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2 years ago

What is the boiling point of cyclohexane at 620 mmhg? can you show calculations?

Chemistry

2 answers:

Anton [14]2 years ago

7 0

Boiling point of cyclohexane at 620 mm hg?

Standard atmospheric pressure is 760 mm Hg. Boiling point at 760 mm= 80.74˚C

A liquid boils when its vapor pressure is equal to the atmospheric pressure. The vapor pressure of a liquid is proportional the absolute temperature of the liquid.

As the atmospheric pressure decreases, less vapor pressure is needed to cause the liquid to boil. Less vapor pressure means lower temperature.

The boiling point of cyclohexane at 620 mm Hg is less than 80.74˚C

TRY THAT OR THIS

271.78

vitfil [10]2 years ago

7 0

Answer:

Boiling point of cyclohexane at 620 mm Hg is 440.6 K

Explanation:

According to clausius-clapeyron equation for a liquid-vapour equilibrium-

$ln(\frac{P_{2}}{P_{1}})=\frac{-\Delta H_{vap}}{R}(\frac{1}{T_{2}}-\frac{1}{T_{1}})$

where $P_{2}$ and $P_{1}$ are vapor pressure of liquid at $T_{2}$ and $T_{1}$ temperature respectively

$\Delta H_{vap}$ is the molar enthalpy of vaporization of liquid

Let's assume $\Delta H_{vap}$ is equal to standard molar enthalpy of vaporization of a liquid

For cyclohexane, standard molar enthalpy of vaporization is 32.83 kJ/mol

Here, $P_{2}$ = 620 mm Hg, $T_{1}$ is 450.8 K (boiling point of cyclohexane) and $P_{1}$ is 760 mm Hg

So, $ln(\frac{620mm Hg}{760mm Hg})=\frac{-32.83\times 10^{3}J/mol}{8.314 J/(mol.K)}\times (\frac{1}{T_{2}}-\frac{1}{450.8K})$

So, $T_{2}=440.6K$

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Elza [17]

Answer:

It will be less than 26 °C as water has a relatively higher specific heat than sand.

Explanation:

The specific heat of a substance is the amount of heat energy absorbed by one unit of mass of the substance when its temperature increases one unit.

From that, you can derive the equation for the specific heat of a substance:

specific heat = heat / (mass × ΔT)

Thus, assuming that all the heat provided by the lamp to both samples is the same and, as given, the amount (mass) of both samples is also the same, you have that the specific heat of the samples will be:

specific heat = constant / ΔT

So, specific heat and ΔT are inversely related.

It is known that water has a higher specific heat than sand (that is why the sand on the shore of a beach is, during the day, hotter than the water and your feet get burned when you walk on a sandy beach on a sunny day).

Then, since the specific heat of water is greater than the specific heat of sand, the increase of temperature of water will be lower and, consequently, water will reach a lower final temperature than sand, when equal amounts of water and sand are heated as described in the experiment. This is the second choice: the final temperature of water is less than 26°C as water has a relatively higher specific heat than water.

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Using the following standard reduction potentials, Fe3+(aq) + e- → Fe2+(aq) E° = +0.77 V Ni2+(aq) + 2 e- → Ni(s) E° = -0.23 V ca

lina2011 [118]

Answer: The above reaction is non-spontaneous.

Explanation:

For the given chemical reaction:

$Ni^{2+}(aq.)+2Fe^{2+}(aq.)\rightarrow 2Fe^{3+}(aq.)+Ni(s)$

Here, nickel is getting reduced because it is gaining electrons and iron is getting oxidized because it is loosing electrons.

We know that:

$E^o_{(Fe^{3+}/Fe^{2+})}=0.77V\\E^o_{(Ni^{2+}/Ni)}=-0.23V$

Substance getting oxidized always act as anode and the one getting reduced always act as cathode.

To calculate the $E^o_{cell}$ of the reaction, we use the equation:

$E^o_{cell}=E^o_{cathode}-E^o_{anode}$

$E^o_{cell}=-0.23-0.77=-1.0V$

Relationship between standard Gibbs free energy and standard electrode potential follows:

$\Delta G^o=-nFE^o_{cell}$

As, the standard electrode potential of the cell is coming out to be negative for the above cell. Thus, the standard Gibbs free energy change of the reaction will become positive making the reaction non-spontaneous.

Hence, the above reaction is non-spontaneous.

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