Chem I Homework Page, Exam 5 Material
This page has all of the required homework for the material covered in the fifth exam of the first semester of General Chemistry. The textbook associated with this homework is CHEMISTRY The Central Science by Brown, LeMay, et.al. The last edition I required students to buy was the 12th edition (CHEMISTRY The Central Science, 12th ed. by Brown, LeMay, Bursten, Murphy and Woodward), but any edition of this text will do for this course.
Note: You are expected to go to the end of chapter problems in your textbook, find similar questions, and work out those problems as well. This is just the required list of problems for quiz purposes. You should also study the Exercises within the chapters. The exercises are worked out examples of the questions at the back of the chapter. The study guide also has worked out examples.
These are bare-bones questions. The textbook questions will have additional information that may be useful and that connects the problems to real life applications, many of them in biology.
Ideal Gas Law (Ch. 10)
During a normal breath, our lungs can have about 1.0 atm pressure and take up 2.0 L at some instant. If the gas inside of the lung is maintained at normal body temperature (37 °C), how many moles of gas are in the lung during this instant?
- PV=nRT n = PV/RT
= [(1 atm)(2 L)]/[(0.082 L-atm/K-mole)(310 K)] = 0.079 mole of gas
A gas initially in a 5 L container at 20 °C and 2 atm is placed into a new 2 L container at the same temperature (20 °C). What is the new pressure?
n = (2 atm)(5 L)/[(0.082 L-atm/K-mole)(293 K)]
|V||5 L||2 L|
|T||293 K||293 K|
= 0.416 moles
P2 = (0.416 moles)(0.082 L-atm/K-mole)(293 K)/(2 L)
= 5 atm
A 44 g sample of CO2 is in a 3 L container at 21 °C. (a) What is the pressure of the gas? (b) What will the volume be at STP?
(a) P = nRT/V = [(1 mole)(0.082 L-atm/K-mole)(294 K)]/(3 L) = 8.04 atm
(b) V = nRT/P = [(1 mole)(0.082 L-atm/K-mole)(273 K)]/(1 atm) = 22.4 L
A gaseous mixture is made from 5 g N2 and 5 g Cl2 is placed in a 12 L vessel at 21 °C. (a) What is the partial pressure of each gas? (b) What is the total pressure in the vessel?
(a) (5 g N2)[(1 mole N2)/(28 g N2)] = 0.179 mole N2
PN2 = [(0.179 mole N2)(0.082 L-atm/K-mole)(294 K)]/(12 L) = 0.36 atm
(5 g Cl2)[(1 mole Cl2)/(70.9 g Cl2)] = 0.071 mole Cl2
PCl2 = [(0.071 mole Cl2)(0.082 L-atm/K-mole)(294 K)]/(12 L) = 0.14 atm
(b) Ptot = 0.36 atm + 0.14 atm = 0.5 atm
A quantity of NH3(g) originally held at 2.5 atm in a 1.2 L container at 23 °C is transferred to a 6 L container at 15 °C. A quantity of CH4(g) originally at 5.0 atm and 10 °C in a 3 L container is transferred to the same 6 L container at 15 °C. What is the total pressure in the 6 L container?
nNH3 = [(2.5 atm)(1.2 L)]/[(0.082 L-atm/K-mole)(296 K)] = 0.1236 mole NH3
PNH3 = [(0.1236 mole NH3)(0.082 L-atm/K-mole)(288 K)]/(6 L) = 0.4865 atm
nCH4 = [(5 atm)(3 L)]/[(0.082 L-atm/K-mole)(283 K)] = 0.6464 mole CH4
PCH4 = [(0.6464 mole CH4)(0.082 L-atm/K-mole)(288 K)]/(6 L) = 2.544 atm
Ptot = 0.4865 atm + 2.544 atm = 3 atm
Using total moles, 0.1236 + 0.6464 = 0.77 moles:
Ptot = [(0.77 moles)(0.082 L-atm/K-mole)(283 K)]/(6 L) = 3 atm
Solid nickel reacts with hydrochloric acid as follows:
Ni(s) + 2HCl(aq) → NiCl2(aq) + H2(g)
If the H2 is collected over water at 22 °C and takes up a volume of 180 mL with a total pressure of 740 torr, how many grams of Ni have been consumed? The vapor pressure of water at 22 °C is 19.83 torr.
PH2 = 740 torr - 19.83 torr = 720.17 torr
(720.17 torr)[(1 atm)/(760 torr)] = 0.948 atm
nH2 = [(0.948 atm)(0.18 L)]/[(0.082 L-atm/K-mole)(295 K)] = 0.0071 mole H2
(0.0071 mole H2)[(1 mole Ni)/(1 mole H2][58.69 g Ni)/(1 mole Ni)] = 0.42 g Ni
The combustion of methane is shown below:
CH4(g) + 2O2(g) → 2H2O(g) + CO2(g)
How many liters of methane gas at 180 °C and 3 atm are required to react with 10.0 mole of oxygen gas in this reaction?
The number of moles of methane needed to react with 10 moles of oxygen is 5 moles of methane.
(10 moles O2)(1 mole CH4)/(2 mole O2) = 5 mole CH4
VCH4 = (5 mole)(0.082 L-atm/K-mole)(453 K)/(3 atm) = 61.91 L