200mL of 12.0M HNO3 is prepared from stock solution of 16.0M HNO3. A, What volume of stock solution of NH3 is used? And b, What volume of water should be added to the stock

a. 150mL of the stock solution was added

b. 50mL of water was added

Explanation:

The solution was diluted from 16.0M to 12.0M, that is:

16.0M / 12.0M = 1.333 times was diluted.

As the volume of the diluted solution is 200mL, the amount of the 16.0M stock solution is:

200mL / 1.333 =

As the volume of the solution is 200mL and 150mL were added of the stock solution, the volume of water is:

200mL - 150mL =

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