Question

4. A Noble Gas has a density of 2.3013 g/L at 1185.6 torr and 330.0 K....

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.)

Homework Answers

Answer #1

4)

Molarity of the gas (n/V ) is calculated from ideal gas equation

Ideal gas equation is

PV = nRT

rearranging equation

n/V = P/RT

where,

P = Pressure, 1185.6 torr = 1.56atm

R = gas constant , 0.082057(L atm /mol K)

T = Temperature, 330.0K

n/V = 1.56atm/(0.082057(L atm / mol K) × 330.0K )

n/V = 0.0576 mol/L

No of moles of gas in 1litee of gas = 0.0576mol

From densisity we know that mass of 1L of gas = 2.3013g

molar mass = mass/no of moles

molar mass of the gas = 2.3013g/0.0576mol = 39.953g/mol

Therefore,

the noble gas is Argon

Sympol of the element is Ar

5)

According to boyl's law at constant temperature and amount of gas , pressure of the gas is inversely proportional to the volume of the gas

P1V1 = P2V2

P2 = P1V1/V2

P2 = 125ml × 684K/188ml

P2 = 455 mmHg

Therefore,

The final pressure = 455 mmHg

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