Domanda del libro di testo
A hollow aluminum sphere with outer diameter 10.0 cm has a mass of 690 g. What is the sphere's inner diameter?
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Ch 18: A Macroscopic Description of Matter
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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TOC.tableOfContents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Non è quello che usi tu?Cambia libro di testo
Capitolo 18, Problema 5
How many atoms are in a 2.0 cm×2.0 cm×2.0 cm cube of aluminum?
Guida verificata passo dopo passo1
Step 1: Calculate the volume of the aluminum cube. Use the formula for the volume of a cube, which is \( V = a^3 \), where \( a \) is the length of one side of the cube. Here, \( a = 2.0 \, \text{cm} \).
Step 2: Convert the volume from cubic centimeters (\( \text{cm}^3 \)) to cubic meters (\( \text{m}^3 \)) because the density of aluminum is typically given in \( \text{kg/m}^3 \). Use the conversion factor \( 1 \, \text{cm}^3 = 10^{-6} \, \text{m}^3 \).
Step 3: Use the density of aluminum, \( \rho = 2700 \, \text{kg/m}^3 \), to calculate the mass of the aluminum cube. The formula is \( m = \rho \cdot V \), where \( \rho \) is the density and \( V \) is the volume in \( \text{m}^3 \).
Step 4: Determine the number of moles of aluminum in the cube. Use the molar mass of aluminum, \( M = 26.98 \, \text{g/mol} \), and the formula \( n = \frac{m}{M} \), where \( m \) is the mass in grams (convert \( \text{kg} \) to \( \text{g} \) by multiplying by 1000).
Step 5: Calculate the number of atoms in the cube using Avogadro's number, \( N_A = 6.022 \times 10^{23} \, \text{atoms/mol} \). Multiply the number of moles \( n \) by \( N_A \) using the formula \( N = n \cdot N_A \).
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Atomic Structure
Atoms are the basic building blocks of matter, consisting of protons, neutrons, and electrons. In the context of aluminum, each atom has a specific arrangement of these subatomic particles, which determines its properties. Understanding atomic structure is essential for calculating the number of atoms in a given volume of a material.
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Percorso guidato
01:06
Charge of Atom
Density
Density is defined as mass per unit volume and is a critical property for determining how many atoms are present in a specific volume of a substance. For aluminum, the density is approximately 2.7 g/cm³. By knowing the density, one can calculate the mass of the aluminum cube and subsequently the number of atoms it contains.
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8:13Intro to Density
Avogadro's Number
Avogadro's number, approximately 6.022 x 10²³, is the number of atoms or molecules in one mole of a substance. This concept is crucial for converting the mass of aluminum into the number of atoms. By using the molar mass of aluminum (about 27 g/mol), one can determine how many moles are in the cube and then multiply by Avogadro's number to find the total number of atoms.
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Percorso guidato
07:19Moles & Avogadro's Number
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