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pashok25 [27]
3 years ago
11

The heat of vaporization of 1-pentanol is 55.5 kJ/mol, and its entropy of vaporization is 148 J/K.mol. What is the approximate b

oiling point of 1-pentanol? 100 oC 375 oC 0 oC 25 oC
Chemistry
1 answer:
____ [38]3 years ago
6 0

Answer:

Approximately 100 °C.

Explanation:

Hello,

In this case, since the entropy of vaporization is computed in terms of the heat of vaporization and the temperature as:

\Delta S_{vap}=\frac{\Delta H_{vap}}{T}

We can solve for the temperature as follows:

T=\frac{\Delta H_{vap}}{\Delta S_{vap}}

Thus, with the proper units, we obtain:

T=\frac{55500J/mol}{148J/(mol*K)} =375K\\\\T=102 \°C

Hence, answer is approximately 100 °C.

Best regards.

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⦁answer Calculate the density in g/L of 478 mL of krypton at 47° C and 671 mm Hg. ⦁ Determine the molar mass of a gas that has a
STALIN [3.7K]

Answer:

The correct answers are:

- Krypton: density= 2.8 g/L

- Molar Mass= 63.99 g/mol

- Mass of O₂= 15.29 g

Explanation:

The general equation of an ideal gas is the folllowing:

P x V = n x R x T

Where: P= pressure (in atm), V= volume; n= number of moles, R= gas constant (0,082 L.atm/K.mol) and T= temperature (in K).

<u>For krypton</u>:

P= 671 mmHg = 0,882 atm

V= 478 ml x 1000 ml/1 L= 0,478 L

T= 47ºC= 320 K

MM= 83.8 g/mol (from Periodic Table, Kr is an inert gas so it is a monoatomic gas)

P x V = n x R x T

Since the number of moles of a compound can be calculated by dividing the mass of compound (m) into its molar mass (MM):

n= m/MM

We can replace the expression in the first equation to obtain:

P x V= \frac{m}{MM} x R x T

m/V= \frac{P x MM}{R x T}

Density (d) is equal to the mass per volume (m/V), so we can directly calculate the density:

d= m/V= \frac{P x MM}{R x T}=

           = (0.882 atm x 83.8 g/mol)/(0.082 L.atm/K.mol x 320 K)

           = 2.81 g/L

<u>For the gas:</u>

d= 2.18 g/L

T= 66ºC= 339 K

P= 720 mmHg= 0.947 atm

d= \frac{P x MM}{R x T}

⇒MM = \frac{dx R x T}{P}

         = (2.18 g/L x 0.082 L.atm/K.mol x 339 K)/(0.947 atm)

         = 63.99 g/mol ≅ 64 g/mol

<u>For the O₂</u>:

V= 5.60 L

P= 1.75 atm

T= 250 K

MM(O₂) = 2 x Atomic Mass O= 2 x 16 g/mol= 32 g/mol

We can use the second equation:

P x V= \frac{m}{MM} x R x T

⇒  m = \frac{P x V x MM}{R x T}= (1.75 atm x 5.6 L x 32 g/mol)/(0.082 L.atm/K.mol x 250 K)

                         = 15.29 g≅ 16 g

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