When a lead acid car battery is recharged by the alternator, it acts essentially as an electrolytic cell in which solid lead(II) sulfate PbSO4 is reduced to lead at the cathode and oxidized to solid lead(II) oxide PbO at the anode. Suppose a current of 17.0A is fed into a car battery for 15.0 seconds. Calculate the mass of lead deposited on the cathode of the battery. Be sure your answer has a unit symbol and the correct number of significant digits.

0.273504 grams

Explanation:

Charge Deposited = Current ×Time

q=It

Given: I= 17.0A, t= 15.0 seconds

Q= 17×15= 255 C

Since, 1 mole contains 96500C of charge deposition

Therefore, number of moles = 255/96500 = 0.00264 moles

Moreover, since Pb changes to Pb2+,

Hence number of moles of Pb deposited = 0.00264/2

= 0.00132

Also, Molar mass of Lead = 207.2 gm/mol

Therefore, the mass of lead deposited = 0.00132×207.2 = 0.273504 grams

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when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml