Suppose you digested 0.25 g of a copper ore samplein 15 mL of solution. One mL of this solution is transferred to a 10 mL volumetric flask, which is filled to the calibration level. The resulting concentrationwas 0.0087M Cu2+. What is the concentration of copper in the ore sample in units of percentage?

Explanation:

10 mL = 10⁻² L

1 mL of solution contains .0087 x 10⁻² mole of Cu

15 mL of solution will contain .1305 x 10⁻² mole of Cu

.1305 x 10⁻² mole of copper = .1305 x 63.5 x 10⁻² gm of Cu

= .08286 gm

concentration of copper in the ore sample

= .08286 x 100 / .25

= 33 % approx .

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