If 50.00 ml of 1.10 m naoh is added to 25.00 ml of 1.90 m hcl, with both solutions originally at 24.72 ∘c, what will be the final solution temperature? (assume that no heat is lost to the surrounding air and that the solution produced in the neutralization reaction has a density of 1.02 g/ml and a specific heat of 3.98 jg−1∘c−1.)

d

explanation:

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