Homework Help is Here – Start Your Trial Now!
Science
Chemistry
What volume in milliliters of 0.0200 M Ca(OH), is required to neutralize 75.0 mL of 0.0300 M HCI? mL 1 4 C 7 8 +/- х 100 2. 5

What volume in milliliters of 0.0200 M Ca(OH), is required to neutralize 75.0 mL of 0.0300 M HCI? mL 1 4 C 7 8 +/- х 100 2. 5

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...
See similar textbooks
icon
Related questions
Question
**Question:** What volume in milliliters of 0.0200 M Ca(OH)₂ is required to neutralize 75.0 mL of 0.0300 M HCl? **Response Area:** ____ mL **Calculator Interface:** - Numeric keypad with numbers 1 to 9 - Decimal point (.) - Clear button (C) - Positive/negative button (+/-) - Multiplication by 10 button (x 10) - Addition button (+) - Submit button **Description:** This question involves calculating the volume of a calcium hydroxide solution needed to neutralize a given volume of hydrochloric acid. The user is expected to perform stoichiometric calculations to determine the answer. A calculator interface is provided for inputting numerical values and performing calculations.
Transcribed Image Text:**Question:** What volume in milliliters of 0.0200 M Ca(OH)₂ is required to neutralize 75.0 mL of 0.0300 M HCl? **Response Area:** ____ mL **Calculator Interface:** - Numeric keypad with numbers 1 to 9 - Decimal point (.) - Clear button (C) - Positive/negative button (+/-) - Multiplication by 10 button (x 10) - Addition button (+) - Submit button **Description:** This question involves calculating the volume of a calcium hydroxide solution needed to neutralize a given volume of hydrochloric acid. The user is expected to perform stoichiometric calculations to determine the answer. A calculator interface is provided for inputting numerical values and performing calculations.
**Neutralization Reaction Problem** **Question:** What volume (in milliliters) of 0.200 M HNO₃ are required to neutralize 90.0 mL of 0.180 M LiOH? **Help Guide:** This is a stoichiometry problem involving a neutralization reaction between nitric acid (HNO₃) and lithium hydroxide (LiOH). **Steps to Solve:** 1. **Write the Balanced Chemical Equation:** - HNO₃ + LiOH → LiNO₃ + H₂O 2. **Determine Moles of LiOH:** - Molarity (M) = Moles/Volume (L) - Moles of LiOH = 0.180 mol/L * 0.0900 L = 0.0162 mol 3. **Use Stoichiometry to Find Moles of HNO₃:** - The mole ratio of HNO₃ to LiOH is 1:1 from the balanced equation. - Moles of HNO₃ required = Moles of LiOH = 0.0162 mol 4. **Calculate the Volume of HNO₃ Needed:** - Volume = Moles/Molarity - Volume of HNO₃ = 0.0162 mol / 0.200 mol/L = 0.081 L = 81.0 mL **Calculator Interface:** The question is accompanied by a built-in digital calculator interface. You can use this to input numbers and perform calculations to verify your result: - Numerical keypad - Function keys: clear (C), delete (x), and operations like addition, multiplication, etc. **Conclusion:** To neutralize 90.0 mL of 0.180 M LiOH, 81.0 mL of 0.200 M HNO₃ is required.
Transcribed Image Text:**Neutralization Reaction Problem** **Question:** What volume (in milliliters) of 0.200 M HNO₃ are required to neutralize 90.0 mL of 0.180 M LiOH? **Help Guide:** This is a stoichiometry problem involving a neutralization reaction between nitric acid (HNO₃) and lithium hydroxide (LiOH). **Steps to Solve:** 1. **Write the Balanced Chemical Equation:** - HNO₃ + LiOH → LiNO₃ + H₂O 2. **Determine Moles of LiOH:** - Molarity (M) = Moles/Volume (L) - Moles of LiOH = 0.180 mol/L * 0.0900 L = 0.0162 mol 3. **Use Stoichiometry to Find Moles of HNO₃:** - The mole ratio of HNO₃ to LiOH is 1:1 from the balanced equation. - Moles of HNO₃ required = Moles of LiOH = 0.0162 mol 4. **Calculate the Volume of HNO₃ Needed:** - Volume = Moles/Molarity - Volume of HNO₃ = 0.0162 mol / 0.200 mol/L = 0.081 L = 81.0 mL **Calculator Interface:** The question is accompanied by a built-in digital calculator interface. You can use this to input numbers and perform calculations to verify your result: - Numerical keypad - Function keys: clear (C), delete (x), and operations like addition, multiplication, etc. **Conclusion:** To neutralize 90.0 mL of 0.180 M LiOH, 81.0 mL of 0.200 M HNO₃ is required.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 2 steps with 1 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Knowledge Booster
Concentration Terms
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.
Similar questions
Recommended textbooks for you
Chemistry
Chemistry
Chemistry
ISBN:
9781305957404
Author:
Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:
Cengage Learning
Chemistry
Chemistry
Chemistry
ISBN:
9781259911156
Author:
Raymond Chang Dr., Jason Overby Professor
Publisher:
McGraw-Hill Education
Principles of Instrumental Analysis
Principles of Instrumental Analysis
Chemistry
ISBN:
9781305577213
Author:
Douglas A. Skoog, F. James Holler, Stanley R. Crouch
Publisher:
Cengage Learning
Organic Chemistry
Organic Chemistry
Chemistry
ISBN:
9780078021558
Author:
Janice Gorzynski Smith Dr.
Publisher:
McGraw-Hill Education
Chemistry: Principles and Reactions
Chemistry: Principles and Reactions
Chemistry
ISBN:
9781305079373
Author:
William L. Masterton, Cecile N. Hurley
Publisher:
Cengage Learning
Elementary Principles of Chemical Processes, Bind…
Elementary Principles of Chemical Processes, Bind…
Chemistry
ISBN:
9781118431221
Author:
Richard M. Felder, Ronald W. Rousseau, Lisa G. Bullard
Publisher:
WILEY
SEE MORE TEXTBOOKS