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Write "true" or "false" for each statement. (a) We balance chemical equations as we do because energy must be conserved. (b) If the reaction 2 \(\mathrm{O}_{3}(g) \rightarrow 3 \mathrm{O}_{2}(g)\) goes to completion and all \(\mathrm{O}_{3}\) is converted to \(\mathrm{O}_{2},\) then the mass of \(\mathrm{O}_{3}\) at the beginning of the reaction must be the same as the mass of \(\mathrm{O}_{2}\) at the end of the reaction. (c) You can balance the "water-splitting" reaction \(\mathrm{H}_{2} \mathrm{O}(I) \rightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\) by writing it this way: \(\mathrm{H}_{2} \mathrm{O}_{2}(l) \rightarrow \mathrm{H}_{2}(g)+\mathrm{O}_{2}(g)\)

Short Answer

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\( \textbf{False, True, False} \)

Step by step solution

01

Statement (a) - Balancing Chemical Equations and Energy Conservation

We balance chemical equations to ensure that the number of atoms of each element remains the same on both the reactant and product side. This is in line with the law of conservation of mass, which states that mass cannot be created nor destroyed in chemical reactions. So, the reason we balance chemical equations is not directly related to the conservation of energy but the conservation of mass. Therefore, the statement is: \( \boxed{\textbf{False}} \)
02

Statement (b) - Comparison of Masses before and after the Reaction

Given the balanced chemical equation: \( 2\,\mathrm{O}_{3}(g) \rightarrow 3\,\mathrm{O}_{2}(g) \) This equation tells us that every two molecules of ozone (\(\mathrm{O}_{3}\)), upon going to completion, are converted into three molecules of molecular oxygen (\(\mathrm{O}_{2}\)). According to the law of conservation of mass, the mass of the reactants should be equal to the mass of the products. Given the stoichiometry of the reaction, the conservation of mass should hold true. Hence, the statement is: \( \boxed{\textbf{True}} \)
03

Statement (c) - Balancing the Water-Splitting Reaction

The given water-splitting reaction reads: \( \mathrm{H}_{2}\,\mathrm{O}(l) \rightarrow \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \) This reaction is not balanced as written, as there are only two hydrogen atoms and one oxygen atom on the reactant side, while there are two oxygen atoms in the product \(\mathrm{O}_{2}(g)\). To balance the equation, we can adjust the coefficients as follows: \( 2\, \mathrm{H}_{2}\,\mathrm{O}(l) \rightarrow 2\, \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \) The suggested balanced equation: \( \mathrm{H}_{2}\,\mathrm{O}_{2}(l) \rightarrow \mathrm{H}_{2}(g) + \mathrm{O}_{2}(g) \) is incorrect as it introduces a new molecule (hydrogen peroxide) that is not involved in the actual reaction, creating a mismatch in terms of the structural and stoichiometric properties of the balanced equation. Thus, the statement is: \( \boxed{\textbf{False}} \)

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Balancing Chemical Equations
Balancing chemical equations is crucial in chemistry. When you look at a chemical reaction, it represents how atoms rearrange to form new substances. However, even as the reactants turn into products, the total number of atoms must remain the same. This is known as the Law of Conservation of Mass.
To ensure that the reaction equation reflects this law, chemists balance the equation. They adjust the coefficients before each chemical formula. This way, each type of atom is equal on both sides of the equation.
  • The aim is to make sure that the number of atoms of each element is the same on both sides.
  • This does not mean atoms are lost or gained, only rearranged.
  • Coefficients are only changed, not the subscripts within the formulas.

Remember, correctly balancing an equation is not just an academic exercise; it provides insight into the quantities involved in a chemical reaction.
Stoichiometry
Stoichiometry is a fundamental concept in chemistry that follows the principle of the Law of Conservation of Mass. It refers to the calculation of reactants and products in chemical reactions. It allows chemists to predict the quantities of substances consumed and produced.
Stoichiometry brings balance to chemistry by using balanced equations to provide a quantitative relation between reactants and products. When you balance a chemical equation, stoichiometry tells you the ratios of molecules involved.
  • Helps predict the amount of product produced from given reactants.
  • Useful for scaling reactions up or down.
  • Essential for designing industrial processes, ensuring efficiency and cost-effectiveness.

In the real world, stoichiometry provides a bridge between the theoretical yield and what can be achieved practically. Understanding stoichiometry is key to mastering chemistry.
Chemical Reactions
Chemical reactions are processes whereby reactants convert into products. They involve breaking old bonds and forming new ones, resulting in substances with new properties. Chemical reactions are written as equations, depicting the transformation of materials.
There are various types of chemical reactions, such as synthesis, decomposition, single replacement, and combustion reactions, to name a few. It's important to identify the type of reaction to understand the chemical processes involved.
  • Involves breaking and making chemical bonds.
  • Reactions can release or absorb energy.
  • Identified by changes like color change, temperature change, gas production, or precipitate formation.

By understanding chemical reactions, you gain insights into the continuous and dynamic changes that happen in the world around us, from digesting food to developing new materials.

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Most popular questions from this chapter

Vanillin, the dominant flavoring in vanilla, contains \(\mathrm{C}, \mathrm{H}\) , and \(\mathrm{O} .\) When 1.05 \(\mathrm{g}\) of this substance is completely combusted, 2.43 \(\mathrm{g}\) of \(\mathrm{CO}_{2}\) and 0.50 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\) are produced. What is the empirical formula of vanillin?

Epsom salts, a strong laxative used in veterinary medicine, is a hydrate, which means that a certain number of water molecules are included in the solid structure. The formula for Epsom salts can be written as \(\mathrm{MgSO}_{4} \cdot x \mathrm{H}_{2} \mathrm{O},\) where \(x\) indicates the number of moles of \(\mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{MgSO}_{4}\) When 5.061 \(\mathrm{g}\) of this hydrate is heated to \(250^{\circ} \mathrm{C},\) all the water of hydration is lost, leaving 2.472 g of MgSO \(_{4} .\) What is the value of \(x ?\)

When hydrocarbons are burned in a limited amount of air, both CO and \(\mathrm{CO}_{2}\) form. When 0.450 g of a particular hydrocarbon was burned in air, 0.467 \(\mathrm{g}\) of \(\mathrm{CO}, 0.733 \mathrm{g}\) of \(\mathrm{CO}_{2},\) and 0.450 \(\mathrm{g}\) of \(\mathrm{H}_{2} \mathrm{O}\) were formed. (a) What is the empirical formula of the compound? (b) How many grams of O \(_{2}\) were used in the reaction? (c) How many grams would have been required for complete combustion?

Determine the empirical formula of each of the following compounds if a sample contains \((\mathbf{a})0.104 \mathrm{mol} \mathrm{K}, 0.052 \mathrm{mol}\) \(\mathrm{C},\) and \(0.156 \mathrm{mol} \mathrm{O} ;(\mathbf{b}) 5.28 \mathrm{g} \mathrm{Sn}\) and \(3.37 \mathrm{g} \mathrm{F} ;(\mathbf{c}) 87.5 \% \mathrm{N}\) and 12.5\(\% \mathrm{H}\) by mass.

(a) What is the mass, in grams, of 1.223 mol of iron( III) sulfate? (b) How many moles of ammonium ions are in 6.955 g of ammonium carbonate? (c) What is the mass, in grams, of \(1.50 \times 10^{21}\) molecules of aspirin, \(\mathrm{C}_{9} \mathrm{H}_{8} \mathrm{O}_{4} ?\) (d) What is the molar mass of diazepam (Valium') if 0.05570 mol has a mass of 15.86 \(\mathrm{g} ?\)

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