Textbook Question
Two small aluminum spheres, each having mass kg, are separated by cm. How many electrons does each sphere contain? (The atomic mass of aluminum is g/mol, and its atomic number is .)
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Ch 21: Electric Charge and Electric Field
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552
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Table of contents
- Ch 01: Units, Physical Quantities & Vectors37
- Ch 02: Motion Along a Straight Line57
- Ch 03: Motion in Two or Three Dimensions45
- Ch 04: Newton's Laws of Motion23
- Ch 05: Applying Newton's Laws54
- Ch 06: Work & Kinetic Energy11
- Ch 07: Potential Energy & Conservation6
- Ch 08: Momentum, Impulse, and Collisions15
- Ch 09: Rotation of Rigid Bodies37
- Ch 10: Dynamics of Rotational Motion44
- Ch 11: Equilibrium & Elasticity27
- Ch 12: Fluid Mechanics27
- Ch 13: Gravitation34
- Ch 14: Periodic Motion26
- Ch 15: Mechanical Waves22
- Ch 16: Sound & Hearing28
- Ch 17: Temperature and Heat28
- Ch 18: Thermal Properties of Matter35
- Ch 19: The First Law of Thermodynamics29
- Ch 20: The Second Law of Thermodynamics18
- Ch 21: Electric Charge and Electric Field28
- Ch 22: Gauss' Law21
- Ch 23: Electric Potential31
- Ch 24: Capacitance and Dielectrics18
- Ch 25: Current, Resistance, and EMF24
- Ch 26: Direct-Current Circuits29
- Ch 27: Magnetic Field and Magnetic Forces16
- Ch 28: Sources of Magnetic Field10
- Ch 29: Electromagnetic Induction22
- Ch 30: Inductance14
- Ch 31: Alternating Current20
- Ch 33: The Nature and Propagation of Light17
- Ch 34: Geometric Optics29
- Ch 35: Interference13
- Ch 36: Diffraction19
- Ch 37: Special Relativity19
- Ch 38: Photons: Light Waves Behaving as Particles12
- Ch 39: Particles Behaving as Waves25
- Ch 40: Quantum Mechanics I: Wave Functions20
- Ch 41: Quantum Mechanics II: Atomic Structure21
- Ch 42: Molecules and Condensed Matter17
- Ch 43: Nuclear Physics13
- Ch 44: Particle Physics and Cosmology17
Young & Freedman Calc15th EditionUniversity PhysicsISBN: 9780135159552Not the one you use?Change textbook
Chapter 21, Problem 8c
Two small aluminum spheres, each having mass kg, are separated by cm. What fraction of all the electrons in each sphere does this represent?
Verified step by step guidance1
First, determine the number of aluminum atoms in each sphere. Use the formula for the number of moles: \( n = \frac{m}{M} \), where \( m \) is the mass of the sphere (0.0250 kg) and \( M \) is the molar mass of aluminum (approximately 27 g/mol or 0.027 kg/mol).
Convert the mass of the sphere from kilograms to grams to match the units of the molar mass: \( m = 0.0250 \text{ kg} \times 1000 \text{ g/kg} = 25 \text{ g} \).
Calculate the number of moles of aluminum in each sphere: \( n = \frac{25 \text{ g}}{27 \text{ g/mol}} \).
Determine the number of atoms in each sphere using Avogadro's number \( N_A = 6.022 \times 10^{23} \text{ atoms/mol} \): \( N = n \times N_A \).
Since each aluminum atom has 13 electrons, calculate the total number of electrons in each sphere: \( N_e = N \times 13 \). This gives the total number of electrons in one sphere, and you can compare this to the number of electrons involved in the interaction to find the fraction.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Atomic Structure of Aluminum
Aluminum is an element with atomic number 13, meaning each atom has 13 protons and, typically, 13 electrons. Understanding the atomic structure is crucial for calculating the total number of electrons in a given mass of aluminum, as it allows us to determine the number of atoms present.
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Pouring Hot Water in an Aluminum Cup
Avogadro's Number
Avogadro's Number, approximately 6.022 x 10^23, is the number of atoms or molecules in one mole of a substance. This concept is essential for converting the mass of aluminum spheres into moles, which then allows us to calculate the total number of electrons by multiplying the number of moles by Avogadro's Number and the number of electrons per atom.
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Fraction Calculation
To find the fraction of electrons involved, one must compare the number of electrons in the interaction to the total number of electrons in the spheres. This involves calculating the total number of electrons using atomic and molecular data, and then determining what portion of these electrons are relevant to the problem at hand.
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Related Practice
Textbook Question
Two small spheres spaced cm apart have equal charge. How many excess electrons must be present on each sphere if the magnitude of the force of repulsion between them is N?
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Textbook Question
Two small plastic spheres are given positive electric charges. When they are cm apart, the repulsive force between them has magnitude N. What is the charge on each sphere if the two charges are equal?
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Textbook Question
Two small aluminum spheres, each having mass kg, are separated by cm. How many electrons would have to be removed from one sphere and added to the other to cause an attractive force between the spheres of magnitude N (roughly ton)? Assume that the spheres may be treated as point charges.
Textbook Question
The nuclei of large atoms, such as uranium, with protons, can be modeled as spherically symmetric spheres of charge. The radius of the uranium nucleus is approximately m. What magnitude of electric field does it produce at the distance of the electrons, which is about m?
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Textbook Question
Two small plastic spheres are given positive electric charges. When they are cm apart, the repulsive force between them has magnitude N. What is the charge on each sphere if one sphere has four times the charge of the other?
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