How many ml of 0.01 M NaOH are required to neutralize 50 ml of 0.02 M HCl

100 ml

Explanation:

The neutralization reaction is:

NaOH(aq) + HCl(aq) → NaCl(aq) + H₂O(l)

Since 1 mol NaOH reacts with 1 mol HCl, at the equivalence point of the titration, the number of moles of NaOH is equal to the number of moles of HCl. We calculate the moles as the product of molarity (M) and volume (V):

moles NaOH = moles HCl

M(NaOH) x V(NaOH) = M(HCl) x V(HCl)

Thus, we calculate the volume of NaOH needed to completely neutralice HCl:

V(NaOH) = (M(HCl) x V(HCl)) /M(NaOH)

               = (0.02 M x 50 ml)/ (0.01 M)

               = 100 ml

Therefore, 100 ml of 0.01 M NaOH are required to neutralize 50 ml of 0.02 M HCl.

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