Excess alcohol is removed by Vacuum filtration.
<h3>What is vacuum filtration?</h3>
- Vacuum filtration is a methodology where a tension differential is kept up with across the channel medium by clearing the air beneath the channel paper.
- Vacuum filtration gives a power on the arrangement notwithstanding that of gravity and builds the pace of filtration.
- Vacuum filtration is the standard strategy utilized for isolating a strong fluid blend when the objective is to hold the strong (for instance in crystallization).
<h3>How do you dry a product after vacuum filtration?</h3>
- The strong can be dried rapidly by washing the strong with an unpredictable fluid (methanol, CH3)2CO, or ether are normal) that the strong won't break up in. In the event that the filtrate is to be gathered, change the getting jar prior to washing.
To learn more about vacuum filtration from the given link
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the moles of CO2(g) =2.5moles
I hope this help
Answer:
ΔGreaction < 0
ΔSuniverse > 0
ΔHreaction < 0
Explanation:
A spontaneous process is one which can proceed without additional input of energy releasing free energy in the process and then moves to a lower more stable thermodynamical state.
For an isolated system, a spontaneous process proceeds with an increase in entropy.
The conditions for a spontaneous process at constant temperature and pressure, can be determined using the change in Gibbs free energy, which is given by: ∆G = ∆H - T∆S
Where ∆G is change in free energy; ∆H is change in enthalpy or Heat content; ∆S is change in entropy, T is temperature.
For a process to be spontaneous, the following conditions are necessary:
1. ∆G < 0; must be negative
2. ∆S > 0; there must be an increase in entropy
3. ∆H < 0; enthalpy change must be negative such that heat is lost to the surroundings
The above conditions ensures that ∆G is negative and the process is spontaneous.
Answer:
2.30 liters.
Explanation:
- The balanced equation of the reaction is:
<em>Na₂O₂ + CO₂ → Na₂CO₃ + 1/2O₂,</em>
- It is clear that 1.0 mole of Na₂O₂ reacts with 1.0 mole of CO₂ to produce 1.0 mole of Na₂CO₃ and 0.5 mole of O₂.
- The no. of moles of CO₂ in (4.60 L) reacted can be calculated from the relation: <em>PV = nRT</em>.
P is the pressure of the gas (P = 1.0 atm at STP),
V is the volume of the gas (V = 4.60 L),
R is the general gas constant (R = 0.082 L.atm/mol.K),
T is the temperature of the gas (T = 273.0 K at STP).
∴ n = PV/RT = (1.0 atm)(4.6 L) / (0.082 L.atm/mol.K)(273.0 K) = 0.205 mol.
<u><em>Using cross multiplication:</em></u>
1.0 mole of CO₂ produces → 0.5 mole of O₂, from the stichiometry.
0.205 mole of CO₂ produces → ??? mole of O₂.
- The no. of moles of O₂ produced from 4.60 L of CO₂ = (0.5 mole)(0.205 mole) / (1.0 mole) = 0.103 mole.
- ∴ The volume of O₂ produced from 4.60 L of CO₂ = nRT/P = (0.103 mol)(0.082 L.atm/mol.K)(273.0 K) / (1.0 atm) = 2.30 liters.
As you can see, the unit of heat of vaporization is in kJ/mol, while the unit for entropy of vaporization is in J/mol·K. Since it is vaporization, this occurs at the boiling point. Thus, the formula would be:
Boiling Point = ΔHvap/ΔSvap
Make sure the units are consistent.
Boiling point = 55.5 kJ/mol * 1000 J/kJ / 148 J/mol·K = <em>375 K or 102°C</em>
