Concept explainers
(a)
Interpretation: The difference in the behavior of the sample of gas and liquid should be determined if the sample is transferred from one container to a large one.
Concept introduction:
Interconversion of physical
For example, at high temperatures and at a certain pressure, solid changes to a liquid, and liquid changes to a gas.
(a)

Answer to Problem 5.1P
When a sample of gas is transferred from one container to a large one, the volume of the gas increases to the volume of the larger container whereas the volume of the liquid remains constant.
Explanation of Solution
The gaseous state of matter has a large intermolecular distance between particles with the weakest intermolecular force between them. Due to high kinetic energy, the particles can move randomly so they occupy the volume of the container. On the contrary, liquids have fixed volume due to stronger intermolecular forces between particles compared to gases. When a sample of gas is transferred from one container to a large one, the volume of the gas increases to the volume of the larger container whereas the volume of the liquid remains constant.
(b)
Interpretation: The difference in the behavior of the sample of gas and liquid should be determined if the sample is heated in an expandable container without any change in the state of matter.
Concept introduction:
Interconversion of physical states of matter refers to the application of temperature and pressure to change one physical state of matter into another.
For example, at high temperatures and at a certain pressure, solid changes to a liquid, and liquid changes to a gas.
(b)

Answer to Problem 5.1P
When a sample of gas is heated in an expandable container without a change of physical state, then the volume of the container will increase whereas the volume of a sample with liquid does not change on heating.
Explanation of Solution
The gaseous state of matter has a large intermolecular distance between particles with the weakest intermolecular force between them. On the contrary, liquids have fixed volume due to stronger intermolecular forces between particles compared to gases. When a sample of gas is heated in an expandable container without a change of physical state, then the volume of the gas will increase whereas the volume of a sample with liquid does not change on heating.
(c)
Interpretation: The difference in the behavior of the sample of gas and liquid should be determined if the sample is placed in a cylinder with a piston and external force is applied.
Concept introduction:
Interconversion of physical states of matter refers to the application of temperature and pressure to change one physical state of matter into another.
For example, at high temperatures and at a certain pressure, solid changes to a liquid, and liquid changes to a gas.
(c)

