The answer is D!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
Answer:
because they cannot be seen by our naked eyes to make them visible we must use microscope.
Answer:
-138.9 kJ/mol
Explanation:
Step 1: Convert 235.8°C to the Kelvin scale
We will use the following expression.
K = °C + 273.15 = 235.8°C + 273.15 = 509.0 K
Step 2: Calculate the standard enthalpy of reaction (ΔH°)
We will use the following expression.
ΔG° = ΔH° - T.ΔS°
ΔH° = ΔG° / T.ΔS°
ΔH° = (-936.92kJ/mol) / 509.0K × 0.51379 kJ/mol.K
ΔH° = -3.583 kJ (for 1 mole of balanced reaction)
Step 3: Convert -9.9°C to the Kelvin scale
K = °C + 273.15 = -9.9°C + 273.15 = 263.3 K
Step 4: Calculate ΔG° at 263.3 K
ΔG° = ΔH° - T.ΔS°
ΔG° = -3.583 kJ/mol - 263.3 K × 0.51379 kJ/mol.K
ΔG° = -138.9 kJ/mol
Answer:
The problem is solvable
Explanation:
The information that is provided, together with some equations are enough to solve the problem:
- The inlet states are totally defined.
- A heat balance under adiabatic assumption, allows to calculate the outlet temperature (both outlets stream are in equilibrium, so at the same temperature).
- The rule of phases implies that por the two-phase equilibrium, there is only one pressure for each temperature.
- The mass balance and equilibrium relationships allows to calculate the res of properties for the outlet streams.
Ka expression for any substance:
Ka = [H+] [A-] / [HA]
In this case,
Ka = [H+][C₃H₅O₃-]/[C₃H₆O₃]
1.4 x 10⁻⁴ = [H+] [C₃H₅O₃-] / [C₃H₆O₃]
pKa = -log(Ka)
pH = -log([H+])
The difference between pKa and pH is that pH is the negative logarithm of only the concentration of hydrogen ions, while pKa is the negative logarithm of the ratio of the product of the concentrations of hydrogen ions and concentration of base to the concentration of acid.
