CHE140Chapter 4 Homework (1).docx - Chapter 4 Homework Group 1 Catherine Coble Jasmine Decker Austin Anderson Comment Instruction Comment on who

CHE140Chapter 4 Homework (1).docx - Chapter 4 Homework...

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Chapter 4 Homework Group # 1 Catherine Coble Jasmine Decker Austin Anderson Comment: Instruction: Comment on who participated among the members of the group Include detailed calculations on quantitative problems to obtain partial/full credit 4.2 Match the key terms with the descriptions provided. (a)the mass of 1 mol of any substance, in units of grams per mole molar mass (b) an expression of the percent of the total mass due to each element in a compound percent composition by mass (c)the substance doing the dissolving; usually, that component of a solution that is present in the larger amount solvent (d) the relative amounts of solute and solvent in a solution concentration (e) a solution that contains a relatively small concentration of solute dilute solution (f) the process of adding more solvent to give a solution of lesser concentration dilution 4.6 In addition to its more well-known use, a common erectile dysfunction drug has been shown to be effective in treating a rare pulmonary disorder. A 20.0-mg dose of the drug contains 3.54 mg of nitrogen. What is the percent nitrogen in this drug? (3.54 mg/20.0 mg ) x 100 = 17.7 % 4.22 How many oxygen atoms are in 0.2 mol of SO 2 ? 0.2 mol SO 2 x 6.022 x 10 23 atoms/mol = 1 x 10 23 SO 2 atoms 4.28 Calculate the molar mass of each of the following compounds.
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(a)K 2 SO 4 39.098x 2 + 32.066+ 16.00 x 4 = 174.3 g/mol (b)NiCl 2 6H 2 O NiCl 2 = 129.59 H 2 O = 18.02 18.02 X 6 = 108.12 129.59 X 108.12 = 14011.27 (c)C 2 H 4 Cl 2 12.01 x 2 = 24.02 1.01 x 4 = 4.04 35.45 x 2 = 70.90 24.02+4.04+70.90 = 98.96 (d)Mg(NO 3 ) 2 Mg = 24.30 N = 14 O 3 = (15.99 x 3 ) =47.97 (14+47.97) x 2 = 123.94 123.94+24.30 = 148.24 4.32 What is a mole? Why do chemists need to use the concept? A mole is the amount of substance that contains the same number of particles (atoms, molecules or formula units) as there are atoms in exactly 12 g of 12 C, which is Avogadros # 6.022 X 10 23 .
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