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vivado [14]
3 years ago
14

80 liters of oxygen is collected over water at 50.0 degrees Celsius. The atmospheric pressure during the collection was 96.00 kp

a. What is the partial pressure of the oxygen?
Chemistry
1 answer:
vlabodo [156]3 years ago
3 0

Answer:

The value of partial pressure of oxygen P_{O_{2} } = 83.66 K pa

Explanation:

Volume of oxygen = 80 liters

Temperature = 50°c

The total pressure = 96 K pa

The partial pressure of water at 50°c is calculated from the tables.

P_{H_{2}O } = 12.344 K pa

The total pressure is given by

P_{total} = P_{H_{2}O } + P_{O_{2} }

Put all the values in the above equation we get

96 = 12.344 + P_{O_{2} }

P_{O_{2} } = 96 - 12.344

P_{O_{2} } = 83.66 K pa

This is the value of partial pressure of oxygen.

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The answer is 1.56L. Avogadro's Law states that the volume of a gas is directly proportional to the number of moles (or a number of particles) of gas when the temperature and pressure are held constant.

V∝n

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When 3.51 g of phosphorus was burned in chlorine, the product was a phosphorus chloride. Its vapor took 1.77 times as long to ef
kumpel [21]

Answer:

<em>molar mass of the phosphorus chloride = 138.06 g/mol</em>

<em></em>

Explanation:

<em>mass of phosphorus will be the same as mass of CO2, since it is stated that they are of equal amount.</em>

mass = 3.51 g

<em>lets assume that it took the CO2 1 sec to effuse, then the time taken by the phosphorus chloride will be 1.77 sec</em>

From this we can say that

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rate of effusion of the phosphorus chloride = 3.51/1.77 = 1.98 g/s

<em>From graham's equation of effusion</em>,

\frac{Rc}{Rp} = \sqrt{\frac{Mp\\}{Mc} }

Rc = rate of effusion of CO2 = 3.51 g/s

Rp = rate of effusion of phosphorus chloride = 1.98 g/s

Mc = molar mass of CO2 = 44.01 g/mol

Mp = molar mass of the phosphorus chloride = ?

Imputing values into the equation, we have

\frac{3.51}{1.98} = \sqrt{\frac{Mp\\}{44.01} }

1.77 = \frac{\sqrt{Mp} }{6.64}

11.75 = \sqrt{Mp}

Mp = 11.75^{2}

Mp = <em>molar mass of the phosphorus chloride = 138.06 g/mol</em>

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