The change in the volume of 22.4 L rigid box, when a sample of 1.00 mol of argon gas vented into the rigid box which already contains a 1.00 mol of nitrogen gas. (a) Decrease the total gas pressure in the box by 50%. (b) Occupy the entire 22.4 L volume of the box. (c) Increase the total gas pressure in the box by less than 50%. (d) Spread out into the box but the actual volume occupied by the gas cannot be known without pressure information. (e) Occupy only a volume of 11.2 L of the box, since there is already 1.0 mol of nitrogen present in the container. Concept Introduction: Effusion : The diffusion of gas molecules through the small holes into an emptied space. So, the gas molecules diffuse through the holes rather than colliding with it. The molecules free enough to make a move from the positions, so the faster the movement of molecule the more possibly they encounter the hole. The rate of effusion depends on the average molecular speed Graham law of effusion : At constant pressure, constant temperature the rate of effusion of the gas is inversely proportional to square root of the molar mass of the gas. Rate 1 Rate 2 = Molar mass 1 Molar mass 2
The change in the volume of 22.4 L rigid box, when a sample of 1.00 mol of argon gas vented into the rigid box which already contains a 1.00 mol of nitrogen gas. (a) Decrease the total gas pressure in the box by 50%. (b) Occupy the entire 22.4 L volume of the box. (c) Increase the total gas pressure in the box by less than 50%. (d) Spread out into the box but the actual volume occupied by the gas cannot be known without pressure information. (e) Occupy only a volume of 11.2 L of the box, since there is already 1.0 mol of nitrogen present in the container. Concept Introduction: Effusion : The diffusion of gas molecules through the small holes into an emptied space. So, the gas molecules diffuse through the holes rather than colliding with it. The molecules free enough to make a move from the positions, so the faster the movement of molecule the more possibly they encounter the hole. The rate of effusion depends on the average molecular speed Graham law of effusion : At constant pressure, constant temperature the rate of effusion of the gas is inversely proportional to square root of the molar mass of the gas. Rate 1 Rate 2 = Molar mass 1 Molar mass 2
Solution Summary: The author explains that the rate of effusion depends on the hole cross-sectional area, molecules per unit volume, and average molecular speed.
The change in the volume of 22.4 L rigid box, when a sample of 1.00 mol of argon gas vented into the rigid box which already contains a 1.00 mol of nitrogen gas.
(a) Decrease the total gas pressure in the box by 50%.
(b) Occupy the entire 22.4 L volume of the box.
(c) Increase the total gas pressure in the box by less than 50%.
(d) Spread out into the box but the actual volume occupied by the gas cannot be known without pressure information.
(e) Occupy only a volume of 11.2 L of the box, since there is already 1.0 mol of nitrogen present in the container.
Concept Introduction:
Effusion:
The diffusion of gas molecules through the small holes into an emptied space. So, the gas molecules diffuse through the holes rather than colliding with it. The molecules free enough to make a move from the positions, so the faster the movement of molecule the more possibly they encounter the hole. The rate of effusion depends on the average molecular speed
Graham law of effusion:
At constant pressure, constant temperature the rate of effusion of the gas is inversely proportional to square root of the molar mass of the gas.
What are the IUPAC Names of all the compounds in the picture?
1) a) Give the dominant Intermolecular Force (IMF) in a sample of each of the following
compounds. Please show your work. (8) SF2, CH,OH, C₂H₂
b) Based on your answers given above, list the compounds in order of their Boiling Point
from low to high. (8)
19.78 Write the products of the following sequences of reactions. Refer to your reaction road-
maps to see how the combined reactions allow you to "navigate" between the different
functional groups. Note that you will need your old Chapters 6-11 and Chapters 15-18
roadmaps along with your new Chapter 19 roadmap for these.
(a)
1. BHS
2. H₂O₂
3. H₂CrO4
4. SOCI₂
(b)
1. Cl₂/hv
2. KOLBU
3. H₂O, catalytic H₂SO4
4. H₂CrO4
Reaction
Roadmap
An alkene 5. EtOH
6.0.5 Equiv. NaOEt/EtOH
7. Mild H₂O
An alkane
1.0
2. (CH3)₂S
3. H₂CrO
(d)
(c)
4. Excess EtOH, catalytic H₂SO
OH
4. Mild H₂O*
5.0.5 Equiv. NaOEt/EtOH
An alkene 6. Mild H₂O*
A carboxylic
acid
7. Mild H₂O*
1. SOC₁₂
2. EtOH
3.0.5 Equiv. NaOEt/E:OH
5.1.0 Equiv. NaOEt
6.
NH₂
(e)
1. 0.5 Equiv. NaOEt/EtOH
2. Mild H₂O*
Br
(f)
i
H
An aldehyde
1. Catalytic NaOE/EtOH
2. H₂O*, heat
3. (CH,CH₂)₂Culi
4. Mild H₂O*
5.1.0 Equiv. LDA
Br
An ester
4. NaOH, H₂O
5. Mild H₂O*
6. Heat
7.
MgBr
8. Mild H₂O*
7. Mild H₂O+
Chapter 5 Solutions
Student Solutions Manual for Ebbing/Gammon's General Chemistry, 11th
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