Calculate the number of moles of NaOH in an aqueous solution of 36 cm3 of 0.1 mol dm−3.

0.0036 moles of NaOH are present

Explanation:

A solution of 0.1mol/dm³ contains 0.1 moles of solute (In this case, NaOH), per 1 dm³ of solution.

36cm³ are in dm³ (1000cm³ / 1dm³):

36cm³ * (1dm³/1000cm³) = 0.036dm³

The moles are:

0.036dm³ * (0.1mol/dm³) = 0.0036 moles of NaOH are present

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