Question

2. What is the temperature in Kelvin for 0.112 moles of an ideal gas at 0.888 atm in a container that has a volume of 675 mL? (Your answer should have three significant figures).

3.What is the total pressure in atm for a 1807 mL container at
28.05º C that contains 5.00 grams each of CO_{2},
H_{2}O, and N_{2}? (The answer should have three
sig figs.)

4. A Noble Gas has a density of 2.3013 g/L at 1185.6 torr and 330.0 K. What is the elemental symbol for that gas?

5. A fixed sample of gas at 45.5 K is initially at 125 mL and 684 mm Hg. What would the final pressure be in mm Hg if the volume was expanded to a final volume of 188 mL? (The answer should have three sig figs.)

Answer #1

2)

Given:

P = 0.888 atm

V = 675.0 mL

= (675.0/1000) L

= 0.675 L

n = 0.112 mol

use:

P * V = n*R*T

0.888 atm * 0.675 L = 0.112 mol* 0.08206 atm.L/mol.K * T

T = 65.218 K

Answer: 65.2 K

3)

Molar mass of CO2,

MM = 1*MM(C) + 2*MM(O)

= 1*12.01 + 2*16.0

= 44.01 g/mol

Molar mass of H2O,

MM = 2*MM(H) + 1*MM(O)

= 2*1.008 + 1*16.0

= 18.016 g/mol

Molar mass of N2 = 28.02 g/mol

n(CO2) = mass/molar mass

= 5.0/44.01

= 0.1136

n(H2O) = mass/molar mass

= 5.0/18.016

= 0.2775

n(N2) = mass/molar mass

= 5.0/28.02

= 0.1784

n(CO2),n1 = 0.1136 mol

n(H2O),n2 = 0.2775 mol

n(N2),n3 = 0.1784 mol

Total number of mol = n1+n2+n3

= 0.1136 + 0.2775 + 0.1784

= 0.5696 mol

Given:

V = 1807.0 mL

= (1807.0/1000) L

= 1.807 L

n = 0.5696 mol

T = 28.05 oC

= (28.05+273) K

= 301.05 K

use:

P * V = n*R*T

P * 1.807 L = 0.5696 mol* 0.08206 atm.L/mol.K * 301.05 K

P = 7.7872 atm

Answer: 7.79 atm

Only 1 question at a time please

4. A Noble Gas has a density of 2.3013 g/L at 1185.6 torr and
330.0 K. What is the elemental symbol for that gas?
5. A fixed sample of gas at 45.5 K is initially at 125 mL and
684 mm Hg. What would the final pressure be in mm Hg if the volume
was expanded to a final volume of 188 mL? (The answer should have
three sig figs.)

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