How many milliliters of a 1: 5000 w/v solution of the preservative benzalkonium chloride can be made from 125 ml of a 0.2% w/v solution of the preservative? a. 1250 ml b. 2500 ml c. 625 ml d. 1875 ml

a. 1250 mL

Explanation:

We use the C1V1=C2V2 equation to solve this problem, with:

C1=1/5000=0,0002 w/v,             V1=?

C2=0,2%=0,2/100=0,002 w/v     V2=125 mL

Then:

V1=C2V2/C1=0,002*125/0,0002=125O mL

Thus, 1250 mL of the preservative 1:5000 w/v can be made from 125 mL of a 0,2% w/v solution.

a:

by the combination of an unlimited number of elements

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