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Leto [7]
4 years ago
11

A 135 g sample of carbon disulfide requires 43.2 kj of heat to vaporize completely. what is the enthalpy of vaporization for car

bon disulfide?
a. 1.77 kj/mol
b. 24.4 kj/mol
c. 76.2 kj/mol
d. 0.320 kj/mol
e. 3.13 kj/mol
Chemistry
1 answer:
Sergio [31]4 years ago
8 0
Molecular weight of CS2 = 76.14 g

number of moles of CS2 = \frac{weight.of.CS2}{molecular.weight}
                                        = \frac{135}{76.14}
                                        = 1.773

Now, 1.733 mol requires 43.2 kj of heat to vaporize.
∴  1 mol will require \frac{43.2}{1.77} = 24.4 kj/mol

Thus, correct answer is option B

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Answer:

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8 0
4 years ago
When 3.0 grams of H2 is reacted with excess C at constant pressure, the reaction forms CH4 and releases 53.3 kJ of heat. C(s) +
Murrr4er [49]

Answer:

THE ENTHALPY OF REACTION IN KJ/MOL OF CH4 IS 7.07 KJ/MOL.

Explanation:

Mass of H2 = 3 g

Molar mass of H2 = 2 g/mol

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Equation of the reaction:

C(s) + 2H2(g) -------> CH4(g)

First:

Calculate the number of moles of H2 that was used:

Number of moles = mass / molar mass

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Number of moles = 1.5 moles

So therefore, when 53.3 kJ of heat was released from the reaction, 1.5 moles of hydrogen was used.

From the equation of the reaction, one mole of carbon reacts with two moles of hydrogen to form one mole of methane.

For 3 g of hydrogen, 1.5 mole of hydrogen is involved.

It means:

1.5 moles of hydrogen reacts with 0.75 moles of carbon and produces 0.75 moles of methane. This is so because the reaction occurs in 1: 2: 1 in respect to carbon, hydrogen and methane respectively.

So we can say that the production of 0.75 mole of methane will evolve 53.3 kJ of heat.

0.75 mole of methane releases 53.3 kJ of heat.

1 mole of methane will release ( 53.3 kJ * 1 / 0.75 )

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In conclusion, the enthalpy of the reaction in kJ/ mole of CH4 is 71.07 kJ/mol.

7 0
4 years ago
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