Explanation To calculate the density of the solution Record the given data Volume of H 3 PO 4 = 100 .0 mL Mass of H 3 PO 4 = 10 .0 g Density of water = 1 .00g/cm 3 100 mL of water has a mass of 100 g since the density of water is 1 .00g/cm 3 Density of solution = 10 .0 g H 3 PO 4 +100 .0 g H 2 O 104 mL = 1 .06 g cm -3 Density of solution = 1 .06 g cm -3 To calculate the mole of H 3 PO 4 and Water: Record the given data Mass of H 3 PO 4 = 10 .0 g Mass of Water = 100 .0 g Molar mass of H 3 PO 4 = 97 .99 g Molar mass of Water = 18 .01 g Moles of H 3 PO 4 = 10 .0 g× 1 mol 97 .99 g = 0 .102 mol Moles of Water = 100 .0 g× 1 mol 18 .02 g = 5 .55 mol Moles of H 3 PO 4 = 0

2nd Edition

ISBN: 9781305079243

Author: Steven S. Zumdahl, Susan A. Zumdahl

Publisher: Cengage Learning

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Chapter 10, Problem 29E

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The density, molality, molarity and mole fraction of solution has to be calculated.

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Chapter 10, Problem 28Q

Chapter 10, Problem 30E

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The SN 1 mechanism starts with the rate-determining step which is the dissociation of the alkyl halide into a carbocation and a halide ion. The next step is the rapid reaction of the carbocation intermediate with the nucleophile; this step completes the nucleophilic substitution stage. The step that follows the nucleophilic substitution is a fast acid-base reaction. The nucleophile now acts as a base to remove the proton from the oxonium ion from the previous step, to give the observed product. Draw a curved arrow mechanism for the reaction, adding steps as necessary. Be sure to include all nonzero formal charges. Cl: Add/Remove step G Click and drag to start drawing a structure.

Chapter 10 Solutions

Chemistry: An Atoms First Approach

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The density, molality, molarity and mole fraction of solution has to be calculated.

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Interpretation: The density, molality, molarity and mole fraction of solution has to be calculated.

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Chapter 10 Solutions

Interpretation:

The density, molality, molarity and mole fraction of solution has to be calculated.