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A watt is a measure of power (the rate of energy change) equal to 1 J/s. (a) Calculate the number of joules in a kilowatt-hour.
Ch.1 - Introduction: Matter, Energy, and Measurement
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232
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TOC.tableOfContents
- Ch.1 - Introduction: Matter, Energy, and Measurement177
- Ch.2 - Atoms, Molecules, and Ions231
- Ch.3 - Chemical Reactions and Reaction Stoichiometry216
- Ch.4 - Reactions in Aqueous Solution189
- Ch.5 - Thermochemistry175
- Ch.6 - Electronic Structure of Atoms168
- Ch.7 - Periodic Properties of the Elements150
- Ch.8 - Basic Concepts of Chemical Bonding177
- Ch.9 - Molecular Geometry and Bonding Theories200
- Ch.10 - Gases179
- Ch.11 - Liquids and Intermolecular Forces113
- Ch.12 - Solids and Modern Materials154
- Ch.13 - Properties of Solutions157
- Ch.14 - Chemical Kinetics199
- Ch.15 - Chemical Equilibrium128
- Ch.16 - Acid-Base Equilibria164
- Ch.17 - Additional Aspects of Aqueous Equilibria163
- Ch.18 - Chemistry of the Environment105
- Ch.19 - Chemical Thermodynamics164
- Ch.20 - Electrochemistry171
- Ch.21 - Nuclear Chemistry122
- Ch.22 - Chemistry of the Nonmetals82
- Ch.23 - Transition Metals and Coordination Chemistry79
- Ch.24 - The Chemistry of Life: Organic and Biological Chemistry16
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Brown 14th EditionCh.1 - Introduction: Matter, Energy, and MeasurementProblema 38c
Brown 14th EditionCh.1 - Introduction: Matter, Energy, and MeasurementProblema 38cCapitolo 1, Problema 38c
(c) A spherical ball of lead has a diameter of 5.0 cm. What is the mass of the sphere if lead has a density of 11.34 g>cm3? (The volume of a sphere is 14>32pr3, where r is the radius.)
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Convert the diameter of the sphere to the radius by dividing the diameter by 2. The radius r = \(\frac{5.0 \text{ cm}\)}{2}.
Use the formula for the volume of a sphere: V = \(\frac{4}{3}\) \(\pi\) r^3. Substitute the radius into this formula to find the volume.
Calculate the volume of the sphere using the value of \(\pi\) \(\approx\) 3.14159.
Use the density formula: \(\text{Density}\) = \(\frac{\text{Mass}\)}{\(\text{Volume}\)}. Rearrange this to find the mass: \(\text{Mass}\) = \(\text{Density}\) \(\times\) \(\text{Volume}\).
Substitute the density of lead (11.34 \(\text{ g/cm}\)^3) and the calculated volume into the mass formula to find the mass of the lead sphere.
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Density
Density is defined as mass per unit volume and is a key property of materials. It is typically expressed in grams per cubic centimeter (g/cm³) for solids and liquids. In this problem, the density of lead is given as 11.34 g/cm³, which will be used to calculate the mass of the lead sphere once its volume is determined.
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Percorso guidato
01:56
Density Concepts
Volume of a Sphere
The volume of a sphere can be calculated using the formula V = (4/3)πr³, where r is the radius of the sphere. The radius is half of the diameter, so for a sphere with a diameter of 5.0 cm, the radius would be 2.5 cm. This volume will be essential for finding the mass of the lead sphere using its density.
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Percorso guidato
02:35Constant-Volume Calorimetry
Mass Calculation
Mass can be calculated using the formula mass = density × volume. Once the volume of the sphere is calculated using the radius, and knowing the density of lead, this formula allows us to find the mass of the lead sphere. This relationship is fundamental in chemistry for converting between different properties of substances.
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Percorso guidato
03:12Molar Mass Calculation Example
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