In which of the titrations described below will the first (or only) equivalence point be reached upon the addition of 25.0 ml of 0.100 m naoh? 1) the titration of 25.0 ml of 0.100 m hcl (hydrochloric acid) 2) the titration of 50.0 ml of 0.100 m h2c2o4 (oxalic acid) 3) the titration of 25.0 ml of 0.100 m hc2h3o2 (acetic acid) 4) the titration of 12.5 ml of 0.200 m hbr (hydrobromic acid) view available hint(s) in which of the titrations described below will the first (or only) equivalence point be reached upon the addition of 25.0 ml of 0.100 m naoh? 1) the titration of 25.0 ml of 0.100 m hcl (hydrochloric acid) 2) the titration of 50.0 ml of 0.100 m h2c2o4 (oxalic acid) 3) the titration of 25.0 ml of 0.100 m hc2h3o2 (acetic acid) 4) the titration of 12.5 ml of 0.200 m hbr (hydrobromic acid) 1 only 1, 3, and 4 both 1 and 3 all four titrations will have their first (or only) equivalence point after the addition of 25.0 ml of 0.100 m naoh

1, 3, and 4

Step-by-step explanation:

We must calculate the volume of NaOH needed for each titration.

1) HCl

HCl + NaOH ⟶ NaCl + H₂O

n(HCl)       = 25.0 mL × (0.100 mmol/1mL) = 2.50 mmol

n(NaOH)  = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

2) H₂C₂O₄

H₂C₂O₄ + NaOH ⟶ NaHC₂O₄

n(H₂C₂O₄) = 50.0 mL × (0.100 mmol/1mL) = 5.00 mmol

n(NaOH)    = 5.00 mmol H₂C₂O₄ × (1 mmol NaOH/1 mmol H₂C₂O₄)

= 5.00 mmol NaOH

V(NaOH)   = 5.00 mmol × (1 mL/0.100 mmol) = 50.0 mL

3) HC₂H₃O₂

HC₂H₃O₂ + NaOH ⟶ NaC₂H₃O₂ + H₂O

n(HC₂H₃O₂) = 25.0 mL × (0.100 mmol/1mL) = 2.50 mmol

n(NaOH) = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl)

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

4) HBr

HBr + NaOH ⟶ NaBr + H₂O

n(HBr) = 25.0 mL × (0.100 mmol/1mL) =2.50 mmol

n(NaOH) = 2.50 mmol HCl × (1 mmol NaOH/1 mmol HCl)

= 2.50 mmol NaOH

V(NaOH) = 2.50 mmol × (1 mL/0.100 mmol) = 25.0 mL

Titrations 1, 3, and 4 reach the first or only equivalence point  at 25.0 mL NaOH.

atomic number of carbon is 6

electronic configuration of carbon is 1s²2s²2p²

orbital diagram of carbon is as shown here in the image.

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