3. Fe(NO3), (aq) +
Rb, PO,(aq)
à
FePO,(s) + RbNO,(aq)
How many moles of iron(III) phosphite will precipitate when 0.473 moles of rubidium phosphite react
with excess of iron(III) nitrate?

the answer is learned

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

i think it's c. if you look search electronic symbol on google and then click on wiki it shows you a