Question

# 2. What is the temperature in Kelvin for 0.112 moles of an ideal gas at 0.888...

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 CO2, H2O, and N2? (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.)

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

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

Only 1 question at a time please

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