Respuesta :
Answer:
The molar concentration of the solution is 4.46 M.
Explanation:
From the given,
Molarity of sulfuric acid = 5.00 M
Volume of sulfuric aci = 1.50L
Amperes of battery = 2.20 A
Time = 9.50 h
[tex]Charge= Current \times time(s)[/tex]
[tex]=2.30 \times 9.50 \times 3600 s = 78660 \, C[/tex]
[tex]Moles\,of\,electrons = \frac{charge}{Faraday\,constant}[/tex]
[tex]= \frac{78660}{96485}= 0.8153 mol[/tex]
[tex]Mole\,of\,H_{2}SO_{4}\,consumed = moles\,of\,electrons = 0.8153\,mol[/tex]
[tex]Initial\,moles\,of\,H_{2}SO_{4}\,=Volume \,\times Initial\,concentration[/tex]
[tex]1.50 \times 5.00 = 7.5\,mol[/tex]
[tex]Final\,moles\,of\,H_{2}SO_{4}=7.5-0.8513=6.6847\,mol[/tex]
[tex]Final\,concentration\,of\,H_{2}SO_{4}= \frac{final\,moles}{Volume}[/tex]
[tex]= \frac{6.6847}{1.50}=4.46M[/tex]
Therefore, The molar concentration of the solution is 4.46 M.
The concentration of H2SO4 is mathematically given as
FC=4.46M
What will be the concentration of H2SO4 in the battery after 2.20 A of current is drawn from the battery for 9.50 hours?
Question Parameter(s):
a fully charged lead-acid battery contains 1.50 L of 5.00 M H2SO4
the battery after 2.20 A of current
the battery for 9.50 hour
Generally, the equation for the Charge is mathematically given as
Charge= Current *time
Therefofre
C=2.30 * 9.50* 3600 s
C= 78660 C
Moles of E=[tex]\frac{charge}{Faraday\ constant}[/tex]
Moles of E= 78660/96485
Moles of E= 0.8153 mol
In conclusion, Initial moles of H2SO4
M=1.50* 5.00
M= 7.5mol
Hence, final moles of H2SO4
FM=7.5-0.8513
FM=6.6847mol
Final concencetration of H2SO4 is now
FC= 6.6847/1.50
FC=4.46M
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