Suppose that a fully charged lead¨Cacid battery contains 1.16 L of 5.00 M H2SO4. What will be the concentration of HSO4- in the battery after 2.83 A of current is drawn from the battery for 6.3 h?

Pb(s) + PbO2(s) + 2H+(aq) + 2HSO4−(aq) ¡ú 2PbSO4(s) + 2H2O(l)

Enter your answer with 3 significant digits.

To find the concentration of HSO4- in the battery after the given time and current, we need to use the equation relating the moles of HSO4- to the current and time:

moles of HSO4- = (current × time) / (2 × F)

Where:
- Current is given as 2.83 A
- Time is given as 6.3 h
- F is the Faraday's constant, which is 96,485 C/mol e-

First, we need to calculate the number of moles of HSO4- present initially in the battery.

To do this, we need to find the number of moles of H2SO4 initially:

moles of H2SO4 = volume × concentration

Where:
- volume is given as 1.16 L
- concentration is given as 5.00 M

moles of H2SO4 = 1.16 L × 5.00 mol/L = 5.80 mol

Since the stoichiometry of the reaction tells us that 2 moles of HSO4- are produced for every 2 moles of H2SO4, the initial number of moles of HSO4- is also 5.80 mol.

Now, we can calculate the number of moles of HSO4- remaining after the given time:

moles of HSO4- remaining = initial moles of HSO4- - (current × time) / (2 × F)

moles of HSO4- remaining = 5.80 mol - (2.83 A × 6.3 h) / (2 × 96485 C/mol e-)

moles of HSO4- remaining = 5.80 mol - (2.83 A × 6.3 h) / (192970 C/mol e-)

Now, we can calculate the concentration of HSO4- remaining:

concentration of HSO4- = moles of HSO4- remaining / volume

Since the volume is not given, we cannot calculate the concentration of HSO4- with the information provided.