in order to prepare 250 mL total volume of your desired solution?
Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) A. What is the molarity of the solution? B. How many mL of the 2M palmitic acid stock solution will you need? Final answer must be rounded off to 2 decimal places, and shall NOT have any units.
You are the public health officer in the water treatment facility for a city of 50,000 people. A concentration of 1.0 ppm fluride in the drinking water is sufficient for the purpoise of helping to prevent tooth decay. The compound normally chosen for fluoridation is the same as is found in some toothpastes, sodium fluoride (NaF). Calculate how many kilograms of sodium fluride you will need to purchase in order to fluoridate the city's water supply for one year, based upong your estimate that the average daily consumption of water is 150 gallons per person You will find the following information helping in solving the problem below. Sodium fluride, NaF, is 45.0% fluride by mass. 1 ppm means "one part per million" by mass, e.g. 1 g of fluoride peron million grams of water 1 gallon = 3.79 L 1 year = 365 days 1 ton = 2000lb 1 lb = 453.6 g Density of water = 1.0 g/mL
Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) You are to prepare 250 mL of the said molar concentration. If you have an available 2M stock solution of Palmitic acid. How many mL of solvent do you need to add to the amount gotten from the 2M Palmitic acid stock solution in order to prepare 250 mL total volume of your desired solution?
Consider Palmitic acid C16H32O2, a common fatty acid used in the manufacture of soap. A solution of palmitic acid is prepared by mixing 112 g palmitic acid with 725 mL of benzene C6H6. The density of the resulting solution is 0.902 g/mL. Palmitic acid (molar mass = 256 g/mol and density = 0.852 g/mL); Benzene (molar mass = 78 g/mol and density = 0.879 g/mL) You are to prepare 250 mL of the said molar concentration. If you have an available 2M stock solution of Palmitic acid. How many mL of the 2M palmitic acid stock solution will you need?