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Mariulka [41]
3 years ago
5

Magnesium has three naturally occurring isotopes with masses of 23.99, 24.99, and 25.98 amu and natural abundances of 78.99%, 10

.00%, and 11.01%. Calculate the atomic mass of magnesium.
Chemistry
1 answer:
Elodia [21]3 years ago
3 0

Answer:

24.309

Explanation:

Let A, B, and C represent the three isotopes.

For isotope A:

Mass number = 23.99

Abundance = 78.99%

For isotope B:

Mass number = 24.99

Abundance = 10.00%

For isotope C:

Mass number = 25.98

Abundance = 11.01%

Atomic mass of Mg =?

Atomic mass = [(Mass of AxA%)/100] + [(Mass of BxB%)/100] + [(Mass of CxC%)/100]

Atomic Mass = [(23.99x78.99)/100] + [(24.99x 10)/100] + [(25.98x11.01)/100]

Atomic number = 18.950 + 2.499 + 2.860

Atomic mass of Mg = 24.309

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A gas occupies a volume of 67.54 liters at 158°C and 4.87 atm pressure. Calculate the number of moles of this gas.
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Answer:

The number of moles of the gas is 9.295 moles or 9.30 moles

Explanation:

We use PV = nRT

Where P = 4.87 atm;

V = 67.54 L

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the number of moles can be obtained by substituting the values in the respective columns and solve for n

n = PV / RT

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n = 328.9198 / 35.3851

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A containing vessel holds a gaseous mixture of nitrogen and butane. Thepressure in the vessel at 126.9 Cis 3.0 atm. At 0 C, the
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A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

A vessel contains a gaseous mixture of nitrogen and butane. At 126.9 °C (400.1 K) the pressure is due to the mixture is 3.0 atm.

We can calculate the total number of moles using the ideal gas equation.

P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{3.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.091 mol/L \times V

At 0 °C (273.15 K), the pressure due to the gaseous nitrogen is 1.0 atm.

We can calculate the moles of nitrogen using the ideal gas equation.

P \times V = n \times R \times T\\\\n = \frac{P \times V}{R \times T} = \frac{1.0 atm \times V}{0.082 atm.L/mol.K \times 400.1 K} = 0.030 mol/L \times V

The mole fraction of nitrogen in the mixture is:

X(N_2) = \frac{0.030 mol/L \times V}{0.091 mol/L \times V} = 0.33

A vessel that contains a mixture of nitrogen and butane has a pressure of 3.0 atm at 126.9 °C and a pressure of 1.0 atm at 0 °C. The mole fraction of nitrogen in the mixture is 0.33.

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