Answer to Problem 5.1P
When a sample of gas is placed in a cylinder with a piston and external force is applied, the volume of the liquid remains constant whereas the volume of the gas is reduced.
Explanation of Solution
The gaseous state of matter has a large intermolecular distance between particles with the weakest intermolecular force between them. On the contrary, liquids have fixed volume due to stronger intermolecular forces and lesser intermolecular space between particles compared to gases. When a sample of gas is placed in a cylinder with a piston and external force is applied, the volume of the liquid remains constant as liquids are not compressible whereas the volume of the gas is reduced because gases are highly compressible due to large intermolecular distance.
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Chapter 5 Solutions
CHEMISTRY: MOLECULAR...(LL) W/ALEKS
- Predict the product of this organic reaction: + NH2 HO A P+ H2O Specifically, in the drawing area below draw the skeletal ("line") structure of P. If there is no reasonable possibility for P, check the No answer box under the drawing area. Click and drag to start drawing a structure. ✓arrow_forward个 Stuc X ctclix ALE X A ALE × A ALE X Lab x (195 × Nut x M Inbx EF 目 → C www-awu.aleks.com/alekscgi/x/Isl.exe/10_u-IgNslkr7j8P3jH-IQ1g8NUi-mObKa_ZLx2twjEhK7mVG6PulJI006NcKTV37JxMpz Chapter 12 HW = Question 27 of 39 (5 points) | Question Attempt: 1 of Unlimited Part: 1/2 Part 2 of 2 Give the IUPAC name. Check 3 50°F Clear ©2025 McGraw Hill L Q Search webp a عالياكarrow_forward个 Stuck x ctc xALE X A ALE × A ALE X Lab x (19: x - G www-awu.aleks.com/alekscgi/x/Isl.exe/10_u-lgNslkr7j8P3jH-1Q1g8NUi-mObka ZLx2twjEhK7mVG6PUUIO06 Chapter 12 HW 三 Question 26 of 39 (4 points) 1 Question Attempt: 1 of Unlimited Answer the following questions about the given alkane. Part: 0 / 2 Part 1 of 2 Give the IUPAC name. Skip Part 2 53°F Clear Check × Q Search hp hp 02arrow_forward
- Calculate the equilibrium constant at 25.0 oC for the following equation. Cd(s) + Sn+2(aq) ↔Cd+2(aq) + Sn(s) Group of answer choices 3.11x104 1.95x1018 9.66x108 1.40x109arrow_forwardWhat is the pH at the cathode for the following cell written in line notation at 25.0 oC with a Ecell = -0.2749 V? Ni(s)|Ni+2(aq, 1.00 M)||H+1(aq, ?M)|H2(g, 1.00 atm)|Pt(s)arrow_forwardCalculate Ecell for a hydrogen fuel cell at 95.0 oC using the following half-reactions with PH2 = 25.0 atm and PO2 = 25.0 atm. O2(g) + 4H+1(aq) + 4e-1 → 2H2O(l) Eo = 1.229 V 2H2(g) → 4H+1(aq) + 4e-1 Eo = 0.00 Varrow_forward
- Calculate Ecell at 25.0 oC using the following half-reactions with [Ag+1] = 0.0100 M and [Sn+2] = 0.0200 M. Ag+1(aq) + 1e-1 Ag(s) Sn+2(aq) + 2e-1 Sn(s)arrow_forwardDone 18:19 www-awu.aleks.com Chapter 12 HW Question 27 of 39 (5 points) | Question Attempt: 1 of Unlimited .. LTE סוי 9 ✓ 20 ✓ 21 × 22 23 24 25 26 27 28 29 30 Answer the following questions about the given alkane. Part: 0 / 2 Part 1 of 2 Classify each carbon atom as a 1º, 2º, 3º, or 4°. Highlight in red any 1° carbons, highlight in blue any 2° carbons. highlight in green any 3° carbons, and leave any 4° carbons unhighlighted. Skip Part Check Save For Later © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use Privacy Center | Accessibility ☑ คarrow_forward< Done 19:22 www-awu.aleks.com Chapter 12 HW Question 4 of 39 (2 points) | Question Attempt: 5 of Unlimited : .. LTE סוי 1 ✓ 2 ✓ 3 = 4 ✓ 5 ✓ 6 ✓ 7 ✓ 8 ✓ 9 = 10 11 ✓ 12 Consider the molecule (CH3)2CHCH2CHCн for the following questions. Part 1 of 2 Which of the following molecules is/are constitutional isomer(s) to (CH3)2CHCH2CH2CH3? Check all that apply. Part 2 of 2 (CH3),C(CH2)2CH3 CH3 H,C-CH-CH-CH, CH 3 None of the above. ☑ Which of the following molecules is/are identical molecules to (CH3)2CHCH2CH2CH₁₂? Check all that apply. CH3 H,C-CH-CH₂-CH2-CH, CH3(CH2)2CH(CH3)2 CH2-CH2-CH3 HỌC-CH=CH, 乂 ☑ а None of the above Check Save For Later Submit Assignment © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center Accessibilityarrow_forward
- 18:11 LTE ا... US$50 off hotels is waiting for you Book now, hotels in Nashville are going fast QUTSLIVII 25 61 69 points) | QuestIVIT ALLēm... now Give the IUPAC name for each compound. Part 1 of 3 Part 2 of 3 X ☑ Х Check Save For Later Submit © 2025 McGraw Hill LLC. All Rights Reserved. TOMS CT US ...vacy Center | Accessibilityarrow_forwardDone 19:17 www-awu.aleks.com Chapter 12 HW Question 29 of 39 (6 points) | Question Attempt: 1 of Unlimited .III LTE סוי 27 28 = 29 30 31 32 = 33 34 35 Consider this structure. CH3CH2CH2 Part 1 of 3 3 CH2 CH2CH3 - C-CH2CH 3 H CH₂ Give the IUPAC name of this structure. 3-ethyl-3,4-dimethylheptane Part: 1/3 Part 2 of 3 Draw the skeletal structure. Skip Part < Check Click and drag to start drawing a structure. Save For Later Submit © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility Хarrow_forward18:57 .III LTE www-awu.aleks.com Chapter 12 HW Question 31 of 39 (8 points) | Question Attem... Give the IUPAC name of each compound. Part 1 of 4 Part 2 of 4 Х Х Check Save For Later Submit © 2025 McGraw Hill LLC. All Rights Reserved. TOMS OF US vacy Center | Accessibilityarrow_forward
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