Answer:
626.7nm
Explanation:
The energy of a photon is defined as:
E = hc / λ
<em>Where E is the energy of the photon, h is Planck constant (6.626x10⁻³⁴Js), c is speed of light (3x10⁸m/s) and </em>λ is the wavelength of light
The energy of 1 photon is:
(191000 J / mol) ₓ (1 mole / 6.022x10²³) = 3.1717x10⁻¹⁹ J
Replacing:
3.1717x10⁻¹⁹ J = <em>6.626x10⁻³⁴Jsₓ3x10⁸m/s / </em>λ
λ = 6.267x10⁻⁷m
as 1nm = 1x10⁻⁹m:
6.267x10⁻⁷m ₓ (1nm / 1x10⁻⁹m) =
<h3>626.7nm</h3>
No se si aun necesitas ayuda o no
Answer:
pH = 1.32
Explanation:
H₂M + KOH ------------------------ HM⁻ + H₂O + K⁺
This problem involves a weak diprotic acid which we can solve by realizing they amount to buffer solutions. In the first deprotonation if all the acid is not consumed we will have an equilibrium of a wak acid and its weak conjugate base. Lets see:
So first calculate the moles reacted and produced:
n H₂M = 0.864 g/mol x 1 mol/ 116.072 g = 0.074 mol H₂M
54 mL x 1L / 1000 mL x 0. 0.276 moles/L = 0.015 mol KOH
it is clear that the maleic acid will not be completely consumed, hence treat it as an equilibrium problem of a buffer solution.
moles H₂M left = 0.074 - 0.015 = 0.059
moles HM⁻ produced = 0.015
Using the Henderson - Hasselbach equation to solve for pH:
ph = pKₐ + log ( HM⁻/ HA) = 1.92 + log ( 0.015 / 0.059) = 1.325
Notes: In the HH equation we used the moles of the species since the volume is the same and they will cancel out in the quotient.
For polyprotic acids the second or third deprotonation contribution to the pH when there is still unreacted acid ( Maleic in this case) unreacted.
