The volume of the 0.15 M LiOH solution required to react with 50 mL of 0.4 M HCOOH to the equivalence point is 133.3 mL
<h3>Balanced equation </h3>
HCOOH + LiOH —> HCOOLi + H₂O
From the balanced equation above,
The mole ratio of the acid, HCOOH (nA) = 1
The mole ratio of the base, LiOH (nB) = 1
<h3>How to determine the volume of LiOH </h3>
- Molarity of acid, HCOOH (Ma) = 0.4 M
- Volume of acid, HCOOH (Va) = 50 mL
- Molarity of base, LiOH (Mb) = 0.15 M
- Volume of base, LiOH (Vb) =?
MaVa / MbVb = nA / nB
(0.4 × 50) / (0.15 × Vb) = 1
20 / (0.15 × Vb) = 1
Cross multiply
0.15 × Vb = 20
Divide both side by 0.15
Vb = 20 / 0.15
Vb = 133.3 mL
Thus, the volume of the LiOH solution needed is 133.3 mL
Learn more about titration:
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The energy range expected is 6.6 × 10^-19 J < E < 7.33 × 10^-19 J
The energy of the photon is given by;
E = hc/λ
E = energy of the photon
h = Plank's constant
c = speed of light
λ = wavelength of light
For the upper boundary range;
E = ?
h = 6.6 × 10^-34 Js
c = 3 × 10^8 m/s
λ = 270 × 10^-9
E = 6.6 × 10^-34 Js × 3 × 10^8 m/s / 270 × 10^-9
E = 7.33 × 10^-19 J
For the lower range;
E = ?
h = 6.6 × 10^-34 Js
c = 3 × 10^8 m/s
λ =300 × 10^-9
E = 6.6 × 10^-34 Js × 3 × 10^8 m/s / 300 × 10^-9
E = 6.6 × 10^-19 J
Hence, the energy range 6.6 × 10^-19 J < E < 7.33 × 10^-19 J
Learn more: brainly.com/question/24857760
