help with this homework please! Module 6Homework- Google Chrome a Secure https//
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help with this homework please!
Module 6Homework- Google Chrome a Secure https//session.masteringchemistry.com/myct/itemView?assignmentProblemlD-90036023 a Acid Concentration in a Lead Battery previous 21 of 28 I nedt ± Acid Concentration In a Lead Battery Part A .50 L of 5.00 M H,SO, What wil be the concentration of H2SO, in the battery after 3.30 A A lead-acid battery uses a redox reaction in which lead(0) and lead IV) are both converted to lead(ll) This reaction is Suppose that a fully charged lead-acid battery contains 1 facilitated by the presence of sulturic acid, H So, as shown by the reaction Express your answer with the appropriate units. Hints ale Units Submit My Answers Give Up 1227 PM ^4x 11/27/2017Explanation / Answer
Answer
4.22M
Explanation
1coloumbs in 1seconds = 1Ambs
3.30A = 3.30coloumbs in seconds
Total running hour = 9.50hours
Total coloumbs = 3.30× 60×60 ×9.50 =112860Coloumbs
Coloumbs for 1mole of electron = 96485
No of mole of electron discharged = 112860/96485=1.1697
now look at the reaction
Oxidation half cell
Pb (aq) + H2SO4(aq) ------> PbSO4(aq) + 2H+(aq) + 2e
Reduction half cell
PbO2(s) + H2SO4(aq) + 2H+(aq) + 2e -----> PbSO4 + 2H2O(l)
Overall reaction
Pb(s) + PbO2(s)+ 2H2SO4(aq)------> 2PbSO4(aq) + 2H2O(l)
So, for 2mol of H2SO4 consumption 2 mole of electron relesed
1.1697mole of electron discharge indicate 1.1697mole of H2SO4 consumption
Initial mole of H2SO4 = (5mol/1000ml)×1500ml = 7.5mol
remaining mole of H2SO4 = 7.5mol - 1.1697mol = 6.3303mol
Concentration of H2SO4 = (6.3303mol/1500ml)×1000ml = 4.22M
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