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snow_lady [41]
3 years ago
10

There are 24 g of carbon in a substance with a mass of 30 g. What is the percent carbon by mass for this substance?

Chemistry
1 answer:
mihalych1998 [28]3 years ago
8 0

Answer:

Explanation:

The composition of the vapor in mass percent.

According to Dalton's law,

where  is the partial pressure of the gas j,  is the mole fraction of the gas j in the gas mixture and  is the total pressure.

To know the composition of the vapor we must first calculate the molar fractions of the components of the mixture in the vapor:

→  

 →  

If we consider 1 mole of solution then we will have 0.56 mol of benzene and 0.44 mol of toluene in the vapor. In this way, the number of grams of each component in the vapor will be,

The percentage by mass of each component in the vapor will be,

% benzene = (g benzene / g total) x 100% = (43.7 g / (43.7 g + 40.6 g)) x 100% → % benzene = 51.8 %

% toluene = 100 % - % benzene → % toluene = 41.2%

So, the composition of the vapor in mass percent is 51.8 % benzene and 41.2% toluene.

Why is the composition of the vapor different from the composition of the solution?

The composition of the vapor will be different from that of the solution, since the more volatile compound will have a larger molar fraction in the vapor phase than in the liquid phase.

In a mixture with different volatile components the compound that volatilizes more easily is the one that will have greater capacity to escape from the solution in the form of vapor and, therefore, will be in a greater composition in the vapor above the solution.

We can measure this by means of vapor pressure, which is a measure of the volatility of a substance, that is, the capacity of the substance to pass from a liquid to a gaseous state. In the question, benzene is in greater proportion than toluene in the vapor mixture since it has a higher vapor pressure (94.2 torr) than toluene (28.4 torr).

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2 years ago
What mass of octane must be burned in order to liberate 5270 kj of heat? δhcomb = -5471 kj/mol?
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As,
                           5471 kJ heat is given by  =  1 mole of Octane
Then,
                    5310 kJ heat will be given by  = X moles of Octane

Solving for X,
                                  X  =  (5310 kJ × 1 mol) ÷ 5471 kJ

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So, 0.970 moles of Octane will liberate 5310 kJ energy. Now changing moles to mass,
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Or,
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Putting values,
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