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.
The number of C2H5OH in a 3 m solution that contain 4.00kg H2O is calculate as below
M = moles of the solute/Kg of water
that is 3M = moles of solute/ 4 Kg
multiply both side by 4
moles of the solute is therefore = 12 moles
by use of Avogadro law constant
1 mole =6.02 x10^23 molecules
what about 12 moles
=12 moles/1 moles x 6.02 x10^23 = 7.224 x10^24 molecules
They are made up of particles that are arranged in a repeating pattern.
Answer:
1.98x10⁻¹² kg
Explanation:
The <em>energy of a photon</em> is given by:
h is Planck's constant, 6.626x10⁻³⁴ J·s
c is the speed of light, 3x10⁸ m/s
and λ is the wavelenght, 671 nm (or 6.71x10⁻⁷m)
- E = 6.626x10⁻³⁴ J·s * 3x10⁸ m/s ÷ 6.71x10⁻⁷m = 2.96x10⁻¹⁹ J
Now we multiply that value by <em>Avogadro's number</em>, to <u>calculate the energy of 1 mol of such protons</u>:
- 1 mol = 6.023x10²³ photons
- 2.96x10⁻¹⁹ J * 6.023x10²³ = 1.78x10⁵ J
Finally we <u>calculate the mass equivalence</u> using the equation:
- m = 1.78x10⁵ J / (3x10⁸ m/s)² = 1.98x10⁻¹² kg
