A 20 mL volume of a hydrochloric acid solution, HCl (aq), was placed into a beaker containing an appropriate acid-base indicator. It was titrated with a solution of sodium hydroxide, NaOH (aq), until the endpoint was
reached. If 25 mL of sodium hydroxide solution with a concentration of 0.20 mol/L was used, what is the
concentration of the acid?
Select one:
a. 2.5 mol/L
O
b. 0.20 mol/L
C. 0.25 mol/L
d. 2.0 mol/L

C- 0.25mol/L

Explanation:

Hcl+Naoh-Nacl+H2O

Ratio =1:1

0.2moles of Naoh in1000cm3 what about 25=(25×0.2)÷1000=

0.005moles

0.005moles in 20cm³ what about in 1000

(0.005×1000)÷20=0.25mol/L

m=9

step-by-step explanation:

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