Answer:
Exam 3 Material
Homework Page Without Visible Answers
This page has all of the required homework for the material covered in the third 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.
Explanation:
Answer:
Its probably none of those.
Explanation:
White dwarf temperatures can exceed 100,000 Kelvin according to NASA (that's about 179,500 degrees Fahrenheit). Despite these sweltering temperatures, white dwarfs have a low luminosity as they're so small in size according to NMSU.
Answer:
51207 torr is the new pressure of the gas
Explanation:
We can solve this question using combined gas law that states:
P1V1T2 = P2V2T1
<em>Where P is pressure, V volume and T absolute temperature of 1, initial state and 2, final state of the gas</em>
<em> </em>
Computing the values of the problem:
P1 = 710torr
V1 = 5.0x10²mL
T1 = 273.15 + 30°C = 303.15K
P2 = ?
V2 = 25mL
T2 = 273.15 + 820°C = 1093.15K
Replacing:
710torr*5.0x10²mL*1093.15K = P2*25mL*303.15K
3.881x10⁸torr*mL*K = P2 * 7.579x10³mL*K
P2 = 51207 torr is the new pressure of the gas
Answer:
1.5 × 10² mL
Explanation:
Step 1: Given data
- Initial pressure of the gas (P₁): 1.9 atm
- Initial volume of the gas (V₁): 80 mL
- Final pressure of the gas (P₂): 1.0 atm (standard pressure)
- Final volume of the gas (V₂): ?
Step 2: Calculate the final volume of the gas
For an ideal gas, we can calculate the final volume of the gas using Boyle's law.
P₁ × V₁ = P₂ × V₂
V₂ = P₁ × V₁/P₂
V₂ = 1.9 atm × 80 mL/1.0 atm
V₂ = 1.5 × 10² mL
Since the pressure decreased, the volume of the gas increased.
Answer:
2.2×10^8
Explanation:
Cu(OH)2(s)<---------> Cu^2+(aq) + 2OH^-(aq) Ksp=2.2 x 10 ^-20
2H3O^+(aq) + 2OH^-(aq) <-------> 4H2O(l). Kw= 1×10^14
Cu^2+(aq) + 4H2O(l) <--------> [Cu(H2O)4]^2+(aq)
Overall ionic reaction:
Cu(OH)2(s) +2H3O^+(aq) <---------> [Cu(H20)4]^2+(aq)
Equilibrium constant for the reaction: Ksp×Kw= 2.2 x 10 ^-20 × (1/(1×10^-14))^2
Keq= 2.2×10^8
Kw= ion dissociation constant of water
