Chemistry Homework Help!?

At STP, the volume of one mole of an ideal gas is 22.4 liter. The mass of one mole of carbon dioxide is 44 grams.

n = 8.75 ÷ 44
This is approximately 0.2 mole.
V = 22.4 * (8.75 ÷ 44) = 196 ÷ 44
The volume is approximately 4.45 liters.


(b) 190°C and 2.00 atm
The temperature must be in ˚K.
T = 273 + 190 = 463˚K
The volume is directly proportional to the change of the temperature. The volume is inversely proportional to the change of the pressure. Since we know the volume at standard temperature and pressure, we can use the following equation to determine the volume at a different temperature and pressure.

V2 = V1 * T2/T1 * P1/P2
V = (196 ÷ 44) * 463˚/273˚ * ½
This is approximately 3.78 liters

(c) 215 K and 115 kPa

Standard pressure is 101.3 kPa.

V = (196 ÷ 44) * 215/273 * 101.3 ÷ 115
This is approximately 3.09 liters. I hope this is helpful for you.

Molar mass of CO₂ = (12.0 + 16.0×2) g/mol = 44.0 g/mol
No. of moles of CO₂, n = (8.75 g) / (44.0 g/mol) = 8.75/44.0 mol


(a)
Pressure, P = 1.00 atm
Temperature, T = 273 K
Gas constant, R = 0.08206 L atm / (mol K)

Gas law: PV = nRT
Volume, V = nRT/P = (8.75/44.0) × 0.08206 × 273 / 1.00 L = 4.46 L


(b)
Pressure, P = 2.00 atm
Temperature, T = (273 + 190) K = 463 K
Gas constant, R = 0.08206 L atm / (mol K)

Gas law: PV = nRT
Volume, V = nRT/P = (8.75/44.0) × 0.08206 × 463 / 2.00 L = 3.78 L


(c)
Pressure, P = 115 kPa
Temperature, T = 215 K
Gas constant, R = 8.314 J / (mol K)

Gas law: PV = nRT
Volume, V = nRT/P = (8.75/44.0) × 8.314 × 215 / 115 L = 3.09 L