Suppose that a fully charged lead-acid battery contains 1.50 L of 5.00 M H2SO4. What will be the concentration of H2SO4 in the battery after 3.50 A of current is drawn from the battery for 5.50 hours ?
Added by Danielle M.
Step 1
Given: Current = 3.50 A Time = 5.50 hours = 5.50 * 60 * 60 seconds Molar mass of H2SO4 = 98.08 g/mol Number of moles = (Current * Time) / (96500 C/mol) Number of moles = (3.50 A * 5.50 * 60 * 60 s) / (96500 C/mol) Number of moles = 0.718 mol Show more…
Show all steps
Your feedback will help us improve your experience
Maryada Jain and 98 other Chemistry 101 educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
Suppose that a fully charged lead-acid battery contains 1.50 L of 5.00 M H2SO4. What will be the concentration of H2SO4 in the battery after 3.60 A of current is drawn from the battery for 5.50 hours ? Express your answer with the appropriate units.
Shaiju T.
Suppose that a fully charged lead-acid battery contains 1.50 L of 5.00 M H2SO4. What will be the concentration of H2SO4 in the battery after 3.10 A of current is drawn from the battery for 5.50 hours? Express your answer with the appropriate units.
Madhur L.
Sulfuric acid is needed in a lead storage battery. A 25.00 -mL sample of "battery acid" taken from a functioning battery requires $53.7 \mathrm{~mL}$ of $4.552 \mathrm{M} \mathrm{NaOH}$ for complete neutralization. What is the molarity of the battery acid?
Recommended Textbooks
Chemistry: Structure and Properties
Chemistry The Central Science
Chemistry
Transcript
Watch the video solution with this free unlock.
EMAIL
PASSWORD