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3 years ago

Look up and record the boiling point of acetic acid, and explain why only some of it evaporates from the reaction mixture.

Chemistry

1 answer:

REY [17]3 years ago

8 0

Answer:

Heating the mixture to a temperature above the boiling point of acetic acid, but below 100°C (the boiling point of water). The vapours from the acetic acid rise, and go into a tube. They are then condensed within the tube, and run off into a separate storage area. Because water can exist as a gas at pretty much any temperature above 0°C, it will result in an impure mixture, but repeatedly doing this will get the acetic acid to the desired purity.

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e-lub [12.9K]

It's back wards lol you need to take it right

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3 years ago

Read 2 more answers

The sodium salt, NaA, of a weak acid is dissolved in water; no other substance is added. Which of these statements (to a close a

ddd [48]

Answer:

Your question is somewhat poorly worded, even so what I can contribute is the statement is false, since the salt is sodium chloride, where its severe chemical formula NaCl, is a SALT not an acid and if it dissolves in water, seriously the solute of a solution, where water plays the role of the solvent

Explanation:

Sodium chloride is a binary salt, very easy to dissolve in water, it is also called the famous table salt, since it can be ingested in food even though it is not so recommended against high blood pressure conditions.

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3 years ago

Rubbing alcohol is less polar than water. Both are liquids at room temperature which One boils first? why?

Sergeeva-Olga [200]

Answer:

because both liquid are made from different substances.

Explanation:

8 0

3 years ago

A sample of gas has a density of 0.53 g/L at 225 K and under a pressure of 108.8 kPa. Find the density of the gas at 345 K under

sukhopar [10]

Answer:

$\rho _2=0.22g/L$

Explanation:

Hello!

In this case, since we are considering an gas, which can be considered as idea, we can write the ideal gas equation in order to write it in terms of density rather than moles and volume:

$PV=nRT\\\\PV=\frac{m}{MM} RT\\\\P*MM=\frac{m}{V} RT\\\\P*MM=\rho RT$

Whereas MM is the molar mass of the gas. Now, since we can identify the initial and final states, we can cancel out R and MM since they remain the same:

$\frac{P_1*MM}{P_2*MM} =\frac{\rho _1RT_1}{\rho _2RT_2} \\\\\frac{P_1}{P_2} =\frac{\rho _1T_1}{\rho _2T_2}$

It means we can compute the final density as shown below:

$\rho _2=\frac{\rho _1T_1P_2}{P_1T_2}$

Now, we plug in to obtain:

$\rho _2=\frac{0.53g/L*225K*68.3kPa}{345K*108.8kPa}\\\\\rho _2=0.22g/L$

Regards!

8 0

3 years ago

Predict the boiling point of water at a pressure of 1.5 atm.

Lina20 [59]

Answer:

100.8 °C

Explanation:

The Clausius-clapeyron equation is:

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

Where 'ΔHvap' is the enthalpy of vaporization; 'R' is the molar gas constant (8.314 j/mol); 'T1' is the temperature at the pressure 'P1' and 'T2' is the temperature at the pressure 'P2'

Isolating for T2 gives:

$T_{2}=(\frac{1}{T_{1}} -\frac{Rln\frac{P_{2}}{P_{1}} }{Delta H_{vap}}$

(sorry for 'deltaHvap' I can not input symbols into equations)

thus T2=100.8 °C

7 0

3 years ago