B and d will work out and a and c will also work out
When dT = Kf * molality * i
= Kf*m*i
and when molality = (no of moles of solute) / Kg of solvent
= 2.5g /250g x 1 mol /85 g x1000g/kg
=0.1176 molal
and Kf for water = - 1.86 and dT = -0.255
by substitution
0.255 = 1.86* 0.1176 * i
∴ i = 1.166
when the degree of dissociation formula is: when n=2 and i = 1.166
a= i-1/n-1 = (1.166-1)/(2-1) = 0.359 by substitution by a and c(molality) in K formula
∴K = Ca^2/(1-a)
= (0.1176 * 0.359)^2 / (1-0.359)
= 2.8x10^-3
Answer:
2812.6 g of H₂SO₄
Explanation:
From the question given above, the following data were obtained:
Mole of H₂SO₄ = 28.7 moles
Mass of H₂SO₄ =?
Next, we shall determine the molar mass of H₂SO₄. This can be obtained as follow:
Molar mass of H₂SO₄ = (1×2) + 32 + (16×4)
= 2 + 32 + 64
= 98 g/mol
Finally, we shall determine the mass of H₂SO₄. This can be obtained as follow:
Mole of H₂SO₄ = 28.7 moles
Molar mass of H₂SO₄ =
Mass of H₂SO₄ =?
Mole = mass / Molar mass
28.7 = Mass of H₂SO₄ / 98
Cross multiply
Mass of H₂SO₄ = 28.7 × 98
Mass of H₂SO₄ = 2812.6 g
Thus, 28.7 mole of H₂SO₄ is equivalent to 2812.6 g of H₂SO₄
The expectancy of how long a product will last divided by two!
Answer:
a) [A⁻]/[HA] = 0.227
b) [A⁻]/[HA] = 0.991
c) [A⁻]/[HA] = 2.667
Explanation:
In the Henderson-Hasselbalch equation, HA stands from an acid an A⁻ stands from its conjugate base, as follows:
pH = pka + Log [A⁻]/[HA]
pH = 4.874 + Log[CH₃CH₂CO₂⁻]/[CH₃CH₂CO₂H]
4.23 = 4.874 + Log [A⁻]/[HA]
-0.644 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.227 = [A⁻]/[HA]
4.87 = 4.874 + Log [A⁻]/[HA]
-0.004 = Log [A⁻]/[HA]
= [A⁻]/[HA]
0.991 = [A⁻]/[HA]
5.30 = 4.874 + Log [A⁻]/[HA]
0.426 = Log [A⁻]/[HA]
= [A⁻]/[HA]
2.667 = [A⁻]/[HA]
