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liraira [26]
4 years ago
5

we can separate the liquids from solutions using distillation because liquids evaporate but dissolved solids do not drag the lab

els to the diagram of sea water distiller ​
Chemistry
1 answer:
Lisa [10]4 years ago
3 0

Answer:

Yes, we can separate the liquids from solutions using distillation.

Explanation:

In distillation, the components or substances are separated from a liquid mixture by selective boiling and condensation. Distillation can result in substantially complete separation (near-pure components), or it can be a partial separation that increases the concentration of the selected components in the mixture. In both cases, the process takes advantage of the differences in the relative volatility of the components of the mix. In industrial chemistry, distillation is a unitary process of universal practical importance, but it is a process of physical separation, not a chemical reaction. Distillation is also be used to separate a solution from two or more liquids or fractions. They are called miscible liquids and must have different boiling points, e.g. B. ethanol and water or fermented fruit juice. A capacitor is used. A mixture is heated to just above 80 degrees Celsius (the boiling point of ethanol is 79 degrees Celsius). Ethanol evaporates and condenses. It can be collected in a cup. The water remains in the balloon.

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Consider the following reaction: CO(g)+2H2(g)⇌CH3OH(g) Kp=2.26×104 at 25 ∘C. Calculate ΔGrxn for the reaction at 25 ∘C under eac
valentinak56 [21]

Answer : The value of \Delta G_{rxn} is, 8.867kJ/mole

Explanation :

The formula used for \Delta G_{rxn} is:

\Delta G_{rxn}=\Delta G^o+RT\ln Q   ............(1)

where,

\Delta G_{rxn} = Gibbs free energy for the reaction

\Delta G_^o =  standard Gibbs free energy

R = gas constant = 8.314 J/mole.K

T = temperature = 25^oC=273+25=298K

Q = reaction quotient

First we have to calculate the \Delta G_^o.

Formula used :

\Delta G^o=-RT\times \ln K_p

Now put all the given values in this formula, we get:

\Delta G^o=-(8.314J/mole.K)\times (298K)\times \ln (2.26\times 10^{4})

\Delta G^o=-24839.406J/mole=-24.83\times 10^3J/mole=-24.83kJ/mole

Now we have to calculate the value of 'Q'.

The given balanced chemical reaction is,

CO(g)+2H_2(g)\rightarrow CH_3OH(g)

The expression for reaction quotient will be :

Q=\frac{(p_{CH_3OH})}{(p_{CO})\times (p_{H_2})^2}

In this expression, only gaseous or aqueous states are includes and pure liquid or solid states are omitted.

Now put all the given values in this expression, we get

Q=\frac{(1.4)}{(1.2\times 10^{-2})\times (1.2\times 10^{-2})^2}

Q=8.1\times 10^{5}

Now we have to calculate the value of \Delta G_{rxn} by using relation (1).

\Delta G_{rxn}=\Delta G^o+RT\ln Q

Now put all the given values in this formula, we get:

\Delta G_{rxn}=-24.83kJ/mole+(8.314\times 10^{-3}kJ/mole.K)\times (298K)\ln (8.1\times 10^{5})

\Delta G_{rxn}=8.867\times 10^3J/mole=8.867kJ/mole

Therefore, the value of \Delta G_{rxn} is, 8.867kJ/mole

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