Part 1. determine the molar mass of a 0.458-gram sample of gas having a volume of 1.20 l at 287 k and 0.980 atm. show your work. part 2. if this sample was placed under extreme pressure, describe how the actual volume would compare to the predicted volume. explain your answer.

Part 1: 9.165 g/mol.

Part 2: The volume will decrease.

Explanation:

Part 1:

PV = nRT,

where, P is the pressure of the gas (atm) (P = 0.98 atm).

V is the volume of the gas (L) (V = 1.2 L).

n is the no. of moles of the gas (mol).

R is the general gas constant (R = 0.082 L.atm/mol.K).

T is the temperature of the gas (K) (T = 287 K).

∴ n = PV/RT = (0.98 atm)(1.2 L)/(0.082 L.atm/mol.K)(287 K) = 0.0499 mol ≅ 0.05 mol.

∵ n = mass/molar mass

∴ molar mass = mass/n = (0.458 g)/(0.05 mol) = 9.165 g/mol.

Part 2:  

The volume of the gas will decrease as the sample is placed under extreme pressure.

Boyle’s Law  states that at constant temperature, the  volume of a given quantity of  a gas varies inversely with its  pressure.

P ∝ 1/V.

so, as the sample is placed under extreme pressure, the volume of the gas will be decreased.

the answer is 1260

you spend 7 hours in school for 1 school day with 5 school days in 1 school week with 36 school weeks in a school year.

7*5*36= 1260 it's all about multiplying