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.

first let us lay out the formula for potential

and kinetic energy:

pe = mgh

ke = (mv^2)/2

where: m=mass, v=velocity, g=gravitational

acceleration, h=height

calculating for mass using the known values of

pe:

340j = m (9.81 m/s^2) (50ft) (1m/3.28ft)

m = 2.27 kg

the law of conservation of energy states that

energy is neither created nor destroyed. therefore, the change in kinetic

energy from 50 ft to 35 ft would be equal to the change in potential energy

from 50 ft to 35 ft. however it would be opposite in signs since one is losing

while other is gaining.

ke(35ft) – ke(50ft) = - [ pe(35ft) – pe(50ft]

ke(50ft) = 0 since the brick is initially at rest

ke(35ft) = 340j - 2.27kg (9.81 m/s^2) (35ft)

(1m/3.28 ft)

ke(35ft)

= 102.38 j