In an experiment, a 45.0 mL sample of 0.15 M triethylamine, (CH3)3N (aq), was placed in a flask and titrated to the equivalence point and beyond using 0.10 M HNO3(aq), according to the balanced equation below: (CH3)3N (aq) + HNO3(aq) →(CH3)3NH+(aq) + NO3−(aq)

Calculate the pH after the following volumes of HNO3have been added. The Kb of triethylamine is equal to 6.4 × 10-5.

1) 0.0 mL
2) 36.0 mL
3) 67.5 mL
4) 70.0 mL

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

answerto predict;