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Answer questions 1 c, d and e only!

Answer questions 1 c, d and e only!

Chemistry
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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Answer questions 1 c, d and e only!
Certainly! Here is a transcription suitable for an educational website: --- **(d)** Using the mmoles of salt produced and the total volume in mL of solution (from the mixing of the acid and base solutions), calculate the molarity of the salt solution produced by this reaction. **(e)** Consider this example problem: If 100. mL of 0.100 M HCl solution is mixed with 100. mL of 0.100 M NaOH, what is the molarity of the resulting salt solution? (Assuming the volumes are additive and ignore the change in H₂O, which is negligible). \[ \begin{array}{c} \text{Rxn ratio:} \\ \text{HCl}_{(\text{aq})} + \text{NaOH}_{(\text{aq})} \rightarrow \text{NaCl}_{(\text{aq})} + \text{H}_2\text{O}_{(\ell)} \\ 1 \, \text{mol} + 1 \, \text{mol} \rightarrow \end{array} \] \[ \begin{array}{ccc} \text{Mols @ Start:} & \quad & \text{100 mL} \times \left( \frac{0.100 \, \text{mol HCl}}{1 \, \text{L}} \right) = 10 \, \text{mmol HCl} \\ \quad & \quad & \text{100 mL} \times \left( \frac{0.100 \, \text{mol NaOH}}{1 \, \text{L}} \right) = 10 \, \text{mmol NaOH} \\ \text{Change:} & -10 \, \text{mmol} & -10 \, \text{mmol} \\ \text{After rxn:} & 0 \, \text{mmol} & 0 \, \text{mmol} \\ \quad & \quad & 10 \, \text{mmol} \\ \end{array} \] The resulting solution contains 10 mmol NaCl in a total volume of 200 mL. The molarity of NaCl solution is: \[ \frac{10 \, \text{mm
Transcribed Image Text:Certainly! Here is a transcription suitable for an educational website: --- **(d)** Using the mmoles of salt produced and the total volume in mL of solution (from the mixing of the acid and base solutions), calculate the molarity of the salt solution produced by this reaction. **(e)** Consider this example problem: If 100. mL of 0.100 M HCl solution is mixed with 100. mL of 0.100 M NaOH, what is the molarity of the resulting salt solution? (Assuming the volumes are additive and ignore the change in H₂O, which is negligible). \[ \begin{array}{c} \text{Rxn ratio:} \\ \text{HCl}_{(\text{aq})} + \text{NaOH}_{(\text{aq})} \rightarrow \text{NaCl}_{(\text{aq})} + \text{H}_2\text{O}_{(\ell)} \\ 1 \, \text{mol} + 1 \, \text{mol} \rightarrow \end{array} \] \[ \begin{array}{ccc} \text{Mols @ Start:} & \quad & \text{100 mL} \times \left( \frac{0.100 \, \text{mol HCl}}{1 \, \text{L}} \right) = 10 \, \text{mmol HCl} \\ \quad & \quad & \text{100 mL} \times \left( \frac{0.100 \, \text{mol NaOH}}{1 \, \text{L}} \right) = 10 \, \text{mmol NaOH} \\ \text{Change:} & -10 \, \text{mmol} & -10 \, \text{mmol} \\ \text{After rxn:} & 0 \, \text{mmol} & 0 \, \text{mmol} \\ \quad & \quad & 10 \, \text{mmol} \\ \end{array} \] The resulting solution contains 10 mmol NaCl in a total volume of 200 mL. The molarity of NaCl solution is: \[ \frac{10 \, \text{mm
Given the balanced equation: \[ \text{HBr}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{NaBr}_{(aq)} + \text{H}_2\text{O}_{(l)} \] and remembering that Molarity = moles/liter OR mmoles/mL. **(i) Calculations:** (a) Calculate the number of mmoles of HBr in 100.0 mL of 0.250 M HBr. \[ \text{mmoles of HBr} = \text{molarity} \times \text{volume} \] \[ = 0.250 \times 100 = 25 \text{ mmoles HBr} \] (b) Calculate the number of mmoles of NaOH in 100.0 mL of 0.250 M NaOH. \[ \text{mmoles of NaOH} = \text{molarity} \times \text{volume} \] \[ = 0.250 \times 100 = 25 \text{ mmoles NaOH} \] (c) When these two solutions are mixed, the acid and base should neutralize one another exactly. This means that all of the acid and base are completely used up; either one could be considered a "limiting reactant". Starting with the moles of either the acid or base, calculate the number of mmoles of salt produced by the reaction. (d) Using the mmoles of salt produced and the total volume in mL of solution (from the mixing of the acid and base solutions), calculate the molarity of the salt solution produced by this reaction. **(e) Consider this example problem:** If 100. mL of 0.100 M HCl solution is mixed with 100. mL of 0.100 M NaOH, what is the molarity of the resulting salt solution? (Assuming the volumes are additive and ignore the change in \( \text{H}_2\text{O} \), which is negligible). \[ \text{HCl}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{NaCl}_{(aq)} + \text{H}_2\text{O}_{(l)} \] (Note: Additional calculations and steps may follow related to this problem for a comprehensive understanding.)
Transcribed Image Text:Given the balanced equation: \[ \text{HBr}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{NaBr}_{(aq)} + \text{H}_2\text{O}_{(l)} \] and remembering that Molarity = moles/liter OR mmoles/mL. **(i) Calculations:** (a) Calculate the number of mmoles of HBr in 100.0 mL of 0.250 M HBr. \[ \text{mmoles of HBr} = \text{molarity} \times \text{volume} \] \[ = 0.250 \times 100 = 25 \text{ mmoles HBr} \] (b) Calculate the number of mmoles of NaOH in 100.0 mL of 0.250 M NaOH. \[ \text{mmoles of NaOH} = \text{molarity} \times \text{volume} \] \[ = 0.250 \times 100 = 25 \text{ mmoles NaOH} \] (c) When these two solutions are mixed, the acid and base should neutralize one another exactly. This means that all of the acid and base are completely used up; either one could be considered a "limiting reactant". Starting with the moles of either the acid or base, calculate the number of mmoles of salt produced by the reaction. (d) Using the mmoles of salt produced and the total volume in mL of solution (from the mixing of the acid and base solutions), calculate the molarity of the salt solution produced by this reaction. **(e) Consider this example problem:** If 100. mL of 0.100 M HCl solution is mixed with 100. mL of 0.100 M NaOH, what is the molarity of the resulting salt solution? (Assuming the volumes are additive and ignore the change in \( \text{H}_2\text{O} \), which is negligible). \[ \text{HCl}_{(aq)} + \text{NaOH}_{(aq)} \rightarrow \text{NaCl}_{(aq)} + \text{H}_2\text{O}_{(l)} \] (Note: Additional calculations and steps may follow related to this problem for a comprehensive understanding.)
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