Domanda del libro di testo
Vanillin, the dominant flavoring in vanilla, contains C, H, and O. When 1.05 g of this substance is completely combusted, 2.43 g of CO2 and 0.50 g of H2O are produced. What is the empirical formula of vanillin?
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Ch.3 - Chemical Reactions and Reaction Stoichiometry
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.3 - Chemical Reactions and Reaction StoichiometryProblema 94
Brown 14th EditionCh.3 - Chemical Reactions and Reaction StoichiometryProblema 94Capitolo 3, Problema 94
The fluoride ion reacts with water to produce HF. (a) Write out the chemical equation for this reaction.
Guida verificata passo dopo passo1
Step 1: Identify the reactants and products. In this case, the reactants are the fluoride ion (F-) and water (H2O). The product is hydrofluoric acid (HF).
Step 2: Write the chemical equation with the reactants on the left and the products on the right. The arrow in the middle represents the direction of the reaction. So, the initial equation would look like this: F- + H2O -> HF
Step 3: Balance the equation. In this case, the equation is already balanced, as there is one atom of each element on both sides of the equation.
Step 4: Include the state of each substance. In this case, the fluoride ion is in aqueous solution (aq), water is a liquid (l), and hydrofluoric acid is also in aqueous solution (aq). So, the final balanced chemical equation is: F-(aq) + H2O(l) -> HF(aq)
Step 5: Note that this is a simplified version of the reaction. In reality, the fluoride ion reacts with water to produce hydrofluoric acid and a hydroxide ion (OH-), so the complete balanced equation would be: F-(aq) + H2O(l) -> HF(aq) + OH-(aq)
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Chemical Reactions
A chemical reaction involves the transformation of reactants into products through the breaking and forming of chemical bonds. In this case, the fluoride ion (F-) reacts with water (H2O) to form hydrofluoric acid (HF). Understanding the nature of reactants and products is essential for writing balanced chemical equations.
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Percorso guidato
03:31
Chemical Properties
Ionic and Molecular Compounds
Fluoride ion (F-) is an example of an ionic species, while hydrofluoric acid (HF) is a molecular compound. Ionic compounds typically form from the transfer of electrons between metals and nonmetals, while molecular compounds result from the sharing of electrons. Recognizing the difference helps in predicting the behavior of substances in reactions.
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Percorso guidato
02:11Ionic Compounds Naming
Acid-Base Chemistry
The reaction of fluoride ion with water can be understood through the lens of acid-base chemistry, where HF acts as a weak acid. In this context, water can act as a base, accepting a proton (H+) from HF. This concept is crucial for understanding the properties of acids and bases in aqueous solutions.
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02:00Arrhenius Acids and Bases
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Cinnamaldehyde is a compound that is responsible for the
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An organic compound was found to contain only C, H, and Cl. When a 1.50-g sample of the compound was completely combusted in air, 3.52 g of CO2 was formed. In a separate experiment, the chlorine in a 1.00-g sample of the compound was converted to 1.27 g of AgCl. Determine the empirical formula of the compound.
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(a) One molecule of the antibiotic penicillin G has a mass of 5.342×10-21 g. What is the molar mass of penicillin G?
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