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(d) the grams of Na2SO4 that can be obtained from 38 mL of 0.060 M NaOH g Na2SO4 the tolerance is +/-2% (e) the volume of 0.240 M H,SO4 needed to react with 34.2 mL of 0.779 M NaOH mL H2SO4 the tolerance is +/-2% (f) the molarity (M) of the NaOH solution when 63.23 mL react with 48.47 mL of 0.273 M H2SO4 M NaOH the tolerance is +/-2%

(d) the grams of Na2SO4 that can be obtained from 38 mL of 0.060 M NaOH g Na2SO4 the tolerance is +/-2% (e) the volume of 0.240 M H,SO4 needed to react with 34.2 mL of 0.779 M NaOH mL H2SO4 the tolerance is +/-2% (f) the molarity (M) of the NaOH solution when 63.23 mL react with 48.47 mL of 0.273 M H2SO4 M NaOH the tolerance is +/-2%

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
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Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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In this chemical equation, we describe a neutralization reaction between sodium hydroxide (NaOH) and sulfuric acid (H₂SO₄). This reaction produces sodium sulfate (Na₂SO₄) and water (H₂O). The balanced chemical equation is: **2 NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2 H₂O(l)** ### Explanation: 1. **Reactants:** - **2 NaOH(aq):** Sodium hydroxide in an aqueous state (dissolved in water). Here we have two moles of sodium hydroxide. - **H₂SO₄(aq):** Sulfuric acid in an aqueous state (dissolved in water). 2. **Products:** - **Na₂SO₄(aq):** Sodium sulfate in an aqueous state (dissolved in water). - **2 H₂O(l):** Water in a liquid state. This reaction produces two moles of water. ### Process: In this reaction, the hydroxide ions (OH⁻) from NaOH and the hydrogen ions (H⁺) from H₂SO₄ combine to form water (H₂O). The sodium ions (Na⁺) from NaOH and the sulfate ions (SO₄²⁻) from H₂SO₄ combine to form the salt sodium sulfate (Na₂SO₄). ### Graphs and Diagrams: There are no graphs or diagrams in the provided image. However, this reaction can be depicted in terms of the ionic equation which simplifies to: **2 OH⁻ (aq) + 2 H⁺ (aq) → 2 H₂O (l)** This showcases the neutralization aspect where hydroxide ions and hydrogen ions react to form water. ### Application: This reaction is commonly used to neutralize acidic solutions in various chemical processes. Understanding such neutralization reactions is fundamental in chemistry, especially for tasks involving pH adjustments and titration experiments.
Transcribed Image Text:In this chemical equation, we describe a neutralization reaction between sodium hydroxide (NaOH) and sulfuric acid (H₂SO₄). This reaction produces sodium sulfate (Na₂SO₄) and water (H₂O). The balanced chemical equation is: **2 NaOH(aq) + H₂SO₄(aq) → Na₂SO₄(aq) + 2 H₂O(l)** ### Explanation: 1. **Reactants:** - **2 NaOH(aq):** Sodium hydroxide in an aqueous state (dissolved in water). Here we have two moles of sodium hydroxide. - **H₂SO₄(aq):** Sulfuric acid in an aqueous state (dissolved in water). 2. **Products:** - **Na₂SO₄(aq):** Sodium sulfate in an aqueous state (dissolved in water). - **2 H₂O(l):** Water in a liquid state. This reaction produces two moles of water. ### Process: In this reaction, the hydroxide ions (OH⁻) from NaOH and the hydrogen ions (H⁺) from H₂SO₄ combine to form water (H₂O). The sodium ions (Na⁺) from NaOH and the sulfate ions (SO₄²⁻) from H₂SO₄ combine to form the salt sodium sulfate (Na₂SO₄). ### Graphs and Diagrams: There are no graphs or diagrams in the provided image. However, this reaction can be depicted in terms of the ionic equation which simplifies to: **2 OH⁻ (aq) + 2 H⁺ (aq) → 2 H₂O (l)** This showcases the neutralization aspect where hydroxide ions and hydrogen ions react to form water. ### Application: This reaction is commonly used to neutralize acidic solutions in various chemical processes. Understanding such neutralization reactions is fundamental in chemistry, especially for tasks involving pH adjustments and titration experiments.
### Chemistry Practice Problems: Titration Calculations #### Problem (d): Calculation of Grams of Sodium Sulfate Calculate the grams of Na₂SO₄ that can be obtained from 38 mL of 0.060 M NaOH. **Enter your answer**: [________] g Na₂SO₄ *The tolerance is +/- 2%* #### Problem (e): Volume of Sulfuric Acid Needed Calculate the volume of 0.240 M H₂SO₄ needed to react with 34.2 mL of 0.779 M NaOH. **Enter your answer**: [________] mL H₂SO₄ *The tolerance is +/- 2%* #### Problem (f): Molarity of Sodium Hydroxide Solution Determine the molarity (M) of the NaOH solution when 63.23 mL react with 48.47 mL of 0.273 M H₂SO₄. **Enter your answer**: [________] M NaOH *The tolerance is +/- 2%* --- In this section, we present a series of titration problems designed to enhance your understanding of volumetric analysis and stoichiometric relationships. Each problem’s answer must be within a 2% tolerance of the exact value. Ensure your calculations are precise and double-check your work for accuracy.
Transcribed Image Text:### Chemistry Practice Problems: Titration Calculations #### Problem (d): Calculation of Grams of Sodium Sulfate Calculate the grams of Na₂SO₄ that can be obtained from 38 mL of 0.060 M NaOH. **Enter your answer**: [________] g Na₂SO₄ *The tolerance is +/- 2%* #### Problem (e): Volume of Sulfuric Acid Needed Calculate the volume of 0.240 M H₂SO₄ needed to react with 34.2 mL of 0.779 M NaOH. **Enter your answer**: [________] mL H₂SO₄ *The tolerance is +/- 2%* #### Problem (f): Molarity of Sodium Hydroxide Solution Determine the molarity (M) of the NaOH solution when 63.23 mL react with 48.47 mL of 0.273 M H₂SO₄. **Enter your answer**: [________] M NaOH *The tolerance is +/- 2%* --- In this section, we present a series of titration problems designed to enhance your understanding of volumetric analysis and stoichiometric relationships. Each problem’s answer must be within a 2% tolerance of the exact value. Ensure your calculations are precise and double-check your work for accuracy.
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