Chemistry, 19.11.2019 01:31 kyliexhill

Part a: three gases (8.00 g of methane, ch_4, 18.0g of ethane, c_2h_6, and an unknown amount of propane, c_3h_8) were added to the same 10.0- l container. at 23.0 degrees c, the total pressure in the container is 5.40 atm. calculate the partial pressure of each gas in the container. part b: a gaseous mixture of o_2 and n_2 contains 37.8 % nitrogen by mass. what is the partial pressure of oxygen in the mixture if the total pressure is 405 mmhg?

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Answer from: leo4687

Explanation:

Part A:

Total pressure of the mixture = P = 5.40 atm

Volume of the container = V = 10.0 L

Temperature of the mixture = T = 23°C = 296.15 K

Total number of moles of gases = n

PV = nRT (ideal gas equation)

$n=\frac{PV}{RT}=\frac{5.40 atm\times 10.0 L}{0.0821 atm L/mol K\times 296.15 K}=2.22 mol$

Moles of methane gas = $n_1=\frac{8.00 g}{16 g/mol}=0.5 mol$

Moles of ethane gas = $n_2=\frac{18.0 g}{30 g/mol}=0.6 mol$

Moles of propane gas = n_3

n=n_1+n_2+n_3

2.22=0.5 mol +0.6 mol+ n_3

n_3= 2.22 mol - 0.5 mol -0.6 mol= 1.12 mol

Mole fraction of methane = $\chi_1=\frac{n_1}{n_1+n_2+n_3}=\frac{n_1}{n}$

$\chi_1=\frac{0.5 mol}{2.22 mol}=0.2252$

Similarly, mole fraction of ethane and propane :

$\chi_2=\frac{n_2}{n}=\frac{0.6 mol}{2.22 mol}=0.2703$

$\chi_3=\frac{n_3}{n}=\frac{1.12 mol}{2.22 mol}=0.5045$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

Partial pressure of methane gas:

$p_1=P\times \chi_1=5.40 atm\times 0.2252=1.22 atm$

Partial pressure of ethane gas:

$p_2=P\times \chi_2=5.40 atm\times 0.2703=1.46 atm$

Partial pressure of propane gas:

$p_3=P\times \chi_3=5.40 atm\times 0.5045=2.72 atm$

Part B:

Suppose in 100 grams mixture of nitrogen and oxygen gas.

Percentage of nitrogen = 37.8 %

Mass of nitrogen in 100 g mixture = 37.8 g

Mass of oxygen gas = 100 g - 37.8 g = 62.2 g

Moles of nitrogen gas = $n_1=\frac{37.8 g g}{28g/mol}=1.35 mol$

Moles of oxygen gas = $n_2=\frac{62.2 g}{32 g/mol}=1.94 mol$

Mole fraction of nitrogen= $\chi_1=\frac{n_1}{n_1+n_2}$

$\chi_1=\frac{1.35 mol}{1.35 mol+1.94 mol}=0.4103$

Similarly, mole fraction of oxygen

$\chi_2=\frac{n_2}{n_1+n_2}=\frac{1.94 mol}{1.35 mol+1.94 mol}=0.5897$

Partial pressure of each gas can be calculated by the help of Dalton's' law:

$p_i=P\times \chi_1$

The total pressure is 405 mmHg.

P = 405 mmHg

Partial pressure of nitrogen gas:

$p_1=P\times \chi_1=405 mmHg\times 0.4103 =166.17 mmHg$

Partial pressure of oxygen gas:

$p_2=P\times \chi_2=405 mmHg\times 0.5897=238.83 mmHg$

Answer from: Quest

a mercury barometer is a device that uses the physical properties of mercury in a tube to measure barometric pressure. mercury barometers are used for : -

1. measurement of atmospheric pressure

2. determining the altitude of a place

3. weather forecasting

4. calibration and checking of aneroid barometers

5. application in fluid mechanics

6. measurement of pressure in aircrafts

7. surface weather analysis

8. preparation of barographs

9. usage in physical sciences like physics, astronomy & chemistry

10. preparation of aircraft altimeters

characteristics of mercury barometer : -

1.used to define and measure atmospheric pressure.

2.on the average at sea level the column of mercury rises to a height of about 760 mm.

3.this quantity is equal to 1 atmosphere.

4.it is also known as standard atmospheric pressure.

5.a barometer in which the weight of a column of mercury in a glass tube with a sealed top is balanced against that of the atmosphere pressing on an exposed cistern of mercury at the base of the mercury column, the height of the column varying with atmospheric pressure.

Write 15 uses of a mercury barometer along with its characterestics

Answer from: Quest

The answer is probably c. xylem

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Part a: three gases (8.00 g of methane, ch_4, 18.0g of ethane, c_2h_6, and an unknown amount of pro...

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