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 96.0A is fed into a car battery for 37.0 seconds. Calculate the mass of lead deposited on the cathode of the battery.

3.73 grams

Explanation:

where Q= quantity of electricity in coloumbs

I = current in amperes = 96.0 A

t= time in seconds = 37.0 sec

The reaction at cathode is:

of electricity deposits 1 mole of  

3552 C of electricity deposits = of Pb

Thus mass of lead deposited is 3.73 g

може ли да зареждате повече информация, моля

answer : (a) volume of the gas = 8.67 ×   l

                (b) second virial coefficient b = -1.21226 l/mole

solution :   given,     t = 300k

                                p = 20atm

                                z = 0.86

                                n = 8.2 mmoles = 8.2 × moles

                                r =   0.082 l atm/k/mol

        find,     (a) volume of gas   and (b) second virial coefficient b at 300k.

formula used :            

                          and            

(a)            

                    = \times 0.082\times 300}{20}[/tex]

                    = 8.67 ×   l

(b)      

                    = (0.00867-1) × 8.659

                    = - 8.5839 l/mole