For the next two problems, consider three 1 L flasks at STP. Flask A contains NH
ID: 701968 • Letter: F
Question
For the next two problems, consider three 1 L flasks at STP. Flask A contains NH, gas, flask B contains NO2 gas, and flask C contains N2 gas- all at the same pressure and temperature A1. Which contains the largest number of atoms? a. Flask A b. Flask B c. Flask C d. They all have the same number of atoms e. Not enough inform to tell A2. In which flask do the molecules have the highest average velocity a. Flask A b. Flask B c. Flask C d. They all have the same number of atoms e. Not enough information to tell B. The energy of gaseous molecules: a. IS b. is distributed over a wide range at a constant temperature d. is the same for all molecules at a constant temperature e. increases with an increase in pressure C. Complete the following statement such that it is true: The higher the molecular weight, the a. higher b. lower c. steadier d. more precise e. more accurate the effusion rate D1. Show that the molar volume of a gas at STP is about 22.4 L. (5 points) D2. A 4.37 gram sample of a certain diatomic gas occupies a volume of 3.00 L at 1.00 atm and a temperature of 45 °C. Determine the molecular weight of the gas. (5 points) D3. Determine the density of N2(g) at 0 °C and 1.00 atm. (5 points)Explanation / Answer
A1
At STP and at the same temperature and pressure
Number of moles remains constant for every gas.
Number of atoms will also remain same.
Option D is the correct answer
A2
Lower the molecular weight of gas, higher the velocity of gas molecules in the flask
NH3 = 17
NO2 = 46
N2 = 28
NH3 has the lowest lowest molecular weight, Flask A have the highest average velocity.
Option A is the correct answer
B
Kinetic energy of a gas depends on the temperature
For constant temperature, the energy is same.
Option B is the correct answer
C
Effusion rate is inversely proportional to the molecular weight
Higher the molecular weight, lower the effusion rate.
Option B is the correct answer
D1
At STP
Gas constant R = 0.0821 L-atm /mol-K
Temperature T = 273 K
Pressure P = 1 atm
V = RT / P
= 0.0821*273/1 = 22.4 L