Answer:
1. 0.121 moles
2. 46.9 liters
3. 12.3 atm
4. 253 K
5. 0.552 g/L.
6. 22.8 g/mol.
7. 0.0494 moles
8. 5370 liters
9. 34.5 grams
10. helium
Step-by-step explanation:
1. Using the ideal gas law equation, we can solve for the number of moles of oxygen:
PV = nRT
n = PV/RT
n = (1.2 atm) * (2.5 L) / [(0.0821 L atm/mol⚫K) * (298 K)]
n = 0.121 moles of oxygen
Therefore, 0.121 moles of oxygen will occupy a volume of 2.5 liters at 1.2 atm and 25°C.
2. Using the ideal gas law equation, we can solve for the volume of nitrogen:
PV = nRT
V = nRT/P
V = (2.0 moles) * (0.0821 L atm/mol⚫K) * (293 K) / (720 Torr)
V = 46.9 L
Therefore, 2.0 moles of nitrogen will occupy a volume of 46.9 liters at 720 Torr and 20°C.
3. Using the ideal gas law equation, we can solve for the pressure of CO:
PV = nRT
P = nRT/V
n = (25 g) / (28.01 g/mol) = 0.892 mol
P = (0.892 mol) * (0.0821 L atm/mol⚫K) * (298 K) / (0.5 L)
P = 12.3 atm
Therefore, 25 g of CO will exert a pressure of 12.3 atm at a temperature of 25°C and a volume of 500 mL.
4. Using the ideal gas law equation, we can solve for the temperature of Cl:
PV = nRT
T = PV/nR
n = (5.00 g) / (35.45 g/mol) = 0.141 mol
T = (900. Torr) * (0.750 L) / (0.141 mol) / (0.0821 L atm/mol⚫K)
T = 253 K
5. Therefore, 5.00 g of Cl will exert a pressure of 900. Torr at a volume of 750 mL at a temperature of 253 K.
Using the ideal gas law equation, we can solve for the density of NH3:
PV = nRT
n/V = P/RT
n/V = (800 Torr) / [(0.0821 L atm/mol⚫K) * (298 K)]
n/V = 0.0324 mol/L
The molar mass of NH3 is 17.03 g/mol, so the density of NH3 is:
density = (0.0324 mol/L) * (17.03 g/mol) = 0.552 g/L
Therefore, the density of NH3 at 800 Torr and 25°C is 0.552 g/L.
6. We can use the ideal gas law to calculate the number of moles of the gas and then divide the mass of the gas by the number of moles to get the molecular mass.
PV = nRT
n = PV/RT
n = (1.2 g/L) / [(0.0821 L atm/mol⚫K) * (293 K) * (745. Torr / 760 Torr)]
n = 0.0526 mol
The mass of the gas is 1.2 g, so the molecular mass is:
molecular mass = 1.2 g / 0.0526 mol = 22.8 g/mol
Therefore, the molecular mass of the gas is 22.8 g/mol.
7. Using the ideal gas law equation, we can solve for the number of moles of nitrogen gas:
PV = nRT
n = PV/RT
n = (6680 Torr) * (0.347 L) / [(0.0821 L atm/mol⚫K) * (300 K)]
n = 0.0494 moles of nitrogen gas
Therefore, 0.0494 moles of nitrogen gas will occupy a volume of 347 mL at 6680 Torr and 27°C.
8. We can use the ideal gas law to calculate the volume of hydrogen:
PV = nRT
V = nRT/P
n = (454 g) / (2.016 g/mol) = 225 mol
V = (225 mol) * (0.0821 L atm/mol⚫K) * (298 K) / (1.05 atm)
V = 5370 L
Therefore, 454 grams (1 lb.) of hydrogen will occupy a volume of 5370 liters at 1.05 atm and 25°C.
9. Using the ideal gas law equation, we can solve for the number of moles of CO:
PV = nRT
n = PV/RT
n = (785 Torr) * (32.5 L) / [(0.0821 L atm/mol⚫K) * (305 K)]
n = 1.23 moles of CO
The molar mass of CO is 28.01 g/mol, so the mass of CO is:
mass = (1.23 moles) * (28.01 g/mol) = 34.5 g
10. Therefore, 34.5 grams of CO will exert a pressure of 785 Torr at a volume of 32.5 L and a temperature of 32°C.
Using the ideal gas law equation, we can solve for the identity of the gas:
PV = nRT
n = PV/RT
n = (758 Torr) * (58.4 L) / [(0.0821 L atm/mol⚫K) * (275.5 K)]
n = 2.93 moles of gas
The mass of the gas is 10.3 g, so the molecular mass is:
molecular mass = 10.3 g / 2.93 mol = 3.52 g/mol
Looking at the periodic table, we see that the only element with a molecular mass close to 3.52 g/mol is helium, which has a molecular mass of 4.00 g/mol. Therefore, the gas is likely helium.